Solution & Colligative Properties | Question Bank for Class 12 Chemistry

Get Solution & Colligative Properties important questions for Boards exams. Download or View the Important Question bank for Class 12 Chemistry. These important questions will play significant role in clearing concepts of Chemistry. This question bank is designed by NCERT keeping in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 12 chemistry chapter wise CBSE.

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Q. Why does molality of a solution remain unchanged with the rise in temperature ?

[AI 2004 C]

Sol. Mass of solvent does not change with temperature.


Q. What is van’t Hoff equation for dilute solution ?

Sol. $\pi V=n R T$ where $n$ is the no. of moles of solute present in $V$ litres of solution, $\pi$ is osmotic pressure, $T$ is temperature and $R$ is gas constant or solution constant.


Q. Mention a large scale use of the phenomenon called ‘reverse osmosis’.

[Delhi 2004; CBSE 2004]

Sol. It is used for desalination of sea water.


Q. What do you understand by ‘colligative properties’?

[Delhi 1999]

Sol. Colligative properties are those properties which depends upon number of particles of solute and not on natures of solute.


Q. Why is the cooking temperature in pressure cooker higher than in open pan ?

[Delhi 2002 C]

Sol. In pressure cooker, high pressure is exerted by steam due to which boiling point becomes higher than in open pan.


Q. What is it that elevation in boiling point of water is not same in following solutions? 0.1 molar $N a C l$ solution and 0.1 molar sugar solution.

Sol. $0.1 M N a C /$ has higher elevation in boiling point because it contains more number of particles than $0.1 M$ solution of glucose. $N a C l \rightarrow N a^{+}+C l^{-}$


Q. What is de-icing agent ? How does it work ?

Sol. Common salt is called de-icing agent because it lowers the freezing point of water to such an extent that it does not freeze to form ice. Hence, it is used to clear snow from roads.


Q. Why is the vapour pressure of a solution of glucose in water lower than that of water ?

[A.I.S.B. 2006]

Sol. At a number of sites, non-volatile glucose molecules will occupy the surface instead of water molecules. Hence, effective surface area for evaporation decreases and so the vapour pressure is lower.


Q. Give one example each of solid in gas and liquid in gas solutions.

[H.P.S.B. 1998]

Sol. Iodine vapour in air, aerated drinks.


Q. What happens when blood cells are placed in pure water ?

[Delhi 2003, 04; D.S.B. 2004 C]

Sol. Due to osmosis, water molecules move into blood cells through the cells walls. As a result, blood cells swell and may even burst.


Q. Write expression for the Raoult’s law for non-volatile solutes.

Sol. $\left(p^{\circ}-p_{s}\right) / p^{\circ}=n_{2} /\left(n_{1}+n_{2}\right)$ where $p^{\circ}=V . P .$ of pure solvent $p_{s}=V . P .$ of solution, $n_{2}=$ no. of moles of solute and $n_{1}=$ no. of moles of solvent.


Q. What is the difference between lowering of vapour pressure and relative lowering of vapour pressure ?

Sol. Lowering of V.P. $=p^{\circ}-p_{s} .$ Relative lowering of V.P. $=\left(p^{\circ}-p_{s}\right) / p^{\circ}$


Q. State any two characteristic of ideal solution.

[Delhi 1999; Foreign 1999]

Sol. Ideal solutions

(i) Follow Raoult’s law

(ii) They can be separated by fractional distillation


Q. On mixing equal volumes of water and ethanol, what type of deviation would you expect from Raoult’s law ?

[Delhi 1999 C]

Sol. It is because of $H$ -bonding between acetone and $C H C l_{3}$, the force of attraction increases, therefore energy is released.


Q. Why are ethers not miscible in water ?

Sol. Because they can not form H-bonds with water.


Q. Why is benzene insoluble in water but soluble in toluene?

[AI 1999 C]

Sol. Benzene is insoluble in water because benzene is non polar where as water is polar. Toluene is non polar solvent therefore, Benzene is soluble in toluene.


Q. Give an example of a compound in which hydrogen bonding results in the formation of a dimer.

Sol. Acetic acid forms dimer due to H-bonding.


Q. Why is freezing point depression 0.1 M sodium chloride solution nearly twice that of 0.1 M glucose solution ?

[A.I.S.B. 2006]

Sol. $\mathrm{NaCl},$ being an electrolytes, dissociates almost completely to give $N a^{+}$ and $\quad C I^{-}$ ions whereas glucose, being nonelectrolyte, does not dissociate. Hence, the number of particles in 0.1 MNaCl solution is nearly double than in

$0.1 M$ glucose solution. Freezing point depression, being a colligative properly, is therefore, nearly twice for solution than for glucose solution of same molarity.


Q. State the formula relating pressure of a gas with its mole fraction in a liquid solution in contact with it.

[DSB2005]

Sol. According to Henry’s law.

Partial pressure of gas above the solution

$=k_{H} \times$ mole fraction of the gas in the solution

or $\quad p_{A}=k_{H} \times x_{A^{\prime}}$ where $k_{H}=$ Henry’s constant.


Q. When fruits and vegetables are dried and placed in water, they slowly swell and return to original form. Why ? Does an increase in temperature accelerate the process ? Explain.

Sol. When fruits and vegetables that have dried and are placed in water osmosis takes place, i.e., water molecules pass through semipermiable membranes present in cell walls, therefore, they swell. If temperature is increased, osmosis will be faster.


Q. Why does the use of pressure cooker reduce cooking time?

Sol. At higher pressure over the liquid (due to weight of the pressure cooker lid), the liquid boils at higher temperature. Therefore, cooking occurs faster.


Q. Sodium chloride solution freezes at lower temperature than water but boils at higher temperature than water. Explain.

Sol. Freezing point of a liquid depresses on the addition of a non-volatile solute and therefore, a solution of sodium chloride freezes at lower temperature than freezing point of water. On he other hand, there is elevation in boiling point on the addition of a non-volatile solute and consequently boiling point of sodium chloride solution is more than that of water.


Q. Addition of $\mathrm{HgI}_{2}$ to aqueous solution of $K I$ shows an increase in vapour pressure. Why?

Sol. HgI$_{2}$ forms a complex with $K I$ and therefore, the number of particles in solution increases.

$H g I_{2}+2 K I \rightarrow K_{2}\left[H g I_{4}\right] \Longrightarrow 2 K^{+}+\left[H g I_{4}\right]^{2-}$

Therefore, the osmotic pressure increases.


Q. Give one example each of miscible liquid pairs showing positive and negative deviations from Raoult’s law. Give one reason each for such deviations.

[Delhi 2000; AI 2000]

Sol. Ethanol and water, methanol and $H_{2} O$ show positive deviation from Raoult’s law because force of attraction between them decreases on mixing. and and, show negative deviation from Raoult’s law because force of attraction between them increases on mixing.


Q. “The solution of a non-volatile solute boils at a higher temperature than the pure solvent.” Show this relationship on a graphic diagram.

[AI 2002 C]

Sol. The diagram shows that solution containing non-volatile solute boils at temperature $T_{2}$ which is higher than boiling point of pure solvent, i.e., $T_{1}$

Variation of vapour pressure with Temperature showing boiling point of solution is higher than that of pure solvent.


Q. Carbon tetrachloride and water are immiscible whereas ethanol and water are miscible in all proportions. Correlate this behaviour with molecular structures of these compounds.

[AI 2003]

Sol. Carbon tetrachloride is a non-polar compounds, it cannot form $H$ -bond with water, that is why $C C l_{4}$ and water do not mix with each other. Ethanol is a polar compound and $H_{2} \mathrm{O}$ is polar solvent, there is $H$ -bond between ethanol and water, therefore, ethanol and water are miscible in all proportions.


Q. Calculate molarity and molality of a $13 \%$ solution (by weight) of sulphuric acid? Its density is $1.020 \mathrm{g} \mathrm{cm}^{-3}$

(Atomic mass $H=1, O=16, S=32 \text { a. } m . u .)$

[Foreign 1999]

Sol. $M=\frac{w_{B}}{M_{B}} \times \frac{1000}{w_{A}}=\frac{13 \times 1000}{98 \times \frac{100}{1.02}}=\frac{13 \times 10 \times 1.02}{98}=\frac{1326}{98}$

$=1.353 \mathrm{mol} L^{-1}$ or $1.353 \mathrm{M}$

$m=\frac{w_{B}}{M_{B}} \times \frac{100}{w_{A}}=\frac{13}{98} \times \frac{1000}{87}=1.524 \mathrm{mol} / \mathrm{kg}_{\mathrm{or}} 1.524 \mathrm{m}$


Q. A solution is prepared by adding 60 g of methyl alcohol to 120 g of water. Calculate the mole fraction of methanol and water.

[H.P. 1995]

Sol. Mass of methanol = 60 g

Moles of methanol $\left.=\frac{60}{32}=1.875 \text { (Molar mass }=32\right)$

Moles of water $=\frac{120}{18}=6.667$

Total No. of moles = 1.875 + 6.667 = 8.542

Mole fraction of methanol $=\frac{1.875}{8.542}=0.220$

Mole fraction of water $=\frac{6.667}{8.542}=0.780$


Q. Vapour pressure of pure water at $35^{\circ} \mathrm{C}$ is $31.82 \mathrm{mm} \mathrm{Hg}$. When $27.0 \mathrm{g}$ of solute is dissolved in $100 \mathrm{g}$ of water (at the same temperature) vapour pressure of the solution thus formed is $30.95 \mathrm{mm}$ Hg. Calculate the molecular mass of the solute.

Sol. $\frac{P_{A}^{\circ}-P_{A}}{P_{A}^{\circ}}=X_{B} \Rightarrow \frac{31.82-30.95}{31.82}=\frac{27 / M}{100 / 18}$

$\Rightarrow \frac{0.87}{31.82}=\frac{27}{M_{B}} \times \frac{18}{100} \Rightarrow M_{B}=\frac{27 \times 18 \times 31.82}{0.87 \times 100}$

$\Rightarrow 177.75 g \mathrm{mol}^{-1}$


Q. In a solution of urea, $3.0 g$ of it is dissolved in $100 \mathrm{ml}$ of water. What will be the freezing point of this solution? State the approximation made if any.

$\left[k_{f} \text { for water }=1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}, \text { molar mass of urea }=60 \mathrm{g}\right.$ $\left.m o l^{-1}\right]$

[Delhi 2002 C]

Sol. $\Delta T_{f}=k_{f} \times m$

$\Delta T_{f}=1.86 \times \frac{W_{B}}{M_{B}} \times \frac{1000}{W_{A}}=\frac{1.86 \times 3}{60} \times \frac{1000}{100}=0.93$

Freezing point of solution $=273 K-0.93=272.07 K$


Q. Give one example each of miscible liquid pairs showing positive and negative deviation from Raoults law. give one reason each for such deviations.

[CBSE 2004]

Sol. Solution of $n$ hexane and ethanol shows positive deviations from Raoults law because $n$ -hexane molecular weaken the H-bonds between ethanol molecules which increases its vapour

Solution of acetone and chloroform shows negative deviations from Raoult’s law because of the formation of the $H$ -bonds between acetone and chloroform molecules.


Q. What is meant by abnormal molecular mass of solute ? Discuss the factors, which bring abnormality in the experimentally determined molecular masses of solutes using colligative properties.

[AI 1999 C]

Sol. The molecular mass obtained with the help of colligative property sometimes is different from normal molecular mass, it is called abnormal molecular mass. The factor which bring abnormality are

(i) Association : When solute particles undergo association, number of particles become less and molecular mass determined with the help of colligative property will be more.

(ii) Dissociation : It leads to increase in number of particles, therefore, increase in colligative property, therefore, therefore, decrease in molecular weight because colligative property is inversely proportional to molecular weight.


Q. Explain why vapour pressure of solvent is lowered by addition of a non-volative solute ?

Sol. It is observed that the presence of a non-volatile solute in a solution reduces the escaping tendency of solvent molecules into vapour phase. It is because some of the solute particles occupy the position of the solvent molecules on the liquid surface and thus dower the vapour pressure of the solvent.


Q. Ristinguish between ideal and non ideal solutions.

[MP 2001, 02, 06]

Sol.


Q. How will you determine the molecular mass of a substance by osmotic pressure method ?

Sol. According to Van’t Hoff equation

$\pi v=n R T$ …..(i)

$\pi=$ osmotic pressure; $T=$ Temperature for a solution

If $w$ gram of solute is dissolved in $v$ litres of he solution and $M$ is the molecular mass of the solute, then

$n=\frac{w}{M}$

Substituting this value in equation (v) we get,

$\pi v=\frac{w}{M} R T \Rightarrow M=\frac{w R T}{\pi v}$ …..(ii)

Thus, measuring the osmotic pressure of a solution containing grams of the solute in litres of the solution, at temperature the molecular man, of the solute can be calculated using equation (ii).


Q. The vapour pressure of pure liquid $A$ and $B$ are $70 \mathrm{mm}$ and $90 \mathrm{mm}$ Hg respectively at $25^{\circ} .$ The mole fraction of ^{ } $A^{\prime}$ in a solution of the two is $0.3 .$ Assuming that $A$ and $B$ form an ideal solution, calculate the partial pressure of each component in equilibrium with the solution.

[AI 1999 C]

Sol.


Q. Assuming complete dissociation, calculate the expected freezing point of a solution prepared by dissolving $6.00 \mathrm{g}$ Glaubers salt, $N a_{2} S O_{4}, 10 H_{2} O$ in. 100 kg of $H_{2} O\left[K_{f}=1.86\right.$ $\left.K \mathrm{kg} \mathrm{mol}^{-1}\right)$

[CBSE Board 1998]

Sol.


Q. $2 \mathrm{g}$ of $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}$ dissolved in $25 \mathrm{g}$ of benzene shows a depression in freezing point equal to $1.62 \mathrm{K} .$ Molar depression constant for benzene is $4.9 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} .$ What is the percentage association of acid if it exists as dimer in solution.

[CBSE 2004]

Sol. Here $W_{A}=25 g, W_{b}=2 g, \Delta T_{f}=1.62 K, K_{f}=4.9 K \mathrm{kg} \mathrm{mol}^{-1}$

Observed molecular mass of benzoic acid,

$M_{B}=\frac{1000 K_{f} \times W_{B}}{\Delta T_{f} \times W_{A}}=\frac{1000 \times 4.9 \times 2}{1.62 \times 25}=242$

Calculated molecular mass of benzoic acid

$=72+5+12+32+1=122$

Van’t Hoff factor

$i=\frac{\text { Calculated molecular mass }}{\text { Observed molecular mass }}=\frac{122}{242}=.504$

Let $\alpha$ be the degree of dissociation

\[ 2 C_{6} H_{5} C O O H \rightleftharpoons\left(C_{6} H_{5} C O O H\right)_{2} \]

Total number of moles after association

$=1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2}$

$\therefore \quad i=\frac{1-\frac{\alpha}{2}}{1}=.504$

or $\quad \alpha=(1-.504) \times 2=.496 \times 2=.992$

percentage of association $=99.2 \%$


Q. With the help of a simple diagram, describe Berkeley and Hartley’s method of determining the osmotic pressure of a dilute solution.

[MP 2002, 04, 06]

Sol. Experimental Measurement of Osmotic Pressure. A number of methods are available for measurement of osmotic pressure. The best out of these is Berkeley and Hartley’s Method.

This method is based upon applying pressure on the solution which is just sufficient to prevent the entry of the solvent into the solution through the semi-permeable membrane. The apparatus used by Berkeley and nartley is shown in Figure.

It consists of a porous tube (open at both ends) containing the semi-permeable membrance of copper ferrocyanide. The porous tube is fitted with a reservoir R on one side and a tube T on the other.

The porous tube is filled with the pure solvent so that the level in the tube T stands at the mark M. The porous tube is fitted into an outer vessel made of gun metal. This vessel has a wide tube at the top which is fitted with a frictionless piston and a pressure gauge as shown in figure. The solution under study is taken in the outer gun metal vessel. As a result of osmosis, the level in the tube tends to fall. The pressure applied on the solution by means of the piston which keeps the level in the tube at is a measure of the osmotic pressure and can be read directly on the pressure gauge.

This method is superior to the other methods because of the following reasons :

(a) In this method, the osmotic pressure is balanced by the external pressure so that there is no strain on the membrane.

(b) The concentration of the solution does not change because the entry of the solvent into the solution is prevented by the external pressure.

(c) The time taken for the measurement of osmotic pressure is much less in this method as compared to the other methods.


Q. Draw a suitable labelled diagram to express the relationships for ideal solution of A and B between vapour pressure and mole fractions of components at constant temperature.

[CBSE 2002; Delhi 2002]

Sol.

Relationships between V.P. and mole fraction for ideal solution


Q. $2 g$ of benzoic acid $\left(C_{6} H_{5} C O O H\right)$ dissolved in $25 g$ of benzene shows a depression in freezing point equal to $1.62 K .$ Molal depression constant for benzene is $4.9 K$ $k g \mathrm{mol}^{-1} .$ What is the percentage association of acid if it forms dimer in solution?

Sol. The given quantities are : $w_{2}=2 g ; K_{f}=4.9 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}$ ;

$w_{1}=25 g, \Delta T_{f}=1.62 K$

Substituting these values in equation

$M_{2}=\frac{k_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}}$

$M_{2}=\frac{4.9 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} \times 2 \mathrm{g} \times 1000 \mathrm{g} \mathrm{kg}^{-1}}{25 \mathrm{g} \times 1.62 \mathrm{K}}=241.98 \mathrm{gmol}^{-1}$

Thus, experimental molar mass of benzoic acid in benzene is $=241.98 \mathrm{g} \mathrm{mol}^{-1}$

Now consider the following equilibrium for the acid:

$2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}$ BHA $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)_{2}$

If $x$ represents the degree of association of the solute then we would have $(1-x)$ mol of benzoic acid left in

unassociated form and correspondingly $\frac{x}{2}$ as associated moles of benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium is:

$1-x+\frac{x}{2}=1-\frac{x}{2}$

Thus, total number of moles of particles at equilibrium equals Van’t Hoff factor $i$

But $i=\frac{\text { Normal molar mass }}{\text { Abnormal molar mass }}=\frac{122 g \mathrm{mol}^{-1}}{241.98 \mathrm{gmol}^{-1}}$

$\Rightarrow \frac{x}{2}=1-\frac{122}{241.98}=1-0.504=0.496$

$\Rightarrow x=2 \times 0.496=0.992$

Therefore, degree of association of benzoic acid in benzene is $99.2 \%$


Solid State | Question Bank for Class 12 Chemistry

Get Solid State important questions for Boards exams. Download or View the Important Question bank for Class 12 Chemistry. These important questions will play significant role in clearing concepts of Chemistry. This question bank is designed by NCERT keeping in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 12 chemistry chapter wise CBSE.

Click Here for Detailed Chapter-wise Notes of Chemistry for Class 12th, JEE & NEET. 

You can access free study material for all three subject’s Physics, Chemistry and Mathematics.

Click Here for Detailed Notes of any chapter. 

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We have transformed classroom in such a way that a student can study anytime anywhere. With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one.
Visit eSaral Website to download or view free study material for JEE & NEET.
Also get to know about the strategies to Crack Exam in limited time period.

Q. What is the C.N of $\mathrm{Ca}^{2+}$ and $\mathrm{F}^{-}$ in $\mathrm{CaF}_{2}$ crystal lattice?

Sol. C.N of $C a^{2+}$ ion $=8,$ C.N. of $F^{-}$ ion $=4$


Q. How many particles are present in a face centred unit cell?

Sol. Four particles are present in each unit cell.


Q. How many atoms are there in a unit cell of a metal crystallising in fcc structure ?

Sol. Four


Q. A compound $\mathrm{AB}_{2}$ possesses the $\mathrm{CaF}_{2}$ type crystal structure. Write the coordination numbers of $\mathrm{A}^{2+}$ and $\mathrm{B}^{-}$ ions in its crystals.

Sol. Coordination no. of $\mathrm{A}=8,$ Coordination no. of $\mathrm{B}=4$


Q. What other element may be added to silicon to make electrons available for conduction of an electric current ?

Sol. Phosphorus.


Q. Define coordination number of a metal ion in an ionic crystal.

Sol. In an ionic crystal, coordination number of a metal ion (+ve ion) is the number of negative ions surrounding the metal ion, i.e., which are present as its nearest neighbours.


Q. At what temperature ferrimagnetic substance changes to paramagnetic ?

Sol. 850 K


Q. What is the co-ordination number of $\mathrm{Na}^{+}$ and $\mathrm{Cl}$ – ions in $\mathrm{NaCl}$ structure?

Sol. Six.


Q. Sodium chloride has a face centred cubic crystal. What is the co-ordination number of sodium in sodium chloride ?

Sol. Six


Q. What are the co-ordination numbers of each of the ions present in the cubic close packed structure of $\mathrm{Na}_{2} \mathrm{O}$ at ordinary temperature and pressure?

Sol. $N a=4, O=8$


Q. A metallic element crystallises into a lattice containing a sequence of layers of ABABAB $\ldots . .$ Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space?

Sol. 26%


Q. AgI crystallises in cubic close packed ZnS structure. What fraction of tetrahedral sites are occupied by Ag” ions?

Sol. In the face-centred unit cell, there are eight tetrahedral voids. Of these, half are occupied by silver cations.


Q. The unit cell of a substance has cations $\mathrm{A}^{+}$ at the corners of the unit cell and the anions $\mathrm{B}^{-}$ in the centre. What is the simplest formula of the substance?

Sol. AB


Q. What is the maximum possible co-ordination number of an atom in an hcp crystal structure of an element?

Sol. 12


Q. In a crystal of zinc sulphide, zinc occupies tetrahedral voids. What is the co-ordination number of zinc ?

Sol. 4.


Q. Which out of $\mathrm{Cd} \mathrm{Cl}_{2}$ and $\mathrm{NaCl}$ will produce Schottky defect if added to $\mathrm{AgCl}$ crystal $?$

Sol. $\mathrm{CdCl}_{2}$ will produce Schottky defect.


Q. What type of stoichiometric defects are noticed in crystals ?

Sol. These are of two types i.e., Frenkel and Schottky defects.


Q. What type of semi-conductors is produced when silicon is doped with arsenic ?

Sol. It is of n-type because one electron of each silicon atom remains non-bonded.


Q. What is the effect of Frenkel defect on the electrical conductivity of crystalline solids ?

Sol. It increases because of the vacancies that are created.


Q. Why does Frenkel defect not change the density of AgCl crystals?

Sol. Due to Frenkel defect, as no ions are missing from the crystal as a whole, so there is no change in density.


Q. Mention one property which is caused due to the presence of Fcentre in a solid.

Sol. F-centre is reponsible for the colour and paramagnetic behaviour of the solid.


Q. What type of crystal defect is produced when sodium chloride is doped with MgCl_?

Sol. It is called impurity defect. A cation vacancy is produced. A substitutional solid solution is formed (because $2 \mathrm{Na}^{+}$ ion are replaced by one Mg $^{2+}$ ion in the lattice site).


Q. What causes the conduction of electricity by semiconductors ?

Sol. Electrons and holes produced by defects.


Q. A cubic solid is made up of two atoms A and B. Atoms A are present at the corners and B at the centre of the body. What is the formula of the unit cell ?

Sol. Contribution by the atoms A present at eight corners = 1

Contribution by the atom B present in the centre of the body = 1

Thus, the ratio of atoms of A : B = 1 : 1

Formula of unit cell = AB


Q. Define the term ‘amorphous’. Give a few examples of amorphous solids.

Sol. An amorphous solid is a substance whose constituent particles do not possess a regular orderly arrangement. Some examples of amorphous solids are glass, plastics, rubber, starch etc.


Q. Calculate the density of silver, which is known to crystallise in face-centred cubic form. distance nearest metal atoms is $287 \mathrm{pm}$ (Molar mass of $\left.\mathrm{Ag}=107.87 \mathrm{g} \mathrm{mol}^{-1} ; \mathrm{N}_{0}=6.022 \times 10^{23} \mathrm{mol}^{-1}\right)$

Sol. Density of unit cell $(\rho)=\frac{Z \times M}{N_{0} \times a^{3} \times 10^{-30}}$

According to available data,

Distance between nearest metal atom $(2 r)=287 \mathrm{pm}$

Edge length for foc crytal $(a)=\sqrt{2} \times 2 r=405.87 \mathrm{pm}$

No. of atoms per unit cell $(z)=4$

Atomic mass of silver $(M)=107.87 \mathrm{gmol}^{-1}$

Avogadro’s number $\left(N_{0}\right)=6.022 \times 10^{23} \mathrm{mol}^{-1}$

Density of silver

$=\frac{4 \times 107.87 g m o l^{-1}}{(405.87)^{3} \times 6.022 \times 10^{23} \mathrm{mol}^{-1} \times 10^{-30} \mathrm{m}^{3}}$

$=10.77 \mathrm{g} \mathrm{cm}^{-3}$


Q. What is the distance between $N a^{+}$ and $C I^{-}$ ions in $N a C l$ crystal if its density is $2.165 \mathrm{g} \mathrm{cm}^{-3} ?$ crystallises in fec lattice?

Sol. Step I. Calculation of Edge Length of unit cell

Let the edge length of unit cell $=a$

Volume of unit cell $=a^{3}$

Gram formula mass of $N a C l=23+35.5=585 g m o l^{1}$

No. of particles in foc type unit cell $(Z)=4$

Mass of unit cell $=\frac{Z \times \text { Gramformulamassof } N a C l}{\text { Avogadro’s Number }\left(N_{0}\right)}$

$=\frac{4 \times\left(58.5 g \mathrm{mol}^{-1}\right)}{\left(6.022 \times 10^{23} \mathrm{mol}^{-1}\right)}=3.886 \times 10^{-22} \mathrm{g}$

Density of unit cell $(\rho)=2.165 g \mathrm{cm}^{-3}$

Now, Density of unit cell $=\frac{\left(3.886 \times 10^{-22} g\right)}{\text { Volume of unit cell }\left(a^{3}\right)}$

volume of unit cell $(a)=\frac{\left(3.886 \times 10^{-22} g\right)}{\left(2.165 g \mathrm{cm}^{-3}\right)}$

$=1.795 \times 10^{-22} \mathrm{cm}^{3}=179.5 \times 10^{-24} \mathrm{cm}^{3}$

Edge length $(\mathrm{a})=\left(179.5 \times 10^{-24} \mathrm{cm}^{3}\right)^{1 / 3}=5.64 \times 10^{-8} \mathrm{cm}$

\[ =564 \times 10^{-10} \mathrm{cm}=564 \mathrm{pm} \]

Step II. Calculation of distance between $N a^{+}$ and $C l^{-}$ ions.

Edge length of $N a^{+} C l^{-}$ unit cell $=2\left(r_{N a^{+}}+r_{C l}^{-}\right)$

$2\left(r_{N a^{+}}+r_{C I^{-}}\right)=564 \mathrm{pm}$

or $\left(r_{N a^{+}}+r_{C l^{-}}\right)=\frac{564}{2}=282 \mathrm{pm}$

$\therefore$ Distance in and ions $=282$ pm


Q. Calculate the value of Avogadro’s number from the following data : Density of $\mathrm{NaCl}=2.165 \mathrm{g} \mathrm{cm}^{-3},$ Distance between $\mathrm{Na}^{+}$ and $\mathrm{Cl}^{-}$ ions in $\mathrm{NaCl}$ crystal $=281 \mathrm{pm}$

Sol. We know that $\rho=\frac{Z \times M}{a^{3} \times N_{0} \times 10^{30}}$ or $\quad N_{0}=\frac{Z \times M}{a^{3} \times \rho \times 10^{-30}}$

Edge length of

$N a C l(a)=2\left(r_{N a^{+}}+r_{C l}\right)=2 . \times 281=562 \mathrm{pm}$

No. of atoms per unit cell $(Z)=4$

$(\because N a C l$ hasfcc structure)

Density of $N a C l(\rho)=2.165 g \mathrm{cm}^{-3}$

Gram molecular mass of $N a C l(M)=23+35.5$

$=58.5 g \mathrm{mol}^{-1}$

$\therefore$ Avogadro’s Number

$\left(N_{0}\right)=\frac{4 \times\left(58.5 g m o l^{-1}\right)}{(562)^{3} \times\left(2.165 g \mathrm{cm}^{-3}\right) \times\left(10^{-30} \mathrm{cm}^{3}\right)}$

$=6.089 \times 10^{23} \mathrm{mol}^{-1}$


Q. How does the electrical conductivity of metallic conductors vary with temperature ?

Sol. Electrical conductivity decreases with increase in temperature because kernels start vibrating and create hinderance in the flow of electrons.


Q. What is the coordination number of

(i) Sodium in sodium oxide $\left(\mathrm{Na}_{2} \mathrm{O}\right)$

(ii) Oxide ion in sodium oxide (Na,O)?

(iii) Calcium in calcium fluoride $\left(\mathrm{CaF}_{2}\right) ?$

(iv) $\operatorname{zinc}$ in zinc blende (ZnS)?

Sol. (i) 4 (ii) 8 (iii) 8 (iv) 4 (v) 4.


Q. In CaF- $_{2}$ crystal $\mathrm{Ca}^{2+}$ ions are present in FCC arrangement. Calculate the number of $\mathrm{F}^{-}$ ions in the unit cell.

Sol. No. of $C a^{2+}$ ions per unit cell $=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4 .$ Hence,

no. of $F^{-}$ ions per unit cell $=2 \times 4=8$


Q. How many atoms can be assigned to its unit cell if an element forms (i) body-centred cubic cell (ii) a face-centred cubic cell.

Sol. (i) 2 (ii) 4.


Q. What are the types of lattice imperfections found in crystals ?

Sol. (a) Stoichiometric defects, viz., Schottky defect and Frenkel defect and (b) Non-stoichiometric defects, viz., metal excess, metal deficiency and impurity defects.


Q. What are non-soichiometric compounds ?

Sol. If the actual ratio of the cations and anions is not same as represented by the ideal chemical formula of the compound, it is called a non-stoichiometric compound.


Q. What is Frenkel defect ?

Sol. When some ions (usually cations) are missing from the lattice sites and they occupy the interstitial sites so that electrical neutrality as well as stoichiometry is maintained, it is called Frenkel defect


Q. Explain with the help of diagrams the structural differ-ences between three types of cubic crystals.

Sol. The crystals belonging to cubic system have three kinds of Bravais lattices. These are :

(i) Simple cubic lattice : There are points only at the corners of each unit.

(ii) Face-centred cubic lattice : There are points at the corners as well as at the centre of each of the six faces of the cube.

(iii) Body-centred cubic lattice : There are points at the corners as well as in the body centre of each cube.


Q. What is the difference between ‘Schottky defects’ and ‘Frenkel detects’ ?

Sol.


Q. Pure silicon is an insulator. Silicon doped with phosphorus is a semiconductor. Silicon doped with gallium is also a semiconductor. What is the difference between the two doped silicon semiconductors ?

Sol. When silicon is doped with phosphorus (a group 15 element) one electron of each phosphorus atom becomes free because it is not involved in any bonding with silicon atom. Thus, silicon containing phosphorus as the impurity is n-type semi-conductor. When silicon is doped with gallium (a group 13 element) only three valence electrons of silicon may be involved in the bonding with the three valence electrons of gallium. Since silicon atom has four valence electrons, this may result in creating vacancies or holes. These holes responsible for electrical conductivity and will result in p-type semi-conductors.


Q. Explain superconductivity.

Sol. There are certain solids called super conducting solids which do not offer any resistance to flow of electric current which means that they have zero resistivity. A few important examples are

$B a_{7} K_{3} B i O_{3}, L a_{1.8} S r_{.2} C u O_{4}, Y B a_{2} C u_{3} O_{7}$ etc.


Q. Sodium crystallises in a body-centred cubic unit cell (bcc) with edge length 4.29 $\hat{\mathrm{A}}$. What is the radius of the sodium atom? What is the length of the body-diagonal of the unit cell?

Sol. For a body centred cubic unit cell (bcc)


Q. Aluminium metal forms a cubic face centred close packed crystal structure. Its atomic radius is

\[ 125 \times 10^{-12} \mathrm{m} \]

(a) Calculate the length of the side of the unit cell.

(b) How many unit cells are there in $1.0 \mathrm{m}^{3}$ of aluminium?

Sol. (a) For a face centred cubic lattice (fcc).

Radius $(\mathrm{r})=\frac{a}{2 \sqrt{2}}$

$a=r \times 2 \sqrt{2}=125 \times 10^{-12} \times 2 \sqrt{2} m$

$=125 \times 2 \times 1.414 \times 10^{-12}=354 \times 10^{-12} \mathrm{m}$

(b) Volume of unit cell $(a)^{3}$

$=\left(354 \times 10^{-12}\right)^{3} m^{3}=4.436 \times 10^{-29} m^{3}$

No. of unit cells in $1.0 \mathrm{m}^{3}$ of $\mathrm{Al}$

$=\frac{\left(1.0 m^{3}\right)}{\left(4.436 \times 10^{-29} m^{3}\right)}=2.25 \times 10^{28}$


Q. The edge length of $\mathrm{NaCl}$ unit cell is 564 pm. What is the density of $\mathrm{NaCl}$ in $\mathrm{g} / \mathrm{cm}^{3} ?$

Sol. Density of unit cell $(\rho)=\frac{Z \times M}{a^{3} \times N_{0} \times 10^{-30}}$

According to available data,

Edge length $(a)=564$ pm

Molar mass of $N a C l(M)=23+35.5=58.5 \mathrm{g} \mathrm{mol}^{-1}$

No. of atoms per unit cell” $(Z)=4$

Avogadro’s number $\left(N_{0}\right)=6.022 \times 10^{23} \mathrm{mol}^{-1}$

Density of unit cell

$=\frac{4 \times\left(58.5 g m o l^{-1}\right)}{(564)^{3} \times\left(6.022 \times 10^{23} \mathrm{mol}^{-1}\right) \times\left(10^{-30} \mathrm{cm}^{3}\right)}$

$=2.16 g \mathrm{cm}^{-3}$


Q. Cesium chloride crystallises as a body centred cubic lattice and has a density of $4.0 \mathrm{gcm}^{-3} .$ Calculate the edge length of unit cell of cesium chloride crystal.

(Moalr mass of $\left.\mathrm{CsCl}=168.5 \mathrm{g} \mathrm{mol}^{-1} ; \mathrm{N}_{\mathrm{A}}=6.02 \times 1023 \mathrm{mol}^{-1}\right)$

Sol. Density of unit cell $(\rho)=\frac{Z \times M}{N_{0} \times a^{3} \times 10^{-30}}$

or $\quad a^{3}=\frac{Z \times M}{\rho \times N_{0} \times 10^{-30}}$

According to available data,

No. of atoms in the b.c.c. unit cell $(Z)=2$

Density of unit cell $(\rho)=4.0 \mathrm{g} \mathrm{cm}^{-3}$

Molar mass of $\operatorname{CsCl}(M)=168.5 \mathrm{g} \mathrm{mol}^{-1}$

Avogadro’s no. $\left(N_{0}\right)=6.02 \times 10^{23} \mathrm{mol}^{-1}$

By substituting the values,

$a^{3}=\frac{2 \times\left(168.5 g m o l^{-1}\right)}{\left(4.0 g c m^{-3}\right) \times\left(6.02 \times 10^{23} \mathrm{mol}^{-1}\right) \times 10^{-30}}$

$=1.399 \times 10^{8}(p m)^{3}$

Edge length $(a)=\left[1.399 \times 10^{8}(p m)^{3}\right]^{1 / 3}$

(.: Value of a is in pm)

$=1.12 \times 10^{2} \mathrm{pm}$


Q. How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell ? Explain

Sol. (a) The number of nearest neighbours in a packing is called coordination number. For example, in hexagonal close packing (HCP) in three dimensions each atom (or constituent particle) is touching 12 other atoms. So its coordination number is 12 .

(b) $\quad$ (i) In cubic close-packed structures each atom is in direct contact with 12 other atoms. Hence, its coordination number is 12

(ii) In body centred cubic structure coordination number is $8 .$


Q. ‘Stability of a crystal is reflected in the magnitude of its melting point.’ Comment. Collect melting point of solid water, ethyl alcohol, diethyl ether, and methane from a data book. What can you say about the intermolecular forces between these molecules.

Sol. A solid having stronger interparticle forces has higher melting point and is more stable. Thus, we can get an idea about stability of a crystal from its melting point. More stable crystals have higher melting points than less stable crystal. Thus, it is correct to say that stability of a crystal is reflected in magnitude of its melting point.

Methane < Diethyl ether < Ethyl alcohol < Water

Water having highest melting point has the strongest interparticle forces whereas methane having lowest melting point has the weakest intermolecular forces.

The intermolecular forces in these molecules are in the order :

Methane < Diethyl ether < Ethyl alcohol < Water


Q. How will you distinguish between the following pairs of terms

(a) Hexagonal close packing and cubic close packing

(b) Crystal lattice and unit cell

(c) Tetrahedral void and octahedral void.

Sol. (a) In hexagonal close packing every third layer is similar to the first layer. Thus there is ABAB ………..arrangement. In cubic close packing every fourth layer is similar to first thus, there is ABC ABC ….. arrangement.

(b) A crystal lattice may be defined as a regular three dimensional arrangement of identical points in space. While a unit cell may be defined as a three dimensional group of lattice points that generates the whole lattice by translation or stacking.

(c) The vacant space between four touching sphere is called tetrahedral void. Centres of these spheres occupy corners of regular tetrahedron.

The interstitial void formed by combination of two triangular voids of the first and second layer is called octahedral void because this is enclosed between six spheres, centres of which occupy corners of a regular octahedron.


Q. What is a semiconductor ? Describe the two main types of semiconductors and contrast their conduction mechanisms.

Sol. Semi-conductors. These are the solids whose conductivity lies in-between those of conductors and insulators. It normally ranges between $10^{-6}$ to $10^{4}$ ohm $^{-1} \mathrm{cm}^{-1} .$ They, infact allow only a portion of applied electric current to pass through them.

Conduction of electricity in semi-conductors

The electric conductivity of semi conductors is explained with the help of band theory. As we know, there is only a small energy gap between valence band (filled band) and conduction band (empty band) in semi-conductors. Therefore, some electrons can jump from valence band to vacant band and electric conductance is thus, possible. It can further increases with the rise in temperature since more electrons can jump to the conduction band.

Intrinsic semi-conductors. The extrinsic semi-conductors are formed when impurities of certain elements are added to the insulators.

This process known as doping makes available electrons or holes for conductivity and thus, leads to two types of semi-conduction. These are called n-type semi-conductors and p-type semi-conductors.

n-type semi-conductors. n-type semiconductors are formed when impurity atoms containing more valence electrons than the atoms of the parent in insulator are introduced in it. These are called electron rich impurities. For example, when traces of phosphorus (a group 15 element) is added to pure silicon (a group 14 element) on electron of each phosphorus atom (has five valence electrons) becomes free because it is not involved in any bonding with silicon atom (has four valence electrons). The unbonded electrons are free to carry the electric current. Thus, silicon containing phosphorus as the impurity is n-type semi-conductor.

(ii) p-type semi-conductors. These are formed when impurity atoms containing lesser number of valence electrons than the atoms of the parent insulator element added to it. These are called electron deficient impurities. For example, when traces of boron. (a group 13 element) are added to pure silicon (a group 14 element) only three valence electrons of silicon may be involved in the bonding with the three valence electrons of boron. Since silicon atom has four valence electrons, this may result in creating vacancies or holes. These holes are responsible for electrical conductivity and will result in p-type semi-conductors.


Q. Classify each of the following as being either a p-type or an n-type semiconductor :

(i) Ge doped with In

(ii) B doped with Si

Sol. (i) Ge is an element of group- 14 and has configuration $4 s^{2} 4 p^{2}$ It has been doped with In, a $13^{\text {th }}$ group element having $5 s^{2} 5 p^{1}$ configuration. All three valence electrons of In get bonded with three out of four electrons of Ge. Thus, the fourth bond of Ge contains only one electron and hence is an electron deficient bond or a hole. Conductivity is due to these holes or electron deficient bonds. Therefore, it is a p-type semiconductor.

(ii) Boron, $\mathrm{B}$ is an element of group- 13 and has $2 s^{2} 2 p^{1}$ configuration. It is doped with $S i,$ an element of group-

14 having $3 s^{2} 3 p^{2}$ configuration. All three electrons of boron get bonded with 3 out of 4 electrons of Si and $4^{\text {th }}$ electron of impurity atom (i.e., Si) is responsible for conductivity. Thus it is a n-type semiconductor.


Q. An element crystallises in a bcc structure. The edge length of its unit cell is 288 pm. If the density of the crystal is $7.3 \mathrm{g} \mathrm{cm}^{-3}$, what is the atomic mass of the element?

Sol. We know that $\rho=\frac{Z \times M}{a^{3} \times N_{0} \times 10^{-30}}$

or $M=\frac{\rho \times a^{3} \times N_{0} \times 10^{-30}}{Z}$

According to available data,

Edge length $(a)=288$ pm

No. of atoms per unit cell $(Z)=2$

$(\because \text { bcc structure })$

Density of the element $=7.3 \mathrm{g} \mathrm{cm}^{-3}$

Avogadro’s No. $\left(N_{0}\right)=6.022 \times 10^{23} \mathrm{mol}^{-1}$

$=52.51 g \mathrm{mol}^{-1}$


Q. The element chromium crystallises in a body centred cubic lattice whose density is $7.20 \mathrm{g} / \mathrm{cm}^{3} .$ The length of the edge of the unit cell is $288.4 \mathrm{pm} .$ Calculate Avogadro’s number (Atomic mass of $\mathrm{Cr}=52)$

Sol. We know that $\rho=\frac{Z \times M}{a^{3} \times N_{0} \times 10^{-30}}$

Or

$N_{0}=\frac{Z \times M}{a^{3} \times \rho \times 10^{-30}}$

Edge length of the unit cell $(\mathrm{a})=288.4 \mathrm{pm}$

No. of atoms per unit cell $(Z)=2$

Density of the unit cell $(\rho)=7.2 \mathrm{g} / \mathrm{cm}^{3}$

Atomic mass of the element $(\mathrm{M})=52 \mathrm{g} \mathrm{mol}^{-1}$

$N_{0}=\frac{2 \times\left(52 g m o l^{-1}\right)}{(288.4)^{3} \times\left(7.2 g \mathrm{cm}^{-3}\right) \times\left(10^{-30} \mathrm{cm}^{3}\right)}$

$=6.04 \times 10^{23} \mathrm{mol}^{-1}$


Q. The density of chromium is $7.2 \mathrm{g} \mathrm{cm}^{-3} .$ If the unit cell is a cubic with edge length of 289 pm, determine the type of the unit cell. (Atomic mass of $\mathrm{Cr}=52$ amu)

Sol. We know that $\rho=\frac{Z \times M}{a^{3} \times N_{0} \times 10^{-30}}$

or $\quad Z=\frac{\rho \times a^{3} \times N_{0} \times 10^{-30}}{M}$

Gram atomic mass of $C r(M)=52.0 \mathrm{gmol}^{-1}$

Edge length of unit cell $(\mathrm{a})=289 \mathrm{pm}$

Density of unit cell $(\rho)=7.2 \mathrm{g} \mathrm{cm}^{-3}$

Avogadro’s number $\left(N_{0}\right)=6.022 \times 10^{23} \mathrm{mol}^{-1}$

$\therefore Z=\frac{\left(7.2 g \mathrm{cm}^{-3}\right) \times\left(289^{3} \times\left(6.022 \times 10^{23} \mathrm{mol}^{1}\right) \times\left(10^{-30} \mathrm{cm}^{3}\right)\right.}{\left(520 \mathrm{gmol}^{1}\right)}=2$

since the unit cell has 2 atoms, it is body centre in nature.


Q. Calculate the efficiency of packing in case of a metal crystal for

(a) Simple cubic (b) Body centred cubic

(c) Face centred cubic

Sol. (a) Packing Efficiency of Simple Cubic Structure

In a simple cubic unit cell there is only one atom per unit cell.

Let a be the edge length of the unit cell and r be the radius of sphere.

Volume of the sphere $=\frac{4}{3} \pi r^{3}$

As the spheres at the corners are touching each other, the edge length a is equal to $2 \mathrm{r}$.

Volume of the cube $=a^{3}=(2 r)^{3}=8 r^{3}$

$\%$ of the space occupied by spheres

$=\frac{\text { Volume of sphere }}{\text { Volume of cube }} \times 100=\frac{\frac{4}{3} \pi r^{3}}{8 r^{3}} \times 100$

Thus, packing efficiency of simple cubic lattice is 52.4%

(b) Packing Efficiency of Body Centred Cubic (BCC) Structure

We know that in a body centred cubic unit cell there are two spheres per unit cell. Let a be the side of the unit cell and r be the radius of sphere.

volume of sphere $=\frac{4}{3} \pi r^{3}$

Volume occupied by two spheres $=2 \times \frac{4}{3} \pi r^{3}=\frac{8}{3} \pi r^{3}$

Face diagonal $A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{2 a^{2}}$

In right angled triangle ACD

Body diagonal $A D=\sqrt{A C^{2}+C D^{2}}=\sqrt{2 a^{2}+a^{2}}=\sqrt{3} a$

In a body centred cubic unit cell, the spheres at the corners are not touching each other but are in contact with the sphere at the centre. As a result the body diagonal AD is equal to the four times the radius of the sphere.

$\therefore A D=\sqrt{3} a=4 r$

$a=\frac{4}{\sqrt{3}} r$

Volume of unit cell $=a^{3}=\left(\frac{4}{\sqrt{3}} r\right)^{3}=\frac{64}{3 \sqrt{3}} r^{3}$

Percentage of space occupied by spheres

$=\frac{\text { Volume of spheres }}{\text { Volume of cube }} \times 100$

$=\frac{\frac{8}{3} \pi r^{3}}{\frac{64}{3 \sqrt{3}} r^{3}} \times 100=\frac{8}{3} \times \frac{22}{7} \times \frac{3 \sqrt{3}}{64} \times 100=68 \%$

Thus, packing efficiency of a bcc arrangement is $68 \% .$

(c) Face centred cubic

In a cubic close packing, the unit cell is face centred cube.

In a face centred cubic unit cell there are 4 spheres per unit cell.

Let r be the radius of sphere and a be the edge length of the cube.

Volume of sphere $=\frac{4}{3} \pi r^{3}$

Volume of four spheres $=4 \times \frac{4}{3} \pi r^{3}=\frac{16}{3} \pi r^{3}$

In a face centred cubic unit cell, the spheres at corners are in contact with sphere at the face centre but are not touching each other. Therefore, face diagonal $\mathrm{AC}$ is equal to four times the radius of sphere.

$A C=4 r$

But in right angled triangle ABC (figure)

$A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{a^{2}+a^{2}}=\sqrt{2} a$

$\therefore \sqrt{2} a=4 r$

$a=\frac{4 r}{\sqrt{2}}$

Volume of the unit cell $=a^{3}=\left(\frac{4 r}{\sqrt{3}}\right)^{3}=\frac{64}{2 \sqrt{2}} r^{3}$

Percentage of space occupied by spheres

$=\frac{\text { Volume of spheres }}{\text { Volume of cube }} \times 100=\frac{\frac{16}{3} \pi^{3}}{\frac{64}{2 \sqrt{2}} r^{3}} \times 100$

$=\frac{16}{3} \times \frac{22}{7} \times \frac{2 \sqrt{2}}{64} \times 100=74 \%$

Thus, packing efficiency of ccp arrangement is 74%. Similarly, packing efficiency of hcp is also 74%.


Q. Explain the following terms with suitable examples :

(i) Schottky defect

(ii) Frenkel defect

(iii) Interstitials

(iv) F-centres

Sol. (i) Schottky Defects. This type of defect is created when one positive ion and one negative ion are missing from their respective positions leaving behind a pair of holes (figure.). Schottky defect are more common in ionic compounds with high co-ordination number, and where the sizes of positive and negative ions are almost equal. For example, $N a C l, K C l, C s C l, A g B r$ and $K B r .$

$\mathrm{NaCl}$ has one defect in $10^{15}$ lattice sites at room temperature. The number of defects increases with increases in temperature.

The number of defects increases to one in $10^{6}$ sites at $775 \mathrm{K}$

and one in $10^{4}$ sites at $1075 \mathrm{K} .$ The presence of large number of Schottky defects in crystal results in significant decrease in its density. This defect is also known as vacancy defect.

(ii) Frenkel Defect. This type of defect is created when an ion leaves its correct lattice site and occupies an interstitial site (fig. Frenkel defects are common in ionic compound which have low co-ordination number and in which there is large difference in size between positive and negative ions. For example,

$Z n S, A g C l, A g B r$ and $A g I$

(iii) Interstitial Defects. This type of defect is caused due to the presence of ions in the normally vacant interstitial sites in the crystal. The ions occupying the interstitial sites are called interstitials. The formation of interstitial defects in determined by the size of the interstitial ion.

(iv) The “holes’ occupied by electrons are called F-centres (or colour centres) and are responsible for the colour of the compound and many other interesting properties. For example, the excess sodium in $\mathrm{NaCl}$ makes the crystal appear yellow, excess potassium in $K C l$ makes it violet and excess lithium in LiClmakes it pink. Greater the number of F-centres, greater is the intensity of colour. Solids containing F-centres are paramagnetic because the electrons occupying the “holes’ are unpaired.


Q. Aluminium crystallizes in a cubic close packed structure. Its metallic radius is $125 \mathrm{pm} .$

(a) What is the length of the side of the unit cell?

(b) How many unit cells are there in $1.00 \mathrm{cm}^{3}$ of aluminium?

Sol. (a) In cubic close packed structure, face diagonal of the unit cell is equal to four times the atomic radius

Face diagonal $=4 \times r=4 \times 125 \mathrm{pm}=500 \mathrm{pm}$

But face diagonal $=\sqrt{2} \times$ edge length

$\therefore$ Edge length $=\frac{\text { Face diagonal }}{\sqrt{2}}=\frac{500}{\sqrt{2}}=354 \mathrm{pm}$

(b) Volume of one unit cell

$=\mathrm{a}^{3}$

$=\left(3.54 \times 10^{-8} \mathrm{cm}\right)^{3}$

No. of unit cells $1.00 \mathrm{cm}^{3}$

$=\frac{1.00}{\left(3.54 \times 10^{-8}\right)^{3}}$

$=2.26 \times 10^{22}$


Chemical Kinetics Solved Problems
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Q. Give the IUPAC names of:

Sol. (a) Diacetone alcohol, (b) Crotonaldehy de.

(a) $\quad 4$ – Hy droxy – 4 -methylpentan-2-one

(b) $\quad$ But- 2 -en- 1 -al.


Q. Name one reagent used to distinguish acetaldehyde from acetone.

Sol. Tollen’s reagent or Fehling’s solution.


Q. Name one reagent used to convert toluene into benzaldehyde.

Sol. Chromyl chloride in $\mathrm{CS}_{2}\left(\mathrm{CrO}_{2} \mathrm{Cl}_{2} / \mathrm{CS}_{2}\right)$ or chromic trioxide in acetic anhydride $\left[\mathrm{CrO}_{3} /\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}\right]$ followed by acid or alkaline hydrolysis.


Q. What type of aldehydes and ketones undergo aldol condensation?

Sol. Aldehydes and ketones containing $\alpha$ -hydrogens.


Q. Which type of aldehydes undergo Cannizzaro reaction?

Sol. Aromatic and aliphatic aldehydes which do not contain $\alpha$ hydrogens


Q. Which alkene on reductive ozonolysis gives acetone as the only product?

Sol.


Q. Write the IUPAC name of

Sol. 2-Methylpentan-3-one


Q. To what oxidation state does ethanal reduce $C u(\mathrm{II}) ?$

Sol. +1 oxidation state


Q. Write the structure and give the IUPAC names of the following compounds. (a) Cinnamic acid (b) Crotonic acid (c) Oxalic acid.

Sol.


Q. Why does benzoic acid not undergo Friedel-Crafts reaction ?

Sol. (a) Due to deactivation of the benzene ring by electronwithdrawing effect of the $-C O O H$ group.

$(b)$ $A I C l_{3}$ gets bonded to the $-C O O H$ group.


Q. Why carboxylic acid do not form oximes ?

Sol. Due to resonance between lone pairs of electrons on the -atom of the group and the carboxyl carbon is less electrophilic than carbonyl carbon in aldedhydes and ketones. Therefore, nucleophilic addition of to the group of carboxylic acids does not occur and hence carboxylic acids do not form oximes.


Q. Give a suitable example of Hell-Volhard Zelinsky reaction.

Sol.


Q. Carbonyl compounds are more polar than alcohols although electronegativity difference between C and O atoms is less than O and H atoms. Explain.

Sol. In the carbonyl group $(>C=O),$ the $\pi$ -electron pair is loosely held and can be readily shifted to the oxygen atom. It is not the case with the alcoholic $(O-H)$ group. Therefore, carbonyl compounds are more polar and have higher dipole moment values $(2.3-2.8 D)$ than the alcohols $(1.6-1.8 D)$


Q. Give the IUPAC name of (a) Isobutyraldehy de (b) Acrolein (c) Valeraldehyde.

Sol.


Q. Which alkene upon reductive ozonolysis will give acetone as the product ?

Sol. 2, 3-dimethylbut-2-ene will form acetone by reductive ozonolysis


Q. Formaldehyde gives cannizzaro’s reaction while acetaldehyde does not. Why ?

Sol. Formaldehyde does not have an $\alpha$ -hydrogen. It therefore, takes part in cannizzaro’s reaction when heated with strong $N a O H$ to form a methyl alcohol and sodium formate mixture. Acetaldehyde fails to react since it has $\alpha$ -hydrogens present.


Q. What happens when ethylbenzene is heated with acidified $K_{2} C r_{2} O_{7} ?$

Sol. Benzoic acid is formed.


Q. Why is benzoic acid a stronger acid than acetic acid?

Sol. The $K_{a}$ value of benzoic acid $\left(6.3 \times 10^{-5}\right)$ is more than that of

acetic acid $\left(1.75 \times 10^{-5}\right) .$ Actually, $C_{6} H_{5}$ group with $-I$

effect in facilitates the release of $H^{+}$ from benzoic acid while

$\mathrm{CH}_{3}$ group with $+I$ effect tends to retard it.


Q. Which out of acetic and peracetic acid is a stronger acid and why ?

Sol. Acetic acid $\left(p K_{a}=4.8\right)$ is stronger than peracetic acid

$\left(p K_{a}=8.2\right) .$ It is because after losing $H^{+}$ ion acetic acid molecule gets stabilised by resonance while the peracetate ion does not.


Q. Carboxylic acids do not give the characteristic reactions of carbonyl group.

Sol. Due to the presence of lone pairs of electrons on the oxygen atom of the OH group, the carboxylic acids are stabilized by resonance.

As a result, the double bond character of the bond in carboxylic acids is greatly reduced as compared to that in aldehydes and ketones. In other words, the carbonyl group in carboxylic acids is not a true carbonyl group as in aldehydes and ketones and hence does not give some of the characteristic reactions of the carbonyl group. For example, unlike aldehydes and ketones, carboxylic acids do not form oximes, hydrazones, semicarbazones etc.


Q. Write reactions and conditions for following conversions :

(a) Phenol to salicylic acid

(b) $\quad 2$ -propanone into 2 -methyl- 2 -propanol.

Sol. (a) By Kolbe’s reaction. Sodium salt of phenol reacts with

$\mathrm{CO}_{2}$ under high pressure of $4-7$ atm, at about $400 \mathrm{K}$ to form salicylic acid.


Q. Write chemical tests to distinguish between formic acid and acetic acid.

Sol.


Q. Although both $>\mathrm{C}=\mathrm{C}<$ and $>\mathrm{C}=\mathrm{O}$ have a double bond, they exhibit different type of addition reactions. Explain.

Sol. The $>\mathrm{C}=\mathrm{C}<$ undergoes electrophilic addition reactions while $>\mathrm{C}=\mathrm{O}$ shows nucleophilic addition reactions. This difference in behaviour is due to the different shapes of their $\pi$ -electron clouds. The electron cloud of bond due to similar electronegativities of the two carbon atoms is almost symmetrical and surrounds both the carbon atoms equally. Consequently, if a reagent is to attack one of the carbon atoms of the bond, it has to pass through the electron cloud. since the -electron cloud consists of loosely held electrons, therefore, it can readily allow electrophiles to react. As a result, the typical reactions of are electrophilc addition reactions. In contrast, the electron cloud of the $\succ \mathrm{C}=0$ is unsymmetrical i.e., shifts towards oxygen due to greater electronegativity of O than C. As a result, the C-atom of the bond acquires a partial charge and hence is readily attacked by nucleophiles. Thus, the typical reactions of are nucleophilic addition reactions.


Q. What is the function of Roschelle salt in Fehling’s solution?

Sol. In alkaline medium, $C u^{2+}$ ions get precipitated as $C u(O H)_{2}$ To keep ions in solution in alkaline medium, Rochelle salt is added. The insoluble first formed goes into solution due to the formation of a soluble complex between ions and tartrate ion as shown below:


Q. Suggest a suitable oxidising agent for the conversion $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCOCH}_{3} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCO}_{2} \mathrm{H}$

Sol. Alkaline $K M n O_{4},$ acidified $K_{2} C r_{2} O_{7}$ or $H N O_{3}$ cannot be used since all of these will cleave the molecule at the site of double bond giving a mixture of ketone/acids. The most suitable reagent for this oxidation is $\mathrm{NaOI} \quad\left(I_{2} / N a O H\right)$ since methyl ketones on treatment with NaOI undergo iodoform reaction to give iodoform along with the salt of a carboxylic acid having one carbon atom less than the starting methyl ketone.


Q. Fluorine is more electronegative than chlorine even than $p-$ fluorobenzoic acid a weaker acid than $p$ -chlorobenzoic acid. Explain

Sol. since halogens are more electronegative than carbon and also possess lone pairs of electrons, therefore, they exert both $-I$ and $+R$ -effects. Now in $F,$ the lone pairs of electrons are present in $2 p$ -orbitals but in $C l,$ they are present in $3 p-$ orbitals.

since $2 p$ -orbitals of $F$ and $C$ are of almost equal size, therefore, the $+R$ -effect is more pronounced in $p-$ fluorobenzoic acid than in -chlorobenzoic acid

Thus, in $p$ -fluorobenzoic acid, $+R$ -effect outweighs the $-I-$ effect but is p-chlorobenzoic acid, it is the I-effect which outweighs the $+$ R-effect. Consequently acid is a weaker acid than p-chlorobenzoic acid.


Q. Give an example of a reaction where a Grignard reagent acts as a reducing agent.

Sol. If the Grignard reagent or the ketone contains branching at the $\alpha$ -carbon, sometimes the usual addition reaction does not occur due to steric hindrance. Instead reduction occur in which a hydride ion is transferred from the Grignard reagent to the ketone via a six-membered cyclic transition state in a manner similar to

M.P.V. reduction. For example, when isopropylmagnesium bromide is added to di-isopropyl ketone, the expected $3^{\circ}$ alcohol (i.e., tri-isopropylcarbinol) is not formed; instead the secondary alcohol, di-isopropylcarbinol is obtained by reduction of the keto group.


Q. Boiling points of carbonyl compounds lie between the parent alkanes and corresponding alcohols. Justify.

Sol. Alkanes are non-polar in nature and the attractive forces are weak vander Waal’s forces. In alcohols, intermolecular hydrogen bonding is present in the molecules. In the carbonyl compounds, only dipolar forces exist in the molecules. It is, therefore, quite obvious that the attractive forces in carbonyl compounds are more than in alkanes and less in comparison to alcohols. This justifies the trend in boiling point values in the members of these families.


Q. Halogen acids readily combine with alkenes to form addition products but fail to react with carbonyl compounds. Discuss.

Sol. Halogen acids (HX) react with alkenes to form haloalkanes but they fail to react with the carbonyl compounds. In fact, the attack does take place but the product is unstable and decomposes to form carbonyl compound along with halogen acid. Thus, the reaction is reversible.


Q. Out of benzaldehyde and propionaldehye which is more reactive towards nucleophilic addition ?

Sol. Propionaldehyde is more reactive than benzaldehyde towards nucleophilic addition. Actually, in benzaldehyde the aldehydic group has (mesomeric) effect. As a result, the electron density on the carbonyl carbon atom tends to increase and the nucleophile attack is more difficult as compared to propanal where alkyl group is attached to the aldehydic group.


Q. Chloral hydrate is a gem-diol but still stable. How will you account for it?

Sol. In chloral hydrate, the presence of three $C l$ atoms with $-I$ effect increases the magnitude of the positive charge on the carbonyl carbon atom. As a result, even a weak nucleophile like $H_{2} O$ attacks to form a hydrate. Moreover, there is weak intermolecular hydrogen bonding in the chloral hydrate which tends to increase its stability.


Q. Why does not formaldehyde take part in aldol condensation?

Sol. Aldol condensation involves the nucleophile attack of carbanion generated by one molecule of a particular carbonyl compound on the other molecule. For this, the carbonyl compound must have atleast one $\alpha$ -hydrogen present. since formaldehyde $(H C H O)$ has no such hydrogen present, it fails to take part in aldol condensation alongwith another carbonyl compound with atleast one -hydrogen atom e.g., formaldehyde and acetaldehyde.


Q. Why is it necessary to control the pH during the reaction of aldehydes and ketones with ammonia derivatives?

Sol. The addition of ammonia derivative $\left(N H_{2}-G\right)$ to aldehydes and ketones is done in weakly acidic medium (pH about 3.5). In case the medium is strongly acidic (pH close to 1 ), then the ammonia derivative will be also protonated and will not be able to act as a nucleophile.


Q. Benzophenone does not react with $\mathrm{NaHSO}_{3} .$ Explain.

Sol. The addition of $N a H S O_{3}$ on benzophenone $\left[\left(C_{6} H_{5}\right)_{2} C=O\right]$ involves the nucleophile attack of bisulphite $\left(H S O_{3}^{-}\right)$ ion. The phenyl groups act as hindrance to the attacking nucleophile. Therefore, benzophenone does not react with $\mathrm{NaHSO}_{3}$.


Q. Arrange the following in decreasing ease of hydration:

(a) $\mathrm{CH}_{3} \mathrm{CHO},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}, \mathrm{HCHO}, \mathrm{CH}_{2} \mathrm{ClCOCH}_{3}, \mathrm{CH}_{2} \mathrm{ClCHO}$

Sol. Aldehydes are more easily hydrated than ketones. Further, the presence of electron withdrawing groups increases the ease of hydration of a carbonyl compound. In the light of this, the correct order is :

(a) $\mathrm{HCHO}>\mathrm{CH}_{2} \mathrm{ClCHO}>\mathrm{CH}_{3} \mathrm{CHO}>\mathrm{CH}_{2} \mathrm{ClCOCH}_{3}>\mathrm{CH}_{3} \mathrm{COCH}_{3}$


Q. Write the mechanism for the following conversion.

Sol.

Phenylacetylene


Q. Carboxylic acids do not give the characteristic reactions of carbonyl group. Explain.

Sol. The carbonyl group in carboxylic acid is not a free group as in aldehydes and ketones. It is involved in resonance.

Therefore, carboxylic acids fail to give the characteristic reactions of the carbonyl groups as given by aldehydes and ketones.


Q. Formic acid reduces Tollen’s reagent while other carboxylic acids do not. Justify.

Sol. $\mathrm{HCOOH}$ has an aldehydic (CHO) group in addition to

carboxyl group $(\mathrm{COOH}) .$ Therefore, it is expected to behave as reducing agent also and reduces Tollens to form a shining mirror.


Q. 2, 4, 6-trimethylbenzoic acid is quite difficult to esterify. Assign reason.

Sol.

In the trimethylbenzoic acid there are three electron releasing $\mathrm{CH}_{3}$ groups present. They tend to increase the electron density on the oxygen atom of $O-H$ bond in carboxylic acid. As a result, the acidic strength is reduced considerably and it is not so easy to esterify $2,4,6$ -trimethyl benzoic acid.


Q. m-Hydroxybenzoic acid is a stronger acid than benzoic acid while p-hydroxybenzoic acid is weaker. Explain.

Sol. The hydroxyl $(O H)$ group has both $-I$ effect and $+R$ effect. Whereas the former helps in the release of $H^{+}$ from the carboxylic group, the latter tends to oppose the same because of resonance. At the para position, the $+\mathrm{R}$ effect dominates the effect. Therefore, p-hydroxy benzoic acid is a weaker acid than benzoic acid. But in case of m-hydroxybenzoic acid, the $+\mathrm{R}$ effect does not operate to the same extent as in para isomer. Thus, the acidic weakening effect is smaller and meta hydroxy benzoic acid is a stronger acid than benzoic acid.


Q. Carbon-oxygen bond lengths in formic acid are different but are the same in sodium formate. Justify.

Sol. In sodium formate, the contributing structures for the anion are equivalent while these are not the same in formic acid.

Thus, carbon-oxygen bond length in formate ion is the same for both the bonds while these are different in formic acid.


Q. Phenate ion has more number of contributing structures than benzoate ion; but still benzoic acid is a stronger acid. Explain.

Sol. In phenate ion, the negative charge available is dispersed only one electrongative oxygen atom while there are two oxygen atoms in benzoate ion to disperse the negative charge.

This means that benzoate ion is more stable than phenate ion and this makes benzoic acid stronger acid than phenol although the contributing structures for phenate ion (five) are more than for benzoate ion (two).


Q. Tertiary butyl benzene does not give benzoic acid when oxidised with $K M n O_{4} .$ Why?

Sol. In order that an alkyl group attached to benzene ring may be oxidised to carboxyl group, the presence of atleast one

$\alpha$ -hydrogen is necessary. For example, toluene is oxidised to benzoic acid. But tertiary butyl benzene has no -hydrogen atom. It is therefore, not oxidised to benzoic acid.


Q. $\mathrm{CH}_{3} \mathrm{COO}^{-}$ ion is more stable than $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}$ ion. Assign reason.

Sol. In acetate ion, the negative charge gets dispersed due to the electron withdrawing nature of carbonyl group. On the other

hand, in ethoxide ion, the $C_{2} H_{5}$ group with $+I$ effect tends to increase the electron density on the ion and destabilise it.


Q. Which out of each pair is expected to be a stronger acid ?

(a) $\quad \mathrm{CH}_{3} \mathrm{COOH}$ or $\mathrm{HCOOH}$

(b) $\quad \mathrm{CH}_{3} \mathrm{COOH}$ or $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$

(c) $\quad C_{6} H_{5} \mathrm{COOH}$ or $\mathrm{HCOOH}$

(d) $\quad C H_{2}(C l) C O O H$ or $C H_{2}(B r) C O O H$

Sol. (a) $\mathrm{HCOOH}$

(b) $\mathrm{CH}_{3} \mathrm{COOH}$

(c) HCOOH

(d) $\mathrm{CH}_{2}(\mathrm{Cl}) \mathrm{COOH}$


Q. How will you prepare 2-methylbutanoic acid from butan-2-ol ?

Sol.


Q. Complete following :

Sol.


Q. How will you synthesise the following by using a Grignard reagent?

(a) $2,2$ -Dimethylpentanoic acid (b) But-3-enoic acid

Sol.


Q. Why is acetyl chloride a better acetylating agent than acetic acid ?

Sol. The acetylation is carried by acetyl carbocation $\left(\mathrm{CH}_{3} \mathrm{CO}^{+}\right)$ The cation is formed by acetyl chloride and not by acetic acid.

Therefore, acetyl chloride is a better acetylating agent than acetic acids.


Q. Draw the structural formula of hex-2-en-4-ynoic acid

Describe the following giving a chemical equation for each

(i) Transesterification,

(ii) Hofmann bromamide reaction

Sol. (i) Transesterification : This reaction involves the nucleophilic acyl substitution of the alkoxy group of an ester with the alkoxy group of an alcohol in the presence of an acid or alkali

$R C O O R^{\prime}+R^{\prime \prime} O H \longrightarrow R C O O R^{\prime \prime}+R^{\prime} O H$

(ii) Hofmann bromamide reaction: This reaction involves the conversion of an amide into primary amine with one carbon atom less than the parent amide, by reaction with bromine and alkali



Q. Predict the products of the following reactions :

Sol.

(c) $\quad \mathrm{NH}_{2}$ attached to $\mathrm{NH}$ is more nucleophile than $\mathrm{NH}_{2}$ attached to $\mathrm{C}=$ O group, therefore, reaction occurs through $\mathrm{NH}_{2}$ attached to NH to give the corresponding semicarbozone, i.e.,


Q. Which acid of each pair shown here would you expect to be stronger ?

Sol.


Q. Although phenoxide ion has more number of resonating structure than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

Sol. In the resonating structure of phenoxide ion the negative charge is dispersed on only one oxygen atom whereas in the carboxylate ion the charge gets dispersed on two oxygen atoms. Thus, the dispersal of charge is more in the case of carboxylate ion and hence the resonance stabilisation is more. This is why the carboxylic acids are more stronger than phenol (s).


Q. Give reasons for the following :

(i) Aldehydes are more reactive than ketones in nucleophilic reactions.

(ii) Most aromatic acids are solids but the acids of acetic acid group are mostly liquids.

Sol. (i) The reactivities of aldehydes is more than that of ketones due to the following reasons:

(a) Inductive effect: The ease with which a nucleophile attacks the carbonyl group depends upon the electron deficiency i.e., the magnitude of the positive charge on carbonyl carbon. since an alkyl group has electron donating inductive effect ( $+$ I effect), therefore greater the number of alkyl groups attached to carbonyl carbon greater is the electron density on the carbonyl carbon and hence lower is its reactivity. Aldehydes are more polar than ketones.

In ketone we have two alkyl groups whereas in aldehydes only one alkyl group is attached to carbonyl carbon. Since there is lesser number of alkyl groups in case of aldehydes so they undergo nucleophillic addition more readily than ketones.

(b) Steric effect : The presence of more alkyl groups in case of ketones hinders the attack of nucleophile on the carbonyl group and so the ketones are less reactive than aldehydes.

(ii) Aromatic acids contain benzene ring and their molecular masses are much higher than aliphatic acids of acetic acid group. Aromatic, acids also involve stronger inter molecular attractions, thus they are solids.


Q. Write IUPAC names of following compounds :

Sol.

$2-(2-$ bromophenyl) ethanal. It can also be named as

$2-(\text { o-bromophenyl })$ ethanal.

(b) 5 -chloro-3-ethyl-2-pentanone.

(c) $3,5$ -dimethyl phenyl ethanoate


Q. Give reasons for the following :

(a) Acetic acid is a weaker acid than chloroacetic acid.

(b) During preparation of ammonia derivatives of aldehydes and ketones, $p H$ of medium is controlled.

Sol. (a) The acidic nature of a carboxylic acid is due to its ability to release the $H^{+}$ (proton). Any factor that stabilise the carboxylate ion would facilitate the release of proton and thus increase the acid strength. The electron withdrawing group (attached to $-\mathrm{COOH}$ gp) increases the acid strength. Due to electron withdrawing effect $(-I \text { effect })$ of chlorine atom, the negative charge of carboxylate ion formed gets dispersed. As such this ion gets stabilised and is, therefore, more easily formed.

The acetate ion obtained from acetic acid gets detabilised due to electron releasing effect of methyl group and is not formed

easily. Consequently dissociation of $\mathrm{CH}_{3} \mathrm{COOH}$ takes place to lesser extent.

(b) Formation of ammonia derivatives (oximes, hydrazone, semi-carbazone etc.) proceeds via the attack of carbonyl carbon with proton to form the conjugate acid.

Therefore, presence of an acid is a must for preparing these

derivatives $(p H<7)$ However, in strongly acid medium, the proton attacks the unshared pair of electrons on nitrogen to form the species

$R N^{+} H_{3}$ which cannot attack the carbonyl carbon.

$\stackrel{+}{H}+: N H_{2} R \longrightarrow N H_{3} R$

In basic medium, there is no protonation of carbonyl group and

hence no reaction.

$>C=O \frac{\text { basic medium }}{\text { medium }} \rightarrow$ No protonation Therefore, preparation of ammonia derivatives requires slightly

acidic medium $(p H=3.5)$ and its careful control is essential.


Q. How does $N H_{3}$ react with

(a) Formaldehyde,

(b) Acetaldehyde

(c) Semicarbazide

Sol. (a) $N H_{3}$ reacts with formaldehyde to form hexamethylene tetramine (also called urotropine).


Q. (a) What is ammonolysis of esters?

(b) How will you convert:

(i) Acetaldehyde to acetamide

(ii) Methanol to acetic acid?

Sol. (a) Esters react with ammonia to form amides. The reaction is called ammonolysis of esters.


Q. Addition of Grignard reagents to dry ice followed by hydrolysis gives carboxylic acids while that of organolithium compounds under similar conditions gives ketones. Explain

Sol. since electronegativity of $L i(\mathrm{E} . \mathrm{N} .=1.0)$ is lower than that of $M g(\mathrm{E.N.}=1.2),$ therefore, organolithium compounds are more nucleophilite than Grignard reagents. As a result, organolithium compounds not only add to the more reactive $\mathrm{CO}_{2}$ but also to the less reactive resonance stabilized lithium salt of carboxylic acid thus formed to yield ketones.

Grignard reagents, on the other hand, being less nucleophilic

add only to the more reactive $\mathrm{CO}_{2}$ but not to the less reactive resonance stabilized magnesium salt of the carboxylic acid from which carbxylic acid can be generated by hydrolysis with mineral acids.


Q. How will you prepare methanol from formaldehyde without using a reducing agent.

Sol. Cannizzaro reaction is a disproportionation reaction in which one molecule of an aldehyde is reduced while the other is oxidised.

Here a conc. solution of $N a O H$ is used which is not a reducing agent.


Q. You are provided with following four reagents. $I_{2} / N a O H$ $\mathrm{NaHSO}_{3},$ LiAlH_{4} Schiff’s reagent. Write which two reagents can be used to distinguish between the compounds in each of the following pairs:

(i) $\quad \mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{CH}_{3} \mathrm{COCH}_{3}$

(ii) $\quad \mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}$

(iii) $\quad C_{6} H_{5} C O C H_{3}$ and $C_{6} H_{5} C O C_{6} H_{5}$

Sol. (i) $\quad \mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ can be distinguished by :

LiAIH$_{4}$ and Schiff’s reagent.

(ii) $\mathrm{CH}_{3} \mathrm{CHO}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}$ can be distinguished by $\mathrm{I}_{2} / \mathrm{NaOH}$ and Schiff’s reagent (Benzaldehy de restores pink colour to Schiff’s reagent very slowly).

(iii) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}_{3}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COC}_{6} \mathrm{H}_{5}$ can be distinguished by

$\mathrm{NaHSO}_{3}$ and $\mathrm{I}_{2} / \mathrm{NaOH}$


Q. An organic compound [A] molecular formula $C_{4} H_{8} O$ when reduced with $\mathrm{NaBH}_{4}$ gives compound [\textrm{B} ] \text { which reacts with } HBr to form compound [C][\mathrm{ optically active } ] \text { Identify } \mathrm { A } , \mathrm { B } , \mathrm { C }

and write the two enantiomers of compound $C$.

Sol. The data suggests that the given compound [A] is an aliphatic

ketone butanone $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3}\right) .$ The reactions involved are listed.


Q. Two moles of an ester $[\mathrm{A}]$ are condensed in the presence of sodium ethoxide to give a $\beta$ -ketoester $[\mathrm{B}]$ and ethanol. On heating in an acidic solution $[\mathrm{B}]$ gives ethanol and a $\beta$ -ketoacid $[\mathrm{C}] .$ On decarboxylation, [C] gives pentan- 3 -one. Identify $[\mathrm{A}],[\mathrm{B}]$ and $[\mathrm{C}]$ with proper reasoning.

Sol. (a) According to available data [C] is a $\beta$ -keto acid which upon decarboxylation gives pentane- 3 -one. The structure of is :

$[\mathrm{C}]$

(b) The compound $[\mathrm{C}]$ along with ethanol has been obtained from compound [B] upon acid hydrolysis. since $[\mathrm{B}]$ is a $\beta$ ketoester, its structure may be given as follows:

(c) since $[\mathrm{B}]$ has been formed by the condensation of two moles of $[\mathrm{A}]$ in the presence of sodium ethoxide, the ester $[\mathrm{A}]$ is ethyl propanoate and it participates in Claisen Condensation as follows:


Q. Give chemical tests to distinguish between the following pairs of compounds :

(a) Propanal and Propanone

(b) Methyl acetate and ethyl acetate

(c) Benzaldehyde and Benzoic acid

Sol. (a) When propanal is added to tollen’s reagent (an ammoniacal solution of silver nitrate), silver oxide is reduced to metallic silver which deposits as a mirre

(b) When ethyl acetate is treated with aqueous $N a O H$ Sodium ethanoate and ethanol is formed ethanol on treatment with $I_{2}$ and $N a O H$ forms yellow ppt of iodoform.

When methyl acetate is treated with aqueous NaOH sodium ethanoate and Methanol is formed. Methanol does not respond to iodoform reaction

(c) When benzoic acid is warmed with ethyl alcohol and a little concentrated $\mathrm{H}_{2} \mathrm{SO}_{4},$ a fragrant odour of ethyl benzoate is obtained.

Benzaldehyde does not respond to this test.


Q. Write structural formulae and names of the four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde served as nucleophile and which as electrophile.

Sol.


Q. An organic compound (A) (molecular formula $\left.C_{8} H_{16} O_{2}\right)$ was hydrolysed with dilute sulphuric acid to give a carboxylic acid

(B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-rene. Write equations for the reactions involved.

Sol.

Dehydration of C gives but-1-ene i.e., a product with four carbon atoms, therefore the other product B is also with four carbon atoms and has a terminal COOH group.

Accordingly, A is $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$

(butyl butanoate)

$\mathrm{B}$ is $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ and

$\mathrm{C}$ is $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$


Q. Arrange the following compounds in increasing order of their property as indicated:

(a) Acetaldehyde, Acetone, Di-tert-butyl ketone, methyl tertbutyl ketone (reactivity towards $H C N)$

(b) $C H_{3} C H_{2} C H(B r) \operatorname{COOH}, C H_{3} C H(B r) C H_{2} C O O H$

$\left(\mathrm{CH}_{3}\right) \mathrm{CHCOOH}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ (acid strength)

(c) Benzoic acid, 4-Nitrobenzoic acid, 3, 4- Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

Sol. (a) The reactivity of aldehydes and ketones toward nudeophillic addition depends upon (i) $+$ I effect (ii) steric hinderance. Hence the order is Di-tert-butyl ketone $<$ methyl-tert-butyl ketone < Acetone $<$ Acetaldehy de.

(b) The acidic strength depends upon (i) nature of $+$ I effect

(ii) nature of atom/group attached (iii) position of substituent on the chain. Hence, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOOH}<\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$

$<\mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{2} \mathrm{COOH}<\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{Br}) \mathrm{COOH}$

(c) $\quad 4$ -Methoxy benzoic acid $<$ Benzoic acid $<4$ Nitrobenzoic acid $<3,4$ -Dinitrobenzoic acid


Q. Give plausibe explanation for each of the following

(a) Cyclohexanone forms cyanohydrin in good yield but 2 ,

$2,6$ -trimethylcyclohexanone does not.

(b) There are two $-N H_{2}$ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.

(c) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the

water or the ester should be removed as fast as it is formed.

Sol. (a) The carbonyl group in cyclohexanone is highly

polarised and nuleophilic addition of $H^{\delta+}-C N^{\delta-}$ at carboxyl

group $\left(>C^{\delta+}=O^{\delta-}\right)$ takes place easily. centre of nucleophilic attack by $C N^{-}$

The presence of three methyl group (which are electron repelling) reduces the polarity of $>C=O$ group on one hand and offer a steric hindrance to nucleophilic attack of $C N^{-}$ at group. Therefore, trimethyl cyclohexanone does not give good yield i.e., it gives very poor yield.

(b) The study of semicarbazide (NH3 derivative of urea,

reveal that two $\mathrm{N}-\mathrm{H}$ bonds of $\mathrm{NH}_{2}$ group away from group in semicarbazide are comparatively weaker than two $\mathrm{N}-\mathrm{H}$ bond of the other $-N H_{2}$ group closer to $>C=O$ group. Thus it

produces $H^{+}$ easily for protonation of group of aldehyde or ketone.

The nature of reaction, esterification of carboxylic acid, is reversible as ester formed react with water and produces reactants and thus an equilibrium is set up between reactants and products. The removal of ester or water shifts the equilibrium to right according to Le-Chatelier’s principle and more ester is formed.


Q. Predict the organic product of the following reactions

Sol.


Q. (a) How will you convert

(i) Benzoyl chloride to benzaldehyde

(ii) Propanone to 2 -propanol

(iii) Benzoic acid to $m$ -nitrobenzoic acid

(Write the reaction and state the reaction conditions in each case.)

(b) Write the names and structures of the products formed in the following reactions

(i) Reaction of semicarbazide $\left(N H_{2} C O N H N H_{2}\right)$ with formaldehyde.

(ii) Oxidation of ethyl benzene with alkaline $K M n O_{4}$.

Sol. (a) (i) Benzoyl chloride is subjected to Rosenmund reduction.


Q. Account for the following

(i) Chloroacetic acid has lower $p K a$ value than acetic acid

(ii) Electrophillic substitution in benzoic acid takes place at meta position.

(iii) Carboxylic acid have higher boiling points than alcohols of comparable molecular masses.

Sol. (i) Chloroacetic acid is a stronger acid than acetic acid and has higher value of dissociation constant $K a$ than that of acetic acid. We know that $p K a=-\log K a,$ it means that chloroacetic acid having higher value $K a$ will have lower value of $p K a .$ Chloroacetic acid is stronger than acetic acid because $C l$ group is electron withdrawing and chloroacetate ion is more stabilised than acetate ion (as methyl group is electron releasing).

(ii) Carboxyl group (-COOH) is electron withdrawing i.e. deactivating the benzene ring and thus electron density becomes very less at ortho and para position in comparison to meta position. Electrophiles (+vely charged species) find it easier to attack at meta position as there is higher electron density thus $-C O O H$ group is meta directing.

As there is positive charge on ortho and para position, electron density is higher at meta position and hence electrophilic substitution takes place at meta position. (Similarly nitro group is meta directing)

(iii) Carboxylic group (-COOH) in acids is highly polar and carboxylic acid, generally, exists as dimers containing two hydrogen bonds each as shown below.

These hydrogen bonds in carboxylic acids are stronger than those in alcohols. It is due to following two factors

(a) The $O-H$ bond of the carboxylic acids are more strongly polarised due to adjacent electron attracting $>C=$ O groups.

(b) The oxygen atom of the group $C=O$ in carboxylic acid is more negative as compared to the oxygen atom of the alcohol.

Thus carboxylic acids posses higher boiling points than corresponding alcohols of similar molecular masses.


Q. Write chemical reactions for preparation of acetone from the following

(i) $\quad$ A secondary alcohol

(ii) Carboxylic acid

(iii) Acid chloride

(iv) Alkyne

(v) Alkene

Sol. (i) Acetone can be prepared by oxidation or catalytic dehydrogenation of isopropyl alcohol.


Q. How will you convert.

(a) Ethanal to lactic acid

(b) Acetaldehyde to acetone

(c) Ethanal to 2 -hydroxybut- 3 -enoic acid

(d) Formaldehyde to acetaldehyde

(e) Acetaldehyde to formaldehyde

(f) Formaldehyde to $n$ -butane

(g) Ethanol to butan- 2 -one

(h) Acetaldehyde to crotonic acid

(i) Propanal to propyne

(j) Acetone to phorone

(k) Propanone to iodoform

(1) Formaldehyde to urotropine

(m) Acetone to tertiary butyl alcohol

(n) Ethanal to propan- 2 -ol

(o) Methyl cyanide to ethanal

(p) Benzalchloride into cinnamaldehyde.

(q) Propene to propanone

(r) Acetophenone to 2 -phenylbutan- 2 -ol

Sol. (a) Ethanal to lactic acid


Q. Write the equations for the reaction of :

reagents:

(i) Heating with excess of $H B r$

(ii) Dilute $K M n O_{4}$ in the presence of alkali

(iii) A dilute solution of $N a_{2} C O_{3}$

(iv) Phenyl hydrazine

(v) Tollen’s reagent

Sol.


Q. An organic compound (A) with molecular formula $\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}$ forms an orange-red precipitate with $2,4$ -DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollen’s or Fehlings’ reagent, nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula $C_{7} H_{6} O_{2} .$ Identify the compounds $(\mathrm{A})$ and $(\mathrm{B})$ and explain the reactions involved.

Sol. (A) forms $2,4$ -DNP derivative. Therefore, it is an aldehyde or a ketone. since it does not reduce Tollen’s or Fehling reagent,

(A) must be a ketone. (A) responds to iodoform test. Therefore, it should be a methyl ketone. The molecular formula of (A) indicates high degree of unsaturation, yet it does not decolourise bromine water or Baeyer’s reagent. This indicates the presence of unsaturation due to an aromatic ring. Compound (B), being an oxidation product of a ketone should be a carboxylic acid. The molecular formula of (B) indicates that it should be benzoic acid and compound (A) should, therefore, be a monosubstituted aromatic methyl ketone. The molecular formula of $(\mathrm{A})$ indicates that it should be phenyl methyl ketone (acetophenone). Reactions are as follows:



Q. What is meant by the following terms? Give an example in each case.

(a) Cyanohydrin

(b) Semicarbazone

(c) Hemiacetal

(d) Ketal

(e) $2,4$ -DNP-derivative

(f) Acetal

(g) Aldol

(h) Oxime

(i) Imine

(j) Schiff’s base

Sol. (a) Cyanohydrin is a compound which contains both OH and CN group. For example, Lactic acid can be obtained by hydrolysis of cyanohydrin.

(b) The product of carbonyl compounds with semicarbazide is known as semicarbazone.

(c) Hemiacetal is a compound which contains an ether as well as alcohol functional group. For example, methoxyethanol is a hemiacetal.

(d) Ketal is a cyclic compound obtained by reaction of acetone with ethylene glycol.

(e) $2,4$ -DNP derivative: see properties of aldehydes/ ketones in section 12.7 of text.

(f) Acetal : Acetal are the product of aldehyde and monohydric alcohol

(g) Aldol is a condensation product of acetaldehyde in the presence of dil. $N a O H$

(h) Oxime : Aldehyde and Ketones react with hydroxylamine to form oximes.

(i) Schiff’s base : Schiff’s reagent is an aqueous solution of magenta or pink coloured rosaniline hydrochloride which has been decolourised by passing $\mathrm{SO}_{2}$ when aldehyde are treated with decolourised solution of schiff’s reagent, its pink or magenta colour is restored. This reaction is used as a test for aldehyde because ketones do not restore the pink colour of schiff reagent.


 

Aliphatic Hydrocarbon | Question Bank for Class 12 Chemistry

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Q. What are the natural sources of hydrocarbons ?

Sol. Coal and petroleum are two natural sources of hydrocarbons.


Q. What is petroleum ?

Sol. Petroleum is a complex mixture of solid, liquid and gaseous hydrocarbons.


Q. What is the main constituent of LPG ?

Sol. LPG mainly contains butane and Isobutane.


Q. Give a scale for measuring the quality of gasoline.

Sol. Octane number.


Q. Arrange the following in order of increasing volatility :

gasoline, kerosene oil and diesel.

Sol. Diesel < kerosene oil < gasoline.


Q. Name the compound that is added to LPG to detect its leakage.

Sol. The compound added to LPG to detect its leakage is ethyl

mercaptan $\left(C_{2} H_{5} S H\right)$


Q. Name the product obtained by the fractional distillation of petroleum which is used as a furnace fuel in metallurgical operations.

Sol. Fuel oil is used as furnace fuel in metallurgical operations.


Q. Define knocking.

Sol. A sharp metallic rattling sound produced in an internal combustion engine, is called knocking.


Q. What is octane number of fuel which contains $95 \%$ iso-octaneand $5 \%$ n-heptane?

Sol. Octane number is 95.


Q. What is meant by anti-knocking compound ? Give its example.

Sol. – Those compounds which reduces knocking and help in smooth combustion of gasoline are called antiknocking compounds, e.g. tetra ethyl lead $\left(C_{2} H_{5}\right)_{4} P b$


Q. Why is dipole moment of trans-1, 2 dichloroethene zero ?

Sol. It is because individual dipoles are equal and opposite such that net dipole moment is zero


Q. Why alkanes and cycloalkanes are called paraffins ?

Sol. They are called paraffins because under normal conditions alkanes are inert towards common reagents [In Latin, Parum = little, affinis =affinity]


Q. Why eclipsed and staggered forms of ethane cannot be isolated at room temperature ?

Sol. It is because the difference in their energy is less.


Q. What is Lindlar’s catalyst ? Give its use.

Sol. Lindlar’s catalyst is $P d / B a S O_{4} .$ It is used to convert Alkynes to

alkenes with the help of $H_{2}$


Q. Convert ethylene to ethane.

Sol. $\mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{H}_{2} \frac{\mathrm{Ni}}{573 \mathrm{K}} \mathrm{CH}_{3}-\mathrm{CH}_{3}$


Q. Geometrical isomers are not possible in alkynes why ?

Sol. Due to their straight linear geometry.


Q. Ethyne treated with ozone followed by treatment with zinc and water, leads to formation of compound ‘X’. What is X ?

Sol. X is glyoxal, i.e., ethane 1, 2-dial.


Q. Write chemical equation for the reaction when ethyne is passed through acetic acid in the presence of $\mathrm{Hg}^{+}$.

Sol. In the presence of $H g^{+},$ ethyne adds acetic acid to give vinyl ester.


Q. What are arenes ? Give general formula of monocyclic arenes.

Sol. – Arenes are aromatic hy drocarbons. $C_{n} H_{2 n-6}$ is general formula for monocyclic arenes.


Q. What is trade name of benzene hexachloride ?

Sol. Gammaxene.


Q. What is electrophile in sulphonation ?

Sol. $\mathrm{SO}_{3} \mathrm{H}^{+}$


Q. In presence of conc. $H_{2} S O_{4},$ conc. $H N O_{3}$ acts as acid or base during nitration of benzene?

Sol. It acts as a base.


Q. Write down the products of ozonolysis of $1,2$ dimethylbenzene as cyclohexa- $1,3,5$ -triene.

Sol. With ozone $o$ -xylene forms triozonide that supports the structure of benzene as cyclohexa- $1,3,5$ triene.


Q. How is petroleum refined using technique of fractional distillation ?

Sol. The crude oil is washed with acidic or basic solution depending upon whether the impurities are basic or acidic in nature. The washed oil is then subjected to fractional distillation by heating it to about $723 K$ and allowing the vapours to pass through a fractionating column. The crude oil is separated into various fractions.


Q. Give the full form of CNG and state its composition.

Sol. CNG is compressed natural gas. It mainly consists of methane, $(85$ to $95 \%$ ). Ethane and propane are also present in small quantities. Natural gas is formed in the earth crust by the decomposition of vegetable-matter lying under water by the action of anaerobic bacteria so mostly it is found along with petroleum deposits.


Q. Discuss the stability of conformers of ethane.

Sol. The staggered conformer of ethane is more stable than eclipsed conformer, because in staggered conformer, the H-atoms are far apart. So, the repulsion is lesser than in eclipsed conformation. The difference in the energy content of staggered and eclipsed

conformation is $12.5 \mathrm{kJ} \mathrm{mol}^{-1}$

The order of stability of conformers is in the following order:

Staggered $>$ gauche (skew) $>$ eclipsed.


Q. Classify the following as Z or E isomers :

Sol. (i) E (ii) E (iii) E (iv) Z


Q. What is cracking ? Describe the methods to complete this process.

Sol. Cracking : “The process of breaking down of hydrocarbons of high molecular mass and higher boiling points into smaller hydrocarbons of low molecular mass and lower boiling points, is termed as cracking.

For Example:

Types of cracking : The process of cracking is of two types:

(i) Thermal cracking.

(ii) Catalytic cracking.

(i) Thermal cracking: When cracking is carried at high temperature, it is called thermal cracking.

(ii) Catalytic Cracking: When cracking is carried at low temperature in the presence of a catalyst such as alumina or silica, is called catalytic cracking.


Q. In the alkane $C H_{3}-C H_{2} C\left(C H_{3}\right)_{2} C H_{2} C H\left(C H_{3}\right)_{2},$ identify $1^{\circ}, 2^{\circ}, 3^{\circ}$ and $4^{\circ}$ carbon atoms and give the total number of $H$ atoms bonded to each one of these.

Sol.

Each $1^{\circ}$ Carbon is attached to 3 hydrogen, Each $2^{\circ}$ carbon is attached to 2 hydrogen atoms. Each tertiary ‘C’ atom is bonded to one and quaternary $^{\epsilon} \mathrm{C}^{\prime}$ is not attached to any hydrogen atom.

All five Carbon atoms are attached to total 15 hydrogen atoms.

TwoCarbon atoms are attached to total 4 hydrogen atoms.

OneCarbon atom is attached to total one hydrogen atom

OneCarbon atom is not attached to any hydrogen atom.


Q. Draw Newman projection formulae of the two staggered Gauch conformations of n-butane.

[NCERT]

Sol.


Q. Between the two conformational isomers of cyclohexane, i.e., chair and boat forms, which one is more stable ?

[NCERT]

Sol. Chair form is more stable than boat form.


Q. Addition of to propene yields bromo-2-propane, while in presence of benzoyl peroxide the same reaction yields

1-bromopropane. Explain and give mechanism.

[NCERT]

Sol.


Q. Why propane has only one eclipsed conformation while butane has three ? Explain and give diagrams.

Sol. Propane has only one eclipsed conformer Butane has three eclipsed forms due to changes in dihedral angles.

(i), (ii) and (iii) Newman eclipsed projection of butane.


Q. Complete the reactions

Sol.


Q. Write IUPAC names for the following molecules :

[NCERT]

Sol. (i) 5 -methyl – hept $1,3,6$ -triene.

(ii) 3 methyl pent $-1$ -ene- 4 -yne.


Q. Draw structure of the six isomeric pentenes, $C_{5} H_{10} .$ Specify as $E$ or $Z$ to each geometrical isomer.

[NCERT]

Sol.


Q. Convert ( i ) propene to 1 -bromopropane (ii) 2 -methyl propene to 2 -chloro $-2$ -methyl propane.

Sol.


Q. Why conjugated dienes undergo 1, 4-additions ? Explain.

Sol. Conjugated diene undergo 1, 4 addition due to more resonance stabilization of intermediate carbocation.


Q. Which of the following polymerizes readily and why ?

(i) Acetylene (ii) Ethene (iii) Buta-1, 3-diene.

[NCERT]

Sol. (iii) Buta-1, 3-diene undergo polymerisation reaction readily because it will form resonance stabilized intermediate.


Q. What are the necessary conditions for any compound to show aromaticity ?

[NCERT]

Sol. A compound is said to aromatic if there is delocalization of $\pi-$ electrons and it has planar structure.

It should follow Huckel’s Rule, i.e., in a conjugated, planar, cyclic system if number of delocalized electrons is $(4 n+2) \pi,^{c} n^{\prime}$ being integer, $i . e ., 1,2,3,$ etc. it will be aromatic, otherwise not, e.g.

Benzene, naphthalene, anthracene have $6,10,14$ electrons in conjugated, planar, cyclic system.


Q. Write down the products of ozonolysis of $1,2$ -dimethyl benzene (o-xylene). How does the result support Kekule structure for benzene?

[NCERT]

Sol.

These products prove the structure I and II (Kekule structures of benzene).


Q. Account for the following order of acidity, Acetylen

e > Benzene> Hexane.

[NCERT]

Sol. $-H-C \equiv C-H$ has $s p$ hybridised carbon which is most electronegative due to $50 \%$ s-character therefore it is most acidic.

In Benzene, each carbon is $s p^{2}$ hybridised carbon $(33 \%$ of character), which is less electronegative than hybridised carbon, therefore, it is less acidic.

In hexane, each $^{c} C^{\prime}$ is hybridised carbon $(25 \%$ s-character), which is least electronegative, therefore, it is least acidic.


Q. Why does benzene undergo electrophilic substitution reaction easily and nucleophilic substitution with difficulty ?

[NCERT]

Sol. Benzene has $\pi-$ electrons, therefore, it can undergo electrophilic substitution reactions easily. Nucleophilic substitution are difficult because nucleophile will be repelled by $\pi$ -electrons.


Q. Write down the products and give mechanism of the following reactions :

[NCERT]

Sol.

since electron density is maximum at $o$ and $p$ -positions $S O_{3}$ will attack at $o$ and $p$ -positions.

$\mathrm{NO}_{2}$ will attack at $o$ and $p$ -positions because electron density is maximum at ortho and $p$ -positions.


Q. Arrange the following set of compounds in order of their decreasing reactivity with an electrophile, $E^{+}$

(i) Chlorobenzene, $2,4$ -dinitrochlorobenzene, p-nitrochlorobenzene

(ii) Toluene, $p-H_{3} C-C_{6} H_{4}-C H_{3}, H_{3} C-C_{6} H_{4}-N O_{2}$ $p-O_{2} N-C_{6} H_{4}-N O_{2}$

[NCERT]

Sol.


Q. What happens when-

(i) Sodium acetate is heated with soda lime.

(ii) Acetylene is passed through ammoniacal silver nitrate solution.

(iii) Chlorobenzene is condensed with chloral in the presence of sulphuric acid?

Sol. (i) When sodium acetate is heated with soda lime, methane gas will be formed

(ii) When acetylene is passed through ammonical silver nitrate

solution, white precipitate of silver acetylide will be formed.

(iii) When chlorobenzene is condensed with chloral, in presence of conc. $H_{2} S O_{4},$ DDT is formed.


Q. (i) Write a chemical equation for Friedel-Craft’s reaction.

(ii) Identify $A, B, C$ and $D

$ in the following reactions:

Sol. (i) Friedel Crafts Reaction


Q. Bring out following conversions:

(i) 2 -Bromo-propane to 1 Bromo-propane

(ii) Methane to Ethane

OR

An unknown alkene $A$ on reductive ozonolysis gives two isomeric carbonyl compounds $B$ and $C$ of molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$. Write the structure of $A, B$ and $C$

Sol.


Q. What is LPG ? State its composition, properties and applications.

Or

Give the full form of LPG and state its composition. Why it is considered a good fuel ?

Sol. Petroleum gas : It is a mixture of butane, propane, propene, isobutane and ethane. The main constituent of petroleum gas is butane.

At ordinary pressure butane $\left(C_{4} H_{10}\right),$ Propane $\left(C_{3} H_{8}\right)$ and ethane $\left(C_{2} H_{6}\right)$ are gases. Under high pressure, they are easily liquefied. The petroleum gas which has been liquefied under pressure is called liquefied petroleum gas (LPG). The domestic gas cylinder contains LPG.

LPG a better domestic fuel : Components of LPG burn easily and produce a lot of heat, so LPG is used as a good domestic fuel. Due to the following qualities, LPG is considered as a better fuel.

(i) $\mathrm{LPG}$ is an efficient fuel as it has high calorific value, burning of 1 gram of LPG in air produces about 50 kilo joule of heat energy.

(ii) LPG burns with a smokeless flame, so causes no pollution.

(iii) It leaves no ash on burning so it is a clean fuel.

(iv) On burning no poisonous gases are formed.

(v) It is easy to handle and convenient in storing.


Q. Describe the process of fractional distillation of petroleum with the help of a labelled diagram of the distillation tower. Name any four/one products obtained. State the temperature at which each one of them is obtained.

Sol. Petroleum : It is a naturally occurring fossil fuel that is dark coloured, viscous and foul smelling oil. It is a mixture of several hydrocarbons and other elements such as sulphur, oxygen and nitrogen. Mostly it occurs deep down the earth between two nonporous rocks, that’s why it is called as petroleum. (petra =rock, oleum $=$ oil ) or crude – oil.

Principle of refining of petroleum :”The process of separating different components of petroleum, is known as refining. It is carried out by the process of fractional distillation which is based on the fact that different components of petroleum have different boiling point ranges.

The crude-oil is heated in a tank to a temperature of about and all the vapours formed are fed into a tall fractionating column near the bottom. Except asphalt, all the other components are in vapour state. As the mixture of hot vapours rises up in the column, it begins to cool The vapours of components having higher boiling point-range condense first in the lower part and vapours of components with lower boiling point-range condense later at the upper part of column. Therefore, the components with the highest boiling point condenses first and is collected in the lowest part of the column and the component having lowest boiling point condenses last and is collected in the top most part of column. The vapours that do not liquify, are taken out from the top of the column. This component is called petroleum-gas. The residual oil or liquid residue which does not vapourise in the column is called residual oil. It is collected in the base of column and subjected to further distillation.

Products of petroleum : The various components obtained at different heights of fractional column, (in order of top to bottom of column) are as follows :


Q. Give the products when the following reagents react with alkenes (i) Sulphuric acid (ii) halogens (iii) oxidation (iv) ozone.

Sol. (i) Reaction with sulphuric acid: Cold concentrated sulphuric

-acid reacts with alkenes to gives addition products

(ii) Reaction with halogen : Halogens (chlorine and bromine only react at ordinary temperature and without exposure to $U V$ light to form vicinal halide.

(iii) Oxidation

(a) Oxidation with dilute aqueous potassium permanganate solution : At low temperature alkenes form vicinal glycols.

[This reaction is called hydroxylation as hydroxyl groups are attached across the double bond. $]$

(b) With alkaline potassium permanganate at higher temperature : In such conditions cleavage of $C=C$ bond occurs forming carboxylic acids.

(iv) Ozone: With ozone alkenes form ozonides which undergo cleavage to produce carbonyl compounds. This method was extensively used to determine the position of double bond in alkenes in past.

Procedure : Ozone is passed through a solution of an alkene in inert solvent such as $\mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CHCl}_{3}$ or $\mathrm{CCl}_{4}$ at temperature $196-200 K .$ Alkenes are oxidised to ozonised which are not usually isolated but reduced in situ, with zinc dust and water or $H_{2} / P d$ to give carbonyl compound. The location of double bond is assigned on the basis of products.

(a) In case of symmetrical alkene, only one aldehyde is formed on reduction with zinc and water:

$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}+\mathrm{O}_{3} \longrightarrow$

(b) If double bond is terminal, methanal is one product and a higher aldehyde is other product.

$\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{O}_{3} \longrightarrow$

(c) If ketone is formed, branching is there at C atom having double bond.


Q. An alkene on ozonolysis gives 2 -butanone and 2 methylpropanal. What products will be obtained when it is treated with hot conc. $K M n O_{4} ?$

Sol. Let us first try to write the structure of alkene. The product of ozonolysis are :


Q. (i) What is Octane number? How does it atfect knocking?

(ii) What type of isomerism is shown by methoxy methane and ethanol?

(iii) Complete the following reactions:

Sol. (i) Octane number is defined as percentage of iso-octane in the mixture of iso-octane and $n$ -heptane which gives same knocking as sample fuel. Higher the octane number, lesser will be knocking.

(ii) Functional isomerism.


Q. Complete the following reactions :

(i) Isopropyl bromide $\frac{\text { alc. KOH }}{\text { Heat }} A \frac{H B r}{\text { Peroxide }} B$

(ii) $\quad n$ -Propyl alcohol $\frac{\text { Conc. } H_{2} S O_{4}}{443 K} A \frac{O_{2}, A g}{\text { Heat }}>B$

(iii) $1,1,2,2,$ -tetrachloroethane $\frac{z n, \text { alcohol }}{\text { Heat }}$

$A \frac{\text { Iron tube }}{675 \mathrm{K}}>B$

(iv) Acetylene $\frac{N a N H_{2}}{\longrightarrow} A \stackrel{C H_{3} C H_{2} B r}{\longrightarrow} B$

(v) Propyne $\frac{H_{2}, P d, B a S O_{4}}{\text { Quinoline }} \rightarrow A \frac{(i) O_{3}}{(i i) Z n, H_{2} O} B$

Sol.


Surface Chemistry | Question Bank for Class 12 Chemistry

Get Surface Chemistry important questions for Boards exams. Download or View the Important Question bank for Class 12 Chemistry. These important questions will play significant role in clearing concepts of Chemistry. This question bank is designed by NCERT keeping in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 12 chemistry chapter wise CBSE.

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You can access free study material for all three subject’s Physics, Chemistry and Mathematics.

Click Here for Detailed Notes of any chapter. 

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Also get to know about the strategies to Crack Exam in limited time period.

Q.

Why are substances like platinum and palladium often used for carrying out electrolysis of aqueous solutions?

Sol. Platinum and palladium form inert electrodes, i.e., they are not attacked by the ions of the electrolyte or the products of electrolysis. Hence, they are used as electrodes for carrying out the electrolysis.


Q. Why are powdered substances more effective adsorbent than their crystalline forms ?

Sol. Powdered substances have greater surface area as compared to their crystalline forms. Greater the surface area, greater is the adsorption.


Q. Why is it necessary to remove CO when ammonia is obtained by Haber’s process ?

Sol. CO acts as a poison for the catalyst used in the manufacture of ammonia by Haber’s process. Hence, it is necessary to remove it.


Q. Why is the ester hydrolysis slow in the beginning and becomes faster after some time ?

Sol. The ester hydrolysis takes place as follows :

The acid produced in the reaction acts as catalyst (autocatalyst) for the reaction. Hence, the reaction becomes faster after some time.


Q. What is the role of desorption in the process of catalysis?

Sol. Desorption makes the surface of the solid catalyst free for fresh adsorption of the reactants on the surface.


Q. Give reason why a finely divided substance is more effective as an adsorbent ?

Sol. Adsorption is a surface phenomenon. Since finely divided substance has large surface area, hence, adsorption occurs to a greater extent.


Q. What is demulsification? Name two demulsifiers.

Sol. The decomposition of an emulsion into constituent liquids is called demulsification.

Demulsification can be done by boiling or freezing.


Q. What do you understand by activity and selectivity of catalysts.

Sol. Activity of a catalyst refers to the ability of catalyst to increase the rate of chemical reaction. Selectivity of a catalyst refers to its ability to direct the reaction to give a specific product.


Q. Why does physisorption decrease with increase of temperature ?

Sol. Physisorption is an exothermic process :

According to Le Chatelier’s principle, if we increase the temperature, equilibrium will shift in the backward direction, i.e., gas is released from the adsorbed surface.


Q. What modification can you suggest in the Hardy Schulze law ?

Sol. According to Hardy Schulze law, the coagulating ion has charge opposite to that on the colloidal particles. Hence, the charge on colloidal particles is neutralized and coagulation occurs. The law can be modified to include the following :

When oppositely charged sols are mixed in proper proportions to neutralize the charges of each other, coagulation of both the sols occurs.


Q. Why is it essential to wash the precipitate with water before estimating it quantitatively ?

Sol. Some amount of the electrolytes mixed to form the precipitate remains adsorbed on the surface of the particles of the precipitate. Hence, it is essential to wash the precipitate with water to remove the sticking electrolytes (or any other impurities) before estimating it quantitatively.


Q. Distinguish between the meaning of the terms adsorption and absorption. Give one example of each.

Sol. In adsorption, the substance is concentrated only at surface and does not penetrate through the surface to the bulk of adsorbent while in absorption, the substance is uniformly distributed through out the bulk of the solid.

A distinction can be made between absorption and adsorption by taking an example of water vapour. Water vapours are absorbed by anhydrous calcium chloride but adsorbed by silica gel.


Q. What role does adsorption play in heterogeneous catalysis?

Sol. In heterogeneous catalysis, the reactants are generally gases whereas catalyst is a solid. The reactant molecules are adsorbed on the surface of the solid catalyst by physical adsorption or chemisorption. As a result, the concentration of the reactant molecules on the surface increases and hence the rate of reaction increase. Alternatively, one of the reactant molecules undergo fragmentation on the surface of the solid catalyst producing active species which react faster. The product molecules in either case have no affinity for the solid catalyst and are desorbed making the surface free for fresh adsorption. This theory is called adsorption theory.


Q. Give two chemical methods for the preparation of colloids.

Sol. Colloidal solutions can be prepared by chemical reactions involving, double decomposition, oxidation, reduction and hydrolysis.

(i) Double decomposition : A colloidal sol of arsenious sulphide is obtained by passing hydorgen sulphide into a solution of arsenious oxide in distilled water.

$A s_{2} O_{3}+3 H_{2} S \rightarrow A s_{2} S_{3}+3 H_{2} O$

(ii) Oxidation : A colloidal solution of sulphur can be obtained by passing hydrogen sulphide into a solution of sulphur dioxide in water through a solution of an oxidizing agent (bromide water, nitric acid etc.)

$\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{S} \rightarrow 3 \mathrm{S}+2 \mathrm{H}_{2} \mathrm{O}$

$H_{2} S+[O] \rightarrow H_{2} O+S$


Q. How are the colloidal solutions classified, on the basis of physical states of the dispersed phase and dispersion medium ?

Sol.


Q. Describe a chemical method each for the preparation of sol of sulphur and platinum in water.

Sol. Preparation of sol of sulphur : A colloidal solution o sulphur can be obtained by passing hydrogen sulphide into a solution of sulphur dioxide in water or through solution of an oxidizing agent (bromine water, nitric acid etc.)

$\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{S} \longrightarrow 3 \mathrm{S}+\mathrm{H}_{2} \mathrm{O}$

$H_{2} S+[O] \longrightarrow H_{2} O+[S]$

Preparation of platinum sol : It is prepared by electrical disintegration or Bredig’s arc method. In this method, electric arc is struck between electrodes of the metal immersed in the dispersion medium. The intense heat produced vaporizes some of the metal, which then condenses to form particles of calloidal size.


Q. Give four examples of heterogeneous catalysis.

Sol. When the catalyst exists in a different phase than that of reactants. It is said to be heterogeneous catalyst and the catalysis is called heterogeneous catalysis.

(i) Manufacture of $N H_{3}$ from $N_{2}$ and $H_{2}$ by Haber’s process using iron as catalyst.

$N_{2}+3 H_{2} \stackrel{F e}{\longrightarrow} 2 N H_{3}$

(ii) Manufacture of $\mathrm{CH}_{3} \mathrm{OH}$ from $\mathrm{CO}$ and using $(\mathrm{a}$ mixture of $C u, Z n O$ and $C r_{2} O_{3}$

$\mathrm{CO}+2 \mathrm{H}_{2} \frac{\mathrm{Cu}+\mathrm{ZnO}+\mathrm{Cr}_{2} \mathrm{O}_{3}}{\mathrm{Heat}} \rightarrow \mathrm{CH}_{3} \mathrm{OH}$

(iii) Oxidation of with using as catalyst in ostwald process.

(iv) Hydrogenation of oils to form vegetable ghee using finely divided nickel.

$\mathrm{Oil}+\mathrm{H}_{2} \stackrel{\mathrm{Ni}}{\longrightarrow}$ Vegetable ghee.


Q. Describe some features of catalysis by zeolites

Sol. Features of catalysis by zeolites. (i) Zeolites are hydrated alumino-silicates which have a three-dimensional network structure containing water molecules in their pores. (ii) To use them as catalyst, they are heated so that water of hydration present in the pores is lost and the pores become vacant. (iii) The size of the pores varies from 260 to $740 \mathrm{pm} .$ Thus, only those molecules adsorbed in these pores and catalysed whose size is small enough to enter these pores hence, they act as molecular sieves or shape selective catalyst. An important catalyst used in petroleum industry is ZSM- 5 (Zeolite sieve of molecular porosity 5 ). It converts alcohol into petrol by first dehydrating them to form a mixture of hydrocarbons.


Q. What is shape selective catalysis.

Sol. The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape-selective catalysis. Zeolites are good shape-selective catalysts because of their honeycomb-like structures. They are microporous aluminosilicates with three dimensional network of silicates in which some silicon atoms are replaced by aluminium atoms giving $A l-O-S i$ framework. The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules as well as upon the pores and cavities of the zeolites. They are found in nature as well as synthesised for catalytic selectivity.

Zeolites are being very widely used as catalysts in petrochemical industries for cracking of hydrocarbons and isomerisation. An important zeolite catalyst used in the petroleum industry is ZSM- – . It converts alcohols directly into gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons.


Q. Give four uses of emulsions.

Sol. Uses of emulsions are :

(1) In medicines : A wide variety of pharmaceutical preparations are emulsions. For example, emulsions of cod liver oil. These emulsified oils are easily acted upon by digestive juices in the stomach and, hence are readily digested.

(2) Digestion of fats : Digestion of fats in the intestines is aided by emulsification.

(3) In disinfectants : The disinfectants such as dettol give emulsions of the oil in water type when mixed with water.

(4) In building roads : An emulsion of asphalt and water is used for building roads. In this way there is no necessity of melting the asphalt.


Q. What are micelles ? Give an example of a micellers system.

Sol. There are some substances which at low concentrations behave as normal strong electrolytes, but at higher concentration exhibit colloidal behaviour due to the formation of aggregates. The aggregated Particles thus formed are called micelles. The formation of micelles takes place only above a particular temperature called kraft temperature $\left(\mathrm{T}_{\mathrm{k}}\right)$ and above a particular concentr-ation called critical micelle concentration (CMC). Surface active agents such as soaps and synthetic detergents belong to this class. For soaps, the CMC is $10^{-4}$ to $10^{-3} \mathrm{moll}^{-1}$. micelles may contain as many as 100 molecules or more.


Q. Comment on the statement that “Colloid is not a substance but state of a substance”.

Sol. The given statement is true. This is because the same substance may exist as a colloid under certain conditions and as a crystalloid under certain other conditions. For example, $N a C l$ in water behaves as a crystalloid while in benzene, it behaves as a colloid. Similarly, dilute soap solution behaves like a crystalloid while concentrated solution behaves as a colloid (called associated colloid). It is the size of the particles which matters, i.e., the state in which the substance exists. If the size of the particles lies in the range 1 nm to 1000 nm to it is in the colloidal state.


Q. What is the difference between physical adsorption and chemisorption?

Sol.


Q. What are the factors which influence the adsorption of gas on a solid?

Sol. Factors influencing the adsorption of gas on solid:

(i) Nature of gas: Under given conditions of temperature and pressure, the easily liquifiable gases like $N H_{3}, H C l, C O_{2},$ etc. are adsorbed to a greater extent than

the permanent gases such as $H_{2}, O_{2}, N_{2}$ etc. It is because the vander waal’s forces or molecular forces are more predominant in the former than in later category.

Effect of Nature of the Adsorbent: Activated charcoal (ii) is the most common adsorbent for the gases which can be easily liquefied. The gases $H_{2}, O_{2}$ and $N_{2}$ are adsorbed on metals like $N i, P d$ etc.

(iii) Specific area of the solid : Specific area of an adsorbing solid is the surface available for adsorption per gram of the adsorbent. Greater the specific area of the solid, greater would be its adsorbent power.

(vi) Effect of pressure : It is expected that extent of adsorption increases with increase of pressure.

(v) Effect of temperature : In physisorption

$x / m$ decreases with rise in temperature. While in chemisorptionslightly increases in the beginning and then decreases as the temperature rises. This initial increase is due to the fact that, like chemical reactions, chemisorption also requires activation energy.


Q. What is an adsorption isotherm? Distinguish between Freundlich adsorption isothermal and Langmuir adsorption isotherm.

Sol. Adsorption isotherm. It is a graph indicating the variation of the mass or the gas adsorbed per gram $(x / m)$ of the adsorbent with pressure $(p)$ at constant temperature. Distinction between Freundlich and Langmuir adsorption isotherms

(i) The mathematical expressions representing adsorption are:

Freundlich adsorption isotherm: $\frac{x}{m}=k p^{1 / n}$

Langmuir adsorption isotherm: $\frac{x}{m}=\frac{a p}{1+b p}$

(ii) Langmuir adsorption isotherm is more general and is applicable at all pressure while Freundlich adsorption isotherm fails at high pressure. The latter can be deduced from the former.

(iii) Langmuir adsorption postulates that adsorption is monomolecular whereas Freundlich adsorption suggests that it is multimolecular.


Q. What do you understand by activation of adsorbent? How is it achieved?

Sol. Activation of adsorbent implies increases in the adsorption power of the adsorbent. It involves increase in the surface area of the adsorbent and is achieved by following methods.

$\bullet$ By finely dividing the adsorbent.

$\bullet$ By removing the gases already adsorbed

$\bullet$ By making the surface of adsorbent rough by chemical or mechanical methods.


Q. Discuss the effect of pressure and temperature on the adsorption of gases by solids.

Sol. Effect of pressure : It is expected that extent of adsorption increases with increase in pressure. The extent of the adsorption is generally expressed as $x / m$ where $m$ is the mass of the adsorbent and $x$ is that of the adsorbate when equilibrium has been attained. A graph drawn between

extent of adsorption $\left(\frac{x}{m}\right)$ and the pressure $p$ of the gas at constant temperature is called adsorption isotherm.

. Effect of temperature : Magnitude of adsorption decrease with increase in the temperature.

A graph drawn between extent of adsorption $\left(\frac{x}{m}\right)$ and temperature ( $T$ ) at constant pressure is called adsorption isobar physical adsorption isobar shows a decrease in If with rise in temperature, the isobar of chemisorption $x / 1$ shows an increase in the beginning and then decreases as the temperature rises. This initial increase is due to the fact that like chemical reactions, chemisorption also requires activation energy.


Q. What are lyophilic and lyophobic sols ? give one example of each type.

Sol. (i) Lyophilic colloids : The word ‘lyophilic’ means liquid-loving. Colloidal sols directly formed by mixing substances like gum, gelatine, starch, rubber, etc., with a suitable liquid (the dispersion medium) are called lyophilic sols. An important characteristic of these sols is that if the dispersion medium is separated from the dispersed phase (say by evaporation), the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols. Furthermore, these sols are quite stable and cannot be easily coagulated.

(ii) Lyophobic colloids : The word ‘lyophobic’ means liquid-hating. Substances like metals, their sulphides, etc., when simply mixed with the dispersion medium do not form the colloids. Their colloidal sols can be prepared only by special methods. Such sols are called lyophobic sols. These sols are readily precipitated (or coagulated) on the addition of small amounts of electrolytes, by heating or by shaking and hence, are not stable. Further, once precipitated, they do not give back the colloidal sol by simple addition of the dispersion medium. Hence, these sols are also called irreversible sols. Lyophobic sols need stabilising agents for their preservation.


Q. What is the difference between multimolecular and macromolecular colloids ? Give one example of each. How are associated colloids different from these two types of colloids ?

Sol. On dissolution, a large number of atoms or smaller mole-cules of a substance aggregate together to form species having size in the colloidal range (diameter < 1 nm). The species thus formed are multimolecular colloids while macromolecules in suitable solvents form solutions in which the size of macromolecules may be in colloidal range.

Difference between associated colloids multimolecular and macromolecules colloids. Multimolecular colloids are formed by the aggregation of a large number of small atoms molecules such as S8. Macromolecular colloids contain molecules of large size like starch the size of molecules in this case have the dimensions of colloids. Their size lies in the colloidal range. The associated colloids are formed by electrolytes which dissociate into ions and these ions associate together to form ionic micelles whose size lies in the colloidal range, soap is common example of this type . The micelle formation occurs above a particular concentration (called critical micellisation concentration) and above a particular temperature, called Kraft temperature.


Q. Explain what is observed when

(i) An electrolyte is added to ferric hydroxide sol

(ii) An emulsion is subjected to centrifugation

(iii) Direct current is passed through a colloidal sol

(iv) A beam of light is passed through a colloidal solution.

Sol. (i) The positively charged colloidal particles of $F e(O H)_{3}$ sol get coagulated by the oppositely charged ion provided by electrolyte.

(ii) The constituent liquids of the emulsion separate out. In other words, demulsification occurs.

(iii) On passing direct current, colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated.

(iv) Scattering of light by the colloidal particles takes place and the path of light becomes illuminated. This is called Tyndall effect.


Q. Explain the following terms :

(1) Peptization, (2) Electrophoresis, (3) Coagulation,

Sol. (1) Peptization : Peptization may be defined as the process of converting a precipitate into colloidal sol by shaking it with dispersion medium in presence of a small amount of electrolyte. The electrolyte used for this purpose is called peptizing agent.

(2) Electrophoresis : In this process colloidal particles move towards oppositely charged electrodes, get discharged and precipitated.

(3) Coagulation : The stability of the lyophobic sol is due to the presence of charge on colloidal particles. If, some how, the charge is removed, the particles will come nearer to each other to form aggregate (or coagulate) and settle down.


Q. Explain the following terms with suitable examples

(1) Gel (2) Aerosol and (3) Hydrosol

Sol. (1) Gel : It is a colloidal dispersion of a liquid in a solid, common example is butter.

(2) Aerosol : It is a colloidal dispersion of a liquid in a gas, common example is, fog.

(3) Hydrosol :It is a colloidal sol of a solid in water as the dispersion medium, common example is starch sol or gold sol.


Q. What are emulsions ? What are their different types ? Give one example of each type.

Sol. These are liquid-liquid colloidal systems, i.e., the dispersion of finely divided droplets in another liquid. If a mixture of two immiscible or partially miscible liquids is shaken, a coarse dispersion of one liquid in the other is obtained which is called emulsion. Generally, one of the two liquids is water. There are two types of emulsions.

(i) Oil dispersed in water (O/W type) and

(ii) Water dispersed in oil (W/O type ).

In the first system, water acts as dispersion medium. Examples of this type of emulsion are milk and vanishing cream. In milk, liquid fat is dispersed in water. In the second system, oil acts as dispersion medium. Common examples of this type are butter and cream.

Emulsions of oil in water are unstable and sometimes they separate into two layers on standing. For stabilisation of an emulsion, a third component called emulsifying agent is usually added. The emulsifying agent forms an interfacial film between suspended particles and the medium. The principal emulsifying agents of O/W emulsions are proteins, gums, natural and synthetic soaps, etc., and for W/O, heavy metal, salts of fatty acids, long chain alcohols, etc.

Emulsions can be diluted with any amount of the dispersion medium. On the other hand, the dispersed liquid when mixed, forms a separate layer. The droplets in emulsions are often negatively charged and can be precipitated by electrolytes. They also show Brownian movement and Tyndall effect. Emulsions can be broken into constituent liquids by heating, freezing, etc.


Q. Action of soap is due to emulsification and micelle formation. Comment.

Sol. Cleansing action of soap: Washing action of soap is due to the emulsification of grease and taking it away in the water along with dirt or dust present on grease.

Explanation : The cleansing action of soap can be explained keeping in mind that a soap molecule contains a non-polar hydrophobic group and a polar hydrophilic group. The dirt is held on the surface of clothes by the oil or grease which is present there. since oil or grease are not soluble in water, therefore, the dirt particles cannot be removed by simply washing the cloth with water. When soap is applied, the non-polar alkyl group dissolves in oil droplets while the polar $-C O O^{-} N a^{+}$ groups remain dissolved in water (Figure). In this way, each oil droplet is surrounded by negative charge. These negatively charged oil droplets cannot coagulate and a stable emulsion is formed. These oil droplets (containing dirt particles) can be washed away with water along with dirt particles.


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Q. What is the coordination number of central metal ion in $\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-2}$

Sol. The coordination number of central metal ion is 6.


Q. Define a ‘ligand’. Give an example also.

Sol. Ligand is an atom or group of atoms which is either negatively charged or have lone pair of electrons to form co-ordinate bond with central metal ion.


Q. What is oxidation state of $\mathrm{Co}$ in complex $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{2}\left(\mathrm{NO}_{2}\right) \mathrm{Cl}\right]$ $\left[A u(C N)_{3}\right] ?$

[AI 1997C]

Sol. Cobalt has $+3$ oxidation state. $[x+0-1-1+1-2=0] \Rightarrow x=3$


Q. Write the name of bidentate ligand.

Sol. Ethylene diamine, oxalate ion.


Q. What is the charge on the complex of $P t^{2+}$ in which the ligands are-one water molecule, one pyridine molecule and one ethylenediamine molecule?

Sol. $\left[P t\left(H_{2} O\right) p y(e n)\right]^{2+} .$ The charge on complex ion is $+2$ because ligands are neutral molecules.


Q. Write IUPAC name of the complex $N a_{3}\left[C r(O H)_{z} F_{4}\right] .$

[Delhi 2003]

Sol. Sodium tetrafluorodihydroxochromate (III)


Q. Fill in the blanks :

The complexes with coordination number 6 having three ……………………. ligands are optically active.

[CBSE Sample Paper 1997]

Sol. Different.


Q. Give the chemical formula of potassium hexacyano ferrate $(\mathbb{I})$

Sol. $K_{4}\left[F e(C N)_{6}\right]$


Q. Write the structures of a pair of complexes showing geometrical isomerism.

Sol. (i) $\operatorname{cis}-\left[P t\left(N H_{3}\right)_{2} C l_{2}\right],$ trans-

                     


Q. Write ionisation isomer of $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right]^{2+} \mathrm{SO}_{4}^{2-}$

Sol. $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right]^{+} \mathrm{Br}^{-} \quad$ is ionisation isomer of

$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right]^{2+} \mathrm{SO}_{4}^{2-}$


Q. Name the type of isomerism that occurs in complexes in which both cation and anion are complex ions.

Sol. Co-ordination isomerism.


Q. What scheme of hybridisation is proposed for $C o$ in $\left[C O\left(N H_{3}\right)_{6}\right]^{3+9} ?$

[AI 1999C]

Sol. $d^{2} s p^{3}$


Q. What is the shape of $\left[N i(C N)_{4}\right]^{2-} ?$

[CBSE Sample Paper 1997]

Sol. It is square planar.


Q. Name the hybridisation and the orbitals involved in the shape of $\left[N i(C N)_{4}\right]^{2-}$

Sol. – The hybridization is $d s p^{2}$ and shape is square planar.


Q. Which of the two is more stable $K_{4}\left[F e(C N)_{6}\right]$ or $K_{3}\left[\text { Fe }(C N)_{6}\right] ?$

Sol. $K_{3}\left[F e(C N)_{6}\right]$ is more stable than $K_{4}\left[F e(C N)_{6}\right]$


Q. Give one use of Ziegler Natta catalyst.

Sol. It is used as catalyst in polymerisation of alkanes as heterogeneous catalyst.


Q. Name the metal present in (i) Chlorophyll (ii) Haemoglobin

(iii) Vitamin $\mathrm{B}_{12}$ (iv) Cis-platin.

Sol. (i) $\mathrm{Mg}$ (ii) Fe (iii) Co (iv) Pt.


Q. What is meant by hexadentate ligand? Give one example. How is such a ligand useful for measuring hardness of water?

Sol. Hexadentate ligand is a ligand which has 6 donor atoms, e.g. EDTA

                 

EDTA forms complex with $C a^{2+}$ and $M g^{2+},$ therefore, it is used for measuring hardness of water.


Q. Define coordination number. What is the coordination number of central metal atom in

(i) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$

(ii) $K_{2}\left[F e C l_{4}\right]$

Sol. Coordination number is defined as number of monodentate ligands surrounding a central metal atom or ion. The coordination number of $\mathrm{Co}$ in ( i ) is 6 and $\mathrm{Fe}$ in ( i ) is 4.


Q. Explain geometrical isomerism with reference to square planar complexes giving one example. How is tetrahedral complexes with simple ligands do not exhibit geometrical isomerism?

Sol. Explain geometrical isomerism with reference to square planar complexes giving one example. How is tetrahedral complexes with simple ligands do not exhibit geometrical isomerism?

                             

Tetrahedral complexes do not show geometrical isomerism because relative position of ligands are the same.


Q. In a complex ion : [Co(NH3)3(H2O)2Cl]+

(a) Identify the ligand’s formula and the charge on each one of them and

(b) Write the geometry of complex ion.

Sol. (a) Ligands are $N H_{3},$ the charge on it is zero. $H_{2} O$ has zero charge. $C l$ has negative charge $(-1)$

(b) It has octahedral shape.


Q. Square planar complexes with a coordination number 4 exhibit geometrical isomerism whereas tetrahedral complexes do not, why ?

Sol. The tetrahedral complexes do not show geometrical isomerism because the relative positions of the atoms with respect to each other will be same. The square planar complexes on the other hand show this kind of isomerism because if same kind of ligands occupy position adjacent to each other, it is called cis-form and if these are opposite to each other, it is called trans-form.


Q. How is the magnitude of $\Delta_{0}$ affected by (i) nature of ligand and (ii) oxidation state of metal ion?

[Delhi 2004]

Sol. (i) $\quad \Delta_{0}$ is affected by nature of ligand. Greater the strength of ligand, greater the value of

(ii) is also affected by oxidation state of central metal ion. Generally, the higher the ionic charge on the central metal

ion, the greater will be the value of $\Delta_{0}$.


Q. Using the valence bond approach, deduce the shape and magnetic character of $\left[\mathrm{Cr}(\mathrm{CO})_{6}\right]$

$[\text { At. no. of } C r=24]$

[Delhi 2003]

Sol. Cr has electronic configuration

$C r(0)$ has electronic configuration $[A r] 4 s^{0} 3 d^{6} \because C O$ will cause pairing of electrons.

                   

$d^{2} s p^{3}$ hybridisation octahedral shape, diamagnetic.


Q. Using valence bond theory of complexes, explain the geometry and diamagnetic nature of the ion $\left[C o\left(N H_{3}\right)_{6}\right]^{3+}$ [At. no. of $C o=27]$

[Delhi 2001]

Sol. $\mathrm{Co}(27) 1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{s}^{2} 3 \mathrm{p}^{6} 4 \mathrm{s}^{2} 3 \mathrm{d}^{7}$

                           


Q. Among $\left[A g\left(N H_{3}\right)_{2}\right] C l,\left[N i(C N)_{4}\right]^{2-}$ and $\left[C u C l_{4}\right]^{2-}$ which

(i) Has square planar geometry?

(ii) Remains colourless in aqueous solutions and why? $[\text { At. no. of } A g=47, N i=28, C u=29]$

Sol. $-\left[N i(C N)_{4}\right]^{2-}$ has square planar geometry.

$\left[A g\left(N H_{3}\right)_{2} C l\right]$ remains colourless in aqueous solution

because $A g^{+}$ has no unpaired electron, therefore it cannot undergo $d-d$ -transition.


Q. Explain the following:

(i) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ is diamagnetic, whereas $\left[\mathrm{CoF}_{6}\right]^{3-}$ is paramagnetic.

(ii) $\left[F e\left(H_{2} O\right)_{6}\right]^{3+}$ is more paramagnetic than $\left[F e(C N)_{6}\right]^{3-}$

[AI 2000]

Sol. (i) $\quad\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ is diamagnetic because $\mathrm{NH}_{3}$ is strong

field ligand, causes pairing of electrons whereas $\left[\mathrm{CoF}_{6}\right]^{3-}$ is

paramagnetic because $F^{-}$ is weak field ligand, does not cause pairing of electrons and there are unpaired electrons.

(ii) $\left[F e\left(H_{2} O\right)_{6}\right]^{3+}$ has more unpaired electron than

$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-},$ therefore, more paramagnetic.


Q. Draw the structures and write the hybridised state of the central atom of each of the following species:

(i) $\mathrm{Fe}(\mathrm{CO})_{5}$ (ii) trans- $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+}$

$[\text { At. no. of } F e=26, C o=27]$

[AI 1999]

Sol.

               


Q. Describe the shape of tetra cyano nicklate (II) ion and account for its magnetic property.

[AI 1996; CBSE Sample Paper 1997]

Sol. $\left[N i C N_{4}\right]^{2-}, C N^{-}$ cause pairing of electrons in $N i^{2+}$ ion.

               


Q. Explain how $\left[P t\left(N H_{3}\right) C l_{2}\right]$ and $\left[P t\left(N H_{3}\right)\right] C l_{2}$ will differ in their electrolytic conductances. Give the hybridisation state of $P t$ in these compounds. [At. no. of Pt is $78]$

Sol. Ionization isomerism and Geometrical isomerism is exhibited

by $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Br}$

The ionisation isomers are $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Br}$ and

$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl} \mathrm{Br}\right] \mathrm{Cl}$ The geometrical isomers are

                 

The central atom has $d^{2} s p^{3}$ hybridization.


Q. How is stability of coordination compounds determined in aqueous solution?

Sol. Stability of co-ordination compound in aqueous solution is determined with the help of stability constant. Higher the value of stability constant, greater will be stability. Smaller the cation, higher the charge on the cation, more stable will be the complex.


Q. Explain why a chelating complex is more stable than unchelated complex.

Sol. Chelating complex is more stable than unchelated complex because there is strong force of attraction between cation and polydentate ligand as compared to monodentate ligand.


Q. Draw structures of geometrical isomers of $\left[F e\left(N H_{3}\right)_{2}(C N)_{4}\right]^{-}$

Sol.

                   


Q. (i) Co-ordination compound has the formula $\mathrm{CoCl}_{3}, 4 \mathrm{NH}_{3}$ It does not liberate ammonia but forms a precipitate with $A g N O_{3} .$ Write the structure and IUPAC name of the complex compound.

(ii) Name a ligand which is bidentate and give an example of the complex formed by this ligand.

[AI 1998]

Sol. $\begin{array}{llll}{-(1)} & {\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}} & {\text { Its }} & {\text { IUPAC }} & {\text { name }} & {\text { is }}\end{array}$

tetraamminedichlorocobalt (III) chloride.

(ii) Ethylene diamine (en) is bidentate ligand. $\left[\mathrm{Co}(e n)_{3}\right]^{3+} .$ Its IUPAC name is tris (ethylene diamine) cobalt (MI) ion.


Q. Write the IUPAC name of $\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right] \mathrm{Cl}$ and draw the structures of all the isomers with this formula of complex.

[Delhi 2004]

Sol. The IUPAC name is Dichlorobis (ethylene diamine) cobalt (III) chloride.

                 


Q. Answer the following:

(i) Differentiate between a bidentate ligand and a monodentate ligand.

(ii) Write the IUPAC name of $\left[P t\left(N H_{3}\right)_{2} C l_{2}\right] C l_{2}$

(iii) Draw the structures of geometrical isomers of

$$ \left[\mathrm{Co}\left(\mathrm{NH}_{4}\right)_{4} \mathrm{Cl}_{2}\right]^{+} $$

[AI 2003C]

Sol. (i) Bidentate ligand has two donor atoms which can form two sigma bonds whereas monodentate ligand can form only one bond

(ii) Diammine dichloro platinum (IV) chloride.

                     


Q. Explain each of the following observations:

(i) Tetrahedral $N i$ (II) complexes are paramagnetic but square planar $N i$ (II) complexes are diamagnetic.

(ii) Only transition metals are known to form $\pi$ -complexes.

[Delhi 2002]

Sol. $-(1) \quad$ Tetrahedral complexes of $N i$ (II) are paramagnetic due to presence of unpaired electrons whereas square planar complexes of $N i^{2+}$ are diamagnetic due to absence of unpaired electrons.

(ii) Iransition metals have vacant $d$ -orbitals as well as pair of electrons in $d$ -orbitals, therefore, they can form $\pi$ bonded complexes.


Q. Mention the geometrical shapes attained by the following type of hybrid orbitals:

(i) $S p^{3}$

(ii) $d s p^{2}$

(iii) $d^{2} s p^{3}$

Give an example of each of the above.

Sol. (i) $s p^{3}$ has tetrahedral shape, e.g., $\left[Z n\left(N H_{3}\right)_{4}\right]^{2+}$ is hybridised.

(ii) has square planar, e.g., is hybridised.

(iii) has octahedral shape, e.g., is hybridised.


Q. (i) Name two main factors that favour a metal ion’s forming complex.

(ii) Give an example of industrial application of formation of coordination complex.

(iii) Write the IUPAC name of $\left[\mathrm{Co}(e n)_{2} \mathrm{Cl}(\mathrm{ONO})\right]^{+}$

[AI 2001C]

Sol. (i) $\quad$ (a) small size of cation and higher charge.

(b) presence of vacant $d$ -orbitals.

(ii) $\quad K\left[A g(C N)_{2}\right]$ is used for silver plating and $K\left[A u(C N)_{2}\right]$ is used for gold plating.

(iii) Chlorobis (ethylene diamine) nitrito cobalt (III) ion.


Q. Formation of complex is exothermic or endothermic process. Explain why. What is the effect of temperature on stability of complex compounds ?

Sol. Formation of complex is an exothermic process because there is force of attraction between central metal ion and ligands.

Stability of complex decreases with the increases in temperature because formation of complex is exothermic process. On heating, coordination bond between central metal ion and ligand will break.


Q. Example of synergic bonding interactions in a carbonyl complex.

                   

Sol. The metal-carbon bond in metal carbonyls possess both $s$ and $p$ character. The $M-C \sigma$ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The $M-C \pi$ -bond is formed by the donation of a pair of electrons from a filled $d$ -orbital of metal into the vacant antibonding $\pi^{*}$ orbital of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between $C O$ and the metal.


Q. The hexaquomanganese (II) ion contains five unpaired electrons while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.

Sol. $-M n$ in $+2$ state has the configuration $3 d^{5} .$ In presence of $H_{2} O$ as ligand, the distribution of these five electrons is $t_{2 g}^{3} e_{g}^{2},$ i.e., all the electrons remain unpaired. In presence of $\mathrm{CN}^{-}$ as ligand, the distribution is $t_{2 g}^{5} e_{g}^{0},$ i.e. , two $t_{2 g}$ orbitals contain paired electrons while the third orbital contains one unpaired electron.


Q. Explain the bonding in coordination compounds in terms of Werner’s postulates.

Sol. The main postulates of Werner’s theory of coordination compounds are as follows:

(i) Metal possess two types of valences called

(a) primary valence which are ionisable.

(b) secondary valence which are non-ionisable.

(ii) Primary valency is satisfied by the negative ions and it is that which a metal exhibits in the formation of its simple salts. For example, in the coordination compound, $\mathrm{CoCl}_{3} \cdot 6 \mathrm{NH}_{3}$ valency between $\mathrm{Co}$ and $\mathrm{Cl}$ is primary valency and valency between $\mathrm{Co}$ and $\mathrm{NH}_{3}$ is secondary. In terms of modern theories based on electronic configuration, the primary valency is now referred to as “oxidation state’ of metal.

(iii) Secondary valencies are satisfied by neutral ligand or negative ligand and are those which metal exercises in the formation of its complex ions. Every cation has a fixed number of secondary valencies and are directed in space about central metal ion in certain fixed directions. The six valencies are regarded as directed to corners of a regular octahedron whereas four may be arranged either in square planar manner or tetrahedral manner. This explains the shape of coordination compound (or complex ion or compound).

For example, in coordination compound $\left[\mathrm{CoCl}_{3} \cdot 6 \mathrm{NH}_{3}\right],$ six ammonia molecules linked to $\mathrm{Co}$ by secondary valencies are directed to six corners of a regular octahedron and thus account for structure of as follows:


Q. Using IUPAC norms write the formulas for the following

(i) Tetrahydroxozincate (II)

(ii) Hexaamminecobalt (III) sulphate

(iii) Potassium tetrachloridopalladate (II)

(iv) Potassium tri (oxalato) chromate (III)

(v) Diaminedichloridoplatinum(II)

(vi) Hexaammineplatinum(IV)

Sol.


Q. Using IUPAC norms write the systematic names of the following:

Sol. (i) Hexammine cobalt (III) chloride

(ii) Tetrammine chloronitro cobalt (III) chloride

(iii) Hexammine nickel (II) chloride

(iv) Diamminechloro (methylamine) platinum (II) chloride

(vi) Tris $(1,2$ -ethane diamine) cobalt (III) ion

(vii) Hexaqua titanium (III) ion


Q. $\left[C r\left(N H_{3}\right)_{6}\right]^{3+}$ is paramagnetic while $\left[N i(C N)_{4}\right]^{2-}$ is diamagnetic. Explain why?

Sol. In $\left[C r\left(N H_{3}\right)_{6}\right]^{3+},$ the chromium ion is in $+3$ oxidation state and has electronic configuration of $3 d^{\beta},$ as shown below :


Q. List various types of isomerism possible for coordination compounds, given an example of each.

Sol. Two principal types of isomerism are known among coordination compounds:

(i) Stereo isomerism:

(a) Geometrical isomerism, (b) Optical isomerism

(ii) Structural isomerism:

(a) Linkage isomerism, (b) Coordination isomerism, (c) Ionisation isomerism, (d) Solvate isomerism

(i) Stereo isomerism:

(a) Geometrical isomerism: This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands.

For example, in a square planar complex of formula $\left[M X_{2} L_{2}\right]$ the two ligands $X$ may be arranged adjacent to each other in a cis isomer, or opposite to each other in a trans isomer.

(b) Optical isomerism: Optical isomers are mirror images that cannot be super imposed on one another. There are called as enantiomers. The two forms are called dextro (d) and laevo $(l)$ depending upon direction they rotate the plane of polarised light in polarimeter.

Optical isomers $(d \text { and } l)$ of $\left[C o(e n)_{3}\right]^{3+}$

(ii) Structural isomerism:

(a) Linkage isomerism: Linkage isomerism arises in a coordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocynate ligand, $N C S^{-}$ which may find through the nitrogen to give $M \leftarrow N C S$ or through sulphur to give $M \leftarrow S C N$

(b) Coordination isomerism: This type of isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex.

Example : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]$ and $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$

(c) Ionisation isomerism: This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become counter ion.

Example : $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}$

(d) Solvate isomerism: This form of isomerism is known as “hydrate isomerism’. Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice.

Example: $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3} \&\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}$


Q. Draw a figure to show splitting of degenerate $d$ -orbital in an octahedral crystal field. How does the magnitude of D_{o} decide the actual configuration of $d$ -orbitals in a complex entity?

[Delhi 2004]

Sol. In $d^{1}$ coordination entity, $d^{1}$ electron occupies $t_{2 g}$ orbitals.

In $d^{2}$ and $d^{3}$ also, electrons occupy orbitals singly in accordance with the Hund’s rule. For $d^{4}$ ions two possibility exist. If is less than $P($ energy) required for electron pairing in a single orbital, we have weak field, high spin situation and the fourth electron enters one of the orbitals giving. Fifth electron also enters orbitals, i.e., . When, we have strong field, low spin situation, and pairing will occur in the level with orbitals remain unoccupied in to ions. Greater the value greater the chances of pairing.


Q. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.

Sol. Ligands have different field strengths and as a result the crystal field splitting, $\Delta_{0}$ or $\Delta_{t}$, depends upon the field produced by the ligand and charge on metal ions. Some ligands are able to produce strong fields in which case the splitting will be large whereas others produce weak fields and consequently result in small splitting of $d$ -orbitals. The arrangement of the ligands in order of increasing field strength is known as spectrochemical series.

$I^{-}<B r^{-}<S C N^{-}<C I^{-}<S^{2-}<F^{-}<O H^{-}<C_{2} O_{4}^{2-}$

$<H_{2} O<\mathrm{EDTA}^{4-}<N H_{3}<e n<C N^{-}<C O$


Q. What is meant by stability of coordination compound in solution? State the factors which govern stability of complexes?

Sol. The stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium. The magnitude of the (stability or formation) equilibrium constant for the association, quantitatively expresses the stability. Thus, if we have a reaction of the type:

$M+4 L \rightleftharpoons M L_{4}$

then the larger the stability constant, the higher the proportion of that exists in solution. Free metal ions rarely exist in the solution so that $M$ will usually be surrounded by solvent molecules which will complete with the ligand molecules, $L,$ and be successively replaced by them. For simplicity, we generally ignore these solvent molecules and write four stability constants as follows:

Where $K_{1}, K_{2},$ etc., are referred to as stepwise stability constants. Alternatively, we can write the overall stability constant thus:

$M+4 L \rightleftharpoons M L_{4} \quad \beta_{4}=\left[M L_{4}\right] /[M][L]^{4}$

The stepwise and overall stability constant are therefore related as follows :

$\beta_{4}=K_{1} \times K_{2} \times K_{3} \times K_{4}$ or more generally.

$\beta_{n}=K_{1} \times K_{2} \times K_{3} \times \ldots \ldots \ldots K_{n}$

Factors affecting the stability of complex :

(1) The nature of central ion

(2) The nature of ligand

(1) The nature of central ion : The term nature means the ratio, more is the stability of a complex.

(2) The nature of ligand :

(i) Basic character of ligand : More basic is a ligand, greater is the ease with which it can donate its electrons and therefore more is the stability of complex.

(ii) For charged ligand : The higher the charge and the smaller their size, the more stable are the complexes.

(iii) Chelate effect : Complexes containing chelate ring are usually more stable than similar complexes containing no rings.


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Q. In general which metals do you expect to occur in the native state in nature ? Give examples.

Sol. Metals such as $C u, A g, A u, P t,$ etc. which lie below hydrogen in the electrochemical series are not readily attacked by oxygen, moisture and $\mathrm{CO}_{2}$ of the atmosphere and hence occur in the native state in nature.


Q. Copper and silver lie below in the electrochemical series and yet they are found in the combined state as sulphides in nature. Comment.

Sol. Due to high polarising power of $C u$ and $A g$ ions, their sulphides are even more stable than the metals.


Q. Why do some metals occur in the native state ?

Sol. Metals like $A u$ and $P t,$ which are not reactive, $i . e .,$ are not attacked by $\mathrm{O}_{2},$ moisture and $\mathrm{CO}_{2}$ of the air, occur in the native state.


Q. Metal sulphides occur mainly in rocks and metal halides in lakes and seas. Explain.

Sol. – Metal halides being soluble in water, get dissolved in rain water and are carried to lakes and seas during weathering of rocks. On the other hand, metal sulphides being insoluble are left behind in the rocks as residue.


Q. Give one important ore of aluminium.

Sol. Bauxite $\left(A I_{2} O_{3} .2 H_{2} O\right)$


Q. What do you mean by benefaction process ?

Sol. The process used to concentrate an ore is called the benefaction process.


Q. What is gangue ?

Sol. The earthy and silicious impurities associated with the ores is called gangue.


Q. What is the role of a stabilizer in froth-flotation process ?

Sol. Chemical compounds like cresols and aniline which tend to stabilize the froth are called froth stabilizers.


Q. In the smelting process for the extraction of tin, excess of lime must be avoided. Assign reason.

Sol. In case lime $(C a O)$ is added in excess, it will not only act as flux to remove acidic impurities but will also combine with tin to form calcium stannate.


Q. Why is zinc not extracted from zinc oxide through reduction using $\mathrm{CO} ?$

Sol. – The standard free energy of formation $\left(\Delta_{f} G^{\circ}\right)$ of $C O_{2}$ from $\mathrm{CO}$ is higher than that of the formation of ZnO from Zn. Therefore, CO cannot be used to reduce $\mathrm{ZnO}$ to $\mathrm{Zn}$.


Q. What types of ores are roasted ?

Sol. Sulphide ores are roasted to convert them into their oxides.


Q. What are fluxes ? How are they useful ?

Sol. Flux is a substance that combines with gangue which may still be present in the roasted or the calcined ore to form an easily fusible material called the slag.


Q. Which is a better reducing agent at $983 K\left(\text { or } 710^{\circ} \mathrm{C}\right),$ carbon or carbon monoxide?

Sol. $\mathrm{CO} .$ Above $983 \mathrm{K}, \mathrm{CO}$ being more stable does not act as a good reducing agent but carbon does.


Q. Indicate the temperature at which can be used as a reducing agent for $F e O$

Sol. Above $1123 \mathrm{K},$ carbon can reduce $\mathrm{FeO}$ to $\mathrm{Fe}$


Q. Which metals are generally extracted by electrolytic processes? Which positions these metals generally occupy in the periodic table?

Sol. Electrolytic process is used for the extraction of active metals like $N a, C a, M g, A l,$ etc. where all other methods fail. Except $A l$ and few other metals, these metals belong to $s$ -block elements.


Q. What is the principle of zone refining ?

Sol. When the molten solution of an impure metal is allowed to cool, the pure metal crystallizes out while the impurities remain in solution.


Q. Magnesium oxide is often used as the lining in steel making furnace. Explain.

Sol. The role of magnesium oxide is to act as flux (basic) and it removes certain acidic impurities in the form of slag.


Q. What is a mineral ? How does it differ from an ore ?

Sol. The natural materials or the chemical compounds in which the metals occur in the earth are called minerals. The mineral from which the metal is conveniently and economically extracted is called an ore. Thus, all ores are minerals but all minerals are not $c$


Q. Pine oil is generally added in the Froth Floatation process. Explain.

Sol. In the froth floatation process, sulphide ore is to be separated from gangue or earthy impurities. The impure ore is mixed with water and pine oil in a tank and steam is passed through it. Now, gangue particles are wetted by water while the ore by oil. With the help of steam, the sulphide ore rises in the form of a foam while the gangue impurities settle at the bottom of the container.


Q. Aluminium metal is generally used for the extraction of chromium and manganese from their oxide ores. Explain.

Sol. Aluminium has a very strong affinity for oxygen. It reduces the oxides of chromium and manganese to the metallic forms. It can also be explained on the basis of Ellingham diagram.


Q. Why partial roasting of sulphide ore is done in the metallurgy of copper ?

Sol. Partial roasting of sulphide ore is carried in order to convert it to the oxide form. The oxide form then reduces the remaining sulphide ore to the metallic form. This is called auto reduction.


Q. What is thermit process ? What is its utility ?

Sol. Aluminium metal is a powerful reducing agent. It reduces the oxides of metals like iron and chromium to the metallic state upon heating.

The reaction is known as thermit reaction. It is highly exothermic in nature. As a result, the metal iron or chromium is in the molten state. It can be used to fill the cracks in the railway tracks and also the cracks in the machine parts. This process is known as thermit process.


Q. The value of $\Delta_{f} G^{\circ}$ for formation of $C r_{2} O_{3}$ is $-540 k J m o r^{-l}$ and that of $A l_{2} O_{3}$ is $-827 k J \mathrm{mol}^{-1} .$ Is the reduction of $\mathrm{Cr}_{2} \mathrm{O}_{3}$ possible with aluminium?

Sol. The two equations are:

$\frac{4}{3} A l(s)+O_{2}(g) \rightarrow \frac{2}{3} A l_{2} O_{3}(s) ; \Delta_{f} G_{A l, A_{2} O_{3}}^{\circ}=-827 k J m o l^{1} \ldots$

$\frac{4}{3} C r(s)+O_{2}(g) \rightarrow \frac{2}{3} C r_{2} O_{3}(s) ; \Delta_{f} G^{\circ}=-540 k J m o l^{1}$

Subtracting Eq. (ii) from Eq. (i), we have,

$\frac{4}{3} A l(s)+\frac{2}{3} C r_{2} O_{3}(s) \rightarrow \frac{2}{3} A l_{2} O_{3}(s)+\frac{4}{3} C r(s) ; \Delta_{r} G^{\circ}$

$$ =-287 \mathrm{kJ} \mathrm{mol}^{-1} $$

since $\Delta_{r} G^{\circ}$ of the combined redox reaction is $-v e$

Therefore reduction of $C r_{2} O_{3}$ by $A l$ is possible.


Q. Zinc and not copper is used for the recovery of silver from the complex $\left[A g(C N)_{2} J^{-}\right.$ Discuss.

Sol. Zinc is a very powerful reducing agent and readily displaces silver present in the complex.

No doubt copper is also more reducing than silver but is not so effective as zinc. Moreover, zinc is cheaper than copper.


Q. What is the role of cryolite in the metallurgy of aluminium ?

Sol. The role of cryolite is two-fold.

(i) It makes alumina a good conductor of electricity.

(i) It lowers the fusion (melting point) temperature of the bath from $2323 K$ to about $1140 K$


Q. What is Van Arkel method of obtaining ultrapure metals ?

Sol. The impure metal (Ti or Zr) is first converted into its volatile ¡odide by heating with $I_{2}$ at some low temperature and then decomposing the volatile iodide at a suitable high temperature to give the pure metal. For example,


Q. What is a slag?

Sol. A slag is an easily fusible material which is formed when gangue still present in the roasted ore the calcined ore combines with the flux For example, in the metallurgy of iron, $\mathrm{CaO}(\mathrm{Flux})$ combines with silica gangue to form an easily fusible calcium silicate $\left(C a S i O_{3}\right)$ slag.

$\mathrm{CaCO}_{3} \rightarrow \mathrm{CaO}+\mathrm{CO}_{2} ; \mathrm{CaO}+\mathrm{SiO}_{2} \rightarrow \mathrm{CaSiO}_{3}(\mathrm{slag})$


Q. What is the significance of leaching in the extraction of aluminium ?

Sol. – Aluminium usually contains silica $\left(\mathrm{SiO}_{2}\right),$ iron oxide and

titanium oxide $\left(T i O_{2}\right)$ as impurities. These impurities can be removed by the process of leaching. During leaching, the powdered bauxite ore is heated with a concentrated $(45 \%)$ solution of $N a O H$ at $473-523 K,$ when alumina dissolves as sodium meta-aluminate and silica $\left(\mathrm{Si} \mathrm{O}_{2}\right)$ as sodium silicate leaving the impurities behind.

The impurities are filtered off and the solution of sodium

meta-aluminate is neutralized by passing $\mathrm{CO}_{2}$ when hydrated alumina separates out while sodium silicate remains in the solution. The hydrated alumina thus obtained is filtered, dried and heated to give back pure alumina.

Thus, the significance of leaching in the extraction of aluminium is to prepare pure alumina from the bauxite ore.


Q. At a site, low grade copper ores are available and zinc and iron scraps are also available. Which of the two scraps will be more suitable for reducing the leached copper ore and why?

Sol. The $E^{\circ}$ value for the redox couple $Z n^{2+} / Z n(-0.76 V)$ is more

negative than that of $F e^{2+} / F e(-0.44 V)$ redox couple. Therefore, zinc is more reactive than iron and hence reduction will be faster in case if zinc scraps are used. But zinc is a costlier metal than iron so using iron scraps would be more economical.


Q. Why is the extraction of copper from pyrite difficult than that from its oxide ore through reduction ?

Sol. Carbon is a poor reducing agent for sulphide ores whereas it is good reducing agent for oxide ores.


Q. Why copper matte is put in silica lined converter ?

Sol. Copper matte contains $\mathrm{Cu}_{2} \mathrm{S}$ and $\mathrm{FeS.}$. In the converter FeS gets converted into $F e O$


Q. Differentiate between “minerals” and “ores”

Sol. . The naturally occurring chemical substances in form of which the metals occur in the earth along with impurities are called minerals. The mineral from which the metal can be extracted conveniently and economically is called an ore. Thus, all ores are minerals but all minerals are not ores. For example, iron is found in the earth’s crust as oxides, carbonates and sulphides. Out of these minerals of iron, the oxides of iron are employed for extraction of the metal. Therefore, oxides of iron are the ores of iron. Similarly, aluminium occurs in earth’s crust in form of two minerals, i.e., bauxite $\left(A l_{2} O_{3} . x H_{2} O\right)$ and

clay $\left(A l_{2} O_{3} .2 \mathrm{SiO}_{2} .2 \mathrm{H}_{2} \mathrm{O}\right) .$ Out of these two minerals, Al can be conveniently and economically extracted from bauxite. Therefore, bauxite is the ore of aluminium.


Q. Write one point of difference between electrolytic reduction and reduction with carbon. Give one example of each.

Sol. – The electrolytic reduction takes place at the cathode by the gain of electrons in electrolysis. At the same time, carbon reduction is carried by heating a metal oxide with coke. For example,


Q. Why is the extraction of copper from pyrite difficult than that from its oxide ore through reduction ?

Sol. – The standard free energy $\left(\Delta_{f} G^{\circ}\right)$ of formation of $C u_{2} S$ is greater than those of $C S_{2}$ and $H_{2} S\left(C S_{2}\right.$ is, infact, an endothermic compound). Therefore, neither carbon nor hydrogen can reduce to $\mathrm{Cumetal.}$

In contrast, $\Delta_{f} G^{\circ}$ of $C u_{2} O$ is much lower than that of $C O_{2}$ and hence carbon can easily reduce to $\mathrm{Cu}_{2} \mathrm{O}$ to $\mathrm{Cu}$

$\mathrm{Cu}_{2} \mathrm{O}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \rightarrow 2 \mathrm{Cu}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$

It is because of this reason that the extraction of copper from pyrite is difficult than from its oxide ore through reduction.


Q. Giving examples, differentiate between ‘roasting,’ and “calcination’.

Sol. The process of converting carbonates and hydroxide ores of metals to their respective oxides by heating them strongly below their melting points either in absence or limited supply of air is called calcination.

For example,

On the other hand, the process of converting a sulphide ore into its metallic oxide by heating strongly below its melting point in excess of air is called roasting. For example,


Q. Why copper matte is put in silica lined converter ?

Sol. Copper matte chiefly consists of $\mathrm{Cu}_{2} \mathrm{S}$ along with some unchanged FeS. when a blast of hot air is passed through molten matte taken in a silica lined converter, FeS present in matte is oxidised to $F e O$ which combines with silica $\left(S i O_{2}\right)$ to form $\mathrm{FeSiO}_{3}$ slag.

When whole of iron has been removed as slag, some of the undergoes oxidation to form which then reacts with more to form copper metal.

Thus, copper matte is heated in silica lined converter to remove FeS present in matte as slag.


Q. Write chemical reactions taking place in the extraction of zinc from zinc blende.

Sol. The various steps involved are

(a) Concentration. The ore is crushed and then concentrated by froth-floatation process.

(b) Roasting. The concentrated ore is then roasted in presence of excess of air at about $1200 \mathrm{K}$ when zinc oxide $(\mathrm{ZnO})$ is formed.

(c) Reduction. Zinc oxide obtained above is mixed with powdered coke and heated to $1673 \mathrm{K}$ in a fire clay retort when it is reduced to zinc metal

At $1673 K, \text { zinc metal being volatile (b.p. } 1180 \mathrm{K}),$ distils over and is condensed.

(d) Electrolytic refining. Impure zinc is made the anode while cathode consists of a sheet of pure zinc. The electrolyte consists of $Z n S O_{4}$ solution acidified with dil. $H_{2} S O_{4} .$ On passing electric current, pure $Z n$ gets deposited on the cathode.


Q. Name the process from which chlorine is obtained as a byeproduct. what will happen if an aqueous solution of $N a C l$ is subjected to electrolysis?

Sol. Sodium metal is prepared by Down’s process. It involves the electrolysis of a fused mixture of $N a C l$ and $C a C l_{2}$ at $873 K$ During electrolysis, sodium is discharged at the cathode while $\mathrm{Cl}_{2}$ is obtained at the anode as a bye-product.

If, however, an aqueous solution of $N a C l$ is electrolysed,

$H_{2}$ is evolved at the cathode while is obtained at the anode. The reason being that the $E^{\circ}$ of redox couple is much lower than that of and hence water is reduced to in preference to ions. $\mathrm{NaOH}$ is, however, obtained in the solution.


Q. What is the role of graphite in the electrometallurgy of aluminium ?

Sol. In the electrometallurgy of alumina, a fused mixture of alumina, cryolite and fluorspar $\left(\mathrm{CaF}_{2}\right)$ is electrolysed using graphite as anode and graphite lined iron as cathode. During electrolysis, $A l$ is liberated at the cathode while $C O$ and $C O_{2}$ are liberated at the anode.

At cathode : $A I^{3+}(\text { melt }) \rightarrow A l(l)$

$\begin{array}{ll}{\text { Atanode: }} & {C(s)+O^{2-}(\text {melt}) \rightarrow C O(g)+2 e^{-}} \\ {} & {C(s)+2 O^{2-}(\text {melt}) \rightarrow C O_{2}(g)+4 e^{-}}\end{array}$

If instead of graphite, some other metal is used as the anode, then $O_{2}$ liberated will not only oxidise the metal of the electrode but would also convert some of the Alliberated at the cathode back to $A l_{2} O_{3}$

since graphite is much cheaper than any metal, therefore, graphite is used as the anode. Thus, the role of graphite in electrometallurgy of Al is to prevent the liberation of at the anode which may otherwise oxidise some of the liberated Al back to $\mathrm{Al}_{2} \mathrm{O}_{3}$


Q. Describe a method for refining nickel.

Sol. Mond process: It is used for refining of nickel. When impure nickel is heated in a current of $C O$ at $330-350 K,$ it forms volatile nickel tetracarbonyl complex leaving behind the impurities. The nickel tetracarbonyl thus obtained is then heated to a higher temperature $(450-470 K)$ when it undergoes thermal decomposition giving pure nickel


Q. Predict conditions under which $A l$ might be expected to reduce $M g O$

Sol. Aluminium can reduce magnesium oxide at a temperature of about $2000 K$ because in this temperature range the line of

$\Delta G^{\ominus}_{\left(M g, M_{g} O\right)}$ lies above the line of $\Delta G_{\left(A l, A l_{2} O_{3}\right)}^{\Theta}$ in the

Elingham diagram. Therefore, under these condition $\Delta_{r} G^{\ominus}$ for reduction of $M g O$ with $A l$ will be negative and hence the reaction, $3 \mathrm{MgO}+2 \mathrm{Al} \rightarrow 3 \mathrm{Mg}+\mathrm{Al}_{2} \mathrm{O}_{3}$ would be feasible.


Q. Explain the following:

(a) Carbon reduction process is not applied for reducing aluminium oxide to aluminium.

(b) Aqueous solution of sodium chloride cannot be used for the isolation of sodium by electrolytic reduction method.

(c) Roasting is quite helpful to convert sulphide ore of a metal to the oxide form.

(d) Thermit process is quite useful for repairing the broken parts of a machine.

Sol. (a) Carbon reduction process is not applied for reducing aluminium oxide to aluminium. Actually, aluminium metal itself a very powerful reducing agent and can easily take up oxygen from the oxides of metals like iron, manganese, chromium etc. Therefore, it cannot be reduced by carbon.

(b) If an aqueous solution of sodium chloride is used, then sodium metal which is formed at cathode will immediately react with water. Since sodium has a very strong affinity for water, the reaction will be highly exothermic and the metal will catch fire.

(c) Roasting is carried by heating the metal sulphide with excess of oxygen or air. As a result the metal sulphide will be oxidised to its oxide. For example,

(d) In the thermit process, oxides of metals like iron are reduced by aluminium. The reaction is highly exothermic and a large amount of heat is evolved in the reaction. As a result, the metal will be in the molten state. For example,

If the molten metal is allowed to fall between the broken parts of a machine, the gaps will be filled up and the machine will be repaired.


Q. Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.

Sol. Following reactions occur in the blast furnace

(i) Zone of combustion: Near the tuyeres, coke burns to form carbon dioxide

$$ C+O_{2} \rightarrow C O_{2} ; \Delta H=-393.3 k_{U} $$

since the reaction is exothermic, lot of heat is produced and the temperature here is around $2170 K$

(ii) Zone of heat absorption: This is lower part of the Furnace and the temperature here is between $1423-1673 \mathrm{K.As} \mathrm{CO}_{2}$ formed near tyres moves up; it meets the descending charge. The coke present in the charge reduces $\mathrm{CO}_{2}$ to $\mathrm{CO}$

$\mathrm{CO}_{2}+\mathrm{C} \rightarrow 2 \mathrm{CO} ; \Delta \mathrm{H}=+163.2 \mathrm{kJ}$

since this reaction is endothermic, therefore the temperature gradually falls to $1423 \mathrm{K}$

(iii) Zone of slag formation: It is the middle part of the furnace. The temperature here is around $1123 \mathrm{K} .$ In this region, limestone decomposes to form $\mathrm{CaO}$ and $\mathrm{CO}_{2} .$ The $\mathrm{CaO}$ thus formed acts as a flux and combines with silica (Present as an impurity) to form fusible calcium silicate slag

(iv) Zone of reduction: This is the upper part of the furnace. The temperature here is around $823 \mathrm{K} .$ Here the ores are reduced to $\mathrm{FeOby} \mathrm{CO}$

But the further reduction of $F e O$ to $F e$ by $C O$ occur around $1123 K$

However, direct reduction of iron ores, i.e. haematite, magnetite etc, which might have escaped reduction around $823 \mathrm{K}$, gets completely reduced to iron by carbon above $1123 \mathrm{K}$

(v) Zone of fusion : This is the lower part of the furnace. Temperature here is in between $1423-1673 K .$ In this region, spongy iron melts and dissolves some $C, S, P, S i, M n,$ etc. slag also melts in this region. Both the molten slag and the molten iron trickle down into hearth where they form two separate layers. The molten slag being lighter forms the upper layer while molten iron being heavier form the lower layer. The two liquids are periodically trapped off. The iron thus obtained from the furnace contains about $4 \%$ carbon and many impurities in smaller amount (e.g., S, P, Si, Mn). This is called pig iron and is cast into variety of shapes. Cast iron is different from pig iron and is made by melting pig iron with scrap iron and coke using hot air blast. It has slightly lower carbon content (about $3 \%$ ) is extremely hard and brittle


Q. How can you separate alumina from silica in bauxite ore associated with silica ? Give equations, if any.

Sol. Pure alumina may be separated from bauxite by Baeyer’s process as discussed below. The bauxite ore associated with silica is heated with a concentrated solution of $N a O H$ at 473 – $523 \mathrm{Kand} 35-36$ bar pressure. Under these conditions, alumina dissolves as sodium meta-aluminate and silica as sodium silicate leaving behind the impurities.

The resulting solution is filtered to remove the undissolved impurities, if any and neutralized by passing $\mathrm{CO}_{2}$ gas. Thereafter, the solution is seeded with freshly prepared samples of hydrated alumina when hydrated alumina gets precipitated leaving sodium silicate in the solution.

The hydrated alumina thus precipitated is filtered, dried and heated to give back pure $A l_{2} O_{3} .$

Alternatively, serpeck’s process may be employed for the purification of white bauxite which contains silica as the impurity. The ore is powdered, mixed with coke and heated to $2073 \mathrm{Kin}$

an atmosphere of $N_{2}$ gas. Alumina combines with to form aluminium nitride while silica is reduced to silicon which volatilizes off at this temperature.

Aluminium nitride thus obtained is hydrolysed by water to form aluminium hydroxide which, when ignited, gives pure alumina

This process has one distinct advantage that ammonia is obtained as a by product.


Q. Outline the principles of refining of metals by the following methods

(i) Zone refining (ii) Electrolytic refining

(iii) Vapour phase refining

Sol. (i) Zone refining : As that of Ans. of Q. no.-4.

(ii) Electrolytic refining: A large number of metals such as copper, silver, gold, lead, nickel, chromium, zinc, aluminium etc are refined by this method.

In this method the impure metal is converted into a block which forms the anode while cathode is made up of a pure strip of the same metal. These electrodes are suspended in an electrolyte which is the solution of a soluble salt of the metal usually a double salt of the metal. When electric current is passed, metal ions from the electrolyte are deposited at the cathode in the form of pure metal while an equivalent amount of metal dissolves from the anode and goes into the electrolyte solution as metal ions. The process takes place as under

Anode : $M \rightarrow M^{n+}+n e^{-}$

Cathode : $M^{n+}+n e^{-} \rightarrow M$

The net result is the transfer of pure metal from the anode to the cathode. The voltage applied for electrolysis is such that the impurities of more basic metals (more electropositive metals) remain in the solution as ions whereas impurities of the less basic metals (Less electropositive metals) settle down under the anode as anode mud or anode sludge

(iii) Vapour – phase refining: In this method, the crude metal is freed from impurities by first converting it into a suitable volatile compound by heating it with a specific reagent at a lower temperature and then decomposing the volatile compound at some higher temperature to give the pure metal. Thus, the two requirements are:

(1) The metal should form a volatile compound with a suitable reagent.

(2) The volatile compound should be easily decomposable

so that the recovery is easy.


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Q. Silver atom has completely filled $d$ -orbitals $\left(4 d^{10}\right)$ in its ground state. How can you say that it is a transition element.

Sol. The outer electronic configuration of $A g(Z=47)$ is $4 d^{10} 5 s^{1} .$ In

addition to $+1,$ it shows an oxidation state of $+2$ (e.g., $A g O$ and

$A g F_{2}$ exist). In $+2$ oxidation state, the configuration is $d^{9},$ i.e. the $d$ -subshell is incompletley filled. Hence, it is a transition element.


Q. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Sol. Oxygen and fluorine have small size and high electronegativity. Hence, they can oxidize the metal to the highest oxidation state.


Q. Which is a stronger reducing agent $C u^{2+}$ or $F e^{2+}$ and why?

Sol. $C r^{2+}$ is a stronger reducing agent than $F e^{2+}$

Reason: $E_{C r}^{\circ} / C r^{2+}$ is -ve $(-0.41 \mathrm{V})$ whereas $E_{F e^{3+} / F e^{2+}}^{\circ}$ is $+v e$ $(+0.77 V) .$ Thus, is easily oxidized to but cannot be easily oxidized to. Hence, is stronger reducing agent than.


Q. Calculate the ‘spin only’ magnetic moment of $M_{(a q)}^{2+}$ ion

$(Z=27)$

Sol. Electronic configuration of $M$ atom with $Z=27$ is $[A r] 3 d^{7} 4 s^{2}$

$\therefore$ Electronic configuration of $M^{2+}$ will be $[A \eta] 3 d^{7},$ i.e.,


Q. Actinoid contraction is greater from element to elements than lanthanoid contraction. Why?

Sol. This is due to poor shielding by $5 f$ electrons in the actionids than

that by $4 f$ electrons in the lanthanoids.


Q. Why do the transition elements exhibit higher enthalpies of atomisation?

Sol. Because of large number of unpaired electrons in their atoms they have stronger interatomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.


Q. Name a transition element which does not exhibit ‘variable’ oxidation states.

Sol. Scandium (Z = 21) does not exhibit variable oxidation states.


Q. How would you account for the increasing oxidising power

in the series $V O_{2}^{+}<C r_{2} O_{7}^{2-}<M n O_{4}^{-} ?$

Sol. This is due to the increasing stability of the lower species to

which they are reduced.


Q. Why is the $E^{y}$ value for the $M n^{3+} / M n^{2+}$ couple much more positive than that for $C r^{3+} / C r^{2+}$ or $F e^{3+} / F e^{2+7} ?$ Explain.

Sol. Much larger ionisation energy of $M n$ (where the required change is to $d^{4}$ ) is mainly responsible for this. This also explains why $d^{5}$ the $+3$ state of $M n$ is of little importance.


Q. Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25

Sol. With atomic number $25,$ the divalent ion in aqueous solution will have $d^{5}$ configuration (five unpaired electrons). The magnetic moment, $\mu$ is $\mu=\sqrt{5(5+2)}=5.92 B M$


Q. Name a member of the lanthanoid series which is well known to exhibit $+4$ oxidation state.

Sol. Cerium $(Z=58)$


Q. What may be the stable oxidation state of the transition element with the following $d$ -electron configurations in the ground state of their atoms : $3 d^{\beta}, 3 d^{8}, 3 d^{8}, 3 d^{4} ?$

Sol. – For, $3 d^{3} 4 s^{2},$ the most stable oxidation state is $+5$

For, $3 d^{5} 4 s^{2},$ the stable oxidation states are $+2$ and $+7$

For, $3 d^{8} 4 s^{2},$ the stable oxidation state is $+2$

For, $3 d^{4} 4 s^{2},$ the stable oxidation states are $+3$ and $+6$


Q. Name of the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Sol. Chromate ion, $C r O_{4}^{2-}$ – Oxidation state $+\mathrm{VI}$

Dichromate ion, $C r_{2} O_{7}^{2-}-$ Oxidation state $+\mathrm{VI}$

Permanganate ion, $M n O_{4}^{-}$ – Oxidation state $+\mathrm{VII}$

Group number of chromium is 6 while that of manganese is 7 .


Q. What are the different oxidation states exhibited by the lanthanoids ?

Sol. The principal oxidation state of lanthanoids is $+3 .$ However, some lanthanoids also exhibit oxidation state of $+2$ and $+4 .$ For example, Eu exhibit oxidation state of $+2$ and $C e$ exhibits oxidation state of $+4$


Q. Which metal in the first series of transition metals exhibits $+1$ oxidation state most frequently and why?

Sol. In the first transition series copper exhibit $+1$ oxidation state most frequently. This is due to the reason that $C u($ I) has stable electronic configuration. $C u(I):[A r] 3 d^{10}$


Q. In the series $S c(Z=21)$ to $\operatorname{Zn}(Z=30),$ the enthalpy of atomisation of zinc is the lowest, i.e., $126 \mathrm{kJ} \mathrm{mol}^{-1} .$ Why?

Sol. – In the series, $S c$ to $Z n,$ all elements have one or more unpaired electrons except zinc which has no unpaired electron as its outer electronic configuration is $3 d^{10} 4 s^{2} .$ Hence, atomic intermetallic bonding (metal-metal bonding) is weakest in zinc. Therefore, enthalpy of atomisation is lowest.


Q. Which of the $3 d$ series of the transition metals exhibits the largest number of oxidation states and why?

Sol. Manganese $(Z=25)$ shows maximum number of oxidation states. This is because its electronic configuration is $3 d^{5} 4 s^{2} .$ As $3 d$ and 4s are close in energy, it has maximum number of electrons to lose or share (as all the electrons are unpaired). Hence, if shows oxidation states from $+2$ to $+7(+2,+3,+4,+5,+6 \text { and }+7)$ which is the $\mathrm{n}$ um number.


Q. How would you account for the irregular variation of ionization enthalpies (first and second) in first series of the transition elements?

Sol. Irregular variation of first ionization enthalpy : As we move from left to right along the first transition series, as effective nuclear change increases, it is expected in general that the first ionization enthalpy should show an increasing trend. However, the trend is irregular because removal of the electron alters the relative energies of $4 s$ and $3 d$ orbitals. Thus, there is a reorganisation energy accompanying ionization. This results into the release of exchange energy which increase as the number of electrons increases in the $d^{n}$ configuration and also from the transference of $s$ -electrons into $d$ -orbitals. $C r$ has low first ionization energy because loss of one electron gives stable electronic configuration $\left(3 d^{5}\right) . Z n$ has very high ionization energy because electron has to be removed from orbital of the stable configuration.

Irregularities of second ionization enthalpy : After the loss of one electron, the removal of second electron becomes difficult. Hence, second ionization enthalpies are much higher and in general increase from left to right. However, and show much higher values because the second electron has to be removed from the stable configurations of and.


Q. The $E^{o}\left(M^{+} / M\right)$ value for copper is positive $(+0.34 \mathrm{V}) .$ What is possibly the reason for this?

Sol. $E^{o}\left(M^{2+} / M\right)$ for any metal is related to the sum of the enthalpy changes taking place in the following steps:

$M(s)+\Delta_{a} H \rightarrow M(g),( \left\Delta_{a} H=$ enthalpy of atomisation) \right.

$M(g)+\Delta_{i} H \rightarrow M^{2+}(g),\left(\Delta_{i} H=\text { ionization enthalpy }\right)$ $M^{2+}(g)+a q \rightarrow M^{2+}(a q)+\Delta_{h y d} H( \left\Delta_{h y d} H=$ hydration \right. enthalpy $)$ Copper has high enthalpy of atomisation (i.e., energy absorbed) and low enthalpy o hydration, (i.e., energy released). Hence, $E^{o}\left(C u^{2+} / C u\right)$ is positive.

The high energy required to transform $C u(s)$ to $C u^{2+}(a q)$ is not balanced by its hydration enthalpy.


Q. Explain why $C u^{+}$ ion is not stable in aqueous solution?

Sol. $\mathrm{Cu}^{2+}(a q)$ is much more stable than $\mathrm{Cu}^{+}(a q) .$ This is because although second ionization enthalpy of copper is large but $\Delta_{h y d} H$

for $C u^{2+}(a q)$ is much more negtive than that for and hence it is more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows:

$2 \mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$


Q. On what ground can you say that scandium (Z= 21) is a transition element but zinc (Z = 30) is not ?

Sol. On the basis of incompletely filled $3 d$ orbitals in case of scandium atom in its ground state $\left(3 d^{1}\right),$ it is regarded as a transition element. On the other hand, zinc atom has completely filled $d$ orbitals $\left(3 d^{10}\right)$ in its ground state as well as in its oxidised state, hence it is not regarded as a transition element.


Q. Why is $C r^{2+}$ reducing and $M n^{3+}$ oxidising when both have $d^{4}$ configuration?

Sol. $C r^{2+}$ is reducing as its configuration changes from $d^{4}$ to $d^{3},$ the

latter having half-filled $t_{2 g}$ level. On the other hand, the change

from $M n^{2+}$ to $M n^{3+}$ results in the half-filled $\left(d^{5}\right)$ configuration which has extra stability.


Q. For the first row transition metals the $E^{\circ}$ values are:

Explain the irregularity in the above values.

Sol. The $E^{\ominus}\left(M^{2+} / M\right)$ values are not regular which can be explained from the irregular variation of ionisation enthalpies $\left(\Delta_{i} H_{1}+\Delta_{i} H_{2}\right)$ and also the sublimation enthalpies which are relatively much less for manganese and vanadium.


Q. What is meant by ‘disproportionation’ of an oxidation state? Give an example.

Sol. When a particular oxidation state becomes less stable relative to cother oxidation states, one lower, one higher, it is said to undergo disproportionation. For example, manganese (VI) becomes unstable Telative to manganese (VII) and manganese (IV) in acidic solution.


Q. Write down the electronic configuration of the following ions

(Atomicnos. are: $C r=24, C u=29, C o=27, M n=25, P m=61$

$C e=58, L u=71, T h=90)$

Sol.


Q. Why are $M n^{2+}$ compound more stable than $F e^{2+}$ towards oxidation to their $+3$ state?

Sol. The electronic configuration of $M n^{2+}$ is $3 d^{5},$ which is stable due to half filled sub shell. Hence, is not easily oxidized to. On the other hand, has electronic configuration $3 \mathrm{d}^{6} .$ After losing one electron it change to $\mathrm{Fe}^{3+}$ which has stable electronic confi-guretion. Hence, $\mathrm{Fe}^{2+}$ is relatively easily oxidized to $\mathrm{Fe}^{3+}$


Q. Explain briefly how $+2$ state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Sol. Except scandium, the most common oxidation state of the first row transition elements is $+2$ which arises from the loss of two $4 s$ electrons. The $+2$ state becomes more and more stable in the first half of first transition elements with increasing atomic number because $3 d$ orbitals acquire only one electron in each of five $3 d$ orbitals (i.e., each $d$ -orbital remains half filled) and electronic repulsion is the least and nuclear charge increases. In second half of first row transition elements, electronstarts pairing up in 3 $d$ orbitals.

$\left(T i^{2+} \text { to } M n^{2+} \text { electronic configuration changes from } 3 d^{2} \text { to } 3 d^{5}\right.$ but in second half i.e., $F e^{2+}$ to $Z n^{2+}$ it changes from $3 d^{6}$ to $\left.3 d^{10}\right)$


Q. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with example.

Sol. The stability of an oxidation state is determined to a large extent by the electronic configuration of the element, in case of transition elements. Those oxidation states in which the element acquires estable configuration such as noble gas configuration, exactly halffilled or fully filled $d$ -subshell, are exceptionally stable. For example,

$S c^{3+}, M n^{2+}$ and $Z n^{2+}$ are relatively more stable.


Q. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Sol. The atomic radii and ionic radii of tripositive lanthanoidions $\left(M^{3+}\right)$ show a steady and gradual decrease in moving from La to Lu. Although the atomic radii do show some irregularities but ionic radii decreases steadily from $L a$ to $L u .$ The steady decrease in size of lanthanoid ions with the increase in atomic number is called lanthanoid contraction.

Consequences of Lanthanoid Contraction :

(i) Similarity of second and third transition series : The atomic radii of second row of transition elements are almost similar to those of the third row of transition elements.

(ii) Variation in basic strength of hydronides : The basic strength of hydroxides decreases from to . Due to lanthanoid contraction size of ions decreases and there is increase in the covalent character in MOH bond.

(iii) Separation of lanthanoids : Separation of lanthanoids is also possible due to lanthanoid contraction. All the lanthanoid have quite similar properties and due to this reason they are difficult to be separated. However, because of lanthanoid contraction their properties vary slightly.


Q. In what way is the electronic configuration of the transition elements different from that of the non-transition elements?

Sol. Transition elements have partially filled d-subshell belonging to penultimate energy level whereas non-transition elements do not have any partially filled d-subshell. In non-transition elements, the last electron enters the s or p-subshell whereas in transition elements the last electron enters the d-subshell of penultimate energy level.


Q. Explain giving reason :

(i) Transition metals and many of their compounds show paramagnetic behaviour.

(ii) The enthalpies of ionisation of the transition metal are high.

(iii) The transition metals generally form coloured compounds.

(iv) Transition metals and their many compounds act as good catalyst.

Sol. (i) The transition metal ion generally contain one or more unpaired electrons in them and hence their complexes are generally paramagnetic.

(ii) The ionization enthalpies of transition elements are higher than those of s-block elements but lower than p-block elements. In a particular transition series, ionization enthalpy increases gradually as we move from left to right. The increase in ionization enthalpy is particularly due to increase in nuclear charge.

(iii) The d-orbitals in the transition elements do not have same energy in their complexes under the influence of the ligands attached, the d-orbitals split into two sets of orbitals having slightly different energies ‘In the transition elements which have partly filled’. d-orbitals, the transition of elements can take place from one of the lower d-orbitals to some higher d-orbital within the same sub-shell. The energy required for this transition falls in the visible region. When light falls on these complexes they absorb a particular colour from the radiation for the promotion of electron and the remaining colours are emitted.

(iv) The catalyst activity of transition metals is attributed to the following reasons

(a) Because of their variable valencies transition metals sometimes form unstable intermediate compounds and provide a new path with lower activation energy for the reaction.

(b) In some cases transition metals provide a suitable surface of the reaction to take place. The reactants are adsorbed on the surface of the catalyst where reactions occurs.


Q. What are interstitial compounds? Why are such compounds well known for transition metals ?

Sol. Interstitial compounds are those which are formed when small atoms like H, C, N, B etc. are trapped inside the crystal lattices of metals. They are generally non-stoichiometric.

Most of transition metals form interstitial compounds with small non-metal atoms such as hydrogen, boron, carbon and nitrogen. These small atoms enter into the interstitial voids between the lattice of transition metals and from chemical bonds with transition metals. For examples, steel and cast iron becomes hard by forming interstitial compound with carbon. The existence of vacant d-orbitals in transition elements and their ability to make bonds with trapped small atoms is the main cause of interstitial compound formation.


Q. How is the variability in oxidation states of transition metal different from that of the non-transition metals? illustration with examples.

Sol. In case of transition elements, the oxidation states of differ from each other by unity whereas in case of non-transition elements oxidation states normally differ by units of two. For example, Fe exhibits oxidation states of $+2$ and $+3 .$ Similarly, $C u$ exhibits oxidations states of $+1$ and $+2 .$ On the other hand, $S n$ and $P b$ exhibit oxidation states of $+2$ and $+4$


Q. Describe the preparation of $K_{2} C r_{2} O_{7}$ from iron chromite

ore. What is the effect of increasing $p H$ on a solution of.

Sol. Step I : Preparation of sodium chromate

$4 \mathrm{FeCr}_{2} \mathrm{O}_{4}+16 \mathrm{NaOH}+7 \mathrm{O}_{2} \longrightarrow$

$8 \mathrm{Na}_{2} \mathrm{CrO}_{4}+2 \mathrm{Fe}_{2} \mathrm{O}_{3}+8 \mathrm{H}_{2} \mathrm{O}$

Step II : Conversion of sodium chromate into sodium dichromate

$2 \mathrm{Na}_{2} \mathrm{CrO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}$

Step III : Conversion of sodium dichromate into potassium dichromate $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{KCl} \longrightarrow \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{NaCl}$

Effect of increasing $p H$ on a solution of $\boldsymbol{K}_{2} \mathbf{C r}_{2} \mathbf{O}_{7}$

On increasing $p H$ of solution dichromate ions, change into chromate ion.


Q. Describe the oxidising action of potassium dichromate and write the ionic equation for its reaction with

(i) Iodide

(ii) Iron

(II) solution $\quad$ (iii) $H_{2} S$

Sol. Potassium dichromates, $K_{2} C r_{2} O_{7}$ is a strong oxidising agent. In acidic solution, its oxidising action can be represented as follows:


Q. Describe the preparation of potassium permangnate. How does the acidified permanganate solution reacts with (i) iron (II) ion (ii) Oxalic acid? Write the ionic equations for reactions.

Sol. $-K M n O_{4}$ is prepared from pyrolusite $\left(M n O_{2}\right)$ Pyrolusite is fused with $K O H$ in the presence of oxygen or an oxidizing agent like potassium nitrate or potassium chlorate to give potassium manganate, $K_{2} M n O_{4}$


Q. For $M^{2+} / M$ and $M^{3+} / M^{2+}$ systems the $E^{\mathrm{E}}$ values for some metals are as follows:

Use this data to comment upon:

(i) The stability of $F e^{3+}$ in acid solution as compared to that of $C r^{3+}$ or $M n^{3+}$

(ii) The ease with which iron can be oxidised as compared to the similar process for either chromium or manganese metal.

Sol. (i) We know that higher the reduction potential of species, larger is the tendency for its reduction to take place. Thus, $M n^{3+}$ with

highest reduction potential would be most readily reduced to $M n^{2+}$ and hence is least stable.

From values of reduction potential, it is clear that the stability of $F e^{3+}$ in acid solution is more than that but less than that of.

(ii) The lower the reduction potential of a species, greater is the ease with which its oxidation will take place.

$\therefore M n$ is oxidised most readily to and $F e$ is oxidised to least readily among given metals $M n, C r$ and $F e .$ Thus, order of their tendency to undergo oxidation is $\mathrm{Mn}>\mathrm{Cr}>\mathrm{Fe}$


Q. Predict which of the following will be coloured in aqueous solution?

$T i^{3+}, V^{3+}, C u^{+}, S c^{3+}, M n^{2+}, F e^{3+}, C o^{2+}$ and $M n O_{4}^{-}$

Give reason for each.

Sol. – Any ion that has partially filled $d$ -orbitals is coloured due to $d-d$

transition in visible light. $T i^{3+}(d), V^{3+}\left(d^{2}\right), M n^{2+}\left(d^{5}\right)$

$F e^{3+}\left(d^{6}\right), C o^{2+}\left(d^{8}\right)$ are coloured. $M n O_{4}^{-}$ has purple colour due to change transfer. $C u^{+}\left(d^{10}\right)$ and $S c^{3+}\left(d^{0}\right)$ are white.b


Q. Compare the stability of $+2$ oxidation state for the elements of the first transition series.

Sol. In the first transition series the stability of $+2$ oxidation state of elements, in general, decreases from left to right. This is due to increase in the value of sum of $I E_{1}$ and $I E_{2} .$ However, $M n$ and $Z n$ show exceptional behaviour. The greater stability of $+2$ oxidation states of $M n$ and $Z n$ are due to half filled $d$ -subshell $\left(d^{5}\right)$ in $M n^{2+}$ and fully filled $d$ -subshell $\left(d^{10}\right)$ in $Z n^{2+}$


Q. How would you account for the following:

(1) Of the $d^{4}$ species, chromium (II) is strongly reducing while manganese (III) is strongly oxidising.

(ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing regents it is easily oxidised.

(iii) The $d^{1}$ configuration is very unstable in ions.

Sol. (i) For chromium, $+$ III state is more stable as compared to $+\Pi$ state. Therefore, $C r^{2+}$ readily change into $C r^{3+}$ and behaves as a strong reducing agent. On the other hand, for manganese, + II state is more stable than $+$ III state. Hence, $M n^{3+}$ readily changes into $M n^{2+}\left(3 d^{5}\right)$ by gaining an electron and behave as an strong oxidising agent.

(ii) In the presence of ligands $\mathrm{Co}(\mathrm{II})$ is oxidised to $\mathrm{Co}(\mathrm{III})$ which has $d^{6}$ configuration. Most of the ligands are strong enough to cause spin paring giving rise to diamagnetic octahedral complexes which are very stable and have very large crystal field stabilization energy.

(iii) The ions in $d^{1}$ configuration have great tendency to acquire more stable $d^{0}$ configuration by losing the lone $d$ -electron and act as reducing agent. In some cases, the ion in configuration undergoes disproportionation. For example,

$3 \mathrm{MnO}_{4}^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_{4}^{-}+\mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$


Q. What is meant by ‘disproportionation’? Give two examples of disproportiantion reaction in aqueous solution.

Sol. Sometimes a particular oxidation state becomes less stable relative to other oxidation states, one lower and the other higher. In such a situation a part of the species undergoes oxidation while a part undergoes reduction. Such a species is said to undergo disproportionation.

Thus, $C r(V)$ undergoes disproportionation to $C r(V I)$ and $C r(\mathrm{III})$ while $M n(\mathrm{VI})$ undergoes disproportionation to $M n(\mathrm{VII})$ and $M n(\mathrm{IV})$


Q. Calculate the number of unpaired electrons in following

gaseous ions : $M n^{3+}, C r^{3+}, V^{3+},$ and $T i^{3+} .$ Which one of these is the most stable in aqueous solution?

Sol. $M n^{3+}:[A r] 3 d^{4} ;$ four unpaired electrons

$C r^{3+}:[A r] 3 d^{3} ;$ three unpaired electrons

$V^{3+}:[A r] 3 d^{2} ;$ two unpaired electrons

$T i^{3+}:[A r] 3 d^{1} ;$ one unpaired electron.

Out of these $C r^{3+}$ is most stable in aqueous solution.


Q. Give example and suggest reasons for the following features of the transition metal chemistry :

(i) The lowest oxide of transition metal is basic, the highest is acidic.

(ii) A transition metal exhibits higher oxidation states in oxides and fluorides.

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

Sol. (i) Transition metal oxides in lowest oxidation state are basic and in higher oxidation states become acidic in nature. This is illustrated by following example of various oxides of manganese:

In low oxidation state of metal some of its electrons are not involved in bonding and hence its effective nuclear charge is not very high and its oxide can denote electrons and behave as a base. On the other hand when the metal is in higher oxidation state, its effective nuclear charge is very high and hence its oxide has tendency to gain electrons and behaves as an acid.

(ii) It is due to the fact that fluorine and oxygen are the two most electronegative elements.

(iii) The highest oxidation state of transition element is seen in oxides and these oxides. Dissolve in acids and bases and form oxoanions of transition metal. Thus oxoanions of transition element show highest oxidation state of transition metal. For example, the highest oxidation state of $M n$ is $+7$ and it is observed in its oxoanion

$M n O_{4}^{-}$ (permanganate ion).

The highest oxidation state of $C r$ is $+6$ and it is observed in its

oxoanions, $C r O_{4}^{2-}$ (chromate ion) and $C r_{2} O_{7}^{2-}$ (dichromate ion).


Q. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Sol. An alloy is a homogeneous mixture of a metal with other metals or non-metals. A large number of alloys of transition metals are known and extensively used in modern industries. An important alloy which contains some of lanthanoid metals is “misch-metal”

Uses of Misch-Metal:

(i) Addition of about $3 \%$ misch metal to magnesium increases its strength and thus is used in making jet engine parts.

(ii) It is pyrophoric and is used in cigarette and gas lighters, tracer bullets, shells, etc.


Q. What are inner transition elements? Describe which of the following atomic numbers are the atomic numbers of the inner transition elements: $29,59,74,95,102,104$

Sol. – Inner transition elements are the elements which have partly filled $f$ -orbitals. These are also called $f$ -block elements. There are two series of inner transition elements.

(i) Lanthanoids ( $4 f$ series) : These are the 14 elements from atomic number $58-71$

(ii) Actinoids: ( $5 f$ series ): These are the 14 elements from atomic number $90-103$

Among the given atomic numbers, only 5995 and 102 are the atomic number of inner transition elements.


Q. The chemistry of the actinoid element is not so smooth as that of the lanthanoids. Justify this statement by giving some example from the oxidation state of these elements.

Sol. – Among the actinoids, there is a greater range of oxidation states as compared to lanthanoids. This is in part due to the fact that $5 f, 6 d$ and $7 s$ sub-shells are of comparable energies and the frequent electronic transition among these three sub-shells is possible. The $6 d-5 f$ transition and large number of oxidation states among actinoids make their chemistry more complicated. All actinoids in general exhibit $+3$ oxidation state. The elements in the first half of the series frequently exhibit higher oxidation state. The maximum oxidation state increases from $+4$ in $T h$ to $+5,+6$ and $+7$ respectively in $P a, U$ and $N p$ but decreases in the succeeding elements. since the distribution of oxidation states among the actionoids is so uneven and so different for the earlier and later elements, as a result their chemistry is not so smooth as that of lanthanoids.


Q. Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

Sol. – Lawrencium (Lr) is the last element in the series of the actinoids. Its electronic configuration is

$L r:[R n] 5 f^{14} 6 d^{1} 7 s^{2}$

The possible oxidation state of $L r$ is $+3$ which has stable

electronic configuration $[R n] 5 f^{14}$


Q. Use Hund’s rule to derive the electronic configuration of $C e^{3+}$ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.

Sol. – The electronic configurations of $C e$ and $C e^{3+}$ ion are:

$C e(Z=58)=_{54}[X e] 4 f^{1} 5 d^{1} 6 s^{2}$

$\mathrm{Ce}^{3+}=_{54}[\mathrm{Xe}] 4 \mathrm{f}^{1}$

It has, one unpaired electron.

‘Spin only’ formula for magnetic moment of a species,

$\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)} \mathrm{B} \cdot \mathrm{M}$

Where $n=$ no. of unpaired electrons

$\therefore$ Magnetic moment of $\mathrm{Ce}^{3+}, \mu=\sqrt{1(1+2)}=\sqrt{3} \mathrm{B} \cdot \mathrm{M} \cdot=1.732 \mathrm{B} \cdot \mathrm{M}$


Q. Name the members of the lanthanoid series which exhibit $+4$ oxidation states and those which exhibit $+2$ oxidation states. Try to correlate this type of behaviour with the electronic configuration of these elements.

Sol. $+4$ oxidation state : Cerium $(C e),$ Preseodymium (Pr), Neodymium $(N d),$ Terbium $(T b),$ Dysprosium $(D y)$ $+2$ oxidation state: Neodymium (Nd), Samarium (Sm), Europium $(E u),$ Thulium $(T m)$ and Ytterbium $(Y b)$ $+4$ and $+2$ oxidation states of lanthnoid elements arise mainly from the extra, stability of empty, half filled and fully filled $f$ -subshell.

For example $C e(I V)$ is $4 f^{0}$ and has noble gas configuration.

Similarly $T b(\mathrm{IV})$ and $E u(\mathrm{II})$ have $4 f^{7}$ configuration. $Y b(\mathrm{II})$ has

$4 f^{14}$ configuration.


Q. Write the electronic configurations of the elements with the atomic numbers $61,91,101$ and 109

Sol. Atomic number $(61)=[X e] 4 f^{5} 6 s^{2}$

Atomic number $(91)=_{86}[R n] 5 f^{2} 6 d^{1} 7 s^{2}$

Atomic number $(101)=_{86}[R n] 5 f^{13} 7 s^{2}$

Atomic number $(109)=_{86}[R n] 5 f^{14} 6 d^{7} 7 s^{2}$


Q. Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

(i) Electronic configurations

(ii) Oxidation states

(iii) Ionisation enthalpies

(iv) Atomic sizes

Sol. (i) The elements of first transition series involve progressive filling of 3 dorbitals whereas that of second and third series involve filling of $4 d$ and $5 d$ subshell respectively.

(ii) For the elements of first transition series $+2$ and $+3$ oxidation states are common and these elements form many stable complexes in these oxidation states. For the elements of second and third series higher oxidation states are more important. These elements form more stable compounds in higher oxidation states are more important. These elements form more stable compounds in higher oxidation states. For example, $\left[\mathrm{CrCl}_{6}\right]^{3-}$ is very stable whereas no equivalent complexes of $M o$ and $W$ are known. On the contrary

$\mathrm{O} \mathrm{SO}_{4}$ and $\mathrm{PtF}_{6}$ are quite stable and no corresponding compounds

for first transition series are known.

(iii) The metals of second and third transition series have higher Ionisation enthalpies than the elements of first series. The elements of second and third series form many compounds with $M-M$ bond. (iv) The atomic radii of elements of second and third transition series are larger than those of the elements of first series. Because of lanthanoid contraction, the radii of third series are almost equal to those of second row.


Q. Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

Sol. Some points of difference between properties of elements of the first transition series and those of hevier transition elements are given below:

(i) For elements of first series $+2$ and $+3$ oxidation states are more common while for the heavier transition elements higher oxidation states are more common.

(ii) $M-M$ bonding is rare in the elements of first series but is quite common in heavier transition elements.

(iii) The elements of first transition series do not form complexes with coordination no. 7 or 8 whereas heavier transition elements do so.

(iv) The elements of first transition series form low spin or high spin complexes depending upon strength of the ligand field. On the other hand heavier transition elements form low spin complexes irrespective of the strength of the ligand field


Q. What can be inferred from the magnetic moment values of the following complex species?

Sol. Calculate the unpaired electron by applying formula $\mu=\sqrt{n(n+2)}$

(i) The magnetic moment of 2.2 BM corresponds to $n=1 .$ Thus, in $K_{4}\left[M n(C N)_{6}\right]$ there is only one unpaired electron in $3 d$ subshell.

Thus, the distribution of five $3 d$ electrons in $M n(\mathrm{II})$ is $\left(t_{2 g}\right)^{5}$

(ii) This indicates that there are four unpaired electrons in the complex. Thus, the six $3 d$ electrons in $F e(\Pi)$ are distributed as $\left(t_{2 g}\right)^{4}\left(e_{g}\right)^{2}$

(iii) This indicates that there are five unpaired electrons in the complex. Hence, the five $3 d$ electrons in $M n($ II ) are distributed as $\left(t_{2 g}\right)^{3}\left(e_{g}\right)^{2}$ in the given complex.


Q. What are the characteristics of the transition elements and why are they called transition elements? Which of the $d-$ block elements may not be regarded as the transition elements?

Sol. The important characteristics of transition metals are:

(i) All transition elements are metallic in nature.

(ii) These metals exhibit variable oxidation states.

(iii) Transition metal atoms or ions generally form the complexes with neutral, negative and positive ligands.

(iv) Compounds of transition metals are usually coloured.

(v) The compounds of these metals are usually paramagnetic in nature.

(vi) Transition metals and their compounds act as good catalysts.

(vii) These metals form various alloys with other metals of the series

(viii) These metals form interstitial compounds with C, N, B and H

The presence of partially filled $d$ -orbitals in the electronic configuration of atomic and ionic species of these elements is responsible for the characteristic properties of transition elements. They are called transition elements because of their position in the periodic table. On one side of transition metals are highly reactive metals of s-block whereas on the side are less reactive metals of $p-$

block. A transition element may be defined as a element whose atom or at least one of its simple ions contains partially filled $d-$ orbitals, e.g., iron, copper, chromium, nickel etc. The elements zinc, cadmium and mercury are generally not regarded as transition elements as their atoms and all ions formed do not have partially filled $d$ -orbitals in atomic state or common oxidation state $\left(Z n^{2+}, C d^{+}, H g^{2+}\right)$


Q. Compare the chemistry of actinoids with that of the lanthanoids with special reference to

(i) Electronic configuration

(ii) Oxidation state

(iii) Atomic and ionic sizes

(iv) Chemical reactivity

Sol. (i) Electronic configuration: All the lanthanoids have electronic configuration with $6 s^{2}$ common with variable occupancy of $4 f$ and $5 d$ sub-shells. All actinoids have the electronic configuration of $7 s^{2}$ and variable occupancy of $5 f$ and $6 d$ sub-shell.

(ii) Oxidation state : Lanthanoids exhibit a principal oxidation state of $+3 .$ Cerium and terbium also exhibit oxidation state

of $+4 . \mathrm{Sm}^{2+}, \mathrm{Eu}^{2+}$ and $\mathrm{Yb}^{2+}$ ions also exist in aqueous solutions. The common oxidation state of these elements is $+3 .$ However, they also exhibit oxidation state of $+2,+4,+5$ $+6$ and $+7 .$ Thus, actinoids exhibit greater range of oxidation states.

(iii) Atomic and ionic sizes : The atomic radii and ionic radii of

tripositive lanthanoid ions $\left(M^{3+}\right)$ show a steady and gradual decrease in moving from La to Lu this is due to poor shielding of $4 f$ -electrons. In actinoid, the steady decrease in ionic radii with increase in atomic number referred to as actinoid contraction. The contraction is greater from element to element in this series. This is due to poor shielding by $5 f$ -electrons (iv) Chemical reactivity : The earlier member of the series are quite reactive similar to calcium, however with increase in atomic number their reactivity decreases and they behave more like aluminium. Many lanthanoids react with carbon to form salt like carbides and with hydrogen to give salt-like hydrides.

Actinoids tarnish in air due to formation of oxide coating. They react with not water. The actinoids react with most non-metals at moderate temperatures.


Q. Indicate the steps in the preparation of:

(i) $\quad K_{2} C r_{2} O_{7}$ from chromite ore (ii) $K M n O_{4}$ from pyrolusite ore.

Sol. (i) $\quad \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ from chromite ore:

Following various steps are involved in the preparation of from chromite ore.

(a) Preparation of sodium chromate: The powdered ore is heated with molten alkali in the presence of air in a reverberatory furnance to produce sodium chromate.

(b) Conversion of sodium chromate in to sodium dichromate:

Sodium chromate is extracted with water and acidified with sulphuric acid to get sodium dichromate.

$2 \mathrm{Na}_{2} \mathrm{CrO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}$

On cooling, sodium sulphate separates out as $\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}$

and the solution contains sodium dichromate in it from this solution

orange $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \cdot 2 \mathrm{H}_{2} \mathrm{O}$ can be crystallized.

(c) Conversion of sodium dichromate into potassium dichroamte:

The solution containing sodium dichromate is treated with potassium chloride.

$\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{KCl} \longrightarrow \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{NaCl}$

Sodium chloride being least soluble separates out and is removed by filtration. On cooling potassium dichromate crytallizes out in the form of orange crystals. potassium dichromate is fairly soluble in water.

(ii) $\mathrm{KMnO}_{4}$ from pyrolusite ore:

Pyrolusite is fused with $\mathrm{KOH}$ in the presence of atmospheric oxygen or an oxidizing agent like potassium nitrate or potassium

chlorate to give potassium manganate $\mathrm{K}_{2} \mathrm{MnO}_{4}$

The green mass is extracted with water and oxidized to potassium permanganate, electrolytically or by passing chlorine or-ozone into solution.

Electrolytic oxidation:

During electrolysis manganate ions are oxidized to permanganate at anode.

At cathode hydrogen gas is liberated

$2 \mathrm{H}_{2} \mathrm{O}+2 e^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}$

oxidation by chlorine

$2 \mathrm{K}_{2} \mathrm{MnO}_{4}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{KMnO}_{4}+2 \mathrm{KCl}$

$2 M n O_{4}^{2-}+C l_{2} \longrightarrow 2 M n O_{4}^{2-}+2 C l$

or oxidation by ozone

$2 \mathrm{K}_{2} \mathrm{MnO}_{4}+\mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{KMn} \mathrm{A}+2 \mathrm{KOH}+\mathrm{O}_{2}$

$2 M n O_{4}^{2-}+O_{3}+H_{2} O \longrightarrow 2 M n O_{4}^{2-}+2 O H^{-}+O_{2}$

As a result of oxidation, the green colour of solution changes into

purple. The purple solution containing $K M n O_{4}$ is concentrated

by evaporation which on cooling gives crystals of $\mathrm{KMnO}_{4}$.


Q. Write down the number of $3-d$ electrons in each of the following ions $: T i^{2+}, V^{2+}, C r^{3+}, M n^{2+} \quad F e^{2+}, F e^{3+}, C o^{2+}, N i^{2+}$

and $C u^{2+} .$ Indicate how would you expect the five $3 d$ orbitals to be occupied for these hydrated ions (octahedral).

Sol. (i) $\quad T i^{2+}:[A r] 3 d^{2}$ Under the influence of an octahedral ligand field the $d$ -orbitals split into two groups of different energies.

These are $t_{2 g}$ orbitals $(d x y, d x z \text { and } d y z)$ and $e_{g}$ orbitals $\left(d z^{2} \text { and } d x^{2}-y^{2}\right),$ orbitals are of lower energies than orbitals. Thus, the two $d$ -electrons in would be present in two of the three orbitals. Thus there would be two unpaired electrons.

(ii) The three $d$ -electrons in would be present in three different orbitals. Thus, there would be three unpaired electrons.

(iii) $-$ Three unpaired electrons

(iv) – Five unpaired electrons

(v) $\quad$ – Four unpaired electrons

(vi) – Five unpaired electrons

(vii) $-$ Three unpaired electrons

(viii) – Two unpaired electrons

(ix) – One unpaired electron


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Q. Why is $N_{2}$ less reactive at room temperature?

Sol. Due to the presence of a triple bond between the two nitrogen

atoms, the bond dissociation energy of $N_{2}\left(941.4 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ is very high. Therefore, $N_{2}$ is less reactive at room temperature.


Q. How does ammonia react with a solution of $C u^{2+}$ ion $?$

Sol. Ammonia forms a soluble complex with $C u^{2+}$ ion. $C u^{2+}+4 \ddot{N} H_{3} \rightarrow\left[C u\left(N H_{3}\right)_{4}\right]^{2+}$


Q. What happens when white phosphorus is heated with concentrated $\mathrm{NaOH}$ solution in an inert atmosphere of $\mathrm{CO}_{2} ?$

Sol. White phosphorus reacts with $\mathrm{NaOH}$ to form phosphine.


Q. What is the basicity of $H_{3} P O_{4} ?$

Sol. – The structure of $H_{3} P O_{4}$ is tetrahedral. It has three $P-O H$ and

one $P=O$ bond as shown below :

since it has three ionizable $P-O H$ bonds, therefore $H_{3} P O_{4}$ is tribasic as shown below:

since it is tribasic, it forms three series of salts with $N a O H$

viz, $\mathrm{NaH}_{2} \mathrm{PO}_{4}, \mathrm{Na}_{2} \mathrm{HPO}_{4}$ and $\mathrm{Na}_{3} \mathrm{PO}_{4}$


Q. Which of the following does not react with oxygen directly? $Z n, T i, P t, F e$

Sol. Platinum is a noble metal. The sum of its first four ionization enthalpies is very large. Therefore, it does not react with oxygen directly. In contrast $Z n, T i$ and $F e$ are active metals and hence directly react with oxygen to form their respective oxides.


Q. Sea is the greatest source of halogens. Comment.

Sol. Sea water contains chlorides, bromides and iodides of sodium, potassium, magnesium and calcium, but is mainly sodium chloride solution $(2.5 \%$ by mass). Dried up sea beds contain sodium chloride and carnallite, $K C l . M g C l_{2} .6 H_{2} O .$ Certain seaweeds contain upto $0.5 \%$ of iodine as sodium iodide and chile saltpetre $\left(N a N O_{3}\right)$ contain upto $0.2 \%$ of sodium iodate. Thus, sea is the greatest source of halogens.


Q. Give the reason for the bleaching action of $C l_{2}$

Sol. In presence of moisture or in aqueous solution, $C l_{2}$ liberates nascent oxygen.

This nascent oxygen brings about the oxidation of coloured substances present in vegetable and organic matter to colourless substances.

Coloured substance $+O \rightarrow$ Colourless substance

Thus, the bleaching action of is due to oxidation. Since the bleaching effect of is permanent, it is used to bleach cotton and textiles and wood pulp for manufacture of paper and rayon.


Q. Name two poisonous gases which can be prepared from chlorine gas.

Sol. poisonous gases which can be prepared from $C l_{2}$ are:

(i) Phosgene $\quad$

(ii) $\quad$ Mustard gas.

These are prepared as follows:

(ii) is passed through boiling $S,$ when is formed. This when reacted with ethene gives mustard gas.


Q. Why is $I C l$ more reactive than $I_{2} ?$

Sol. ICl is more reactive than $I_{2}$ because $I-C l$ bond is weaker than $I-I$ bond. Consequently, $I C l$ breaks easily to form halogen atoms which readily bring about the reastions.


Q. Why is helium used in diving apparatus?

Sol. – Because of its low solubility (as compared to $\left.N_{2}\right)$ in blood, a mixture of oxygen and helium is used in diving apparatus used by deep sea divers.


Q. Balance the following equation: $X e F_{\delta}+H_{2} O \rightarrow X e O_{2} F_{2}+H F$

Sol. $\mathrm{XeF}_{6}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{XeO}_{3}+6 \mathrm{HF}$


Q. Why has it been difficult to study the chemistry of radon?

Sol. – Because radon is a radioactive element with a short half life of 3.82 days. This makes the study of chemistry of radon difficult.


Q. Write the reaction of thermal decomposition of sodium azide.

Sol. Sodium azide decomposes to form sodium metal and $N_{2}$ gas.


Q. Why does $N H_{3}$ act as a Lewis base?

Sol. N is $N H_{3}$ has one lone pair of electrons which is available for donation. Therefore, it donates an electron pair to a proton to form $N H_{4}^{+}$ ion and hence it acts as a Lewis base.


Q. Why does $N O_{2}$ dimerise?

Sol. $\mathrm{NO}_{2}$ is an odd electron $(7+2 \times 8=23)$ molecule. In the valence shell, $N$ has seven electrons and hence is unstable. To become stable by having even ( 8) number of electrons in the valence shell, it undergoes dimerization to form $N_{2} O_{4}$


Q. In what way it can be proved that $P H_{3}$ is basic in nature?

Sol. $P H_{3}$ reacts with acids like $H I$ to form $P H_{4} I .$ This shows that is basic in nature.

Due to the presence of a lone pair of electrons on $P H_{3}$ is acting as a Lewis base in the above reaction.


Q. Why does $P C l_{3}$ fume in moisture?

Sol. In presence of moisture $\left(H_{2} \mathrm{O}\right), P C l_{3}$ undergoes hydrolysis giving fumes of $H C l$


Q. $H_{2} S$ is less acidic then $H_{2} T e$ why?

Sol. – As the size of the element increases down the group, $E-H$ bond distance increases and hence $E-H$ bond dissociation energy decreases. In other words, $H-S$ bond dissociation energy is higher than that of $H-T e$ bond dissociation energy and hence $H-S$ bond

breaks less easily than $H-$ Te bond. Therefore, $H_{2} S$ is a weaker

acid than $\mathrm{H}_{2} \mathrm{Te}$


Q. When $H C /$ reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Why?

Sol. Fe reacts with $\mathrm{HCl}$ to form $\mathrm{FeCl}_{2}$ and $\mathrm{H}_{2}$

$$ F e+2 H C l \rightarrow F e C l_{2}+H_{2} $$

thus liberated prevents the oxidation of to


Q. Unlike phosphorus, nitrogen shows little tendency for catenation?

Sol. Due to smaller size, the lone pair of electrons on the $N$ -atoms repel the bond pair of $N-N$ bond. In contrast, because of comparatively larger size of $P,$ the lone pair of electrons on $P$ atoms do not repel the bond pair of the $P-P$ bond to the same extent. As a result, $N_{-}$ $N$ single bond is weaker than $P-P$ single bond. Consequently nitrogen shows little tendency for catenation while $P$ has a distinct tendency for catenation.


Q. Give the disproportionation reaction of $H_{3} P O_{3}$

Sol. Phosphorus acid $\left(H_{3} P O_{3}\right)$ when heated at $473 K$ disproportionate to form orthophosphoric acid and phosphine.


Q. Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.

Sol. – Oxygen atom can form $H$ bonds whereas chlorine does not. The tendency for $H$ bonding depends upon

(i) Small size and

(ii) High electronegativity values Although the electronegativity of $O$ and $C l$ is nearly same yet chlorine does not form $H$ bond due to its larger size (99 pm) as compared to oxygen $(66 \mathrm{pm})$


Q. How can you prepare $\mathrm{Cl}_{2}$ from $\mathrm{HCl}$ and $\mathrm{HCl}$ from $\mathrm{Cl}_{2} ?$ Write reactions only.

Sol. (i) Chlorine can be obtained from $H C l$ by heating with oxidising

agents like $M n O_{2}, K M n O_{4}, K_{2} C r_{2} O_{7}, O_{3}$ etc.

(ii) $\quad H C l$ can be obtained from chlorine by burning it in excess of hydrogen $H_{2}+C l_{2} \rightarrow 2 H C l$


Q. With what neutral molecule is $\mathrm{ClO}^{-}$ isoelectronic ? Is that molecule a Lewis Base?

Sol. $\mathrm{ClO}^{-}$ species has 26 electrons $(17(\mathrm{Cl})+8(\mathrm{O})+1 \bar{e}$ (charge). The meutral molecule which is isoelectronic with it is $C I F(17+9=26 \bar{e})$

ClF can act as a lewis base due to presence of lone pair of electrons.


Q. Which one of the following does not exist?

Sol. (ii) $\quad$ NeF $_{2}$ cannot exist because Neon $(z=10)$ do not contain vacant $d$ -orbitals in its valence shell and hence cannot extend its valency.


Q. Why are pentahalides more covalent than trihalides ?

Sol. The elements of group 15 have five electrons (two in the $s$ -and three in the $p \text { -orbitals })$ in their respective valence shells. since it is

difficult to lose all the three electrons to form $E^{3+}$ or even more difficult to lose all the five valence electrons (two $s$ -and three $p-$ ) to $E^{5+}$ ions, therefore, higher elements have no tendency to form ionic compounds. Instead they form covalent compounds by sharing of electrons. since elements in the $+5$ oxidation state have less tendency to lose electrons than in the $+3$ oxidation state, therefore, elements in the $+5$ oxidation state have more tendency to share electronts in the the $+3$ oxidation state. Thus, elements in the $+5$ oxidation state are $_{j}$ more covalent than in the $+3$ oxidation state. In other words, pentahalides are more covalent than trihalides.


Q. Why is $B i H_{3}$ the strongest reducing agent amongst all the hydrides of Group 15 elements?

Sol. As we move down the group, the size of the element increases and, therefore, the length of the $B-H$ bond increases and its strength decreases. In other words, as we move down the group, the $B-H$ bond can break more easily to evolve $H_{2}$ gas which acts as the reducing agent. Thus, $B i-H$ bonds is the weakest amongst the hydrides of elements of group 15 and hence $B i H_{3}$ is the strongest reducing agent.


Q. Mention the conditions to maximise the yield of ammonia.

Sol. Ammonia is prepared by the Haber’s process.

In accordance with Le Chatelier’s principle, to maximize the yield,

a high pressure of $200 \times 10^{5} \mathrm{Pa}$ is used. To increase the rate of the reaction, a temperature of around $700 K$ is used and iron oxide

mixed with some $K_{2} O$ and $A l_{2} O_{3}$ is used as a catalyst. Sometimes, Mo is also used as a promoter to increase the efficiency of the Fe catalyst.


Q. What is the covalency of nitrogen in $N_{2} O_{5} ?$

Sol. Covalency depends upon the number of shared pairs of electrons.

Now in $N_{2} O_{5},$ each nitrogen atom has four shared paired of electrons as shown below:

Therefore, the covalency of $N$ in $N_{2} O_{5}$ is 4


Q. Bond angle in $P H_{4}^{+}$ is higher than in $P H_{3} .$ Why?

Sol. $P$ in $P H_{3}$ is $s p^{3}$ -hybridized. It has three bond pairs and one lone pair around $P .$ Due to greater lone pair-bond pair repulsion than bond pair-bond pair repulsion, the tetrahedral angle decreases from $109^{\circ}-28^{\prime}$ to $93.6^{\circ} .$ As a result, is pyramidal. However, when it reacts with a proton, it forms which has four bond pairs and no lone pair. Due to the absence of lone pair-bond pair repulsion and

presence of four identical bond pair-bond pair interactions $\mathrm{PH}_{4}^{+}$ assumes tetrahedral geometry with a bond angle of $109^{\circ}-28^{\prime} .$ This

explains why the bond angle in $\mathrm{PH}_{4}^{+}$ is higher than in $\mathrm{PH}_{3}$


Q. What happens when $P C l_{5}$ is heated?

Sol. $-P C l_{5}$ has three equatorial $(202 \text { pm long) and two axial }(240\text { pm }$ long) bond. since axial bonds are weaker than equatorial bonds,

therefore, when $\mathrm{PCl}_{5}$ is heated, the less stable axial bonds break

to form $\mathrm{PCl}_{3}$


Q. Write a balanced equation for the hydrolytic reaction of

$P C l_{5}$ with heavy water.

Sol. It reacts with heavy water to form phosphorus oxychloride $\left(P O C l_{3}\right)$ and deuterium chloride $(D C l)$


Q. What happens when $H_{3} P O_{3}$ is heated?

Sol. The oxidation state of $P$ in $H_{3} P O_{3}$ is $+3 .$ since this value is intermediate between the highest $(+5)$ and lowest $(-3)$ oxidation states of $P,$ therefore, when is heated, it undergoes disproportionation to form and with oxidation states of $-3$ and $+5$ respectively.


Q. List the important sources of sulphur.

Sol. Sulphur mainly occurs in the earth’s crust in the combined state primarily in form of sulphates and sulphides.

Sulphates : Gypsum, $\quad \mathrm{CaSO}_{4} .2 \mathrm{H}_{2} \mathrm{O} \quad ;$ epsom salt

$M g S O_{4} \cdot 7 H_{2} O ;$ baryte, $B a S O_{4},$ etc.

Sulphides : Galena, $P b S ;$ zinc blende, $Z n S$; copper pyrites,

CuFeS_{2}

Traces of sulphur occur as $H_{2} S$ and in organic materials such as eggs, proteins, garlic, onion, mustard, hair and wool.


Q. Write the order of thermal stability of the hydrides of group 16 elements.

Sol. – As the size of the element increases down the group, the $E-H$ bond dissociation energy decreases and hence $E-H$ bond breaks more easily. Thus, the thermal stability of the hydrides of group 16 elements decreases down the group,

$H_{2} O>H_{2} S>H_{2} S e>H_{2} T e>H_{2} P_{O}$


Q. Why is $H_{2} O$ liquid and $H_{2} S$ is a gas?

Sol. Due to greater electronegativity of $O$ than $S, H_{2} O$ under goes extensive intermolecular $H$ -bonding. As a result, exists as an associated molecule in which each $O$ is tetrahedrally surrounded by four water molecules. Quite a larger amount of energy is required to break these molecules. Therefore, is a liquid at room temperature.


Q. Complete the following reactions:

(i) $\quad C_{2} H_{4}+O_{2} \rightarrow$

(ii) $4 A l+3 O_{2} \rightarrow$

Sol. (i) $C_{2} H_{4}$ undergoes combustion to form $C O_{2} \& H_{2} O$

(ii) $A l$ combines with $O_{2}$ to form alumina


Q. Why does ozone act as a powerful oxidising agent ?

Sol. $\mathrm{O}_{3}$ is an endothermic compound. On heating, it readily decomposes to give dioxygen and nascent oxygen.

since nascent oxygen is very reactive, therefore, $O_{3}$ acts as a powerful oxidising agent.


Q. How is $O_{3}$ estimated quantitatively?

Sol. When $O_{3}$ is treated with excess of $K I$ solution buffered with

berate buffer (pH 9.2), I $_{2}$ is liberated quantitatively.

$2 I^{-}(a q)+H_{2} O(l)+O_{3}(g) \rightarrow 2 O H^{-}(a q)+I_{2}(s)+O_{2}(g)$

The thus liberated is titrated against a standard solution of sodium thiosulphate using starch as an indicator.


Q. What happens when sulphur dioxide is passed into aqueous solution of $F e(\mathrm{III})$ salt $?$

Sol. $\mathrm{SO}_{2}$ acts as areducing agent and hence reduces an aqueous solution

-of $F e(\text { III ) salt to } F e(I I)$ salt.


Q. Comment upon the nature of two $S-O$ bonds formed in $S O_{2}$ molecule? Are the two $S-O$ bonds in this molecule equal?

Sol. In $S O_{2}, S$ is $s p^{2}$ -hybridized. Two of the three -orbitals form two $\sigma$ -bonds while the third contains the lone pair of electrons. $S$ is now left with one half-filled $p$ -orbital and one half-filled $d$ -orbitals. These form one and one double bond with oxygen atom. Thus, has bent structure with $O S O$ bond angle of $119.5^{\circ} .$ Due to resonance, the two bonds are equal $(143 \mathrm{pm})$


Q. How is the presence of $S O_{2}$ detected?

Sol. $\mathrm{SO}_{2}$ is a pungent smelling gas. It can be detected by the following two tests :

(i) turns the pink violet colour of $\mathrm{KMnO}_{4}$ to colourless due

to reduction of $\mathrm{MnO}_{4}^{-}$ to $\mathrm{Mn}^{2+}$ ions.

(ii) $\mathrm{SO}_{2}$ turns acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ green due to reduction of $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ to $\mathrm{Cr}^{3+}$ ions.


Q. Mention three areas in which $H_{z} S O_{4}$ plays an important role.

Sol. (i) $\quad H_{2} \mathrm{SO}_{4}$ is used in the manufacture of fertilizers such as ammonium sulphate, calcium super phosphate.

(ii) It is used as an electrolyte in storage batteries.

(iii) It is used in petroleum refining, detergent industry and in the manufacture of paints, pigments and dyes.


Q. Write the conditions to maximise the yield of $H_{2} S O_{4}$ by contact process.

Sol. The key step in the production of $H_{2} S O_{4}$ is the oxidation of

The reaction is exothermic, reversible and the forward reaction proceeds with decrease in volume. Therefore, in accordance with

Le chatelier’s principle, to maximize the yield of $S O_{3}$ and hence

of $H_{2} S O_{4},$ a low temperature $(720 \mathrm{K}$ to have optimum rate), a

high pressure $( \left2 \text { bar) and } V_{2} O_{5}$ is used as a catalyst (to increase \right. the rate at $720 \mathrm{K}$ ).


Q. Why is $K_{a_{2}}$ less than $K_{a_{I}}$ for $H_{2} S O_{4}$ in water?

Sol. $H_{2} S O_{4}$ is a dibasic acid, it ionizes in two stages and hence has two dissociation constant.

$K_{a_{2}}$ is less than $K_{a_{1}}$ because the negatively charged $H S O_{4}^{-}$ ion has much less tendency to donate a proton to $H_{2} O$ as compared to neutral to donate a proton to water.


Q. Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration energy, compare the oxidising power of $F_{2}$ and $C l_{2}$

Sol. The electrode potential of $F_{2}(+2.87 V)$ is much higher than that of

$\mathrm{Cl}_{2}(+1.36 \mathrm{V}),$ therefore, $\mathrm{F}_{2}$ is a much stronger oxidising agent

than $C l_{2} .$ Now, electrode potential depends upon three factors:

(i) Bond dissociation energy,

(ii) Electron gain enthalpy and

(iii) Hydration energy. Although electron gain enthalpy of fluorine is less negative $\left(-333 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ than that of chlorine

$\left(-349 \mathrm{kJ} \mathrm{mol}^{-1}\right),$ the bond dissociation energy of $\mathrm{F}-\mathrm{F}$ bond is

much lower $\left(158.8 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ than that of $\mathrm{Cl}-\mathrm{Cl}$ bond

$\left(242.6 \mathrm{kJ} \mathrm{mol}^{-}\right)$ and hydration energy of $\mathrm{F}^{-} \mathrm{ion}\left(515 \mathrm{kJ} \mathrm{mol}^{-1}\right)$

is much higher than that of $C l^{-}$ ion $\left(381 k J m o l^{-1}\right) .$ The later two factors more than compensate the less negative electron gain

enthalpy of $F_{2} .$ As a result, electrode potential of is higher than that of chlorine and hence is a much stronger oxidising agent than


Q. Give two examples to show the anomalous behaviour of fluorine.

Sol. The anomalous behaviour of fluorine is due to its

(i) Small size,

(ii) Highest electronegativity,

(iii) Low $F-F$ bond dissociation enthalpy and

(iv) Non-availability of $d$ -orbitals in its valence shell. The two examples are:

1. Due to non-availability of $d$ -orbitals in its valence shell, fluorine cannot expand its octet, therefore, fluorine shows only an oxidation state of $-1$ while all other halogens due to the presence of $d$ -orbitals, show positive oxidation states of $+1,+3,+5$ and $+7$ besides oxidation state of $-1$

2. Due to its small size, the three lone pair of electrons on each $F$ atom in $F-F$ molecule, repel the bond pair. As a result, $F-F$ bond dissociation energy is lower than that of $C l-C l$ bond.


Q. Though nitrogen exhibits $+5$ oxidation state, it does not form pentahalide, Give reason or Nitrogen does not form pentahalide. Explain.

Sol. The electronic configuration of nitrogen is $1 s^{2} 2 s^{2} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1} .$ It has three half-filled $p$ -orbitals and hence can form a trihalide. To make a pentahalide, we need five half filled orbitals. Since nitrogen with $n=2$ can have only $s$ -and $p$ -orbitals and no $d$ -orbitals, it cannot expand its valence shell to show a covalency of $5 .$ At the same time, $N$ has a vacant $3 s$ orbital. If one of the $2 s$ -electron gets promoted to $3 s$ orbital, we can still have five half-filled orbitals to

form $\mathrm{NCl}_{5}$. since energy required to promote one 2s electron to 3s orbital is more than the energy released during the formation of two additional bonds, therefore, such on excitation is thermodynamically not feasible. That is why nitrogen does not form a pentahalide.


Q. $P H_{3}$ has lower boiling point than $N H_{3} .$ Why?

Sol. The electronegativity of $N(3.0)$ is much higher than that of $P(2.1)$

Therefore, $N H_{3}$ undergoes extensive intermolecular $H$ -bonding and hence it exists as an associated molecule. To break these $H-$ bonds, a large amount of energy is needed. On the other hand, $P H_{3}$ does not undergo $H$ -bonding and thus exists as discrete molecules. Therefore, the boiling point of is much lower than that of


Q. Are all the five bonds in $P C l_{5}$ equivalent? Justify.

Sol. $P C l_{5}$ has trigonal bipyramid structure. It has three equatorial bonds inclined at angle of $120^{\circ}$ and two axial bonds inclined at angle of $90^{\circ}$

As a result, axial bond pairs suffer greater repulsion than equatorial bond pairs. Consequently, axial bonds are longer $(240 p m)$ as compared to equatorial bonds $(202 p m) .$ Thus, all the five $P-C l$ bonds in $P C l_{5}$ are not equivalent.


Q. How do you account for the reducing behaviour of $\mathrm{H}_{3} \mathrm{PO}_{2} ?$

Sol. The structure of $H_{3} P O_{2}$ has two $P-H$ bonds. Due to the presence

of these $P-H$ bonds $\mathrm{H}_{3} \mathrm{PO}_{2},$ acts as a strong reducing agent. For example, it reduces $\mathrm{AgNO}_{3}$ to $A g$ and arene diazonium salts to

arenes.


Q. Elements of group 16 generally show lower value of first ionization enthalpy compared to the corresponding periods of group $15 .$ Why?

Sol. This is because of the following two reasons:

(i) Due to extra stability of the exactly half-filled $p$ -orbitals electronic configuration of group 15 elements, they require larger amount of energy to remove an electron as compared to group 16 elements.

(ii) After removal of an electron, the group 16 elements yield species which have more stable exactly half-filled $p$ -orbitals electronic configuration while this is not so in case of group 15 elements.


Q. Which form of sulphur shows paramagnetic behaviour?

Or

sulphur in the vapour state exhibits paramagnetism.

Sol. In the vapour state $(-1000 \mathrm{K}),$ sulphur partly exists as $S_{2}$ molecule

which has two unpaired electrons in the antibonding $\pi^{*}$ -molecular

orbitals like $O_{2}$ and hence, $\mathrm{S}_{2}$ exhibits paramagnetism.


Q. What happens when :

(i) Concentrated $H_{2} S O_{4}$ is added to calcium fluoride:

(ii) $\quad S O_{3}$ is passed through water?

Sol. (i) $\quad H_{2} \mathrm{SO}_{4}$ being less volatile displaces more volatile $H F$ from $C a F_{2}$

$\mathrm{CaF}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{CaSO}_{4}+2 \mathrm{HF}$

(ii) $\mathrm{SO}_{3}$ dissolves in water to form $\mathrm{H}_{2} \mathrm{SO}_{4}$

$$ \mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4} $$


Q. Halogens have maximum negative electron gain enthalpy. Explain why?

Sol. Halogens have the smallest size (except noble gases) in their respective periods and, therefore, have high effective nuclear charge. As a result, they readily accept one electron to acquire the stable electronic configuration of the nearest noble gas. In other words, large amount of energy is released when a halogen atom accepts an electron to form the corresponding halide ion and thus halogens have maximum negative electron gain enthalpy.


Q. Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidising agent than chlorine. Why?

Sol. – The oxidising capability of a substance can be determined by its electrode potential. Higher the electrode potential, stronger is the oxidising agent. The electrode potential, in turn, depends of three factors :

(i) Bond dissociation enthalpy

(ii) Electron gain enthalpy

(iii) Hydration energy Although electron gain enthalpy of fluorine is less negative than that of chlorine but

(i) Enthalpy of dissociation of $F-F$ bond is lower than that of $C l-C l$ bond

(ii) Hydration energy of $F^{-}$ ion is higher than that of $C l^{-}$ ion.

Because of these two reasons, electrode potential of $F_{2}(+2.87 \mathrm{V})$

is muchhigher than that of $C l_{2}(+1.36 V)$ and hence $F_{2}$ is astronger

oxidising agent than $C l_{2}$


Q. Fluorine exhibits only $-1$ oxidation state whereas other halogens exhibit $+1,+3,+5$ and $+7$ oxidation states. Explain.

Sol. Fluorine, like other halogens has seven electrons in the valence shell. since it is the most electronegative element, therefore, it readily accepts one electron to acquire the stable electronic configuration of the nearest noble gas. Therefore, like other halogens, it also exhibits an oxidation state of – 1

Further, since fluorine is the most electronegative element and does not have $d$ -orbitals in its valence shell, therefore, it cannot expand its octet and hence does not show positive oxidation states. On the other hand, other halogens have $d$ -orbitals and, therefore, ¡can expand their octets and thus show positive oxidation states of $+1,+3,+5$ and $+7$ in addition to oxidation state of $-1$ can expand their octets and thus show positive oxidation states of $+1,+3,+5$ and $+7$ in addition to oxidation state of $-1$


Q. Write the balanced chemical equation for the reaction of $C l_{2}$ with hot and concentrated $\mathrm{NaOH}$ ? Is this reaction a disproportionation reaction ? Justify.

Sol. This is an example of a disproportionation reaction because the

oxidation state of chlorine decreases from zero in $C l_{2}$ to $-1$ in $N a C l$ and increases from zero in to $+5$ in

#tag#


Q. Deduce the molecular shape of $B r F_{3}$ on the basis of $V S E P R$ theory.

Sol. – No. of electrons in the valence shell of central $B r$ atom $=7$

No. of electrons provided by three $F$ atoms $=3 \times 1=3$

Total no. of electrons around the central atom $(B r)=7+3=10$

$\therefore$ Total no. of electron pairs around the central $B r$ atom $=10 / 2=5$

But the no. of bond pairs $=3(\because \text { there are three } B r-F \text { bonds })$

$\therefore$ No. of lone pairs $=5-3=2$

On the basis of $V S E P R$ theory, a molecule with three bond pairs and two lone pairs must be $T$ -shaped (i.e. distorted trigonal bipyramidal) as shown above.

The two lone pairs will occupy equatorial positions to minimize lone pair-lone pair and lone pair-bond pair repulsions which are greater than bond pair-bond pair repulsions. Further, the axial $F$ atoms will be slightly bent towards the equatorial $F$ atom to minimize

the lone pair-lone pair repulsions. Therefore, the shape of $B r F_{3}$ would be slightly bent $T$ as shown above.

#tag#


Q. Why are elements of group 18 known as noble gas?

Sol. The elements of group 18 have their valence shell orbitals completely filled. As a result, they react with only a few elements (oxygen and fluorine) only under certain conditions. Therefore, they are called noble gases.

#tag#


Q. Noble gases have very low boiling points. Why ?

Sol. Noble gases are monoatomic. Their atoms are held together by weak dispersion forces and hence can be liquefied at very low temperatures. Therefore, they have low boiling points.

#tag#


Q. Does the hydrolysis of $X e F_{6}$ lead to a redox reaction?

Sol. Complete hydrolysis of $\mathrm{XeF}_{6}$ gives $\mathrm{XeO}_{3}$

On the other hand, partial hydrolysis of $\mathrm{XeF}_{6}$ gives $\mathrm{XeOF}_{4}$

and $\mathrm{XeO}_{2} \mathrm{F}_{2}$

Since the oxidation states of all the elements in the products of hydrolysis and reactants is remain the same, therefore, it is not a redox reaction.

#tag#


Q. Why does the reactivity of nitrogen differ from phosphorus?

Sol. The reactivity of nitrogen is different from phosphorus because of the following reasons :

(i) Nitrogen has a small size, high electronegativity, high ionisation enthalpy as compared to phosphorus.

(ii) Nitrogen does not contain vacant $d$ -orbitals in its valence shell whereas phosphorus contains vacant $d$ -orbitals in its valence shell.

(iii) Nitrogen has ability to form $(N \equiv N)$ triple bond as a result of which its bond enthalpy $\left(941.4 \mathrm{kJmol}^{-1}\right)$ is very high making it less reactive.

#tag#


Q. Discuss the trends in chemical reactivity of group 15 elements.

Sol. The elements of group 15 differ from one another appreciably in their chemical reactivity. Nitrogen has a very high dissociation energy $\left(941.4 \mathrm{kJmo}^{-1}\right)$ and is practically inert, which is why it has accumulated in large amounts in the atmosphere. Phosphorus, in one of its allotropic forms i.e. white phosphorus is extremely reactive. The strained structure of $P_{4}$ is responsible for high chemical activity. It catches fire when exposed to air forming $P_{4} O_{10} .$ The other allotrope, red phosphorus is stable in air at room temperature, though it reacts on heating. The heavier elements, $A s, S b$ and $B i$ are less reactive. Arsenic is stable in dry air. When heated in air it sublimes at $615^{\circ} \mathrm{C}$ forming $\mathrm{As}_{4} \mathrm{O}_{6}$

Antimony is less reactive and stable towards water and air. On heating in air it forms $S b_{4} O_{6}, S b_{4} O_{8}$ or $S b_{4} O_{10} .$ Bismuth is also known to form $B i_{2} O_{3}$ on heating.

#tag#


Q. Why does $N H_{3}$ form hydrogen bond but $P H_{3}$ does not?

Sol. In $\mathrm{NH}_{3},$ the nitrogen atoms forms hydrogen bond because of the following reasons.

(i) Small size of nitrogen.

(ii) High electronegativity of nitrogen ( 3.0) Due to more difference of electronegativity between $N$ and $H$ atom then $N-H$ bond is polar forming $H$ bond. On the contrary, phosphorus due to its larger size and lesser electronegativity ( 2.1) does not form hydrogen bond because $P-H$ bond is almost purely covalent and no hydrogen bond is formed.

#tag#


Q. How is nitrogen prepared in laboratory? Write chemical equations of the reactions involved.

Sol. Dinitrogen is prepared in laboratory by treating an aqueous solution of ammonium chloride with sodium nitrite. $N H_{4} C l(a q)+N a N O_{2}(a q) \rightarrow N_{2}(g)+2 H_{2} O(l)+N a C l(a q) \quad$ T $\mathrm{h}$ e impurities of $N O$ and $H N O_{3}$ obtained in small amounts is removed by passing the gas through aqueous sulphuric acid containing potassium dichromate.

#tag#


Q. Illustrate how copper metal can give different products on reaction with $\mathrm{HNO}_{3}$

Sol. The reaction of copper with nitric acid produces different products depending upon the concentration of nitric acid.

#tag#


Q. Give the resonating structures of $N O_{2}$ and $N_{2} O_{S}$

Sol. The resonating structure of $N O_{2}$ and $N_{2} O_{5}$ are as follows:

#tag#


Q. The $H N H$ angle value is higher than $H P H, H A s H$ and $H S b H$ angles. Why?

[Hint. Can be explained on the basis of $s p^{3}$ hybridisation in $N H_{3}$ and only $s-p$ bonding between hydrogen and other elements of the group]

Sol. The bond angle of the hydrides of group 15 are given below.

The $H N H$ angle in ammonia is higher as compared to other hydrides of the group. It is because nitrogen in is hybridised. However due to lone pair of electrons the bond angle contracts from $109^{\circ} 28^{\prime}$ to $106.5^{\circ} .$ The decreased bond angle in other hydrides is because of the fact that the hybridisation becomes less and less distinct with increasing size of the central atom i.e. pure $p$ -orbitals are utilised in $M-H$ bonding or in simple words the $s$ -orbital of $H$ atom overlaps with orbital having almost pure $p$ -character.

#tag#


Q. Why does $R_{3} P=O$ exist but $R_{3} N=O$ does not $(R=\text { alky } 1$ group)?

Sol. – Compounds like $R_{3} N=O$ does not exist because of the absence of $d$ -orbitals in the valence shell of nitrogen atom. As a result of this nitrogen cannot expand its covalency beyond 4 (lack of formation of $d \pi-p \pi \text { bond }) .$ On the contrary $R_{3} P=O$ exists because phosphorus has vacant $d$ -orbitals in its valence shell and can expand its covalency beyond 4 by forming bond.

#tag#


Q. Explain why $N H_{3}$ is basic while $B i H_{3}$ is only feebly basic?

Sol. The hydrides of group 15 are basic in character because of the presence of lone pairs of electrons on the nitrogen atom (Lewis

bases $) . N H_{3}$ is strongly basic in character because it can easily donate its electron pair due to small size of nitrogen atom as a result of which electron density of lone pair is concentrated over a

small region. $B i H_{3}$ on the contrary is feebly basic because of its larger size. Due to increase in size the electron density gets diffused over a larger region and hence the ability to donate the electron pair decreases.

#tag#


Q. Nitrogen exists as diatomic molecule and phosphorus as $P_{4}$ why?

Sol. Nitrogen exists as a diatomic molecule $\left(N_{2}\right)$ because due to small size of nitrogen atom and absence of vacant $d$ -orbitals in its valence

shell it has strong ability to form multiple bonds $(N \equiv N) .$ The other members of group 15 i.e. phosphorus, Arsenic and antimony all exist as discrete tetraatomic tetrahedral molecules, $v i z, P_{4},$ As $_{4}$

and $S b_{4}$ because these are not capable offorming multiple bonds due to their bigger size and repulsions between non-bonded electrons of the inner core. Now since the angle $P-P-P$ is $60^{\circ},$ the $p \pi-p \pi$ bonding is therefore not possible.

#tag#


Q. Write main differences between the properties of white phosphorus and red phosphorus.

Sol. The difference in properties and structure of white and red phosphorus is given in following table :

Structure of white and red phosphorus are shown below :

#tag#


Q. Can $P C l_{3}$ act as an oxidising as well as a reducing agent? Justify.

Sol. P C l_{3} can act both as an oxidising agent as well as reducing agent.

(a) As a reducing agent. The following reactions support the reducing character $P C l_{3}+S O_{2} C l_{2} \rightarrow P C l_{5}+S O_{2}$

$P C l_{3}+S O_{3} \rightarrow P O C l_{3}+S O_{2}$

(b) As an oxidising agent. It oxidises metals to their respective chlorides

#tag#


Q. Why is dioxygen a gas but sulphur a solid ?

Sol.

#tag# Dioxygen is a gas because due to small size of oxygen atom and absence of $d$ -orbitals in its valence shell it has strong tendency to form $p \pi-p \pi-p \pi$ bonds multiple bonds $(O=O)$ so as to complete its octet and hence can exist as discrete $\mathrm{O}_{2}$ molecule. On the contrary sulphur is a solid and exists as staggered 8 -atom ring. It is because the tendency of $S=S$ bond formation is missing in sulphur atom due to its larger size and low sulphur-sulphur double bond energy.

As a result sulphur atoms complete their octet through formation of $S-S$ bonds in $S_{8}$ molecule. This leads to increase the forces of attraction and hence its physical state is solid.


Q. Knowing the electron gain enthalpy values for $O \rightarrow O^{-}$ and

$O \rightarrow O^{2-}$ as $-141$ and $702 k J$ mol $^{-1}$ respectively, how can you account for the formation of a large number of oxides having $Q^{2-}$ species and not $Q^{-} ?$

Sol. The second electron gain enthalpy $\left(O^{-}+e^{-} \rightarrow O^{-2}\right)$ is positive $\left(702 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ yet a large number of oxides have $\mathrm{O}^{2-}$ species. This is attributable to high lattice energy released during formation of such oxides. As a result of this the lattice energy released by the process compensates for the second electron gain enthalpy.

#tag#


Q. Which aerosols deplete ozone?

Sol. The aerosols responsible largely for the depletion of ozone layer are

(i) Nitric oxide emitted from the exhaust systems of supersonic jet aeroplanes

Freons (chlorofluoro hydrocarbons) which are used in aerosol sprays and refrigerants.

#tag#


Q. How is $S O_{2}$ an air pollutant?

Sol. $-\mathrm{SO}_{2}$ acts as an air pollutant because of the following reasons:

(i) is strongly irritating to the respiratory tract. at a concentration of 5 ppm causes throat and eye irritation (resulting into cough, tears and redness in eyes). It causes breathlessness and affects larynx, i.e., voice box.

(ii) Even at a low concentration of 0.03 ppm, it has a very damaging effect on the plants. If exposed for a long time, $i . e .,$ a few days or weaks, it slows down the formation of chlorophyll resulting in injury to the leaves including loss of green colour. This is called chlorosis.

(iii) dissolves in moisture present in air to form which damages building materials especially marble.

It corrodes metals particularly iron and steel. It also brings about fading and deterioration of fabrics, leather, paper, etc. and affecting the colour of paints.

#tag#


Q. Why are halogens strong oxidising agents ?

Sol. Halogens act as strong oxidising agents because of their high electronegativity values and high electron affinity values. The oxidising power of the halogens is comparable in terms of their reduction potential values given below :

As the reduction potential values decrease from fluorine to iodine, the oxidising power also decreases. The reduction potential value depends upon various energy terms as shown below :

#tag#


Q. Explain why fluorine forms only one oxoacid. HOF.

Sol. Fluorine is known to form only one oxoacid, $H O F$ which is highly unstable. Other halogens form oxoacids of the type $H O X, H X O_{2}$ $H X O_{3},$ and $H X O_{4}(X=C l, B r, I) .$ Fluorine due to its small size and high electronegativity cannot act as central atom in higher oxoacids and hence do not form higher oxoacids.

#tag#


Q. Write two uses of $\mathrm{ClO}_{2}$

Sol. (i) $\quad \mathrm{ClO}_{2}$ is a powerful oxidising agent and chlorinating agent. Large quantities of are used for bleaching wood pulp and cellulose and for purifying drinking water.

(ii) It is an excellent bleaching agent. Its bleaching powder is about 30 times higher than that of and is used for bleaching fluor to make white bread.

#tag#


Q. Why are halogen coloured ?

Sol. All the halogens are coloured. The colour of halogens deepen with rise in atomic number from fluorine to iodine.

The colour is due to absorption of energy from visible light by their molecules for excitation of outer electrons of higher energy levels (the gap of energy between valence shell of halogens and higher energy shells is less). Fluorine absorbs violet portion of the light and appears yellow while iodine absorbs yellow and green portions of the light and thus appears violet. Change in colour on moving from $F$ to $I$ is called blue shift or bathochromic shift.

#tag#


Q. Write the reactions of $F_{2}$ and $C l_{2}$ with water.

Sol.

#tag# (i) Fluorine is highly reactive and decomposes water very readily even at low temperature and in dark forming amixture of $O_{2}$ and $O_{3}$

(ii) Chlorine decomposes water in the presence of sunlight forming halogen acid and oxoacid


Q. What inspired $N .$ Bartlett for carrying out reaction between $X e$ and $P t F_{6} ?$

Sol.

#tag# N. Bartle $H,$ in $1962,$ prepared a compound by reacting oxygen with $P t F_{6},$ a powerful oxidising agent. The $X$ -ray examination of solid compound, $O_{2} P t F_{6},$ showed that it consisted of $O_{2}^{+}$ and $P t F_{6}^{-}$ ions. Bartle $H$ thought that a similar compound could be prepared with Xenon because the ionisation enthalpy of Xenon is quite close to the ionisation enthalpy of oxygen $\left(1314 \mathrm{kJ} \mathrm{mol}^{-1}\right) .$ Accordingly, he reacted Xenon with $\mathrm{PtF}_{6}$ and got a red solid compound.


Q. What is the oxidation states of phosphorus in the following :

Sol. The oxidation state of phosphorus in the compounds is :

#tag#


Q. How are $\mathrm{XeO}_{3}$ and $\mathrm{XeF}_{4}$ prepared ? Describe their molecular shapes.

Sol.

#tag# (i) $\quad 6 \mathrm{XeF}_{4}+12 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{Xe}+2 \mathrm{XeO}_{3}+24 \mathrm{HF}+3 \mathrm{O}_{2}$

$$ \mathrm{XeF}_{6}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{XeO}_{3}+6 \mathrm{HF} $$

(ii) Partial hydrolysis of $\mathrm{XeF}_{6}$ give $\mathrm{XeOF}_{4}$

$$ \mathrm{XeF}_{6}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{XeOF}_{4}+2 \mathrm{HF} $$

Molecular shape (or structure)

$\mathrm{XeO}_{3}$ has a trigonal pyramidal shape due to presence of one lone

pair of electrons at one corner of tetrahedron. Similarly, $X e O F_{4}$ has square pyramidal shape. Lone pair of electrons lies opposite to oxygen atom.


Q. Give the formula and describe the structure of a noble gas species which is isostructural with:

Sol. (i) Structure of $I C I_{4}^{-}:$ In $I C I_{4}^{-},$ the central $I$ atom has in all 8 electrons ( 7 valence electrons + one due to negative charge). Four of these form single bonds with four $C l$ atoms (four bond pairs) while the remaining four constitute two lone pairs. Thus, $I$ in has four bond pairs and two lone pairs. Therefore, according to $V S E P R$ theory, it should be square planar as shown.

Now, $\mathrm{ICl}_{4}^{-}$ has $(7+4 \times 7+1)=36$ valence electrons. A noble

gas specieshaving 36 valence electrons in $\mathrm{XeF}_{4}(8+4 \times 7=36)$ Thus, like, is also square planar.

(ii) Structure of $I B r_{2}^{-}$

In $, \mathrm{IBr}_{2}^{-}$ the central I atom has 8 electrons (7 valence electrons + one due to negative charge). Two of these form two single bonds (bond pairs) with two $B r$ atoms, while the remaining six constitute three lone pairs. Thus, Iin has two bond pairs and three lone pairs. Therefore, according to $V S E P R$ theory, it should be linear.

Now, $\mathrm{IBr}_{2}^{-}$ has $22(7+2 \times 7+1)$ valence electrons. A noble

gas species having 22 valence electrons is $\mathrm{XeF}_{2}(8+2 \times 7=22)$

Thus, like, $\mathrm{IBr}_{2}^{-}, \quad \mathrm{XeF}_{2}$ is also linear.

(iii) Structure of $\mathrm{BrO}_{3}^{-}$

The central atom $B r$ has seven electrons. Four of these electrons form two double bonds or coordinate bonds with two oxygen atoms while the fifth electron forms a single bond with The remaining two electrons form one lone pair. Thus, in all there are three bond

pairs and one lone pair around Br atom in $\mathrm{BrO}_{3}^{-}$. Therefore,

according to $V S E P R$ theory $\mathrm{BrO}_{3}^{-}$ should be pyramidal.

Now, $\mathrm{BrO}_{3}^{-}$ has $26(7+3 \times 6+1=26)$ valence electrons. A noble gas species having 26 valence electrons is $\mathrm{XeO}_{3}(8+3 \times 6=26) .$ Thus, like $, \mathrm{BrO}_{3}^{-}, \quad \mathrm{XeO}_{3}$ is also pyramidal.

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Q. Why do noble gases have comparatively large atomic sizes?

Sol.

#tag# The atomic size, in the case of noble gases, is expressed in terms of vander Waal’s radii whereas the atomic size of other members of the period is either metallic radii or covalent radii. As the Vander Waal’s radii is larger than both metallic as well as covalent radii, therefore the atomic size of noble gas is quite large.

Among the noble gases, the atomic size increases down the group due to addition of new electronic shells


Q. List the uses of neon and argon gases.

Sol. (i) Uses of neon

(a) It is used in neon discharge lamps and signs which are used for advertising purposes.

(b) It is used in safety devices for protecting electrical instruments because it has a property of carrying exceedingly high currents under high voltage.

(ii) Uses of Argon

(a) It is widely used for filling in candescent metal filament electric bulbs.

(b) It is used for filling radio-valves, rectifiers and fluorescent tubes.

(c) It is used in producing inert atmosphere during welding and extraction of various metals.

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Q. Discuss the general characteristics of group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.

Sol. Electronic Configuration :

The electronic configurations of the elements of group 15 are given in table.

The general valence shell electronic configuration of the elements of group 15 is $n s^{2} n p^{3},$ where $n=2$ to 6

The three electrons in $p$ -orbitals are distributed as $p_{x}^{1}, p_{y}^{1}, p_{z}^{1}$ in

accordance with Hund’s rule.

Electronic configuration of elements of group 15

(2) Oxidation States

(i) Negative oxidation states : All the elements of this group

have five electrons in the valence shell $\left(n s^{2} n p^{3}\right)$ and thus require three more electrons to acquire the nearest noble gas configuration. Although gain of three electrons from more metallic elements to form $M^{3-}$ ions requires large amount of energy yet it takes place only with nitrogen because it is the smallest and the most electronegative element of this group. Thus, nitrogen forms

$N^{3-(\text { nitride })}$ ion and shows an oxidation state of $-3$ in nitrides of some highly electropositive metals such as $M g_{3} N_{2}, C a_{3} N_{2},$ etc. Other elements of this group form covalent compounds even with metals and show a formal oxidation state of- $-3 .$ For example, calcium phosphide $\left(\mathrm{Ca}_{3} P_{2}\right),$ sodium arsenide $\left(\mathrm{Na}_{3} \mathrm{As}\right),$ zinc stibide $\left(Z n_{3} S b_{2}\right)$ and magnesium bismuthide $\left(M g_{3} B i_{2}\right)$

Besides $-3, N$ and $P$ also show oxidation states of $-2$ in hydrazine $\left(N H_{2} N H_{2}\right)$ and diphosphine $\left(P_{2} H_{4}\right)$ respectively. Nitrogen also shows an oxidation state of $-1$ in hydroxylamine $\left(N H_{2} O H\right)$ but phosphorus does not.

(ii) Positive oxidation states : All the elements of this group show positive oxidation states of $+3$ and $+5 .$ However, on moving down the group, the stability of $+5$ oxidation state decreases while that of $+3$ oxidation state increases due to inert pair effect. Thus,

$+5$ oxidation state of $B i$ is less stable than $+5$ oxidation state of $S b$ The only stable compound of bismuth having $+5$ oxidation state is $B i F_{s} .$ Further due to large amount of energy needed to lose all the five valence electrons, $M^{5+}$ ions cannot be formed. In other words, all the compounds of group 15 elements having $+5$ oxidation state (i.e. $\left.P F_{5}, P C l_{5}, S b F_{5}, B i F_{5}\right)$ are essentially covalent. However,

elements of this group form both ionic ( $i . e . B i F_{3}, S b F_{3}$ ) and covalent

compounds (i.e $\left.\mathrm{NC}_{3}, \mathrm{PC}_{3}, \mathrm{AsC}_{3}, \mathrm{SbC}_{3}\right)$ in $+3$ oxidation state. In nut shell, the covalent character decreases in the order:

$N>P>A s>S b>B i$

It may, however, be pointed out here that nitrogen because of its small size, high electronegativity and strong tendency to form $p \pi-p \pi-p \pi$ multiple bonds, shows all the oxidation states from $-3$ to $+5$ as shown below:

(3) Atomic and ionic radii :

(i) The atomic (covalent) and ionic radii (in a particular oxidation state) of the elements of nitrogen family (group 15) are smaller than the corresponding elements of carbon family (group 14).

Explanation : This is because on moving from left to right, i.e., from group 14 to 15 in a given period, the nuclear charge increases while the new electron enters the same shell. Further, the electrons in the same shell do not screen each other. Therefore, the effective nuclear charge increases and hence the electrons are more strongly attracted towards the nucleus. This results in decrease in covalent radii. Same is true of ionic radii.

(ii) On moving down the group, the covalent and ionic radii (in a particular oxidation state) increase with increase in atomic number. There is considerable increase in covalent radius from $N$ to $P .$ However, from $A s$ to $B i,$ only a small increase is observed.

Explanation : Down the group, the covalent radii increase primarily due to the addition of a new principal energy shell in each succeeding element. The considerable increase in covalent radius from $N$ to $P$ is not only due to the addition of a new energy shell but is also due to strong shielding effect of the $s-$ and $p$ -electrons present in the inner shells. However, the small increase in covalent radii from $A s$ to $B i$ is due to the poor shielding of the valence electrons by the $d$ -and/or $f$ -electrons present in the inner shells of these heavier elements. As a result, the effective nuclear charge increases which reduces the effect of addition of a new energy shell to some extent. As a result, the increase in covalent radius is small from $A s$ to $B i .$ Same is true of ionic radii.

(4) Ionization enthalpy :

The ionization enthalpies of the elements of group 15 are much higher than the corresponding elements of group 14 Down the group, the values of the ionization enthalpies decrease regularly.

Explanation : Because of increased nuclear charge, reduced atomic radii and stable half-filled electronic configurations, the electrons of group 15 elements are strongly attracted by the nucleus and hence they have less tendency to lose electrons. As a result, ionization enthalpies of the elements of the nitrogen family are much higher as compared to the elements of carbon family. The decrease in their values as we move down the group is due to gradual increase in the atomic size which reduces the force of attraction on the electrons by the nucleus.

As expected, successive ionization energies of these elements increase in the order : $\Delta_{i} H_{1}<\Delta_{i} H_{2}<\Delta_{i} H_{3}$

(5) Electronegativity :

Group 15 elements are more electronegative than group 14 elements. Electronegativity of elements of group 15 shows a gradual decrease on moving down the group from $N$ to $B i .$

Explanation: As the elements of group 15 have smaller atomic size and need smaller number of electrons to attain noble gas configuration as compared to elements of group $14,$ they are more electronegative. Due to the gradual increase in the atomic size, attraction by the nucleus for the electrons decreases. Hence, their electronegativity values decrease down the group. However, the decrease is not regular.

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Q. How is ammonia manufactured industrially ?

Sol. On a commercial scale, ammonia is manufactured by Haber’s process.

This reaction is reversible, exothermic and occurs with decrease in volume. Therefore, according to Le Chatelier’s principle, the favourable conditions for the manufacture of ammonia are:

(i) Low temperature: since the forward reaction is exothermic, low temperature will favour the formation of ammonia. However, at low temperatures, the rat of the reaction will be slow. The optimum temperature for the reaction has been found to be around $700 K$

(ii) High pressure : since the forward reaction occurs with decrease in volume, high pressure will favour the formation of ammonia. The reaction is usually carried out a pressure of about $200 \times 10^{5} \mathrm{Pa}$ or 200 atmospheres.

(iii) Catalyst: The rate of reaction is fairly low around $700 \mathrm{K}$. It is increased by using iron oxide as catalyst with small amounts of

$K_{2} O$ and $A l_{2} O_{3} .$ Sometimes, molybdenum is used as a promoter (which increases the efficiency of the catalyst). The flow chart for production of ammonia is in figure.

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Q. Justify the placement of $O . S, S e, T e$ and $P o$ in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Sol. Electronic Configuration :

The elements of group 16 have six electrons in the valence shell

and hence their general outer electronic configuration is $n s^{2} n p^{4}$

The four $p$ -electrons are arranged in three $p$ -orbitals as $p_{x}^{2} p_{y}^{1} p_{z}^{1}$

in accordance with Hund’s rule. The names, electronic configurations and common oxidation states of group 16 elements are given in table.

(2) Oxidation States :

since all the elements of this group have $n s^{2} n p^{4}$ configuration in their valence shell (outer-most shell), they can attain noble gas configuration $v i z . n s^{2} n p^{6}$ either by gaining or by sharing electrons. These elements, therefore, show two types of oxidation states.

(i) Negative oxidation states : oxygen the first element of this group, has high electronegativity. Therefore, it preferably completes its octet by gaining of electrons. As a result, all metal oxides are ionic and contain $\mathrm{O}^{2-}$ ions in which oxygen shows an oxidation

state of $-2 .$ In addition, oxygen shows an oxidation state of $-1$ in peroxides such as $H_{2} \mathrm{O}_{2}$ zero in $\mathrm{O}_{2}$ and $\mathrm{O}_{3},+1$ in $\mathrm{O}_{2} \mathrm{F}_{2}$ and $+2$ in $O F_{2}$

since the electronegativities decrease as we move down the group, the tendency of these elements to show $-2$ oxidation state decreases from sulphur to polonium. Hence, there is much less probability of the formation of dinegative ions in case of $S,$ Se and $T e$. The least electronegative element, polonium, in fact, does not exhibit negative oxidation state at all. Rather it shows positive oxidation states only.

(ii) Positive oxidation states: Oxygen does not show positive oxidation states except in and The other elements of this group show positive oxidation states of $+2,+4$ and $+6$ due to promotion of electrons to vacant $d$ -orbitals as explained in the diagram shown below:

In the ground state, these elements have two unpaired electrons and hence can form two bonds. This explains their $+2$ oxidation state. In the first excited state, one of the paired $p$ -electron goes to the vacant $d$ -orbital of the same shell, thus making four unpaired electrons available for chemical bonding. This accounts for their $+4$ oxidation state. On further excitation, one of the s-electrons also gets promoted to $d$ -orbital, thus making six unpaired electrons available for bond formation. This explains their $+6$ oxidation state. However, due to inert pair effect, the stability of $+6$ oxidation state decreases down the group. Thus, $+6$ oxidation state is most stable in case of $S$ and least stable in case of $P O$.

In general, the compounds of $S,$ Se, Te and Po with oxygen are tercovalent $(+4 \text { oxidation state). These }+4$ compounds show both oxidising and reducing properties.

since fluorine is the strongest oxidising agent, therefore, an element shows its maximum oxidation state in its compound with fluorine.

Thus, the compounds of $S,$ Se, Te and Po with fluorine show an oxidation state of $+6$

These compounds in $+6$ oxidations state show only oxidising properties.

The behaviour of oxygen is different, since it does not have $d-$ orbitals in its valence shell. Therefore, in case of oxygen, the $2 p-$ electrons on excitation have to go to $3 s$ -orbital. But, since too much energy is required to excite an electron to a higher shell (in this case from $\mathrm{K} \text { to } \mathrm{L}-\text { shell }),$ the electrons in oxygen do not get unpaired. Therefore, oxy gen behaves as a divalent element only.

3. Reactivity towards Hydrogen (Formation of Hydrides).

(i) All the elements of group 16 form hydrides of the general

formula, $\quad H_{2} E \quad$ where $\quad E=O, S, S e, T e, P o$

$H_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{S}, \mathrm{H}_{2} \mathrm{Se}, \mathrm{H}_{2} \mathrm{Te}$ and $\mathrm{H}_{2} \mathrm{Po}$

Preparation. The hydrides of $S,$ Se and Te are prepared by the action of acids on metal sulphides, selenides and tellurides respectively. You are already familiar with the Kipp’s apparatus used for generation of $H_{2} S$ in the laboratory by the action of dil,

$H_{2} \mathrm{SO}_{4}$ on $\mathrm{FeS}$

$F e S(s)+2 H_{3} O^{+}(a q) \rightarrow F e^{2+}(a q)+H_{2} S(g)+2 H_{2} O(l)$

Structure. All these hydrides have angular shape (Fig.)

involving $s p^{3}$ -hybridization of the central atom. the bond angles,

however, decrease from $H_{2} O$ to $H_{2}$ Te as shown below:

Explanation. Due to stronger lone pair-lone pair than bond pairbond pair repulsions, the bond angle in water decreases from the tetrahedral value of $109.28^{\prime}$ to $104.5^{\circ} .$ As we move down the group from $O$ to $T e,$ the size of the central atom goes on increasing and its electronegativity goes on decreasing. As a result, the position of the two bond pairs shifts away and away from the central atom as we move from $H_{2} O$ to $H_{2}$ Te. Consequently the repulsions between the bond pairs decreases as we move from to and, therefore, the bond angle decreases in the same order:

$\mathrm{H}_{2} \mathrm{O}>\mathrm{H}_{2} \mathrm{S}>\mathrm{H}_{2} \mathrm{Se}>\mathrm{H}_{2} \mathrm{Te}$

Oxygen forms another important hydride, Some properties of hydrides are given in Table.

TABLE : Properties of Hydrides of Group 16 Elements

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Q. Describe the manufacture of $H_{2} S O_{4}$ by contact process?

Sol. $H_{2} S O_{4}$ is mostly prepared by contract process. The acid produced by this process is free from arsenic impurities and is of high purity. The process involves the following steps:

(i) Preparation of $\mathrm{SO}_{2}$ by burning of $S$ or roasting pyrites.

$S_{8}(s)+8 O_{2}(g) \rightarrow 8 S O_{2}$

$4 \mathrm{FeS}_{2}+11 \mathrm{O}_{2} \rightarrow 2 \mathrm{F}_{2} \mathrm{O}_{3}+8 \mathrm{SO}_{2}$

(ii) Sulphurdioxide gas is purified by passing through the purifying unit.

(iii) The pure gas is preheated to $673-723 K$ and passed into catalytic chamber where is oxidised to

(iv) gas is then absorbed in conc. in the absorption tower to form oleum which can be diluted with water to get of desired concentration.

Favourable conditions for the maximum yield of are:

(a) High conc. of oxygen

(b) High pressure

(c) Low temperature. In about 2 atm. practice, the reaction is carried out at about $625-725 K$

(d) Use of catalyst. To increase the rate of reaction, finely divided platinum or (vanadium pentaoxide) is used as catalyst;

(e) Purity of gases. For maximum yield of the most important condition is that reacting gases and must be free from arsenic and other impurities as the impurities act as catalytic poison.

Flow diagram of plant is given below :

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Q. Write balanced equations for the following :

(i) $\quad$ NaCl is heated with sulphuric acid in the presence of $$ M n O_{2} $$

(ii) Chlorine gas is passed into a solution of $N a I$ in water

(iii) $\quad \mathrm{SiO}_{2}$ is treated with $H F$

(iv) $\quad \mathrm{NaClO}_{2}$ is treated with $\mathrm{SO}_{2}$

(v) Iodine is treated with conc. $H N O_{3} .$

Sol. (i) Chlorine gas is produced.

(ii) Iodine is liberated.

$$ 2 \mathrm{NaI}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{NaCl}+\mathrm{I}_{2} $$

(iii) $\quad \mathrm{SiF}_{4}$ is formed.

$$ \mathrm{SiO}_{2}+4 \mathrm{HF} \rightarrow \mathrm{SiF}_{4}+2 \mathrm{H}_{2} \mathrm{O} $$

(iv) $\quad \mathrm{ClO}_{2}$ gas is produced.

(v) $\quad I_{2}$ is oxidised to $H I O_{3}$

$I_{2}+10 H I N O_{3} \rightarrow 2 H I O_{3}+10 N O_{2}+4 H_{2} O$

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Q. How are xenon fluorides $X e F_{2}, X e F_{4}$ and $X e F_{6}$ prepared? Deduce their structure applying $V S E P R$ theory.

Sol. Preparation of $\mathrm{XeF}_{2}, \mathrm{XeF}_{4}$ and $\mathrm{XeF}_{6} .$ All three binary fluorides of $X e$ are formed by direct union of elements under approximate experimental conditions. $\mathrm{XeF}_{2}$ can also be prepared by irradiating a mixture of xenon and fluorine with sunlight or light from a pressure-mercury arc lamp.

Structure of $\mathrm{XeF}_{2} \cdot \mathrm{XeF}_{4}$ and $\mathrm{XeF}_{6}:$ By applying $V S E P R$ theory the shared and unshared pairs of electrons around central atom tends to stay at maximum distance from each other as well as from central atom. On this basis the structures are as follows:

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Q. Arrange the tollowing in the order of property indicated for each set:

(i) $\quad F_{2}, C l_{2}, B r_{2}, I_{2}-$ increasing bond energy.

(ii) $\quad H F, H C l, H B r, H I-$ increasing acid strength.

(iii) $\quad N H_{3}, P H_{3}, A s H_{3}, S b H_{3}, B i H_{3}-$ increasing base strength

(iv) $\quad H_{2} O, H_{2} S, H_{2} S e, H_{2} T e_{3}-$ increasing acid strength

Sol. (i) In the order of increasing bond energy.

$\mathrm{Cl}-\mathrm{Cl}$ bond energy is highest while $I-I$ bond energy is least

$$ I-I<F-F<B r-B r<C l-C l $$