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# Class 11 PHYSICS

Rotational Motion

Gravitation

Simple Harmonic Motion

Circular Motion

Friction

Newton’s Laws Of Motion

Conservation Of Linear Of Momentum

Projectile Motion

Unit And Dimension

Heat And Thermodynamics

General Properties Of Matter

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Wave Optics

Ray Optics

Semiconductor

Modern Physics

Alternating Current

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Capacitor

Current Electricity

# Class 11 CHEMISTRY

General Organic Chemistry

s-block

Chemical Bonding

Mole Concept

Boron Family

Carbon Family

Ionic Equilibrium

Electrochemistry

Periodic Properties

Thermodynamics and Thermochemistry

Chemical Equilibrium

Gaseous State

Atomic Structure

# Class 12 CHEMISTRY

Solution & Colligative Properties

Solid State

Chemical Kinetics

Carboxylic acids & Derivative

Aliphatic Hydrocarbon

Surface Chemistry

Co-ordination Compounds

Metallurgy

d & f block Elements

Noble Gases

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Important Questions of Organic Chemistry Class 11-With Slutions

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Q. Give the IUPAC names of the following compounds :

Sol. 2-bromo- 4-methyl pentan- 3-one

Q. Give the IUPAC name of the alkane having the lowest molecular mass that contain a quaternary carbon.

Sol.

Q. Write the structural formula of (i) sec-butyl and (ii) isobutyl groups.

Sol.

Q. Lassaigne’s test is not shown by diazonium salts, though they contain nitrogen. Why ?

Sol. Diazonium salts $\left(C_{6} H_{5} N_{2}^{+} X^{-}\right)$ readily lose $N_{2}$ on heating

before reacting with fused sodium metal. Therefore, these do

not give positive Lassaigne’s test for nitrogen.

Q. Suggest a suitable technique of separating naphthalene from kerosene present in a mixture.

Sol. By differential extraction.

Q. How will you separate a mixture of o-nitro-phenol and p-nitrophenol ?

Sol. A mixture of o-nitrophenol and p-nitrophenol can be separated by steam distillation. o-nitrophenol being less volatile distils over along with water while p-nitrophenol being non-volatile remains in the flask.

Q. Will $C C l_{4}$ give white precipitate of $A g C l$ on heating with silver nitrate? Give reason for your answer.

Sol. The precipitate of $A g C l$ will not be formed because $C C l_{4}$ is

covalent compound and does not ionise to give $C l^{-}$ ions to react with $A g N O_{3}$. $C C l_{4}+A g N O_{3} \longrightarrow$ No reaction

Q. Which is expected to be more stable, $\mathrm{O}_{2} N \mathrm{CH}_{2} \mathrm{CHO}$ or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}$ and why?

Sol. $\mathrm{O}_{2} \mathrm{N} \mathrm{CH}_{2} \mathrm{CHO}$ is expected to be more stable because each atom has complete its octet and has no charge.

Q. How many $\sigma$ and $\pi$ bonds are present in each of the following molecules?

(i) $\quad \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{CH}$

(ii) $\quad \mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}$

Sol. (i) $\quad \sigma_{C-C}=4, \sigma_{C-H}=6, \pi_{C=C}=3$

(ii) $\quad \sigma_{\mathrm{C}-\mathrm{C}}=4, \sigma_{\mathrm{C}-\mathrm{H}}=6, \pi_{\mathrm{C}=\mathrm{C}}=3$

Q. Which of the following pairs of structures do not constitute resonance structures :

Sol. Pair of structures shown in (i) and (iv) do not constitute resonance structures.

Q. Classify the following molecules/ions as nucleophiles or electrophiles: $H \ddot{S}^{-}, B F_{3}, C H_{3} C H_{2} O^{-}$

$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}, \mathrm{Cl}^{+}, \mathrm{CH}_{3} \mathrm{C}=\mathrm{O}, \mathrm{H}_{2} \mathrm{N}:, \mathrm{NO}_{2}^{+-}$

Sol. Electrophiles : $B F_{3} . C l^{+}, C H_{3}^{+} C=O, N O_{2}^{+}$

Nucleophiles : $H S, C H_{3} C H_{2} O^{-},\left(C H_{3}\right)_{3} \ddot{N}, \dot{H}_{2} \bar{N}$:

Q. In $s p^{3}, s p^{2}$ and sp hybrid orbitals which hybrid orbitals show the electronegativity?

Sol. $s p^{3}$ hybrid orbitals.

Q. What type of hybridization is shown by second carbon atom.

Sol. sp hybridization.

Q. Explain why $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}$ is more stable than $C H_{3}^{+} C H_{2}$ and $^{+} C H_{3}$ is the least stable cation.

Sol.

the nodal plane of the vacant 2p orbitals and hence cannot

overlap with it. Thus lacks hyperconjugative stability.

Q. State bond length and strength of $s p, s p^{2}$ and $s p^{3}$ hybrid

orbitals.

Sol. The sp hybrid orbital contains more $s$ character and hence it is closer to its nucleus and forms shorter and stronger bonds than the $s p^{3}$ hybrid orbital. The $s p^{2}$ hybrid orbital is intermediate between $s p$ and hybrid orbitals, hence the length and enthalpy of the bonds it forms is also intermediate between them.

Q. On the basis of type of hybridization, predict the shape of the following molecules :

(i) $\quad H_{2} C=O \quad$ (ii) $C H_{3} F \quad$ (iii) $\mathrm{HC} \equiv \mathrm{N}$

Sol. (i) $s p^{2}$ hybridised carbon and therefore, shape is trigonal planar.

(ii) $s p^{3}$ hybridised carbon and shape is tetrahedral.

(iii) $s p$ hybridised carbon and shape is linear.

Short Answer (2 or 3 Marks)

Q. Write IUPAC nomenclature of the following :

Sol. (i) 2-ethyl 3-methyl pentan-1-ol

(ii) 1-chloro propan-2-one

(iii) Hexa 1, 3-diene-5-yne

(iv) 2, 4, 6 – tri bromo phenol

Q. Give IUPAC name for the following :

Sol. (i) 2 methyl propan-1-ol (ii) 4-ethyl-2-methyl aniline

Q. Give IUPAC name of the following organic molecules :

Sol. (i) 5-methoxy-2-nitrophenol

(ii) 4-ethyl-2-methyl anisol

(iii) 2-chloro-4-methoxy benzoic acid

(iv) 2-ethoxy-4-methyl aniline

Q. Write the IUPAC names of :

Sol. (i) Trans-2-hexene

(ii) Cis-2-hexene

(iii) (E)-3-methyl-2-pentene

(iv) (Z)-3-methyl-2-pentene

(v) Trans-1, 4-dimethyl cyclo hexane.

Q. Give the IUPAC name of the following compounds :

Sol.

Q. Which of the following represents the correct IUPAC name for the compounds concerned :

(i) 2, 2–dimethylpentane or 2–dimethylpentane.

(ii) 2, 3–dimethyl pentane or 3, 4–dimethyl pentane.

(iii) 2, 4, 7–trimethyloctane or 2, 5, 7–trimethyloctane,

(iv) 2 –chloro–4–methylpentane or 4–chloro 2–methylpentane,

(v) But–3–yn-1-ol or But–4–ol–1–yne.

[NCERT]

Sol. (i) 2, 2-dimethyl pentane

(ii) 2, 3-dimethyl pentane

(iii) 2, 4, 7-trimethyl octane

(iv) 2-chloro-4-methyl pentane

(v) But-3-yne-1-ol

Q. Write IUPAC name of the following :

Sol. (i) 1-ethoxy propan-2-ol

(ii) Ethyl benzoate

(iii) 2-phenyl propanal

(iv) 6-chloro-3-methyl-hexan-2-one

Q. Write bondline formulae for : tert–Butylcyclopentane, Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one, Cyclohexanone.

Sol.

Q. Draw formulae for the first five members of each homologous series beginning with the following compounds.

(i) $\mathrm{HCOOH},$ (ii) $\mathrm{CH}_{3} \mathrm{COCH}_{3},\left(\text { iii) } \mathrm{CH}_{2}=\mathrm{CH}_{2}\right.$

Sol.

Q. Identify the functional groups in the following compounds:

Sol.

(iii) N-Substituted amide and acid chloride

Q. For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate products as free radical, carbocation and carbanion.

[NCERT]

Sol. (i) Homolysis, Free radicals are formed.

Q. Explain the order of stability of primary, secondary and tertiary carbonium ions.

Sol. $3^{\circ}>2^{\circ}>1^{\circ}$ is order of stability of carbocation.

3 ” carbocation are most stable due to maximum positive inductive effect of three alkyl groups.

$2^{\circ}$ carbocation is morestable than $1^{\circ}$ carbocation due togreater inductive effect of two alkyl groups as compared to $1^{\circ} C$ carbocation in which there is one alkyl group.

Q. 0.378 g of an organic acid gave on combustion 0.264 g of carbon dioxide and 0.162 g of water vapours. Calculate the percentage of C and H.

[NCERT]

Sol. Mass of organic compound $=0.378 \mathrm{g}$

Mass of $\mathrm{CO}_{2}$ formed $=0.264 \mathrm{g}$

Mass of $\mathrm{H}_{2} \mathrm{O}$ formed $=0.162 \mathrm{g}$

(i) Percentage of carbon

$44 \mathrm{g}$ of contains carbon

$0.264 \mathrm{g}$ of contains carbon $=12 \mathrm{g}$

$=\frac{12}{44} \times 0.264=0.072 \mathrm{g}$

Percentage of carbon $=\frac{0.072}{0.378} \times 100=19.04 \%$

$18 g$ of contains hydrogen $=2 g$

$0.162 g$ of contains hydrogen $=\frac{2}{18} \times 0.162=0.018 \mathrm{g}$

Percentage of hydrogen $=\frac{0.018}{0.378} \times 100=4.76 \%$

(ii) Percentage of hydrogen

Q. During nitrogen estimation of an organic compound by Kjeldahl’s method, the ammonia evolved by $0.5 g$ of the

compound neutralised $10 \mathrm{mL}$ of $1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} .$ Calculate the percentage of nitrogen in the compound.

Sol. $1 \mathrm{M}$ of $10 \mathrm{mL} \mathrm{H}_{2} \mathrm{SO}_{4} \equiv 1 \mathrm{M}$ of $20 \mathrm{mL}$ of $\mathrm{NH}_{3}$

$1000 \mathrm{mL}$ of $1 \mathrm{M}$ ammonia contains $=14 \mathrm{g}$ nitrogen

$20 \mathrm{mL}$ of $1 \mathrm{M}$ ammonia contain $=\frac{14 \times 20}{1000} \mathrm{g}$ nitrogen

$\therefore$ Percentage of nitrogen $=\frac{14 \times 20}{1000 \times 0.5} \times 100=56.0 \%$

Q. (i) In sulphur estimation, 0.157 g of organic compound gave $0.4813 \mathrm{g}$ of $\mathrm{BaSO}_{4} .$ What is the percentage of sulphur in organic compound?

(ii) $0.092 g$ of organic compound on heating in carius tube and subsequence ignition gave $0.111 g$ of $M g_{2} P_{2} O_{7}$. Calculate the percentage of phosphorus in organic compound.

Sol. (i) Mass of $B a S O_{4}=0.4813 \mathrm{g}$

Mass of organic compound $=0.157 \mathrm{g}$

$\% \mathrm{S}=\frac{32 \times W_{B a S O_{4}} \times 100}{233 \times W_{\mathrm{Substance}}}=\frac{32 \times 0.4813 \times 100}{233 \times 0.157}=42.10$

(ii) Mass of organic compound $=0.092 \mathrm{g}$

Mass of $\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}=0.111 \mathrm{g}$

$\mathrm{g}_{\mathrm{bof}} \mathrm{P}=\frac{62 \times \mathrm{W}_{\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}}}{222 \times \mathrm{W}_{\mathrm{Substance}}} \times 10 \mathrm{C}=\frac{62 \times 0.111 \times 100}{222 \times 0.092}=33690024 Q. (i) A mixture contains benzoic acid and nitrobenzene. How can this mixture be separated into its constituents by the technique of extraction using an appropriate chemical reagent? (ii) Write the formula of iron (III) hexa cyanoferrate (II). [NCERT] Sol. (i) The mixture is shaken with a dilute solution of$\mathrm{NaHCO}_{3}$and extracted with ether or chloroform when nitrobenzene goes into the organic layer. Distillation of the solvent gives nitrobenzene. The filtrate is acidified with dil. HCl when benzoic acid gets precipitated. The solution is cooled and benzoic acid is obtained by filtration (ii)$\quad F e_{4}\left[F e(C N)_{6}\right]$Q. Give hybridization state of each carbon in following compounds :$\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}, \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}, \mathrm{CH}_{2}=\mathrm{CHCN},\mathrm{CH}_{3} \mathrm{CH}_{2}^{-}, \mathrm{CH}_{3} \mathrm{CH}_{2}^{+}, \mathrm{CH}_{3} \mathrm{CH}_{2}$Sol. Q. (i) Giving justification categorise the following species as nucleophile or electrophile :$\begin{array}{ll}{\text { (a) } B F_{3}} & {\text { (b) } C_{2} H_{5} O^{-}}\end{array}$(ii) Write the IUPAC names of the following compounds : Sol. (i)$\quad$(a)$B F_{3}$is electrophile because it is electron deficient, i.e. octet of boron is not complete. (b)$C_{2} H_{5} O^{-}$is nucleophile because it is negatively charged. Q. Indicate the$\sigma$and$\pi$-bonds in the following molecules: (i)$\quad C_{6} H_{6}$(ii)$\quad C_{6} H_{12}$(iii)$C H_{2} C l_{2}$(iv)$\quad \mathrm{CH}_{3} \mathrm{NO}_{2}$(v)$\quad H C O N H C H_{3}$[NCERT] Sol. (i)$\quad C_{6} H_{6}$is benzene and structural formulae is: Q. Draw the complete structures of bromomethane, bromoethane. 2–bromopropane and tert–butylbromide. Arrange them in order of decreasing steric hindrance. [NCERT] Sol. Q. Draw the resonating structures for the following compounds. Show the electron shift using curved arrow notation. (i)$\quad C_{6} H_{5} O H$(ii)$\quad C_{6} H_{5} N O_{2}$(iii)$C_{6} H_{5} C^{\oplus} H_{2}$(iv)$\quad \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCHO}$(v)$\quad \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}^{\oplus}$(vi)$\quad C_{6} H_{5} \mathrm{CHO}$(vii)$C H_{2}=C H O C H_{3}$[NCERT] Sol. Q. Explain why alkyl groups act as electron donors when attached to a$\pi-$system. [NCERT] Sol. Alkyl group has$s p^{3}$hybridization whereas$\pi$-bond atom is$s p^{2}$hybridized which is more electronegative therefore alkyl group acts as electron donors. Q. Classify the reagents shown in bold in the following equation as nucleophiles or electrophiles. Use curved-arrow notation to show the electron movement. (i)$\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{HO}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O}$(ii)$\mathrm{CH}_{3} \mathrm{COCH}_{3}+^{-} \mathrm{NC} \longrightarrow \mathrm{CH}_{3} \mathrm{C}(\mathrm{CN}) \mathrm{OHCH}_{3}$(üii)$\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{CH}_{3} \mathrm{C}^{+} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}_{3}$[NCERT] Sol. (i)$\quad O H^{-}$is nucleophile Q. Classify the following reactions in one of the reaction type: (i)$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{HS}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}+\mathrm{Br}^{-}$(ii)$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{HCl} \longrightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{ClC}-\mathrm{CH}_{3}$(iii)$\left.\mathrm{(CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{HBr} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CBr}_{2} \mathrm{CH}_{3}$(iv)$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{HO} \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}_{2}$[NCERT] Sol. (i) Nucleophilic Substitution reaction (ii) Addition reaction (iii) Nucleophilic Substitution reaction (iv) Elimination reaction. Q. Explain with the help of examples that Geometrical isomerism is different from conformation. Sol. Conformation are obtained by rotation around$\sigma$-bond, e.g., eclipsed and staggered conformation of ethane. Q. What is the relationship between the members of following pairs of structures ? Are they identical, structural or geometrical isomers, or resonance contributors ? Sol. (i) Structural Isomers. (ii) Geometrical Isomers. (iii) Resonating structures. (iv) Geometrical Isomers. Long Answer (5 Mark) Q. (i) Given the IUPAC name for the amine. (ii) Discuss the hybridization of carbon atoms in allene$\left(C_{3} H_{4}\right)$and show the$\pi$– orbital overlaps. Sol. (i)$3, N, N$-Trimethylpentan-$-3$-amine. (ii)$\quad$The structure of allene$\left(C_{3} H_{4}\right)$is The carbon atoms 1 and 3 are$s p^{2}$-hybridized since each one of them is joined by a double bond. In contrast, carbon atom 2 is -hybridized since it has two double bonds. Thus, the two$\pi-$bonds in allene like in acetylene are perpendicular to each other as shown below : Whereas$H_{c}$and$H_{d}$lie in the plane of the paper while$H_{a}$and$H_{b}$lie in a plane perpendicular to the plane of the paper. Q. Give IUPAC names of the following compounds : Sol. In all the questions given above the ring phenyl group. In case the phenyl group is further substituted, the carbon atoms of the ring are separately numbered starting from the carbon atom of the ring directly attached to the parent chain in such a way that substituent on ring gets lowest possible position. According the IUPAC name are : (i) 2-(4-chlorophenyl) propanoic acid (iv) If the compound contains more than one similar complex radicals/substituents, then the numerical prefixes such as di, tri, tetra, etc. are replaced by bis,tris, tetrakis, etc. respectively. Hence, the IUPAC name is : 1, 1, 1-Trichloro-2, 2-bis-(chlorophenyl) ethane. Q. (i) A mixture contains two components A and B. The solubilities of A and B in water near its boiling point are 10 grams per 100 ml and 2 g per 100 ml respectively. How will you separate A and B from this mixture? (ii) Explain why an organic liquid vapourizes at a temperature below its boiling point in steam distillation? (iii) Which technique can be used to separate naphthalene from kerosene oil present in its mixture. [NCERT] Sol. (i) A and B from the mixture can be separated by using Fractional crystallization. When the saturated hot solution of this mixture is allowed to cool, the less soluble component B crystallizes out first leaving the more soluble component A in the mother liquor. (ii) A liquid boils when its vapour pressure becomes equal to the atmospheric pressure. During steam distillation, the mixture boils when sum of the vapour pressure of water and organic liquid becomes equal to the atmospheric pressure. Since the vapour pressure of water is appreciably higher than that of organic liquid, therefore, the organic liquid will vapourize at a temperature much lower than its normal boiling point. (iii) Since naphthalene and kerosene have large difference in their boiling points. Hence they can be separated by simple distillation. Q. (i) The$R_{f}$value of A and B in a mixture determined by TLC in a solvent mixture are 0.65 and 0.42 respectively. If the mixture is separated by column, chromatography using the same solvent mixture as a mobile phase, which of the two components, A or B, will elute first ? Explain. (ii) A mixture contains 71 percent calcium sulphate and 29 percent camphor. Name a suitable technique of separation of the components of this mixture. (iii) Without using column chromatography, how will you separate a mixture of camphor and benzoic acid? [NCERT] Sol. (i) Since the$R_{f}$value of A is 0.65, therefore, it is less strongly adsorbed as compared to compound B with value of 0.42. Therefore, on extraction of the column, A will elute first. (ii) Sublimation process can be used for the separation of camphor and calcium sulphate. Camphor sublimes where as is non-volatile. (iii) Sublimation cannot be used since both camphor and benzoic acid sublime on heating. Therefore, a chemical method using solution is used when benzoic acid dissolves leaving camphor behind. The filtrate is cooled and then acidified with dil. HCl to get benzoic acid. Q. What are reactive intermediates ? How are they generated by bond fission? [NCERT] Sol. Generally, the organic reactions complete through the involvement of certain chemical species which are short lived ($10^{-6}$seconds to a few seconds) and highly reactive. The short lived and highly reactive species are called reactive intermediates such as carbocations, carboanions, free radicals, carbenes and nitrenes. Heterolytic fission of covalent bond results in carbocations and carboanions. Free radicals are formed by homolytic fission. Carbenes are neutral carbon species in which carbon atom is bonded to two monovalent atoms or groups and also contains one non-bonding pair of electrons. They are produced in photolysis (irradiation with UV light or thermolysis or pyrolysis (action of heat) on diazoalkanes or ketones.) Q. What is the relationship between the members of following pairs of structures ? Are they identical, structural or geometrical isomers, or resonance contributors ? Sol. (i) Structural Isomers. (ii) Geometrical Isomers. (iii) Resonating structures. (iv) Geometrical Isomers. Q. What are the different types of structural isomerism ? Describe them with suitable examples. Sol. Structural isomerism is of following types : (i) Chain isomerism : “When two or more compounds have same molecular formula but different carbon skeleton are said to have chain isomerism.” For example,$\mathrm{C}_{5} \mathrm{H}_{12}$shows the following three isomers: (ii) Position Isomerism : ‘When two or more compounds have same molecular formula but different position of substituent atoms or groups on the carbon skeleton, are said to have position isomerism.’ For example,$C_{3} H_{8} O$shows the following two position isomers of alcohol : (iii) Functional group isomerism : ‘When two or more compounds have same molecular formula but different functional groups, are said to have functional group isomerism.’ For example,$C_{4} H_{10} O$represents the following two compounds. Q. Why do alkenes and cycloalkanes show cis-trans isomerism? Or What is the origin of geometrical isomerism in alkenes? Sol. Presence of carbon-carbon double bond, in a molecule is the cause of geometrical isomerism. This is because when a carbon-carbon double bond is present, the molecule cannot rotate freely about the double bond. It is called hindered or restricted rotation of the molecule about the double bond. During the formation of double bond, if two similar atoms or groups are present on the same side of the plane of the double bond it is called a cis isomer and if present on the opposite sides, it is called a trans isomer. Because of restriction in free rotation one form cannot change to the other. Therefore presence of C = C bond is the origin of geometrical isomerism . This restriction in rotation may be present in rigid of cyclic molecules. Therefore cyclic compounds also shows geometrical isomerism. s block Elements Class 11 Important Questions with Answers Get s-block elements important questions with detailed answers for class 11 exams preparation. View the Important Question bank for Class 11 and class 12 Chemistry complete syllabus. These important questions and answers will play significant role in clearing concepts of Chemistry class 11. This question bank is designed keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 chemistry chapters. Learn the concepts of s block elements and other topics of chemistry class 11 and 12 syllabus with these important questions and answers along with the notes developed by experienced faculty. Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, 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. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. 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. Very Short Answer (1 Mark) Q. . Why first group elements are called alkali metals ? Sol. Group I elements are highly reactive and with water (moisture in the atmosphere) form strong alkalies, so they are called alkali metals. Q. . Write the chemical name and formula of washing soda. Sol. Washing soda is sodium carbonate$\left(N a_{2} C O_{3}\right)$. Q. . Why do alkali metals not occur in free state ? Sol. They are highly reactive, therefore, they occur in combined state and do not occur in free state. Q. . Write the important minerals of lithium. Sol. The important minerals of lithium are: (i) Lipidolite$(L i . N a . K)_{2} A l_{2}\left(S i O_{3}\right)_{3} \cdot F(O H)$(ii) Spodumene$L i A I S i_{2} O_{3}$(iii) Amblygonite$\operatorname{LiAl}\left(P O_{4}\right) F$Q. . How sodium hydroxide is prepared at large scale ? Sol. At large scale, sodium hydroxide is prepared by castner kellner cell. Q. . Why$\mathrm{KHCO}_{3}$is not prepared by Solvay process ? Sol. Because solubility of$\mathrm{KHCO}_{3}$is fairly large as compared to$\mathrm{NaHCO}_{3}$. Q. . What is chemical composition of the Plaster of Paris ? Sol. Plaster of paris is calcium sulphate hemihydrate :$\mathrm{CaSO}_{4} \cdot \frac{1}{2} \mathrm{H}_{2} \mathrm{O}$or$\left(\mathrm{CaSO}_{4}\right)_{2} \cdot \mathrm{H}_{2} \mathrm{O}$Q. . Why do alkali metals have low density ? Sol. Due to weak metallic bonds and large atomic size, their density is low. Q. . Why is first ionization energy of alkali metals lower than those of alkaline earth metals ? Sol. Alkali metals have bigger atomic size, therefore they have lower first I.E. than group 2 elements. Q. . First group elements are strong reducing agents, why ? Sol. Because they have a strong tendency to lose outer most electron. Q. . Explain why ? LiI is more soluble than KI in ethanol. [NCERT] Sol.$K I$In the chemical bond is ionic in character. On the other hand due to small size of lithium ion and its high polarising power the bond in is predominently covalent in character. Hence LiI is more soluble than in ethanol. Q. . LiH is more stable than$\mathrm{NaH}$Explain. Sol. Both$L i^{+}$and$H^{-}$have small size and their combination has high lattice energy. Therefore LiH is stable as compared with$\mathrm{NaH}$Q. . Why is$B e C l_{2}$soluble in organic solvents ? Sol.$B e C l_{2}$is covalent, therefore soluble in organic salvents. Q. . Name the metal which floats on water without any apparent reaction with water. [NCERT] Sol. Lithium floats on water without any apparent reaction with it. Q. . Name an element which is invariably bivalent acid and whose oxide is soluble in excess of NaOH and its dipositive ion has a noble gas core. Sol. The element is beryllium, its oxide$B e O$is soluble in excess of$\mathrm{NaOH}B e O+2 N a O H \longrightarrow N a_{2} B e O_{2}+H_{2} O$Its dipositive ion has electronic configuration$\left(B e^{2+}=1 s^{2}\right)$Q. . State reason for the high solubility of$B e C l_{2}$in organic solvents. Sol. Because$B e C l_{2}$is covalent compound. Q. . What is the cause of diagonal relationship ? [NCERT] Sol. The charge over radius ratio, i.e., polarizing power is similar, that is the cause of diagonal relationship. [esquestion]. Name the alkali metals which forms superoxide when heated in excess of air. Q. . Which out of K, Mg,Ca & Al form amphoteric oxide ? Sol. form amphoteric oxide, i.e., acids as well as basic in nature. Q. . Explain the following : Sodium wire is used to dry benzene but cannot be used to dry ethanol. Sol. Sodium metal removes moisture from benzene by reacting with water. However, ethanol cannot be dried by using sodium because it reacts with sodium.$2 \mathrm{Na}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONa}+\mathrm{H}_{2}$Q. . Why is$\mathrm{CaCl}_{2}$added to$N a C l$in extraction of$N a$by Down cell ? Sol.$\mathrm{CaCl}_{2}$reduces melting point of$N a C l$and increases electrical conductivity. Q. . Carbon dioxide is passed through a suspension of limestone in water. Write balanced chemical equation for the above reaction. Sol.$\mathrm{CaCO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \longrightarrow \mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$Q. . What do we get when crystals of washing soda exposed to air ? Sol. We get amorphous sodium carbonate becouse it loses water molecules. Q. .$M g_{3} N_{2}$when react with water gives off$\mathrm{NH}_{3}$but$H C l$is not obtained from$M g C l_{2}$on reaction with water at room temperature. Sol.$M g_{3} N_{2}$is a salt of a strong base,$M g(O H)_{2}$and a weak acid$\left(N H_{3}\right)$and hence gets hydrolysed to give$N H_{3} .$In contrast,$M g C l_{2}$is a salt of a strong base, and a strong acid, and hence does not undergo hydrolysis to give Q. . Why caesium can be used in photoelectric cell while lithium cannot be ? Sol. Caesium has the lowest while lithium has the highest ionization enthalpy among all the alkali metals. Hence, caesium can lose electron very easily while lithium cannot. Q. . Which nitrates are used in pyrotechnics ? Sol. Strontium and Barium nitrates are used in pyrotechnics for giving red and green flames. Q. .What are s-block elements ? Write their electronic configuration. Sol. The elements in which the last electron enters the -orbital of their outermost energy level are called -block elements. It consists of Group 1 and Group 2 elements. Their electronic configuration is Q. . Name the metals which are found in each of the following minerals : (i) Chile Salt Petre (ii) Marble (iii) Epsomite (iv) Bauxite. Sol. (i)$\quad N a$(ii)$\mathrm{Ca}$(iii)$M g \quad$(iv)$\quad A l$Q. . What is composition of Portland cement ? What is average composition of good quality cement ? [NCERT] Sol.$\mathrm{CaO}=50$to$60 \% \quad \mathrm{SiO}_{2}=20$to$25 \%, \quad \mathrm{Al}_{2} \mathrm{O}_{3}=5$to$10 \%M g O=2$to$3 \%, \quad F e_{2} O_{3}=1$to$2 \%, \quad S O_{2}=1 \quad$to$2 \% \quad$is The ratio of$S i O_{2}$(silica) to alumina$\left(A l_{2} O_{3}\right)$should be between 2.5 and 4.0 and the ratio of lime$(\mathrm{CaO})$to total oxides of silicon$\mathrm{SiO}_{2}, \mathrm{Al}_{2} \mathrm{O}_{3}$and$\mathrm{Fe}_{2} \mathrm{O}_{3}$should be as close to 2 as possible. Q. . Write chemical reactions involved in Down process for obtaining Mg from sea water. Sol. Q. . What is the mixture of$\mathrm{CaCN}_{2}$and carbon called ? How is it prepared ? Give its use. Sol. Q. . State the difficulties in extraction of alkaline earth metals from the natural deposits. Sol. Like alkali metals, alkaline earth metals are also highly electropositive and strong reducing agents. Same difficulties, we face in the extraction of these metals. Therefore, these metals are extracted by electrolysis of their fused metal halides. [esquestion]. Starting with sodium chloride how would you proceed to prepare (state the steps only) : (i) sodium metal (ii) sodium hydroxide (iii) sodium peroxide (iv) sodium carbonate #tag# [NCERT] Q. .Write three general characteristics of the elements of s-block of the periodic table which distinguish them from the elements of the other blocks. [NCERT] Sol. (i) They do not show variable oxidation states. (ii) They are soft metals having low melting and boiling point. (iii) They are highly electropositive and most reactive metals. Q. . Why is$L i F$almost insoluble in water whereas$L i C l$is soluble not only in water but also in acetone ? [NCERT] Sol.$L i F$is an ionic compound containing small ions and hence has very high lattice enthalpy. Enthalpy of hydration in this case is not sufficient to compensate for high lattice enthalpy. Hence, is insoluble in water. has partial covalent character due to polarization of chloride ion by ion. Thus, has partial covalent character and partial ionic character and hence is soluble in water as well as less polar solvents such as acetone. Q. . When an alkali metal dissolves in liquid ammonia the solution acquires different colours. Explain the reasons for this type of colour change. [NCERT] Sol. When an alkali metal is dissolved in liquid ammonia it produces a blue coloured conducting solution due to formation of ammoniated cation and ammoniated electron as given below : When the concentration is above the colour of solution is copper-bronze. This colour change is because the concentrated solution contains clusters of metal ions and hence possess metallic lustre. Q. . Arrange the following in the decreasing order of the property mentioned : Sol. [esquestion]. Alkali metals have low ionisation enthalpies.Why is it so? Q. . Which out of$L i, N a, K, B e, M g, C a$has lowest ionisation enthalpy and why ? Sol.$K^{\prime}$has lowest ionisation energy due to larger atomic size among these elements. The force of attraction between valence electron and nucleus is less, therefore it can loose electron easily. Q. . What is responsible for the blue colour of the solution of alkali metal in liquid ammonia ? Give chemical equation also. Sol. The solvated electron,$e\left(N H_{3}\right)_{x}$or ammoniated electron is responsible for blue colour of alkali metal solution in It absorbs light from visible regions and radiates complimentary colour:$2 N a(s)+2 N H_{3}(l) \longrightarrow 2 N a N H_{2}(s)+H_{2}(g)+e\left(N H_{3}\right)_{x}$Q. . The alkali metals follow the noble gases in their atomic structure. What properties of these metals can be predicted from the information ? Sol. (i) They form unipositive ions. (ii) Their second ionisation energy is very high. (iii) They have weak metallic bonds due to larger size and only one valence electron. Q. . Comment on each of the following observations : (i) The mobilities of the alkali metal ions in aqueous solution are$L i^{+}<N a^{+}<R b^{+}<C s^{+}$(ii) Lithium is the only alkali metal to form directly a nitride. Sol. (i)$L i^{+}$ions are smallest in size therefore most hydrated, that is why they have lowest mobility in aqueous solution.$C s^{+}$ions are largest in size, least hydrated, therefore have highest mobility. Size of hydrated cation decreases, therefore, mobility of ions increases down the group. (ii)$L i$is smallest in size and best reducing agent, therefore, it forms nitride with$N_{2}6 L i+N_{2} \longrightarrow 2 L i_{3} N$(Lithium nitride) (iii) It is due to less difference in their standard reduction potentials which is resultant of sublimation energy, ionisation energy and hydration energy. (iv) is least ionic as compared to other fluorides of alkali metals. It has high lattice energy, therefore, it is least soluble. [esquestion]. Why do alkali metals impart characteristic colours to the flame of a bunsen burner ? What is the colour imparted to the flame by each of the following metals ? Lithium, Sodium and Potassium. Q. . Commercial aluminium always contains some magnesium. Name two such alloys of aluminium. What properties are imparted by the addition of magnesium in these alloys ? [NCERT] Sol. Duralium and Magnaliam are alloys of$\mathrm{Al}, \mathrm{Mg}$is lighter in density than$A l$therefore, it makes the alloys lighter. These alloys are used in automobile engines and aeroplanes. Q. . Arrange the (i) hydroxides and (ii) Sulphates of alkaline earth metals in order of decreasing solubilities., giving a suitable reason for each. Sol.$-B e(O H)_{2}<M g(O H)_{2}<C a(O H)_{2}<S r(O H)_{2}<B a(O H)_{2}$Solubility of hydroxide goes on increasing down the group because hydration energy dominates over lattice energy.$B e S O_{4}>M g S O_{4}>C a S O_{4}>S r S O_{4}>B a S O_{4}$Solubility of sulphate goes on decreasing down the group because lattice energy dominates over hydration energy. Q. . State the properties of beryllium different than other elements of the group. Sol. Properties of beryllium different than other elements of the group (i) Beryllium is harder than other elements. (ii) Melting and boiling points are higher than that of other elements. (iii) It does not react with water but other elements do. (iv) It does not react with acids to form hydrogen. (v) It forms covalent compounds but others form ionic compounds. (vi) Beryllium oxide is amphoteric but other oxides are basic. Q. . Mention the general trends in Group 1 and in Group 2 with increasing atomic number with respect to (i) density (ii) melting point (iii) atomic size (iv) ionization enthalpy. [NCERT] Sol. Q. . How do the following properties change on moving from Group 1 to Group 2 in the periodic table ? (i) Atomic size (ii) Ionization enthalpy (iii) Density (iv) Melting points. Sol. (i) Atomic size decrease from group 1 to group 2 due to increase in effective nuclear charge. (ii) First ionisation enthalpy increases from group 1 to group 2 due to decrease in atomic size. (iii) Density increases from group 1 to 2. (iv) Melting points increase from group 1 to 2. Q. . Compare and contrast the chemistry of Group 1 metals with that of Group 2 metals with respect to (i) nature of oxides (ii) solubility and thermal stability of carbonates (iii) polarizing power of cations (iv) reactivity and reducing power. Sol. (i) Group 1 metals form oxides of strong basic nature (except$L i_{2} O$. Group 2 metal form oxide of less basic nature. (ii) Carbonates : Carbonates of Group 1st are soluble and stable, solubility in case of Group 2nd decreases down the group. (iii) Polarizing power of cations increases. (iv) Reactivity and reducing power decreases. Q. . Explain what happens when ? (i) Sodium hydrogen carbonate is heated (ii) Sodium amalgam reacts with water (iii) Fused sodium metal reacts with ammonia. Sol. (i) On heating sodium bicarbonate it forms sodium carbonate $2 \mathrm{NaHCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$ (ii) When sodium amalgam,$N a / H g,$is formed the vigourosity of reaction of sodium with water decreases. $2 \mathrm{Na} / \mathrm{Hg}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NaOH}+\mathrm{H}_{2}$ (iii) Sodium reacts with ammonia to form amide. $2 \mathrm{Na}+2 \mathrm{NH}_{3} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaNH}_{2}+\mathrm{H}_{2}$ Q. . State as to why ? (i) An aqueous solution of sodium carbonate give alkaline tests.$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-}+\mathrm{HCO}_{3}^{-}$(ii) Sodium is prepared by electrolytic method and not by chemical method. Sol. (i) An aqueous solution of sodium carbonate has a large concentration of hydroxyl ions making it alkaline in nature. (ii) Sodium is a very strong reducing agent therefore it cannot be extracted by the reduction of its ore (chemical method). Thus the best way to prepare sodium is by carrying electrolysis of its molten salts containing impurities of calcium chloride. Q. . Explain why : (i) Lithium on being heated in air mainly forms the monoxide and not peroxide. (ii)$K, R b$and$C s$on being heated in the presence of excess supply of air form superoxides in preference to oxides and peroxides. (iii) An aqueous solution of sodium carbonate is alkaline in nature. Sol. (i)$L i^{+}$ions is smaller in size. It is stabilized more by smaller anion, oxide ion$\left(O^{2-}\right)$as compared to larger anion, peroxide ion$\left(O_{2}^{2-}\right)$. (ii)$\quad K^{+}, R b^{+}, C s^{+},$are large cations. A large cation is more stabilized by large anions. since superoxide ion,$O_{2}^{-}$is quite large,$K, R b$and$C s$form superoxides in preference to oxides and peroxides. (iii) In aqueous solution sodium carbonate undergoes hydrolysis forming sodium hydroxide.$\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NaHCO}_{3}+\mathrm{NaOH}$[esquestion]. Write balanced equations or reactions between : Q. . Name an alkali metal carbonate which is thermally unstable and why ? Give its decomposition reaction. Sol.$L i_{2} C O_{3}$is thermally unstable because it is covalent. It decomposes to form$L i_{2} O$and$C O_{2}L i_{2} C O_{3} \stackrel{\Delta}{\longrightarrow} L i_{2} O+C O_{2}$[esquestion]. Why are ionic hydrides of only alkali metals and alkaline earth metals are known ? Give two examples. Q. . Why does the following reaction : proceed better with$K F$than with$N a F ?$Sol. It is because$K F$is more ionic than$N a F$Q. . The enthalpy of formation of hypothetical$\operatorname{CaCl}(s)$theoretically found to be equal to$188 \mathrm{kJ} \mathrm{mol}^{-1}$and the$\Delta H_{f}^{\circ}$for$C a C l_{2}(s)$is$-795 k J m o l^{-1} .$Calculate the$\Delta H^{\circ}$for the disproportionation reaction. Sol. Q. . Why is it that the s-block elements never occur in free state in nature ? What are their usual modes of occurrence and how are they generally prepared ? [NCERT] Sol. s-block elements are highly reactive, therefore they never occur in free state rather occur in combined state in the form of halides, carbonates, sulphates. They are generally prepared by electrolysis of their molten salts . Q. . Name the chief form of occurrence of magnesium in nature. How is magnesium extracted from one of it ores ? [NCERT] Sol. mg occurs in the form of$M g C l_{2}$in sea water from which it can be extracted. Sea water containing$\mathrm{MgCl}_{2}$is concentrated under the sun and is treated with$\mathrm{Ca}(\mathrm{OH})_{2} \cdot \mathrm{Mg}(\mathrm{OH})_{2}$is thus precipitated, filtered and heated to give the oxide. The oxide is treated with$\mathrm{C}$and$\mathrm{Cl}_{2}$to get$\mathrm{MgCl}_{2}$.$\mathrm{MgO}+\mathrm{C}+\mathrm{Cl}_{2} \stackrel{\text { heat }}{\longrightarrow} \mathrm{MgCl}_{2}+\mathrm{CO}\mathrm{MgCl}_{2}$is fused with$\mathrm{NaCl}$and$\mathrm{CaCl}_{2}$at$970-1023 \mathrm{K}$and molten mixture is electrolysed. Magnesium is liberated at cathode and Chlorine is evolved at the anode. At Cathode :$\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}$At Anode :$2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+2 \mathrm{e}^{-}$A steam of coal gas is blown through the cell to prevent oxidation of Mg metal. Q. . How is pure magnesium prepared from sea water ? What happens when mg is burned in air ? Write chemical equations of reactions involved. Sol. Sea water contains$M g C l_{2}$which is concentrated under the sun and is treated with calcium hydroxide,$\mathrm{Ca}(\mathrm{OH})_{2}$, magnesium hydroxide is thus precipitated, filtered and heated to give oxide. The oxide is treated with carbon and$C l_{2}$to get$M g C l_{2}$.$M g O+C+C l_{2} \longrightarrow M g C l_{2}+C O$is mixed with sodium chloride so as to reduce its melting point and increase its electrical conductivity. Molten mixture is electrolysed using steel cathode and carbon anode. A steam of coal gas is blown through the cell to prevent oxidation. of metal. At cathode Mg’t$+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}$At anode$\quad 2 \mathrm{Cl}^{-}-2 \mathrm{e}^{-} \longrightarrow \mathrm{Cl}_{2}$mg obtained in liquid state is further distilled to give pure mg When burns in air, it forms magnesium oxide and magnesium nitride.$3 M g+N_{2} \longrightarrow M g_{3} N_{2} ; 2 \mathrm{Mg}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{MgO}$[esquestion]. (i) Draw a neat and labelled diagram of Castner-Kellner cell for the manufacture of caustic soda. (ii) Give chemical equations of the reaction of caustic soda with (a) ammonium chloride, and (b) carbon dioxide. Q. . What happens when : (i) sodium metal is dropped in water. (ii) sodium metal is heated in free supply of air. (iii) sodium peroxide dissolves in water. Sol. (i)$2 N a+2 H_{2} O \longrightarrow 2 N a O H+H_{2}$; sodium hydroxide is formed with evolution of$H_{2}(g)$The hydrogen gas catches fire due to highly exothermic process. (ii)$\quad 2 \mathrm{Na}+\mathrm{O}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{O}_{2}$sodium peroxide is formed. (iii) Sodium hydroxide and hydrogen peroxide are formed.$\mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2} \mathrm{O}_{2}$Q. . The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain. Sol. The lattice enthalpies of hydroxides and carbonates of magnesium and calcium are very high due to the presence of divalent cations. The enthalpy of hydration cannot compensate for the energy required to break the lattice in these compounds. Hence, they are sparingly soluble in water. On the otherhand, the lattice enthalpies of hydroxides and carbonates of sodium and potassium are low due to the presence of monovalent cations. The enthalpy of hydration in this case is sufficient to break the lattice in these compounds. Hence, hydroxides and carbonates of sodium and potassium are easily soluble in water.[esquestion]. What happens when : (i) magnesium is burnt in air (ii) quicklime is heated with silica (iii) chlorine reacts with slaked lime (iv) calcium nitrate is heated #tag# [NCERT] Q. . Like lithium in Group 1, beryllium shows anomalous behaviour in Group 2. Write three such properties of beryllium which make it anomalous in the group. Sol. (i) Be forms amphoteric oxide whereas others form basic acids. (ii)$B e C l_{2}$is covalent, others form ionic halides. (iii) does not react even with hot water where as others react easily with water. has smallest atomic size, highest ionisation energy, high polarising power which makes it anomalous in this group.[esquestion]. Beryllium exhibits some similarities with aluminium. Point out three such properties. [NCERT] Q. . Compare the solubility and thermal stability of the following compounds of the alkali metals with those of the alkaline earth metals : (i) nitrates (ii) carbonates (iii) sulphates. Sol. (i) Nitrates of alkali metals are more stable than that of alkaline earth metals. All nitrates are soluble in water. (ii) Carbonates of alkali metals are more stable and more soluble in water than that of alkaline earth metals. (iii) Sulphates of alkali metals are more stable and more soluble in water than that of alkaline earth metal.[esquestion]. Discuss the anomalous behaviour of Lithium. Give its diagonal relationship with Magnesium. Q. . State, why : (i) A solution of$\mathrm{Na}_{2} \mathrm{CO}_{3}$is alkaline. (ii) Alkali metals are prepared by electrolysis of their fused chlorides. (iii) Sodium is found more useful than potassium. Sol. (i)$\mathrm{Na}_{2} \mathrm{CO}_{3}$is alkaline due to its hydrolysis, it forms$O H^{-}$more than$H^{+}$because$H_{2} C O_{3}$is weak acid, therefore, is alkaline. (ii) Alkali metals are strong reducing agents and highly reactive with water, therefore, they are prepared by electrolysis of their fused Chlorides. (iii) Sodium is more useful than potassium because sodium is less reactive and found in more abundance than K Q. . Contrast the action of heat on the following : (i)$\quad \mathrm{Na}_{2} \mathrm{CO}_{3}$and$\mathrm{CaCO}_{3}$(ii)$\quad M g C l_{2} \cdot 6 H_{2} O$and$C a C l_{2} \cdot 6 H_{2} O$(iii)$\quad \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$and$\mathrm{NaNO}_{3}$Sol. (i) Sodium carbonate does not decompose whereas calcium carbonate decomposes on heating.$\quad \mathrm{CaCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}$(ii)$\mathrm{MgCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}+2 \mathrm{HCl}+5 \mathrm{H}_{2} \mathrm{O}\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CaCl}_{2}+6 \mathrm{H}_{2} \mathrm{O}$(iii)$2 \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{CaO}+4 \mathrm{NO}_{2}+\mathrm{O}_{2}2 \mathrm{NaNO}_{3} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaNO}_{2}+\mathrm{O}_{2} |$Q. . What happens when : (i) Carbon dioxide gas is passed through an aqueous solution of sodium carbonate. (ii) Potassium carbonate is heated with milk of lime. (iii) Lithium nitrate is heated. Give chemical equation for the reactions involved. Sol. Q. . What happen when (i) Magnesium is burnt in air (ii) Quick lime is heated with silica (iii) Chlorine reacts with slaked lime (iv) Calcium nitrate is heated Sol. Q. . ‘The chemistry of beryllium is not essentially ionic’. Justify the statement by making a reference to the nature of oxide, chloride and fluoride of beryllium. [NCERT] Sol. Be predominantly forms covalent compounds due to smaller size, higher ionisation energy and high polarising power.$B e O, B e C l_{2}$and$B e F_{2}$are covalent and get hydrolysed by water. BeO is least soluble in water due to covalent character.$B e O+H_{2} O \longrightarrow B e(O H)_{2}B e C l_{2}+2 H_{2} O \longrightarrow B e(O H)_{2}+2 H C lB e F_{2}+2 H_{2} O \longrightarrow B e(O H)_{2}+2 H F$They are less soluble in water but more soluble in organic solvents, which shows they are covalent in nature. Q. . Compare and contrast the chemistry of group 1 metals with that of group 2 metals with respect to : (i) nature of oxides (ii) solubility and thermal stability of carbonates (iii) polarizing power of cations (iv) reactivity and reducing power. Sol. (i) Oxides of group 1 elements are more basic than that of group 2. (ii) Solubility and thermal stability of carbonates of group 1 is higher than that of group 2. (iii) Polarizing power of cations of group 2 is higher than that of group 1. (iv) Reactivity and reducing power of group 1 elements is higher than corresponding group 2 elements. Q. . Describe the importance of the following in different areas : (i) limestone (ii) cement (iii) plaster of paris. Sol. (i) Lime stone : (a) It is used in manufacture of glass and cement. (b) It is used as flux in extraction of iron. (ii) Cement : (a) It is used as building material. (b) It is used in concrete and reinforced concrete, in plastering and in construction of bridges, dams and buildings. (iii) Plaster of Paris : (a) It is used for manufacture of chalk. (b) It is used for plastering fractured bones. (c) It is used for making casts and moulds. (d) It is also used in dentistry, in ornaments work and for taking casts of statues. Q. . Describe the general characteristics of group1 elements. Sol. (i) Group-I consists of lithium, sodium, potassium, rubidium, caesium and francium. (ii) Elements have electronic configuration$n s^{1}$(iii) Elements have 1 electron in the outermost s-orbital and have a strong tendency to lose this electron, so : (a) These are highly electropositive metals. (b) Never found in free state due to their high reactivity. (c) They form$M^{+}$ion. (iv) Atomic radii : Atomic radii of alkali metals are largest in their respective periods. (v) Density : Their densities are quite low. Lithium is the lightest known metal. (vi) Oxidation state : The alkali metal atoms show only +1 oxidation state. (vii) Reducing agents : Due to very low value of ionisation energy, alkali metals are strong reducing agents and reducing character increase$N a$to$C s$but$L$but is the strongest reducing agent. (viii) When alkali metals are heated in air, Lithium forms normal oxide$\left(L i_{2} O\right)$sodium forms peroxide$\left(N a_{2} O_{2}\right)$potassium, rubidium and caesium form superoxides with peroxides. (ix) The alkali metals form hydrides of type reacting with hydrogen at about$673 K \quad \text { (Li forms hydride at } 1073 K)$forms hydride at Ionic character of hydrides increases from to Q. . Describe the manufacture process of sodium-carbonate. Sol. Sodium carbonate$\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$or washing soda is manufactured by solvay-process. Principle of process : Carbondioxide gas is passed through brine solution (about 28%$N a C l$saturated with ammonia, sodium carbonate is formed. Plant used for the manufacture of washing soda. Process : It completes in the following steps : (i) Saturation of brine with ammonia : Lime stone (Calcium carbonate is strongly heated to form carbon dioxide. Ammonia and carbondioxide mixture is passed through a tower in which saturated brine is poured down. Ammoniated brine is filtered to remove impurities of calcium and magnesium carbonate. (ii) The milky solution is removed and filtered with the help of a vacuum pump. (iii) Ammonia recovery tower : The filtrate is step 2 is mixed with calcium hydroxide and heated with steam. Ammonia obtained is recycled with carbondioxides. Potassium carbonate cannot be prepared by Solvay process as the solubility of$\mathrm{KHCO}_{3}$is fairly large as compared to$\mathrm{NaHCO}_{3}$Q. . Give chemical equations for the various reactions taking place during the manufacture of washing soda by Solvay’s process. What are the raw materials used in this process ? What is the by-product in this process ? Sol. The raw materials used in this process are sodium chloride and lime stone Calcium chloride is the by-product in this process. Q. . Describe three industrial uses of caustic soda. Describe one method of manufacture of sodium hydroxide. What happens when sodium hydroxides reacts with (i) aluminium metal (ii)$\mathrm{CO}_{2}$(iii)$\mathrm{SiO}_{2} ?$Sol. Industrial uses of caustic soda. (i) It is used for manufacture of soap. (ii) It is used in paper industry. (iii) It is used in textile industries. Sodium hydroxide can be prepared by electrolysis of saturted solution of brine$(N a C l)$(iii) [esquestion]. (i) How is plaster of Paris prepared ? Describe its chief property due to which it is widely used. (ii) How would you explain ? (a)$\quad$BeO is insoluble but$B e S O_{4}$in soluble in water. (b)$\quad B a O$is soluble but$B a S O_{4}$is insoluble in water. (c)$\quad$Lil is more soluble than$K I$. (d)$\quad \mathrm{NaHCO}_{3}$is known in solid state but$\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$is not isolated in solid state. \ #tag# [NCERT] Q. . (i) Name an element which is invariable bivalent and whose oxide is soluble in excess of$N a O H$and its dipositive ion has a noble gas core. (ii) Differentiate between (a) quick-lime (b) lime-water (c) slaked-lime. Sol. (i) Beryllium : It is invariably divalent. Its oxide is soluble in and its dipositive ion has a noble gas core. (ii) Quick lime : It is calcium oxide$(C a O)$It is produced by heating$\mathrm{CaCO}_{3}$Lime water : The clear aqueous solution of Calcium hydroxide in water is called lime water. It is formed when$\mathrm{Ca}(\mathrm{OH})_{2}$is dissolved in excess of$H_{2} O$Slaked Lime : Calcium hydroxide solid is known as Slaked Lime. It is produced when water is added to$\mathrm{CaO}$Q. . What is the effect of heat on the following compounds ? (Write equations for the reactions). (i) Calcium carbonate. (ii) Magnesium chloride hexahydrate. (iii) Gypsum. (iv) Magnesium sulphate heptahydrate. Sol. (i) Action of heat on calcium carbonate : When calcium carbonate is heated, a colourless gas carbon dioxide is given out and a white residue of calcium oxide is left behind. (ii) Action of heat on magnesium chloride hexahydrate : It loses water and hydrogen chloride and yields a residue of magnesium oxide. (iii) Action of heat on gypsum : It forms a hemi-hydrate, called plaster of paris. (iv) Action of heat on magnesium sulphate heptahydrate : It loses water of crystallization and forms anhydrous salt. Q. . (i) Describe two important uses of each of the following: (a) caustic soda (b) sodium carbonate (c) quicklime. Sol. (i) (a) Caustic soda is used in prepartion of pure fats and oils. It is also used for preparation of rayon (artificial silk). (b) Sodium carbonate is used for manufacture of glass. It is used in softening of hard water. (c) Quick lime is used for white washing. It is used for manufacturing of glass and cement. [Polymeric structure of$\left.B e C l_{2}\right]$in solid state Chemical Bonding Class 11 Questions with Answers-Important for Exams Get Chemical Bonding important questions for class 11 exams. View the Important Question bank for Class 11& 12 Chemistry syllabus. These important questions will play significant role in clearing concepts of Chemistry. These questions are designed keeping NCERT class 11 chemical bonding topic in mind for various exams. The questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 chemistry chapters. Learn all the concepts of Chemical Bonding class 11 chemistry with these questions and answers. Also, you can get notes of eact topic of Chemistry class 11 & 12 o our website. Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, 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. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. 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. Arrange the following molecules in increasing order of ionic character of their bonds$L i F, K_{2} O, N_{2}, S O_{2}, C I F_{3}$Sol.$\mathrm{N}_{2}<\mathrm{SO}_{2}<\mathrm{ClF}_{3}<\mathrm{K}_{2} \mathrm{O}<\mathrm{LiF}$Q. Arrange the following bonds in increasing order of ionic character$C-H, F-H, B r-H, N a-I, K-F, L i-C l$Sol.$\mathrm{C}-\mathrm{H}<\mathrm{Br}-\mathrm{H}<\mathrm{Li}-\mathrm{Cl}<\mathrm{Na}-\mathrm{I}<\mathrm{K}-\mathrm{F}$Q. Write Lewis dot symbols for the following elements:$M g, N a, B, O, N, B r, B e$[NCERT] Sol. Q. Write Lewis symbol for the following atoms and ions :$S$and$S^{2-}, \quad A l$and$A I^{3+}, \quad H$and$H^{-}$[NCERT] Sol. Q. Write Lewis dot symbols for the following ions:$L i^{+}, C l^{-}, O^{2-}, M g^{2+}$and$N^{3-} . \quad$[NCERT] Sol. Q. Why is$\sigma$-bond stronger than$\pi$-bond? Sol. It is because extent of overlapping is more in$\sigma$-bond than$\pi$bond. Q. What is the total number of sigma and$p i$bonds in the following molecules: (i)$\mathrm{C}_{2} \mathrm{H}_{2}$(ii)$C_{2} H_{4}$[NCERT] Sol. Q. Considering Z-axis as the inter nuclear axis, which out of the following will form a sigma bond : (i)$1 \mathrm{s}$and$1 \mathrm{s}$(ii)$1 \mathrm{s}$and$2 \mathrm{p}_{\mathrm{z}}$(iii)$2 p_{x}$and$2 p_{x}$(iv)$2 p_{x}$and$2 p_{y}

(\mathrm{v}) 2 p_{y}$and$2 p_{y}$[NCERT] Sol. (i) and (ii) Q. The skeletal form of structure of$C H_{3} C O O H$as shown below is correct, but some of the bonds are wrongly shown. Write the correct Lewis structure for acetic acid, Sol. Q. Correct Lewis structure of$C H_{3} C O O H$is Sol. Q. Which type of hybridisation explain the trigonal bipyramidal shape of$S F_{4} ?$[NCERT] Sol.$s p^{3} d$hybridisation. Q. Explain why$B e H_{2}$molecule has a zero dipole moment although the$B e-H$bonds are polar.$\quad$Sol. BeH_$_{2}$molecule is linear due to presence of sp-hybridization in$B e .$Therefore, dipole moments due to the polar$B e-H$bond get cancel out. Therefore, molecule has zero dipole moment. Q. Predict which out of the following molecule will have higer dipole moment$\mathrm{CS}_{2}$and$\mathrm{OCS}$. Sol.$-O C S$will have higher dipole moment because bond moment of$O=-C$and that of$C=S$do not cancel each other Q.$H_{3} P O_{3}$can be represented by structure$I$and$I I$as shown below. Can these two structures be taken as the canonical forms of the resonance hybrid of$H_{3} P O_{3} .$If not, state reason for the same. Sol. No, these cannot be taken as canonical forms because the position of atoms have changed. Q. Use molecular orbital theory to explain why$B e_{2}$molecule does not exist. Sol. Molecular orbital configuration of$B e_{2}$is$\left(\sigma_{1 s}\right)^{2},\left(\sigma^{*}_{1 s}\right)^{2}\left(\sigma_{2 s}\right)^{2},\left(\sigma_{2 s}\right)^{2} .$Bond order of $B e_{2}=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{1}{2}(4-4)=0$ Q. Compare the relative stability of the following species and indicate their magnetic (diamagnetic or paramagnetic) properties :$C_{2}^{2-}$and$O_{2}^{-}$Sol.$C_{2}^{-2}(B . O=3)$Diamagnetic is more stable than$\quad O_{2}^{-}\left(B . O=1 \frac{1}{2}\right)$paramagnetic Q. Why is it that in the$S F_{4}$molecule, the lone pair of electrons occupies an equatorial position in the overall trigonal bi-pyramidal arrangement in preference to an axial position. [NCERT] Sol. In$S F_{4}$molecule, lone pair of electron occupy an equatorial position because$l . p .-b . p .$repulsions are less in that case. As a result, energy is less which leads to more stability. Q. Define hydrogen bond. Is it weaker or stronger than the vander Waals forces ? Sol. Hydrogen bond is attractive force which binds H-atom of one molecule with electronegative atom of another molecule. It is stronger than vander waals force. Q. How can one non-polar molecule induce a dipole in a nearby non-polar molecule ? [NCERT] Sol. In a non-polar molecule, there may be instantaneous dipole created by specific position of electrons. This instantaneous dipole can induce a dipole in a nearby non-polar molecules. Q. Define H-bond. Sol. H-bond is force of attraction between hydrogen and most electronegative element like F, O and N. It is a weak bond. Q. List two main conditions for forming hydrogen bonds. Sol. (i) High electronegativity of atom bonded to hydrogen. (ii) Small size of electronegative atom. Q. Arrange the following as indicated :$O, F, S, C l, N$in the order of increasing strength of hydrogen bonding (X……H-X). Sol. S < Cl < N < O < F.. Q. Give electron dot formula of (i)$\mathrm{CN}^{-},$(ii)$\mathrm{SO}_{3}^{2-},$(iii)$\mathrm{ClO}_{2}^{-}$(iv)$\mathrm{ClO}_{4}^{-}$Sol. Q. Write Lewis dot symbols for the following atoms and ions :$S \quad$and$\quad S^{2-} ; \quad P$and$P^{3-} ; \quad N a$and$N a^{+} ; \quad A l$and$A l^{3+}$;$H$and$H^{-}$[NCERT] Sol. Q. Write the Lewis dot structure of (i)$\mathrm{Cl}_{4},$(ii)$P H_{3}$and (iii)$B C l_{3} .$Is the octet rule obeyed in these structures? [NCERT] Sol. In$B C l_{3},$Octet rule is not obeyed because boron contains only six electrons around it. Q. What is the formal charge on Nitrite ion ? Sol. Nitrite : The Lewis structure of nitrite ion is Q. Enlist four major differences between$\sigma$(sigma) and$\pi$(pie) bond. Sol. Q. Describe the change in hybridisation of aluminium atom (if any) during the reaction$A l C l_{3}+C l^{-} \longrightarrow A l C l_{4}^{-}$Sol. In$A I C I_{3}$the hybrid state of aluminium is$s p^{2}$but in$A I C I_{4}^{-}$the hybrid state of aluminium is$s p^{3} .$The$A l C l_{4}^{-}$adopts tetrahedral arrangement. Thus hybrid state of$A l$atom changes from$s p^{2}$to$s p^{3}$Q. Draw the shapes of the following hybrid orbitals : Sol. All the hybrid orbitals have same shape,$i . e .,$one lobe larger than the other, however, their sizes are in the order$s p<s p^{2}<s p^{3}$Q. Is there change in the hybridization of the B and N atoms as a result of the following reaction :$B F_{3}+N H_{3} \longrightarrow F_{3} B \leftarrow N H_{3} ?$Sol. In$B F_{3}, B$is$s p^{2}$hybridised and in$N H_{3}, N$is$s p^{3}$hybridised. After the reaction, hybridisation of$B$changes to$s p^{3}$but that of$N$remains unchanged. Hybridisation of Boron changes to involve a vacant orbital to accommodate electron pair coming from$N H_{3} .$Q. Apart from tetrahedral geometry, another possible geometry for$C H_{4}$is square planar with the four$H$atoms at the corners of the square and the$C$atom at its centre. Explain why$C H_{4}$is not square-planar? [NCERT] Sol. In$C H_{4},$central carbon atom is surrounded by four bond pairs of electrons. These bond pairs arrange themselves in tetrahedral manner with maximum possible bond angle of$109^{\circ} 28^{\prime} .$In square planar geometry, the bond angles will be$90^{\circ}$with larger bond pair-bond pair repulsions. Therefore, square planar geometry is not possible. Q. On the basis of VSEPR theory discuss the geometry of (i)$N H_{3}$molecule (ii)$C H_{4}$molecule. Sol. (i) 5 valence electron in nitrogen. 3 bond pairs and one lone pair of electron. Therefore it’s shape is pyramidal. since the Bond pairBond pair repulsion <bond pair-lone pair, repulsion. Therefore, it has pyramidal or distorted tetrahedral shape. (ii)$C H_{4}$is tetrahedral because it has four bonded pair of electrons. Q. Explain why the bond order of$N_{2}$is greater than that of$N_{2}^{+}$but the bond order of$O_{2}$is less than that of$O_{2}^{+}$Explain why$N_{2}$has greater bond dissociation energy than$N_{2}^{+}$whereas$O_{2}^{+}$has greater bond dissociation energy than$O_{2}$Sol. Q. During formation of$N_{2}^{+}$from$N_{2}$the electron is to be removed from bonding molecular orbital$\left(\sigma_{2 p z}\right)$while during formation of$\mathrm{O}_{2}^{+}$from$\mathrm{O}_{2}$the electron is to be removed from anti bonding molecular orbital$\pi_{2 p x}^{*}$or$\pi_{2 p y}^{*} .$Thus, the bond order decreases during formation of$N_{2}^{+}$from 3 to 2.5 while it increases during the formation of$\mathrm{O}_{2}^{+}$from 2 to 2.5 Sol. Q. Sketch the bond moments and resultant dipole moments in the following molecules :$H_{2} \mathrm{O}, P C l_{3}, N H_{3}, N F_{3}$Sol. Q. Out of the following four resonance structures for the$C O_{2}$molecule, which are important for describing the bonding in the molecule and why? Sol. Out of the given structure$1,2$and 3 are important for describing the bonding in$C O_{2}$molecule. These structure are isovalent (same number of bonds) and thus have higher contribution. Structure 4 has unsaturated carbon with only 4 electrons and thus has negligible stability. Q. Write resonance structures for$S O_{3}, N O_{2}$and$N O_{3}^{-}$Sol. Q. How is the bonding molecular orbital in a molecule of hydrogen different from its antibonding molecular orbital? Sol. (i) Bonding molecular orbital has a single lobe and no node whereas antibonding molecular orbital has two lobes which are separated by a node. (ii) Bonding molecular orbital has less energy whereas antibonding has higher energy than atomic orbitals. Q. Explain why the bond order of$N_{2}$is greater than$N_{2}^{+}$but the bond order of$O_{2}$is less than that of Sol. When$N_{2}$change to$N_{2}^{+}$the electron is removed from bonding molecular orbital while when$O_{2}$changes to$O_{2}^{+},$the electron is removed from antibonding molecular orbital. Q. Explain the significance of bond order. Can bond order be used for quantitative comparisons of the strengths of chemical bond ? Calculate the bond order of$N_{2}, O_{2}, O_{2}^{+}$and$O_{2}^{-}$Sol. Bond order cannot be used for quantitative comparison of the strengths of chemical bonds. It can give us only approximate idea about the strength of the bond. Greater the bond order, greater is the strength of the bond.$N_{2}(3), O_{2}(2), O_{2}^{+}(2.5) . O_{2}^{-}(1.5)$Q. Although geometries of$\mathrm{NH}_{3}$and$\mathrm{H}_{2} \mathrm{O}$are distorted tetrahedral, bond angle in water is less than that of ammonia. Explain. Sol.$N H_{3}$molecule has one lone pair while$H_{2} O$has two lone pairs of electrons. Due to the presence of lone pairs the geometries of and are distorted. Due to the presence of stronger$l p-b p$repulsion than$b p-b p$repulsion the bond angle in is reduced from normal tetrahedral bond angle$\left(109.5^{\circ}\right)$to$107^{\circ} .$In case of, two$l p$of electrons force the$O-H$bonds more closely than$N-H$bonds in. So bond angle decreases to a larger extent i.e. to$104.5^{\circ}$Q. Although both$\mathrm{CO}_{2}$and$\mathrm{H}_{2} \mathrm{O}$are triatomic molecules, the shape of$H_{2} O$molecule in bent while that of$C O_{2}$is linear. Explain this on the basis of dipole moment. Sol.$\mathrm{CO}_{2}$has zero dipole moment. This means that molecules is linear so that the two$C=O$bond moments get cancelled giving zero resultant dipole moment. However, molecule has resultant dipole moment showing that it cannot be linear. The two$O-H$bonds are arranged in angular shape and the bond moment of two$O-H$bonds give resultant dipole moment. Q. How do the bonding and anti-bonding MOs formed from a givenpair of AOs compare to each other with respect to: (i) energy (ii) presence of nodes (iii) internuclear electron density? Sol. Q. What is meant by the term bond order in Lewis concept? Calculate the bond order of$N_{2}, O_{2}$and$\mathrm{CO}$. Sol. According to Lewis concept, bond order is number of bonds present between two atoms of a molecule or ion. Bond order in$N \quad N\left(N_{2}\right)$is 3 Bond order in$\mathrm{O} \quad \mathrm{O}\left(\mathrm{O}_{2}\right)$is 2 Bond order in$C \quad O(C O)$is 3 Q. Give the structure of$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}$and$\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Si}\right]_{3} \mathrm{N}$. Are they isostructural? Sol.$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}$is trimethyl amine. It has a pyramidal geometry as shown in structure I.$\left[\left(C H_{3}\right)_{3} \mathrm{Si}\right]_{3} N$is planar as shown is structure II. Thus, the two species are not isostructural. In$\left(C H_{3}\right)_{3} N . N$eatom assumes$s p^{3}$hybrid state whereas in , the$N$atom assumes Inybrid state. Q. Draw Lewis structures for$H_{2} \mathrm{CO}_{3}, \mathrm{SF}_{6}, \mathrm{PF}_{5}, \mathrm{IF}_{7}$and$\mathrm{CS}_{2} .$Is the octet rule obeyed in these cases? Sol. Q. Which hybrid orbitals are used by carbon atoms in the following molecules ? (i)$H_{3} C-C H_{3}$(ii)$H_{3} C-C H=C H_{2}$(iii)$\mathrm{CH}_{3} \mathrm{CHO}$(iv)$\mathrm{CH}_{3} \mathrm{COOH}$(v)$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}$Sol. Q. Discuss the shapes of following molecules using VSEPR model Sol. Q. (i) Explain why$B e C l_{2}$has a zero dipole moment although$B e-C l$bonds are polar. (ii) Explain important aspects of resonance with reference to$\mathrm{CO}_{3}^{2-}$ion. (iii) Why$H_{2} S$is gas and$H_{2} O$is liquid? Sol. (i)$\quad B e C l_{2}$has zero dipole moment because it has linear shape both dipole are equal and opposite such that net dipole moment is zero. (ii) In resonating structure, the position of atom is same. The bond lengths are equal due to resonance. Larger the number of canonical structures, greater will be the stability of molecule / ion. (iii)$\quad H_{2} O$is liquid because water molecule are associated with inter molecular$H$-bonding where as$H_{2} S$is not associated with inter molecular$H$-bonding that is why it is a gas. Q. Write down the$M . O .$configuration of$\mathrm{O}_{2}^{+}, \mathrm{O}_{2}, \mathrm{O}_{2}^{-}$and$\mathrm{O}_{2}^{-2} .$Also compare their stability. Sol. Q. Explain the following: (i)$\mathrm{O}_{2}^{-}$is paramagnetic but$\mathrm{O}_{2}^{2-}$is not. (ii)$N_{2}$has higher bond order than$N O .$Sol. Q. Write down the$M . O .$configuration of$N_{2}, N_{2}^{+}$and$N_{2}^{-}$and compare their stability. Sol. But since, the population of antibonding orbitals is more in case of and thus, is slightly less stable than. Q. Which of the following substances exhibit hydrogen bonding ? Draw the hydrogen bonds between two molecules of the substance where appropriate : Sol. Classification of Elements and Periodicity in Properties Important Questions & Answers Get Classification of Elements and Periodicity in Properties chapter’s important questions & answers. View the Important Question bank for Class 11 Chemistry. These important questions will play significant role in clearing concepts of Chemistry. This question bank is designed keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 chemistry chapters. Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, 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. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. 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. State modern periodic law. [NCERT] Sol. The properties of elements are periodic function of their atomic number. Q. Why do elements in the same group have similar physical and chemical properties ? [NCERT ] Sol. Because they have similar valence shell electronic configuration. Q. In terms of period and group where would you locate the element with Z = 14 ? [NCERT ] Sol. Period – 3, Group – 14. Q. In the modern periodic table, the period indicates the value of (i) Atomic number (ii) Mass number (iii) Principal quantum number (iv) Azimuthal quantum number. [NCERT] Sol. In the modern periodic table, each period begins with the filling with a new shell, therefore, the period indicates the value of principal quantum number. Thus option (iii) is correct. Q. What would be the IUPAC name and symbol of the element with atomic number 120 ? [NCERT] Sol. The roots for$1,2$and 0 are$u n, b i,$and nil respectively.$\therefore$Name of element: Unbinilium Symbol : Ubn. Q. Which element do you think would have been named by (i) Lawrence Berkalay Laboratory (ii) Seaborg’s group? [NCERT] Sol. (i) Lawrencium (Z=103) and Berkelium (Z=97). (ii) Seaborgium (Z=106) Q. How would you react to the statement that electronegativity of on Pauling scale is 3.0 in all its compounds ? [NCERT] Sol. The electronegativity of nitrogen will not be 3.0 in all its compounds. It depends upon other atom attached to it. Q. Arrange the following ions in the order of increasing size:$B e^{2+}, C l^{-}, S^{2}, N a^{+}, M g^{2+}, B r^{-}$[NCERT] Sol.$\mathrm{Be}^{-2}, \mathrm{Mg}^{-2}, \mathrm{Na}^{+}, \mathrm{Cl}^{-}, \mathrm{s}^{-2}, \mathrm{Br}^{-}$Q. Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different ? Justify your answer. [NCERT] Sol. Isotopes are atoms of the same element which have same atomic number but different mass number. Therefore, they have same number of electrons and nuclear charge (protons). Thus, they will have same first ionization enthalpies. Q. What do you understand by isoelectronic species ? Give the formula of a species that will be isoelectronic with the following atoms or ions : [NCERT] Sol. Isoelectronic species are those which have same number of electrons. (i)$F^{-}\left(10 e^{-}\right): O^{2-}$(ii)$A r\left(18 e^{-}\right): C l^{-}$(iii)$M g^{2+}\left(10 e^{-}\right): N a^{+}$(iv)$R b^{+}\left(36 e^{-}\right): K r$Q. Are the oxidation state and covalency of$A l$in$\left[A I C I\left(H_{2} O\right)_{5}\right]^{2+}$same ? [NCERT] Sol. – No, the oxidation state of$A l$is$+3$and covalency is 6 Q. Show by a chemical reaction with water that$N a_{2} O$is basic oxide and$C l_{2} O_{7}$is an acidic oxide. [NCERT] Sol. Q. What is the basic theme of organization of elements in the periodic table. (NCERT) Sol. The basic theme of organisation of elements in the periodic table is to simplify and systematize the study of the properties of all the elements and millions of their compounds. On the basis of similarities in chemical properties, the various elements has now been divided into different groups. This has made the study simple because the properties of elements are now studied in form of groups rather than individually. Q. Write the atomic number of the element in the third period and seventeenth group of the periodic table. [NCERT] Sol. In the third period the filling up of only 3s– and 3p– orbitals occur. Therefore, in this period there are only two s– and six p-block elements. Since third period starts with and ends at , therefore, elements with and are s– block elements. The next six elements with to 18 are p– block elements and belong to groups 13, 14, 15, 16, 17 and 18. Therefore, the elements which will lie in seventeenth group will have Q. In terms of period and group where would you locate the element with Z = 114 ? So, it belongs to$7^{\text {th }}$period and$14^{\text {th }}$group. [NCERT] Sol. The electronic configuration of the element with$Z=114$would be$[R n\} 5 f^{14} 6 d^{10} 7 s^{2} 7 p^{2} .$since it has$n=7$for the valence shell, it belong to the$7^{\text {th }}$period. It receives the last electron in$p$-orbital. therefore, it belongs to$p$-block. The group number will be$10+4$(No. of electrons in$n s \text { and } n p \text { orbitals })=14$So, it belongs to$7^{\text {th }}$period and$14^{\text {th }}$group. Q. How would you justify that there are only 18 elements in the fifth period of the periodic table. [NCERT ] Sol. The fifth period beings with the filling of 5s orbital and continues till the filling of sixth energy level (6s) starts. The sub-shell which follow 5s are .Thus, the elements which involve filling of 5s, 4d and 5p sub-shell are accommodated in the fifth period. These sub-shells have nine orbitals that can accept 18 electrons in all. Hence, there are 18 elements in the fifth period. Q. On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements. [NCERT ] Sol. In the sixth period the subshell filled are and . There is a total of 16 orbitals which can accommodate 32 electrons. Hence there are 32 elements in the sixth period. Q. Consider the following species :$N^{3-}, O^{2-}, F^{-}, N a^{+}, M g^{2+}$and$A l^{3+}$(i) What is common in them ? (ii) Arrange them in the order of increasing ionic radii. [NCERT] Sol. (i) All these ions have same number (10) of electrons. Therefore, there are also called isoelectronic species. (ii) Since the number of electrons are same, the ionic size decrease with increase in nuclear charge. Therefore, the ions can be arranged in increasing order of ionic radii as$A l^{3+}<M g^{2+}<N a^{+}<F^{-}<O^{2-}<N^{3-}$Q. Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell (i) Valence principal quantum number (ii) Nuclear charge (Z) (iii) Nuclear mass (iv) Number of core electrons. [NCERT] Sol. Nuclear mass does not affect the valence shell because nucleus consists of protons and neutrons. Where as protons i.e., nuclear charge affects the valence shell but neutrons do not. Thus option (iii) is wrong. Q. Which of the following statements related to the modern periodic table is incorrect ? (i) The p-block has six columns, because a maximum of 6 electrons can occupy all the orbitals in a p-subshell. (ii) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell. [NCERT] Sol. Statement (ii) is incorrect while other statements are correct. The correct statement (ii) is : the d-block has 10 columns, Statement (ii) is incorrect while other statements are correct. The correct statement (ii) is : the d-block has 10 columns, Q. What is the basic difference in approach between Mendeleev’s Periodic Law and the Modern Periodic Law? [NCERT] Sol. Mendeleev Periodic Law states that the physical and chemical properties of the elements are a periodic function of their atomic weights while Modern Periodic Law states that physical and chemical properties of elements are a periodic function of their atomic number. Thus, the basic difference in approach between Mendeleev’s Periodic Law and Modern Periodic Law is the change in basis of classification of elements from atomic weight to atomic number. Q. How do atomic radii vary in a period and in a group ? How do you explain the variation ? [NCERT] Sol. Within a group, the atomic radius increases down the group. This is because a new energy shell (i.e., principal quantum number increases by unity) is added at each succeeding element while the number of electrons in the valence shell remains to be the same. In other words, the electrons in the valence shell of each succeeding element lie farther and farther away from the nucleus. As a result, the force of attraction of the nucleus for the valence electrons decreases and hence the atomic size increases. In contract, the atomic size decreases as we move from left to right in a period. This is because that within a period the outer electrons remains in the same shell but the nuclear charge increases by one unit at each succeeding element. Due to this increased nuclear charge, the attraction of the nucleus for the outer electrons increases and hence the atomic size decreases. Q. Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character. : Si, Be, Mg, Na, P. [NCERT] Sol. Arranging the elements into different groups and periods in order of their increasing atomic numbers, we have, We know that the metallic character increases down a group and decreases along a period as we move from left to right. Therefore, Na is the most metallic element, followed by Mg and Si, while P is the least metallic element. Among and is more metallic than Be. Therefore, the overall increasing order of metallic character is$P<S i<B e<M g<N a$Q. Write the general electronic configuration of s-, p-, d– and f-block elements : [NCERT] Sol. (i)$s$-Block elements:$n s^{1-2}$where$n=2-7$(ii)$\quad p$-Block elements:$n s^{2} n p^{1-6}$where$n=2-6$(iii)$d$-Block elements:$(n-1) d^{1-10} n s^{0-2}$where$n=4-7$(iv)$f$-Block elements:$(n-2) f^{0-14}(n-1) d^{0-1} n s^{2}$where$n=6-7$Q. Assign the position of the element having outer electronic configuration, (i)$n s^{2} n p^{4}$for$n=3 \quad$(ii)$(n-1) d^{2} n s^{2}$for$n=4$and (iii)$(n-2) f^{7}(n-1) d^{1} n s^{2}$for$n=6$in the periodic table. [NCERT] Sol. (i)$\quad n=3$suggest that the elements belongs to third period. since the last electron enters the$p$-orbital, therefore, the given element is a$p$-block element. Further since the valence shell contains$6(2+4)$electrons, therefore, group number of the elements$=10+$no. of electrons in the valence shell$=10+6=16$The complete electronic configuration of the element is$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4}$and the element is$S($sulphur). (ii)$\quad n=4$suggest that the element lies in the$4^{\text {th }}$Period. since the$d$-orbitals are incomplete, therefore, it is$d$-block element. The group number of the element$=$no. of$d$-electrons$+$no. of$s$electrons$=2+2=4 .$Thus the elements lies in group 4 and$4^{\text {th }}$period. The complete electronic configuration of the element is$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}$and the element in Ti (Titanium). (iii)$\quad n=6$means that the element lies in the sixth period. since the last electron goes to the$f$-orbital, therefore, the element is a$f$-block element. All$f$-block elements lie in group 3. The complete electronic configuration of the element is$[\mathrm{Xe}\} 4 f^{7} 5 d^{1} 6 s^{2} .$The atomic number of the element$54+7+1+2=64$and the element$G d($gadolinium). Q. The elements Z = 117 and 120 have not yet been discovered. In which family group would you place these elements and also give electronic configuration in each case [NCERT] Sol. The electronic configuration of the element with$Z=117$would be$[R n] 4 f^{14} 5 d^{10} 7 s^{2} 7 p^{5} .$It has outermost$n s^{2} n p^{5}$configuration and therefore, it belongs to halogen family or group 17 The electronic configuration of the element with$Z=120$would be [Uuo]$8 s^{2} .$It has outermost$n s^{2}$configuration and therefore, belongs to alkaline earth metals family or group 2. Q. What are major differences between metals and non-metals? [NCERT] Sol. Elements which have a strong tendency to lose electrons to form cations are called metals while those which have a strong tendency to accept electrons to form anions are called non metals. Thus, metals are strong reducing agents, they have low ionization enthalpies, have less negative electron gain enthalpies, low electronegativity, form basic oxides and ionic compounds. Non-metals, on the other hand, are strong oxidising agents, they have high ionization enthalpies, have high negative electron gain enthalpies, high electronegativity, form acidic oxides and covalent compounds. Q. The elements Z=107 and Z=109 have been made recently; element Z=108 has not yet been made. Indicate the group in which you will place the above elements. [NCERT] Sol. The electronic configuration of these elements are :$Z=107 \quad[R n] 5 f^{14} 6 d^{5} 7 s^{2}Z=108 \quad[R n] 5 f^{14} 6 d^{6} 7 s^{2}Z=109 \quad[R b] 5 f^{14} 6 d^{7} 7 s^{2}$These elements will be placed in$d$-block in group$7^{\text {th }}, 8^{\text {th }}$and$9^{\text {th }}$respectively. Q. Which of the elements$N a, M g, S i$and p would have the greater difference between the first and the second ionization enthalpies ? Briefly explain your answer. [NCERT] Sol. – Among these elements,$N a$is an alkali metal and has only one electron in its valence shell$\left(3 s^{1}\right) .$Therefore, its$I E_{1}$is very low. After the removal of one electron, it acquires neon configuration$i . e ., \quad\left(1 s^{2} 2 s^{2} 2 p^{6}\right) .$Therefore, its$I E_{2}$is expected to be very high. Consequently the difference in first and second ionisation enthalpies would be greater in case of$N a$Q. Energy of an electron in the ground state of the hydrogen atom is$-2.18 \times 10^{-18} \mathrm{J}$. Calculate the ionization enthalpy of atomic hydrogen in terms of$J$mol$^{-1}$Sol. Energy of the electron in ground state$=-2.18 \times 10^{-18} \mathrm{J}$Energy required for removal of the electron$=0-\left(-2.18 \times 10^{-18} \mathrm{J}\right)=2.18 \times 10^{-18} \mathrm{J}$This is the energy required to remove the electron from one atom of hydrogen in ground state. The energy required to remove electron from one mole$\left(6.022 \times 10^{23} \text { atoms }\right)$will be$=2.18 \times 10^{-18} \times 6.022 \times 10^{23}=13.12 \times 10^{5} \mathrm{J} \mathrm{mol}^{-1}$Q. Arrange the following elements in the increasing order of metallic character : Si, Be, Mg, Na, P. [NCERT] Sol. We know that the metallic character increases down a group and decreases along a period as we move from left to right. Hence the order of increasing metallic character is$P<S i<B e<M g<N a$Q. (i) Why do Group-I metals have lower ionisation enthalpy than corresponding Group-II metals? (ii) Why is an anion larger in size than its neutral atoms? Sol. (i) Group I elements are largest in size than alkaline earth metals (Group II elements), therefore, There is less force of attraction between nucleus and valence electron, that is why their ionization energy is lower. (ii) Anions are largest than neutral atom because electrons are more than the protons, therefore, effective nuclear charge is less, therefore, distance between centre of nucleus and valence electrons is more. Q. The first ionization enthalpy$\left(\Delta_{i} H\right)$values of the third period elements,$N a, M g,$and$S i$are respectively$496,737$and$786 \mathrm{kJ} \mathrm{mol}^{-1} .$Predict whether the first$\Delta_{i} H$value for$A l$will be more close to 575 or$760 \mathrm{kJ} \mathrm{mol}^{-1}$respectively. Justify your answer. [NCERT] Sol. It will be more close to$575 \mathrm{kJ} \mathrm{mol}^{-1}$because the value of Al should be lower than that of$M g .$This is due to effective shielding of$3 p$electrons from the nucleus by$3 s$-electrons. Q. Explain why cations are smaller and anions are larger in radii than their parent atoms. [NCERT] Sol. The ionic radius of a cation is always smaller than the parent atom because the loss of one or more electrons increases the effective nuclear charge. As a result, the force of attraction of nucleus for the electrons increases and hence the ionic radii decreases. In contrast, the ionic radius of an anion is always larger than its parent atom because the addition of one or more electrons decreases the effective nuclear charge. As a result, the force of attraction of the nucleus for the electrons decreases and hence the ionic radii increases. Q. What is the basic difference between the terms electron gain enthalpy and electronegativity? [NCERT] Sol. Both electrons gain enthalpy and electronegativity refer to the tendency of the atom of an element to attract electrons. Whereas electron gain enthalpy refers to the tendency of an isolated gaseous atom to accept an additional electron to form a negative ion, electronegativity refers to the tendency of the atom of an element to attract the shared pair of electrons towards it in a covalent bond. Q. (i) Predict the position of the element with atomic number 26 in the periodic table. (ii) Why IE of oxygen is less than that of nitrogen ? Sol. (i) It belong to group 8 of periodic table because its electronic configuration is$[\mathrm{Ar}] 4 \mathrm{s}^{2} 3 d^{6}$(ii) It is because nitrogen has stable electronic configuration i.e., half-filled$p$-orbitals$\therefore$its I.E. is more than that of oxygen. Q. The first ionization enthalpy$\left(\Delta_{i} H\right)$values of the third period elements,$\mathrm{Na}, \mathrm{Mg}$and$\mathrm{Si}$are respectively$496,737$and 786 kJ mol$^{-1}$predict whether the$\Delta i H$value for$A l$will be more close to 575 or$760 \mathrm{kJ} \mathrm{mol}^{-1}$. Justify your answer. [NCERT] Sol. Arrange the elements$N a, M g$and$S i$into different groups and periods in order of their increasing atomic numbers, we have, since in case of$A l,$a$3 p$-electron is to be lost while in$M g,$a$2 s$electron is to be lost, therefore, value for$A l$will be lower than that of$M g\left(737 k J \mathrm{mol}^{-1}\right)$because of the effective shielding of the$3 p-$electron from the nucleus by 3 s-electrons. Therefore,$\Delta_{i} H$for$A l$will be more close to$575 \mathrm{kJ} \mathrm{mol}^{-1}$Q. The size of isoelectronic species$-F^{-}, N e$and$N a^{+}$is affected by (i) Nuclear charge$(Z)$(ii) Valence principal quantum number$(n)$(iii) Electron-electron interaction in the outer orbital (iv) None of factors because their size is the same. [NCERT] Sol. – The size of the isoelectronic ions depends upon the nuclear charge$(Z) .$As the nuclear charge increase the size decreases. For example,$F^{-}(+9)>N e(+10)>N a^{+}(+11) .$Therefore, statement (i) is correct while all other statements are wrong. Q. The increasing order of reactivity among group 1 elements is$L i<N a<K<R b<C s$whereas that of group 17 is$F>C D>B r>$I. Explain. [NCERT] Sol. The elements of group 1 have only one electron in their respective valence shell and thus a strong tendency to lose this electron. The tendency to lose electrons. in turn, depends upon the ionization enthalpy. Since the ionization enthalpy decreases down the group, therefore, the reactivity of group 1 elements increases in the same order :$L i<N a<K<R b<C s .$In contrast, the elements of group 17. have seven electrons in their respective valence shells and thus have a strong tendency to accept one more electron. The tendency to accept electrons, in turn depends upon their electrode potentials. since the electrode potential of group 17 elements decreases in the order$F(+2.87 V)>C l(+1.36 V), B r(1.08 V)$and$I(+0.53),$therefore, their reactivities also decrease in the same order:$F>C l>B r>I$Q. Which of the following species has the largest and the smallest size.$M g, M g^{2+}, A l, A l^{3+}$[NCERT] Sol. (i)$\quad M g$and$A l$belong the third period. Across a period, atomic radii decreases due to increased nuclear charge. Therefore, atomic size of Al is smaller than that of$M g .$(ii) Further, cations are smaller than their parent atoms. Therefore,$M g^{2+}$is smaller than$M g$and$A l^{3+}$is smaller than$A l$(iii) and are isoelectronic ions. Among isoelectronic ions, higher the tye charge, smaller the size. Therefore, ionic radius of$A P^{3+}$is smaller than that of From the above discussion, it follows that$M g$has the largest while has the smallest size. Q. What is the significance of the terms ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy ? Sol. (i) Ionization enthalpy is the minimum amount of energy required to move the most loosely bound electrons from an isolated gaseous atom so as to convert it into a gaseous cation is called ionization enthalpy. The force with which an electron is attracted by the nucleus of an atom is appreciably affected by presence of other atoms within its molecule or in the neighbourhood. Therefore, for the purpose of determination of ionization enthalpy, it is essential that these interatomic forces of attraction should be minimum. Since in the gaseous state, the atoms are widely separated, therefore, these interatomic forces are minimum. Further since it is not possible to isolate a single atom for the purpose of determination of its ionization enthalpy, therefore, the interatomic distances are further reduced by carrying out the measurement at a low pressure of the gaseous atom. Due to these reasons, the term isolated gaseous atoms has been included in the definition of ionization enthalpy. (ii) Electron gain enthalpy is the energy released when an isolated gaseous atom in the ground state accepts an extra electron to form the gaseous negative ion. The term ground state here means that the atom must be present in the most stable state, i.e., the ground state. The reason being that when the isolated gaseous atoms is in the excited state, lesser amount of energy will be released when it gets converted into gaseous anions after the accepting an electron. Therefore, for comparison purposes, the electron gains enthalpy of gaseous atoms be determined in their respective most stable state, i.e., ground state. Q. (i) Write the electronic configurations of the elements given below:$A(\text { At. } \mathrm{No} .=9), B(\text { At. } \mathrm{No} .=12), C \quad(\text { At. } \mathrm{No} .=29), D(\mathrm{At}\mathrm{No} .=54), E(\mathrm{At.} \mathrm{No} .=58)$(ii) Also predict the period, group number and block to which they belong. (iii) Classify them as representative elements, noble gases, transition and inner transition elements. Sol. (i) Electronic configuration of elements A, B, C, D and E are as (ii) Element$A$receives the last electron in$2 p-$orbital, therefore, it belongs to$p$-block elements and its group number$=10+\mathrm{No.}$of electrons in the valence shell$=10+7=17 .$Further the period of the element$=$No. of the principal quantum number of the valence shell$=2^{\mathrm{nd}}$Element$B$receives the last electron in$3 s$-orbital, therefore, it belongs to the$s$-block elements and its group number$=$No. of electrons in the valence shell$=2 .$Further the period of the element$=$No. of the principal quantum number of the valence shell$=3^{\text {rd }}$Element$C$receives the last electron in the$3 d$-orbital, therefore, it belongs to$d$block elements and its group number$=$No. of electrons in the penultimate shell and valence shell$=10+1=11 .$Further, the period of the element$=$No. of principal quantum number of the valence shell$=4^{\text {th }}$Element D receives its last electron in the 5 p-orbital, therefore, it belongs to$p$-block elements and its group number$=10+\mathrm{No.}$of electrons in the valence shell$=10+8=18 .$Further, the period of the element$=$No. of the principal quantum number of valence shell$=5^{\text {th }}$Element$E$receives its last electron in the$4 f$-orbital, therefore, it belongs to$f$-block elements. It may be noted here that the filling of 4 f-orbital occurs only when one electron has already entered$5 d$-orbital. Therefore, element$E-$belongs to$f-$block elements and not to$d$-block elements. since it belongs to lanthanide series therefore, as such it does not have any group number of its own but is usually considered to lie in group$3 .$However, its period$=$No. of the principal quantum number of the valence shell$=6^{\text {th }}$(iii) Elements$A$and$B$are representative elements since their last electron enters$p-$and$s$-orbital respectively. Elements$C$is a transition element since it receives its last electron in the$d$– orbital. Element$D$is a$p$-block element with completely filled$s$– and$p$-orbitals of the valence shell. Such a type of$p-$block element is called a noble gas. Element$E$is an inner transition element since it receives its last electron in the f-orbital. Q. Predict the formula of the stable binary compounds that would be formed by the following pairs of elements : (i) Lithium and Oxygen (ii) Magnesium and Nitrogen (iii) Aluminium and Iodine (iv) Silicon and Oxygen (v) Phosphorus and Fluorine (vi) Element 71 and Fluorine. [NCERT] Sol. (i) Lithium belongs to group 1 with a valence of 1 while oxygen belongs to group 16 with a valence of$2 .$Hence, the formula of the compound is$L i_{2} O$. (ii) Magnesium belongs to group 2 with a valence of 2 while nitrogen belongs to group 5 with a valence of 3. Hence, the formula of compound is$M g_{3} N_{2}$(iii) Aluminium belongs to group 13 with a valence of 3 and iodine belongs to group 17 with a valence of$1 .$Hence the formula of the compound is$A l l_{3}$(iv) Silicon belongs to group 14 with valence of 4 and oxygen belongs to group 16 with a valence of$2 .$Hence, the formula of the compound is$S i O_{2}$(v) Phosphorus belongs to group 15 and it has valence 3 or$5,$while fluorine belongs to group 17 with a valence of 1 Hence the formula of the compound may be$P F_{3}$or$P F_{5} .$(vi) Element with atomic number 71 is a lanthanoid called Lutetium$(L u),$its common value is$3 .$Fluorine is a group 17 (halogen) element with a valence of$1 .$Therefore, the formula of the compound formed would be$L u F_{3}$(Lutetium fluoride). Q. Describe the theory associated with the radius of an atom as it (i) gains electron (ii) loses electron. [NCERT] Sol. (i) Gain of electron. When a neutral atom gains one electron to form an anion, its radius increases. The reason being that the number of electrons in the anion increases while its nuclear charge remains the same as the parent atom. Since the same nuclear charge now attracts greater number of electrons, therefore, force of attraction of the nucleus on the electrons of all the shells decreases (i.e., effective nuclear charge decreases) and hence the electron cloud expands. In other words, the distance between the centre of the nucleus and the last shell that contains electrons increases thereby increasing the ionic radius. Thus, (ii) Loss of electrons. When a neutral atom loses one electron to form a cation, its atomic radius decreases. The reason being that the number of electrons in the cation decreases while its nuclear charge remains the same as the parent atom. Since the same nuclear charge now attracts lesser number of electrons, therefore, the force of attraction of the nucleus on the electrons of all the shells-increases (i.e., effective nuclear charges increases) and hence the size of cation decreases. Thus, Q. The first ionization enthalpy values (in$k J$mol$^{-1}$) of group 13 elements are : How will you explain this deviation from the general trend? [NCERT] Sol. On moving the group 13 from$B$to$A l,$the ionization enthalpy decreases as expected due to an increases in atomic size and screening effect which outweigh the effect of increased nuclear charge. However,$\Delta_{i} H$of$G a$is only slightly higher$\left(2 k J m o l^{-1}\right)$than that of$A l$while that of$T i$is much higher than those of$A l, G a$and$I n .$These deviations can be explained as follows: Al follows immediately after$s$-block elements while$G a$and In follows after$d$-block elements and$T i$after$d$– and$f$– block elements. These extra$d-$and$f-$electrons do not shield (or screen) the outer shell-electrons from the nucleus very effectively. As a result, the valence electrons remain more tightly held by the nucleus and hence larger amount of energy is needed for their removal. This explains why$G a$has higher ionization enthalpy than$A l .$Further on moving down the group from$G a$to$I n,$the increased shielding effect (due to presence of additional$4 d$-electrons ) outweighs the effect of increased nuclear charge$(49-31=18$units) and hence the$\Delta_{i} H_{1}$of$\operatorname{In}$is lower than that of the$G a .$Thereafter, the effect of increased nuclear charge$(81-49=32$units ) outweigh the shielding effect due to the presence of additional 4$f$and$5 d$electrons and hence the of$T l$is higher than that of In. Q. The first$\left(\Delta_{i} H_{1}\right)$and the$\left(\Delta_{i} H_{2}\right)$ionization enthalpies$\left(\text { in } k J m o l^{-1}\right)$and the$\left(\Delta_{e g} H\right)$electron gain enthalpy (in$k J$mol$^{-1}$) of a few elements are given below: Which of the above element is likely to be : (i) The least reactive metal (ii) The most reactive metal (iii) The most reactive non-metal (iv) The least reactive non-metal (v) The metal which can form a stable binary halide of the formula$M X_{2}(X=\text { halogen })$(vi) The metal which can form predominantly stable covalent halide of the formula$M X(X=\text { halogen }) ?$[NCERT] Sol. (i) The element$V$has highest first ionization enthalpy$\left(\Delta_{i} H_{1}\right)$and positive electron gain enthalpy$\left(\Delta_{e g} H\right)$and hence it is the least reactive element. since inert gases have positive$\Delta_{e g} H,$therefore, the element-V must be inert gas. The values of$\Delta_{i} H_{1}, \Delta_{i} H_{2}$and$\Delta_{e g} H,$match that of$H e .$(ii) The element II which has the least first ionization enthalpy and a low negative electron gain enthalpy is the most reactive metal. The values of and match that of$K$(potassium). (iii) The element III which has high first ionization enthalpy and a very high negative electrons gain enthalpy is the most reactive non-metal. The values of and match that of (fluorine). (iv) The element IV has a high negative electron gain enthalpy but not so high first ionization enthalpy. Therefore, it is the least reactive non-metal. the values of and match that of$I$(Iodine). (v) The element VIhas low first ionization enthalpy but higher than that of alkalimetals. Therefore, it appears that the element is an alkaline each metal and hence will form binary halide of the formula (where$X=$halogen). The values of and match that of (magnesium).$M g$(vi) The element I has low first ionization but a very high second ionization enthalpy, therefore, it must be an alkali metal. since the metal forms a predominary stable covalent halide of the formula$(X=$halogen), therefore, the alkali metal must be least reactive. athe values of and match that of Li (lithium). Q. What do you understand by isoelectronic species? Name the species that will be isoelectronic with each of the following atoms or ions.$\begin{array}{llll}{\text { (i) } F^{-}} & {\text {(ii) } A r} & {\text { (iii) } M g^{2+}} & {\text { (iv) } R b^{+}}\end{array}$[NCERT] Sol. Ions of different elements which have the same number of electrons but different magnitude of the nuclear charge are called isoelectronic ions. (i)$\quad F^{-}$has$10(9+1)$electrons. Therefore, the species nitride ion,$N^{3-}(7+3) ;$oxide ion;$O^{2-}(8+2),$neon,$N e(10+0)$sodium ion,$N a^{+}(11-1) ;$magnesium ion,$M g^{2+}(12-2)$aluminium ion,$A l^{3+}(13-3)$etc. each one of which contain 10 electrons, are isoelectronic with it. (ii)$\quad$Ar has 18 electrons. Therefore, the species phosphate ion,$P^{3-}(15+3),$sulphide ion;$S^{-2}(16+2) ;$chloride ion;$\mathrm{Cl}^{-}(17+1),$potassium ion,$K^{+}(19-1),$Calcium ion,$\mathrm{Ca}^{2+}(20-2),$etc. each one of which contain 18 electrons, are isoelectronic with it. (iii)$M g^{2+}$has$10(12-2)$electrons, therefore, the species$N^{3-}, O^{2-}, F^{-}, N e, N a^{+}, A l^{3+}$etc. each one of which contains 10 electrons, are isoelectronic with it. (iv)$\quad R b^{+}$has$36(37-1)$electrons. Therefore, the species bromide ion,$B r^{-}(35+1),$krypton,$K r(36+0)$and strontium$S r^{2+}(38-2)$each one of which has 36 electrons, are isoelectronic with it. Structure of Atom Class 11 Questions and Answers-Important for Exams Get Structure of Atom important questions and answers for class 11 Chemistry exams. View the Important Question bank for Class 11 & 12 Chemistry developed by expert faculties from Kota. These important questions will play significant role in clearing concepts of Chemistry. This question bank is designed, keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 chemistry chapter. Learn all the concepts of Structure class 11 chemistry of Atom Chapter through these questions and answers. Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, 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. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. 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. How many protons and neutrons are present in the following nuclei? [NCERT] Sol. (i) 6, 6 (ii) 26, 30 (iii) 38, 50 (iv) 92, 143. Q. The diameter of zinc atom is 2.6 A. Calculate (i) the radius of zinc atom in pm (ii) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise. [NCERT] Sol. Q. A certain particle carries$2.5 \times 10^{-16} \mathrm{C}$of static electric charge. Calculate the number of electrons present in it. [NCERT] Sol. Charge carried by one electron$=1.6022 \times 10^{-19} \mathrm{C}\therefore \quad$Electrons present in particle carrying$2.5 \times 10^{-16} \mathrm{C}$charge$=\frac{2.5 \times 10^{-16}}{-1.6022 \times 10^{-19}}=1560$Q. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge ? (a)$2 s$and$3 s,$(b)$4 d$and$4 f,$(c) [NCERT] Sol. (i)$2 s$is closer to the nucleus than 3 s. Hence 2 swill experience larger effective nuclear charge. (ii)$4 d$(iii)$3 p$(for same reasons). Q. The unpaired electrons in$A l$and$S i$are present in$3 p$orbital. Which electron will experience more effective charge from the nucleus? [NCERT] Sol. Silicon has greater nuclear charge$(+14)$than aluminium$(+13) .$Hence, the unpaired$3 p$electron in case of silicon will experience more effective nuclear charge. Q. Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven, (b) amber light. from traffic signal, (c) radiation from FMradio, (d) cosmic rays from outer space and (e)$X$-rays. [NCERT] Sol. The increasing order of frequency is : Radiation from FM radio < radiation from microwave oven < amber light from traffic signal < X-rays < cosmic rays from outer space. Q. The energy associated with the first orbit of hydrogen atom is$-2.17 \times 10^{-18} \mathrm{J} /$atom. What is the energy associated with the fifth orbit? [NCERT] Sol.$E_{n}=\frac{E_{1}}{n^{2}} ; E_{5}=\frac{-2.17 \times 10^{-18}}{(5)^{2}} \mathrm{J} /$atom$=-8.68 \times 10^{-20} \mathrm{J} /$atom. Q. Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is$5.6 \times 10^{24}$, calculate the power of this laser. [NCERT] Sol.$E=N h v=N h \frac{c}{\lambda} \quad\left(\because v=\frac{c}{\lambda}\right)=\frac{\left(5.6 \times 10^{24}\right)\left(6.626 \times 10^{-34} \mathrm{J} \mathrm{s}\right)\left(3.0 \times 10^{8} \mathrm{ms}^{-1}\right)}{\left(337.1 \times 10^{-9} \mathrm{m}\right)}=3.3 \times 10^{6} J$Q. What is meant by quantization of energy ? Sol. Quantization of energy means the energy of energy levels can have some specific values of energy and not all the values. Q. Write the mathematical expression by which the energies of various stationary states in the hydrogen atom can be calculated. Sol.$\mathrm{E}_{\mathrm{n}}=-\frac{1312 \mathrm{kJ} \mathrm{mol}^{-1}}{\mathrm{n}^{2}}$Q. How many subshells associated with$n=4 ?$How many electrons will be present in the subshell having$m_{s}$value of$-1 / 2$for$n=4 ?$Sol. (a) Subshells in$n=4$are $4 s(l=0), 4 p(l=1), 4 d(l=2), 4 f(l=3)$ The number of electrons having$m_{s}=-\frac{1}{2}$in$n=4$will be 16 Q. Explain the meaning of the symbol$4 d^{6} ?$[NCERT] Sol. It means that 4d sub-shell has 6 electrons. 4 represents fourth energy shell and d is a sub-shell and 6 electrons are present in d orbitals of sub-shell. Q. How does Heisenberg’s Uncertainty principle support concept of orbital? Sol. According to Heisenberg’s Uncertainty principle ‘the exact position of electron cannot be determined, i.e., the exact path of electron cannot be determined but we can determine the region or space where electron spends most of its time, and this region or space is called orbital. Q. Explain why the uncertainty principle is significant only for the motion of subatomic particles but is negligible for the macroscopic objects. Sol. The energy of photon is sufficient to disturb a subatomic particle so that there is uncertainty in the measurement of position and momentum of the subatomic particle. However, the energy is insufficient to disturb a macroscopic object. Q. Can we apply Heisenberg’s uncertainty principle to a stationary electron ? Why or why not? Sol. No, because, velocity = 0 and position can be measured accurately. Q. Which of the four quantum numbers$\left(n, l, m_{l}, m_{s}\right)$determine: (i) the energy of an electron in a hydrogen atom and multi electron atoms (ii) the size of an orbital (iii) the shape of an orbital (iv) the orientation of an orbital in space ? [NCERT] Sol. (i) n (ii) n (iii) l (iv)$m_{l}$Q. What physical meaning is attributed to the square of the absolute value of wave function$\left|\psi^{2}\right| ?$[NCERT] Sol. It measures the electron probability density at a point in an atom. Q. How many electrons in an atom may have the following quantum number ? (i)$n=4, m_{s}=+1 / 2$(ii)$n=3, l=0$[NCERT] Sol. (i)$n=4, m_{\mathrm{s}}=+1 / 2=16$electrons (ii)$n=3, l=0=2$Q. For each of the following pair of hydrogen orbitals, indicate which is higher in energy : (i)$1 s, 2 s$(ii)$2 p, 3 p$(iii)$3 d_{x y}, 3 d_{y z}$(iv)$3 s, 3 d$(v)$4 f, 5 s$[NCERT] Sol. (i)$2 s$(ii)$3 p$(iii) both have same energy (iv) both have same energy (v)$5 \mathrm{s}$Q. The unpaired electrons in$A l$and$S i$are present in$3 p$orbital. Which electrons will experience more effective nuclear charge from the nucleus? [NCERT] Sol. Si Q. An electron is in one of the$3 d$orbitals. Give the possible values of$n, l,$and$m_{l}$for this electron. [NCERT] Sol.$\mathrm{n}=3,1=2, \mathrm{m}_{1}=+2,+1,0,-1,-2$Q. List the quantum numbers ($l$and$m_{l}$) for electrons in 3d orbital. [NCERT] Sol.$1=2,$and$\mathrm{m}_{1}=+2,+1,0,-1,-2$Q. Give the number of electrons in the species$H_{2}^{+}, H_{2}$and$O_{2}^{+}$Sol. Number of electrons :$H_{2}^{+}=1 ; H_{2}=2 ; \quad O_{2}^{+}=15$Q. What atoms are indicated by the following configurations: (i)$[\mathrm{He}] 2 \mathrm{s}^{1}$(ii)$[\mathrm{Ne}] 3 \mathrm{s}^{2} 3 \mathrm{p}^{3} \quad$(iii)$[\mathrm{Ar}] 4 \mathrm{s}^{2} 3 \mathrm{d}^{1}$[NCERT] Sol. (i)$L i$(ii)$P \quad$(iii)$S c$Q.$2 \times 10^{8}$atoms of carbon are arranged side by side. Calculate the radius of carbon atoms if length of this arrangement is$2.4 \mathrm{cm}$Sol. An atom will cover length equal to its diameter Diameter of carbon atom$=\frac{2.4}{2 \times 10^{8}}=1.2 \times 10^{-8} \mathrm{cm}=1.2 \times 10^{-10} \mathrm{m}$Radius of carbon atom$=\frac{1.2 \times 10^{-10}}{2}=0.6 \times 10^{-10} \mathrm{m}=0.06 \times 10^{-9} \mathrm{m}=0.06 \mathrm{nm}$Q. The wavelength range of the visible spectrum extends from = violet$(400 \mathrm{nm})$to red$(750 \mathrm{nm}) .$Express their wave-lengths in frequencies$(H z) .\left(1 n m=10^{-9} m\right)$[NCERT] Sol. For violet light$v=7.50 \times 10^{14} \mathrm{Hz}$For red light$v=4.00 \times 10^{14} \mathrm{Hz}$Q. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol for the ion. [NCERT] Sol. since the ion carries 3 units of positive charge, it will have 3 electrons less than the number of protons. Let number of electrons$=x$No. of protons$=x+3$No. of neutrons$=x+\frac{x \times 30.4}{100}=x+0.304 x=1.304 x$Now, No. of protons$+$No. of neutrons$=56x+3+1.304 x=562.304 x=53x=\frac{53}{2.304}=23$No. of electrons$=23,$No. of protons$=23+3=26$No. of neutrons$=56-26=30$Q. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol. [NCERT] Sol. We know, Mass number$=$No. of protons$+$No. of neutrons$=81$i.e.,$p+n=81$Let number of protons$=x$Number of neutrons$=x+\frac{x \times 31.7}{100}=1.317 x\therefore \quad x+1.317 x=81x=\frac{81}{2.317}=34.96=35$Symbol$=\begin{array}{l}{81} \\ {35}\end{array} B r$Q. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion. [NCERT] Sol. Since the ion carries one unit of negative charge, it means that in the ion the number of electrons is one more than the number of protons. Total number of electrons and neutrons in the ion = 37 + 1 = 38 Let the number of electrons in the ion be = x Number of neutrons in the$=\frac{x \times 111.1}{100}=1.111 x=x+1.111 x=2.111 x2.111 x=38$$x=\frac{38}{2.111}=18$$\therefore$Number of electrons in the ion$=18$Number of protons in the ion$=18-1=17$Thus, atomic number of the elements is 17 which corresponds to chlorine.$\therefore$Symbol of the ion is$_{17}^{37} \mathrm{Cl}^{-}$Q. The threshold frequency$v_{o}$for a metal is$7.0 \times 10^{14} \mathrm{s}^{-1}$Calculate the kinetic energy of an electron emitted when radiation of$v=1.0 \times 10^{15} \mathrm{s}^{-1}$hits the metal. [NCERT] Sol.$K . E .=\frac{1}{2} m v^{2}=\frac{1}{2} m_{e} v^{2}=h\left(v-v_{0}\right)=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(1.0 \times 10^{15} \mathrm{s}^{-1}-7.0 \times 10^{14} \mathrm{s}^{-1}\right)=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(10 \times 10^{14} \mathrm{s}^{-1}-7.0 \times 10^{14} \mathrm{s}^{-1}\right)=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{14} \mathrm{s}^{-1}\right)=1.988 \times 10^{-19} \mathrm{J}$Q. The Vividh Bharti station of All India Radio, Delhi broadcasts on a frequency of 1,368 kHz. Calculate the wavelength of the electromagnetic radiation emitted by the transmitter. Which part of the electromagnetic spectrum does it belong ? [NCERT] Sol. The wavelength,$\lambda,$is equal to$c / v$where$^{t} c^{\prime}$is the speed of electromagnetic radiation in vacuum and ‘$v^{\prime}$is the frequency. Substituting the given values, we have,$\lambda=\frac{3.00 \times 10^{8} \mathrm{ms}^{-1}}{1368 \mathrm{kHz}}=\frac{3.00 \times 10^{8} \mathrm{ms}^{-1}}{1368 \times 10^{3} \mathrm{s}^{-1}}=219.3 \mathrm{m}$This is a characteristic radiowave wavelength. Q. Emission transitions in the Paschen series end at orbit$n=3$and start from an orbit$n$and can be represented as$v=3.29 \times 10^{15}(\mathrm{Hz})\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right)$Calculate the value of$n$if the transition is obtained at$1285 \mathrm{nm} .$Find the region of spectrum. [NCERT] Sol.$v=3.29 \times 10^{15}\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right) s^{-1}\lambda=1285 \mathrm{nm}\therefore \quad v=\frac{c}{\lambda}=\frac{3.0 \times 10^{8} \mathrm{ms}^{-1}}{1285 \times 10^{-9} \mathrm{m}}=2.33 \times 10^{14} s^{-1}2.33 \times 10^{14}=3.29 \times 10^{15}\left(\frac{1}{9}-\frac{1}{n^{2}}\right)$or$\quad \frac{1}{9}-\frac{1}{n^{2}}=\frac{2.33 \times 10^{14}}{3.29 \times 10^{15}}=0.0708|-\frac{1}{n^{2}}=0.0708-\frac{1}{9}=-0.0403\frac{1}{n^{2}}=0.0403$or$n^{2}=24.82\therefore \quad n=5$Q. Electrons are emitted with zero velocity from a metal surface. when it is exposed to radiation of wavelength$6,800$A. Calculate threshold frequency$\left(v_{0}\right)$and work function$\left(W_{0}\right)$of the metal. Sol.$h v=h \theta_{\circ}+\frac{1}{2} m v^{2} \quad$since$v=0h v=h v_{o}+0h v=h v_{0} \Rightarrow v=v_{0}v=\frac{c}{\lambda} \Rightarrow v=\frac{3 \times 10^{8} \mathrm{ms}^{-1}}{6800 \times 10^{-10} \mathrm{m}}=\frac{300}{86} \times 10^{14}=4.41 \times 10^{14} \mathrm{s}^{-1}v=v_{0}=4.41 \times 10^{14} \mathrm{s}^{-1}W_{o}=h v_{o}=6.626 \times 10^{-34} \mathrm{Js} \times 4.41 \times 10^{14} \mathrm{s}^{-1}=2.91 \times 10^{-19} \mathrm{J}$Q. Calculate the energy associated with the first orbit of$H e^{+}$What is the radius of this orbit? [NCERT] Sol. Energy of electron in nth orbit is$E_{n}=-\frac{13.595}{n^{2}} Z^{2} e V$atom$^{-1}$For first orbit of$H e^{+}, Z=2, n=1E_{1}=-\frac{13.595 \times(2)^{2}}{1^{2}}=-54.380 e V$Radius of nth orbit is,$r_{n}=\frac{0.529 n^{2}}{Z} A$Q. Calculate the wave number for the longest wavelength transition in Balmer series of atomic hydrogen. [NCERT] Sol. The longest wavelength corresponds to minimum energy$(\Delta E)$transition. For Balmer series this transition is from$n_{2}=3$to$n_{1}=2v=109677\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] \mathrm{cm}^{-1}=109677\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right] \mathrm{cm}^{-1}=15233 \mathrm{cm}^{-1}=1.5233 \times 10^{4} \mathrm{cm}^{-1}$or$\quad=1.5233 \times 10^{6} \mathrm{m}^{-1}$Q. Show that the circumference of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron moving around the orbit. [NCERT] Sol. According to Bohr postulate of angular momentum,. $m v r=n \frac{h}{2 \pi} \quad \text { or } \quad 2 \pi r=n \frac{h}{m v} \quad \ldots .(\mathrm{i})$ Substituting this value in equation (i), we get$2 \pi r=n \lambda$Thus, the circumference$(2 \pi r)$of the Bohr orbit for hydrogen atoms is an integral multiple of de-Broglie wavelength. Q. Calculate de-Broglie wavelength of an electron (mass$=9.1 \times 10^{-31} \mathrm{kg}$) moving at$1 \%$speed of light.$\left(h=6.63 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1}\right)$[NCERT] Sol. We know that$\lambda=\frac{h}{m v}m=9.1 \times 10^{-31} \mathrm{kg}, h=6.63 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1}v=1 \%$of speed of light$=\frac{1 \times 3.0 \times 10^{8}}{100} m s^{-1}=3.0 \times 10^{6} \mathrm{ms}^{-1}\left(\because \text { speed of light }=3.0 \times 10^{8} \mathrm{ms}^{-1}\right)$Q. The mass of an electron is$9.1 \times 10^{-31} \mathrm{kg} .$If its K.E. is$3.0 \times 10^{-25} \mathrm{J},$calculate its wavelength. [NCERT] Sol.$K . E .=\frac{1}{2} m v^{2}\therefore v=\sqrt{\frac{2 K \cdot E}{m}}=\sqrt{\frac{2 \times 3.0 \times 10^{-25} \mathrm{J}}{9.1 \times 10^{-31} \mathrm{kg}}=812} \mathrm{ms}^{-1}\left(1, J=1 k g m^{2} s^{-2}\right)$By de-Broglie equation,$\lambda=\frac{h}{m v}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{kg}\right)\left(812 \mathrm{ms}^{-1}\right)}$Q. (i) How many sub-shells are associated with$n=4$? (ii) How many electrons will be present in the sub-shells having$m_{s}$value of$-1 / 2$for$n=4 ?$Sol. (i)$n=4, l=0,1,2,3,(4 \text { subshells, viz, } s, p, d \text { and } f)$(ii) No. of orbitals in 4 th shell$=n^{2}=4^{2}=16$Each orbital has one electron with$m_{s}=-1 / 2 .$Hence, there will be 16 electrons with$m_{s}=-1 / 2$Q. Calculate the uncertainty in the velocity of a wagon of mass 2000 kg whose position is known to an accuracy of \pm10$\mathrm{m}$[NCERT] Sol. Mass of wagon$=2000 \mathrm{kg}$Uncertainty in position,$\Delta x=\pm 10 \mathrm{m}$According to Heisenberg uncertainty principle$\Delta x \times \Delta p=\frac{h}{4 \pi}$or$\Delta x \times \Delta v=\frac{h}{4 \pi m}$or$\quad \Delta v=\frac{h}{4 \pi m \Delta x}$Q. If the position of the electron is measured within an accuracy of$\pm 0.002 n m,$calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is$h / 4 \pi \times 0.05 n m,$is there any problem in defining this value. [NCERT] Sol.$\Delta x=0.002 n m=2 \times 10^{-3} n m=2 \times 10^{-12} m\Delta x \times \Delta p=\frac{h}{4 \pi}\therefore \quad \Delta p=\frac{h}{4 \pi \Delta x}=\frac{6.626 \times 10^{-34} \mathrm{kg} m^{2} s^{-1}}{4 \times 3.14 \times\left(2 \times 10^{-12} \mathrm{m}\right)}=2.638 \times 10^{-23} \mathrm{kg} \mathrm{ms}^{-1}$Actual momentum$=\frac{h}{4 \pi \times 0.05 \mathrm{nm}}=\frac{h}{4 \pi \times 5 \times 10^{-11} \mathrm{m}}$It cannot be defined as the actual magnitude of the momentum is smaller than the uncertainty. Q. Using the s, p, d notations, describe the orbital with the following quantum numbers : [NCERT] Sol. Q. How many sub-shells are associated with$n=4$? [NCERT] Sol. For$n=4, l$can have values$0,1,2,3 .$Thus, there are four sub-shells in$n=4$energy level. These four sub-shells are$4 s, 4 p, 4 d$and$4 f$Q. Indicate the number of unpaired electron in (i)$P(\text { ii })$Si ( iii)$C r(\text { iv })$Fe, and (v)$K r$[NCERT] Sol. i)$\quad_{15} P=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1} 3 p_{z}^{1}$No. of unpaired electron$=3$(ii)$\quad_{14} S i=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1}$No. of unpaired electron$=2$(iii)$\quad_{24} C r=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} 4 s^{1}$No. of unpaired electrons$=6$(iv)$\quad 26 F e=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}$No. of unpaired electrons$=4(\text { in } 3 d)$(v)$\quad 36 K r=$Noble gas. All orbitals are filled. Unpaired electron$=0$Q. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element. [NCERT] Sol. No. of protons$=$Number of electrons$=29$Electronic configuration$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1} 3 d^{10}$Q. Consider the following electronic configurations : (i)$1 s^{2} 2 s^{1}$(ii)$1 s^{2} 3 s^{1}$(a) Name the element corresponding to (i) (b) Does (ii) correspond to the same or different element? (c) How can (ii) be obtained from (i) ? (d) Is it easier to remove one electron from (ii) or (i) ? Explain. Sol. (a) The element corresponding to ( i ) is Lithium (Li). (b) This electronic configuration represents the same element in the excited state. (c) By supplying energy to the element when the electron jumps from the lower energy$2 s$-orbital to the higher energy$3 s$orbital. (d) It is easier to remove an electron from (ii) than from (i) since in the former case the electron is present in a$3 s$-orbital which is away from the nucleus and hence is less strongly attracted by the nucleus than an electron in the$2 s$-orbital. Q. What atoms are indicated by the following configuration? Are they in the ground state or excited state ? (i)$1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} 2 p_{z}^{1}$(ii)$1 s^{2} 2 s^{1} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}$(iii)$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1}$(iv)$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1} 3 p_{x}^{1} 3 p_{y}^{1} 3 p_{z}^{1} 3 d^{1}$(v)$[A r] 3 d^{5} 4 s^{2}$Sol. (i)$1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} 2 p_{z}^{1} .$Atomic number is 9 and it is configuration of fluorine in ground state. (ii)$1 s^{2} 2 s^{1} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1} .$Atomic number is 6 and it is configuration of carbon in excited state. (iii)$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1} .$Atomic number is 14 and it is configuration of silicon in ground state. (iv)$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1} 3 p_{x}^{1} 3 p_{y}^{1} 3 p_{z}^{1} 3 d^{1} .$Atomic number is 15 and it is configuration of phosphorus in excited state. (v)$[A r] 3 d^{5} 4 s^{2} .$Atomic number is 25 and it is configuration of manganese in ground state. Q. An atom has 2 electrons in the first$(K)$shell, 8 electrons in the second$(L)$shell and 2 electrons in the third$(M)$shell. Give its electronic configuration and find out the following: (i) Atomic number (ii) Total number of principal quantum numbers (iii) Total number of sub-levels (iv) Total number of$s$-orbitals (v) Total number of$p$-electrons. Sol. The electronic configuration of the atom is :$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$(i) Atomic number$=2+2+6+2=12$(ii) Number of principal quantum numbers$=3$(iii) Number of sub-levels$=4(1 s, 2 s, 2 p, 3 s)$(iv) Number of$s$-orbitals$=3(1 s, 2 s, 3 s)$(v) Total number of$p$-electrons$=6$Q. Account for the following: (i) The expected electronic configuration of copper is$[A r] 3 d^{9} 4 s^{2}$but actually it is$[A r] 3 d^{10} 4 s^{1}$(ii) In building up of atoms, the filling of$4 s$orbitals occurs before$3 d-$orbitals. Sol. (i) It is because half filled and completely filled orbitals are more stable. (ii)$4 s$orbital has lower energy than$3 d$orbital, therefore, it is filled before$3 d$orbital. Q. (i) Calculate the total number of electrons present in 1 mol of methane. (ii) Find (a) the total number and (b) total mass of neutrons in$7 m g$of$^{14} C$. (Assume mass of a neutron $\left.=1.675 \times 10^{-27} \mathrm{kg}\right)$ (iii) Find (a) total number and (b) total mass of protons in$34 m g$of$N H_{3}$at STP. Will the answer change if the temperature and pressure are changed? [NCERT] Sol. (i) 1 molecule of methane contains$=6+4=10$electrons,$6.022 \times 10^{23}$molecules of methane$(1 \mathrm{mol})$contain electrons$=10 \times 6.022 \times 10^{23}=6.022 \times 10^{24}$(ii) One atom of$^{14} C$contains$=8$neutrons,$6.022 \times 10^{23}$atoms of carbon weight$=14 \mathrm{g}14 g$of$C$contain$=6.022 \times 10^{23}$atoms$=6.022 \times 10^{23} \times 8$neutrons $7.0 \times 10^{-3} g \text { of } C \text { contains }$$=\frac{6.022 \times 10^{23} \times 8 \times 7.0 \times 10^{-3}}{14}=2.409 \times 10^{21}$Mass of 1 neutron$=1.675 \times 10^{-27} \mathrm{kg}$Mass of$2.409 \times 10^{21}$neutrons$=1.675 \times 10^{-27} \times 2.409 \times 10^{21}=4.035 \times 10^{-6} \mathrm{kg}$(iii) No. of protons in 1 molecule of$N H_{3}$$=7+1+1+1=10 \text { protons }$$17 g$of$N H_{3}$contains$=6.022 \times 10^{23}$molecules$=6.022 \times 10^{23} \times 10$protons$=6.022 \times 10^{24}$protons$34 \times 10^{-3} g$of$N H_{3}$contains$=\frac{6.022 \times 10^{24} \times 34 \times 10^{-3}}{17}=1.204 \times 10^{22}$Mass of 1 proton$=1.67 \times 10^{-27} \mathrm{kg}$Mass of$1.204 \times 10^{22}$protons$=1.67 \times 10^{-27} \times 1.204 \times 10^{22}=2.01 \times 10^{-5} \mathrm{kg}$No answer with not change by changing temp. and pres., Q. When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of$1.68 \times 10^{5} \mathrm{J} \mathrm{mol}^{-1} .$What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted? Sol. – Energy of a photon of radiation of wavelength 300 nm will be$E=h v=h \frac{c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{8} \mathrm{ms}^{-1}\right)}{\left(300 \times 10^{-9} \mathrm{m}\right)}$$=6.626 \times 10^{-19} \mathrm{J}$$\therefore$Energy of 1 mole of photons$=\left(6.626 \times 10^{-19} \mathrm{J}\right) \times\left(6.022 \times 10^{23} \mathrm{mol}^{-1}\right)=3.99 \times 10^{5} \mathrm{J} \mathrm{mol}^{-1}$As$\quad E=E_{0}+K . E .$of photoelectrons emitted.$\therefore$Miniumu energy$\left(E_{0}\right)$required to remove 1 mole of electrons from sodium$=E-K . E=(3.99-1.68) 10^{5} J m o l^{-1}=2.31 \times 10^{5} \mathrm{J} \mathrm{mol}^{-1}\therefore$Minimum energy required to remove one electron. Q. Following results were observed when sodium metal is irradiated with different wavelengths. Calculate (a) thres-hold wavelength and (b) Planck’s constant. [NCERT] Sol. Suppose threshold wavelength$=\lambda_{0} n m=\lambda_{0} \times 10^{-9} \mathrm{m}$Then$h\left(v-v_{0}\right)=\frac{1}{2} m v^{2} \quad$or$\quad h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m v^{2}$Substituting the given results of the three experiments, we get$\frac{h c}{10^{-9}}\left(\frac{1}{500}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m\left(2.55 \times 10^{6}\right)^{2}\frac{h c}{10^{-9}}\left(\frac{1}{450}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m\left(4.35 \times 10^{6}\right)^{2}\frac{h c}{10^{-9}}\left(\frac{1}{400}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m\left(5.20 \times 10^{6}\right)^{2}$Dividing equation (ii) by equation (i), we get$\frac{\lambda_{0}-450}{450 \lambda_{0}} \times \frac{500 \lambda_{0}}{\lambda-500}=\left(\frac{4.35}{2.55}\right)^{2}$or$\quad \frac{\lambda_{0}-450}{\lambda_{0}-500}=\frac{450}{500}\left(\frac{4.35}{2.55}\right)^{2}=2.619$or$\lambda_{0}-450=2.619 \lambda-1309.5$or$\quad 1.619 \lambda_{0}=859.5 \quad \therefore \quad \lambda_{0}=531 \mathrm{nm}$Substituting this value in equation (iii), we get$\frac{h \times\left(3 \times 10^{8}\right)}{10^{-9}}\left(\frac{1}{400}-\frac{1}{531}\right)=\frac{1}{2}\left(9.11 \times 10^{-31}\right)\left(5.20 \times 10^{6}\right)^{2}$or$\quad h=6.66 \times 10^{-34} \mathrm{Js}$Q. For which hydrogen like ion the wavelength difference between the first lines of the Lyman and Balmer series is equal to 59.3 nm Sol. For a spectral line,$\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$For Lyman line,$\frac{1}{\lambda_{\mathrm{Lyman}}}=R Z^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]=\frac{3 R Z^{2}}{4}$or$\quad \lambda_{L y \operatorname{man}}=\frac{4}{3 R Z^{2}}$For Balmer line,$\frac{1}{\lambda_{\text {Balmer }}}=R z^{2}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=\frac{5 R Z^{2}}{36}$or$\quad \lambda_{\text {Balmer }}=\frac{36}{5 R Z^{2}}\lambda_{\text {Balmer }}-\lambda_{\text {Lyman }}=\frac{36}{5 R Z^{2}}-\frac{4}{3 R Z^{2}}=59.3 \times 10^{-7}=\frac{1}{R Z^{2}}\left[\frac{36}{5}-\frac{4}{3}\right]=59.3 \times 10^{-7}=\frac{1}{109677.8 Z^{2}}\left[\frac{108-20}{15}\right]=59.3 \times 10^{-7}$or$\quad Z^{2}=\frac{1}{109677.8} \times \frac{88}{15 \times 59.3 \times 10^{-7}}=9$or$\quad Z=3 .$This corresponds to$L i^{2+}$ion. Q. Calculate the velocity$\left(\mathrm{cms}^{-1}\right)$of an electron placed in third orbit of the hydrogen atom. Also calculate the number of revolutions per second that this electron makes around the nucleus. Sol. Velocity of electron is given as :$v=\frac{2 \pi e^{2}}{n h}$Now,$e=4.8 \times 10^{-10}$e.s.u.,$n=3, h=6.63 \times 10^{-27}$erg sec$\therefore \quad v=\frac{2 \times 22 \times\left(4.8 \times 10^{-10}\right)^{2}}{7 \times 3 \times 6.63 \times 10^{-27}}=7.27 \times 10^{7} \mathrm{cm} \mathrm{s}^{-1}$No. of revolutions per second$=\frac{\text { Velocity }}{\text { Circumference }}=\frac{v}{2 \pi r}$but$r=\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2}}\therefore$No. of revolutions$=\frac{2 \pi e^{2}}{n h} \times \frac{4 \pi^{2} m e^{2}}{2 \pi n^{2} h^{2}}=\frac{4 \pi^{2} m e^{4}}{n^{3} h^{3}}=\frac{4 \times\left(\frac{22}{7}\right)^{2} \times\left(9.1 \times 10^{-28}\right) \times\left(4.8 \times 10^{-10}\right)^{4}}{3^{3} \times\left(6.62 \times 10^{-27}\right)^{3}}=2.42 \times 10^{14}$Q. Calculate the wavelength in Angstroms of the photon that is emitted when an electron in Bohr orbit$n=2$returns to the orbit$n=1$in the hydrogen atom. The ionisation potential of the ground state of hydrogen atom is$2.17 \times 10^{-11}$ergs per atom. Sol. Q. The angular momentum of an electron in Bohr’s orbit of hydrogen atom is$4.22 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1} .$Calculate the wavelength of the spectral line when the electron falls from this level to the next lower level. Sol. Angular momentum$(m v r)=n \frac{h}{2 \pi}=4.22 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1}$(Given)$\therefore \quad n=4.22 \times 10^{-34} \times \frac{2 \pi}{h}=\frac{2 \times 4.22 \times 10^{-34} \times 3.14}{6.626 \times 10^{-34}}=4$When the electron jumps from$n=4$to$n=3,$the wavelength of the spectral line can be calculated as follows:$\frac{1}{\lambda}=R_{H}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)=109,677 \mathrm{cm}^{-1}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)=109,677 \times\left(\frac{1}{9}-\frac{1}{16}\right)=109,677 \times \frac{7}{144} \mathrm{cm}^{-1}$or$\lambda=\frac{144}{109,677 \times 7} \mathrm{cm}=1.88 \times 10^{-4} \mathrm{cm}$Q. (i) Write the electronic configuration of$C u^{+}$ion$(Z=29).

(ii) Calculate the de-Broglie wavelength of a milligram sized object moving with $1 \%$ speed of light.

Planck’s constant $(h)=6.63 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1}$

Velocity of light $(c)=3.0 \times 10^{8} \mathrm{ms}^{-1}$

Sol. (i) $\quad C u^{+}(29) 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{0} 3 d^{10}$

(ii) $\quad m=1 m g \quad v=1 \% \times 3 \times 10^{8} m s^{-1}$

$m=10^{-6} \mathrm{kg} \quad v=\frac{1}{100} \times 3 \times 10^{8} \mathrm{ms}^{-1}=3 \times 10^{6} \mathrm{ms}^{-1}$

$\lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-34} \mathrm{kg} m^{2} s^{-1}}{10^{-6} \mathrm{kg} \times 3 \times 10^{6} \mathrm{ms}^{-1}}=2.21 \times 10^{-34} \mathrm{m}$

Q. Give the electronic configuration of the following ions:

Sol. During the formation of cations, electrons are lost while in the formation of anions, electrons are added to the valence shell. The number of electrons added or lost is equal to the numerical value of the charge present on the ion. Following this general concept, we can write the electronic configurations of all the ions given in the question.

(i) $\quad C u^{2+}=_{29} C u-2 e^{-}$

$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{1}-2 e^{-}$

$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{9}$

(ii) $\quad C r^{3+}=_{24} C r-3 e^{-}$

$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} 4 s^{1}-3 e^{-}$

$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{3}$

(iii) $\quad F e^{2+}=_{26} F e-2 e^{-}$

$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}-2 e^{-}$

$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}$

$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}$

$F e^{3+}=26 F e-3 e^{-}$

$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5}$

(iv) $\quad H^{-}=_{1} H+1 e^{-}=1 s^{1}+1 e^{1}=1 s^{2}$

(v) $\quad S^{2-}=_{16} S+2 e^{-}$

$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{2} 3 p_{y}^{1} 3 p_{z}^{1}+2 e^{-}$

$=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{2} 3 p_{y}^{2} 3 p_{z}^{2}$

Numericals on Mole Concept Class 11 with Answers

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Q. What is the SI unit of mass ? How is it defined ?

[NCERT]

Sol. The S.I. unit of mass is kilogram. Mass is defined as the amount of matter present in substance. The unit of mass kilogram is defined as equal to the mass of the international prototype of the kilogram.

Q. Express the following in S.I. base units using power of 10 notation. (example $\left.2.54 \mathrm{mm}=2.54 \times 10^{-3} \mathrm{m}\right)$

(i) $1.35 \mathrm{mm}$

(ii) 1 day

(iii) $6.45 \mathrm{mL}$

(iv) $48 \mu g$ (microgram)

(v) 0.0426 inches

[NCERT]

Sol. (i) $1.35 m m=1.35 \times 10^{-3} \mathrm{m}$

(ii) 1 day $=24 \times 60 \times 60 s=86400 s=8.64 \times 10^{4} s$

(iii) $6.45 \times 10^{-6} \mathrm{m}^{3}\left[\because 1 \mathrm{mL}=10^{-6} \mathrm{m}^{3}\right]$

(iv) $48 \mu g=48 \times 10^{-6} g=4.8 \times 10^{-5} g$

0.0426 inches

(v) $\quad$ linch $=2.54 \times 10^{-2} \mathrm{m}$ $=2.54 \times 10^{-2} \mathrm{m} \times 0.0426=1.082 \times 10^{-3} \mathrm{m}$

Q. When $4.2 \mathrm{g}$ of $\mathrm{NaHCO}_{3}$ is added to solution of $\mathrm{CH}_{3} \mathrm{COOH}$ (acetic acid) weighing $10.0 \mathrm{g},$ it is observed that $2.2 \mathrm{g}$ of $\mathrm{CO}_{2}$ is released into atmosphere. The residue left is found to weigh $12.0 \mathrm{g}$. Show that the observations are in agreement with the law of conservation of mass. $\quad$

[NCERT]

Sol.

These observations are in agreement with law of conservation of mass.

Sol. Avogadro’s Law : Equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Q. Give two examples of molecules having molecular formula same as empirical formula

Sol. $\mathrm{CO}_{2}, \quad \mathrm{CH}_{4}$

Q. What is a limiting reactant ? What is its significance during stoichiometric calculations ?

Sol. The limiting reactant is that reactant which is present in minimum amount as required by the stoichiometry of the balanced equation. Since the limiting reactant is completely consumed during the reaction, all the stoichiometric calculations are done keeping the amount of limiting reactant in mind.

Q. Chlorine is prepared according to the reaction :

$4 H C l(a q)+M n O_{2}(s) \longrightarrow 2 H_{2} O(l)+M n C l_{2}(a q)+C l_{2}(g)$

How many grams of $H C l$ react with $5.0 \mathrm{g}$ of manganese dioxide? (At. mass of $M n=55$ ).

[NCERT]

Sol. According to equation mole ratio of $H C l: M n O_{2}=4: 1$

and mass ratio is $36.5 \times 4: 87=146: 87$

$\therefore \quad H C l$ required in $g=\frac{146}{87} \times 5=8.39 \mathrm{g}$

Q. What is the Concentration of sugar $\left(C_{12} H_{22} O_{11}\right)$ in $\operatorname{mol} L^{-1}$ if its $20 \mathrm{g}$ are dissolved in enough water to make a final volume upto $2 L ?$

[NCERT]

Sol. Molarity of solution $\left(\mathrm{mol} L^{-1}\right)$

$=\frac{\text { Mass of solute }(g)}{\mathrm{M} . \text { mass }} \times \frac{1000}{V \mathrm{inmL}}$

Conc. of sugar $=\frac{20}{342} \times \frac{1000}{2000}=0.0292 \mathrm{mol} \mathrm{L}^{-1}$

Q. Calculate the molality of ethanol in water in which the mole fraction of ethanol is 0.040.

[NCERT]

Sol. $X_{\text {ethonol }}=0.040 ; X_{H_{2} O}=0.960$

Molality ( $m)$

$=\frac{1000 \times X_{\text {eth }}}{X_{H_{2} O} \times G M M_{\left(H_{2} O\right)}}=\frac{1000 \times 0.040}{0.96 \times 18}=2.315 m$

Q. If $6.023 \times 10^{23}$ molecules of $N_{2}$ react completely with

$H_{2}$ according to the equation:

$N_{2}(g)+3 H_{2}(g) \longrightarrow 2 N H_{3}(g)$

Calculate the number of molecules of $N H_{3}$ formed.

Sol. $N_{2}(g)+3 H_{2}(g) \longrightarrow 2 N H_{3}(g)$

$6.023 \times 10^{23}$ molecules of $N_{2}$ react completely with $H_{2}$ to save $2 \times 6.023 \times 10^{23}$ molecules $=1.204 \times 10^{24}$ molecules.

Q. Pressure is determined as force per unit area of the surface. The S.I. unit of pressure, pascal, is as shown below :

$1 P a=1 N m^{-2}$

If mass of air at sea level is $1034 g \mathrm{cm}^{-2}$ calculate the pressure in pascal.

[NCERT]

Sol. Pressure is the force (i.e., weight) acting per unit area

Q. Express the result of the following calculation to the appropriate number of significant digits.

$816 \times 0.02456+215.67$

[NCERT]

Sol. $816 \times 0.02456=20.0$ Product rounded off to 3 significant figures because the least number of significant figures in this multiplication is three.

Q. Express the following numbers of four significant figures:

(i) 5.607892

(ii) 32.392800

(iii) $1.78986 \times 10^{3}$

(iv) 0.007837

[NCERT]

Sol. (i) 5.608 (ii) 32.39 (iii) $1.790 \times 10^{3}$ (iv) $7.837 \times 10^{-3}$

Q. Express the following number to three significant figures: .

(i) $6.022 \times 10^{23}$

(ii) 5.359

(iii) 0.04597

(iv) 34.216

[NCERT]

Sol. (i) The last digit to be retained is 2 and the digit to be dropped is 2 which is less than $5 .$ The result will be expressed as $6.02 \times 10^{23}$

(ii) In this case the last digit to be retained is 5 and the digit to be dropped is $9,$ which is greater than $5 .$ Hence, the last digit to be retained is increased by one. The number will be written as 5.36.

(iii) The three digits to be retained in this case are $4,5$ and $9 .$ The digit 7 is to be dropped which is greater than $5 .$ Hence, the last digit to be retained will therefore be increased by one. The number will be rounded off to three significant digits as 0.0460

(iv) In this case the digits 1 and 6 will be dropped. since, the digit following the last digit to be retained is $1,$ the last digit will be kept unchanged and the number is written as $34.2 .$

Q. Calculate the atomic mass (average) of chlorine using the following data :

[NCERT]

Sol.

Q. Which one of the following will have largest number of atoms?

(i) $1 g A u(s)$

(ii) $1 g N a(s)$

(iii) $1 g L i(s)$

(iv) $1 g C l_{2}(g)$

[at. mass $: A u 197, N a=23, \quad L i=7, \quad C l=35.5]$

Sol. No. of atoms can be calculated as:

(i) $1 g A u=\frac{1}{197} \times 6.022 \times 10^{23}=\frac{6.022}{197} \times 10^{23}$

(ii) $\quad 1 g N a=\frac{6.022}{23} \times 10^{23}$

(iii) $1 g L i=\frac{6.022 \times 10^{23}}{7}$

(iv) $1 g C l_{2}=\frac{2 \times 6.022 \times 10^{23}}{71}=\frac{6.022 \times 10^{23}}{35.5}$

It is clear that contains largest number of atoms.

Q. Define the following :

(iii) Molar mass

(i) Average atomic mass

(ii) Mole

(iv) Unit factor

(v) Molarity (vi) Precision and accuracy.

[NCERT]

Sol. (i) Average atomic mass : It is defined as average of the mass of all of the atoms of an elements, e.g., average atomic mass of is 35.5 u.

(ii) Mole is defined as amount of substance that contains as many atoms, molecules and particles as there are atoms in exactly 0.012 kg of Carbon-12 isotope.

(iii) Molar mass : It is mass of 1 mole of substance which contains $6.022 \times 10^{23}$ particles.

(iv) Unit factor : The factor which is used to convert one unit into another is called Unit factor.

(v) Molarity : It is defined as number of moles of solute dissolved per litre of solution.

(vi) Precision and accuracy : Precision means how closely the experimental measurements agree with another.

Accuracy means how close the experimental measurements and exact values are with each other.

Q. In commercial manufacture of $H N O_{3},$ how many moles of $\mathrm{NO}_{2}$ produce $7.33 \mathrm{mol} \mathrm{HNO}_{3}$ in the reaction?

$3 \mathrm{NO}_{2}(g)+H_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$

[NCERT]

Sol. $3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$

2 moles of $H N O_{3}$ need 3 moles of $N O_{2}$

7.33 mole of $\mathrm{HNO}_{3}$ need

$\frac{3}{2} \times 7.33=\frac{21.99}{2}=10.995$ moles of $\mathrm{NO}_{2}$

Q. Use the data given in the following table to calculate the molar mass of naturally occurring argon :

[NCERT]

Sol. Average atomic mass

Q. How many moles and how many grams of $N a C l$ are present in $250 \mathrm{cm}^{3}$ of $0.50 \mathrm{M} \mathrm{NaCl}$ solution? [ Atomic weight of $N a=$ $23 \mathrm{u}, C l=35.5 \mathrm{u}] \quad \quad$

[NCERT]

Sol. $M=n \times \frac{1000}{\text { Vol. of solution in } \mathrm{cm}^{3}} \Rightarrow 0.50=n \times \frac{1000}{250}$

$\Rightarrow n=0.125 \mathrm{mol}$

Amount of $N a C l=n \times M \cdot w t .=0.125 \times 58.5=7.30 \mathrm{g}$

Q. Potassium bromide $K B r$ contains $32.9 \%$ potassium by mass. If $6.4 \mathrm{g}$ of bromine reacts with $3.60 \mathrm{g}$ of $K,$ calculate the number of moles of $^{c} K^{\prime}$ which combine with bromine to form $K B r$

[NCERT]

Sol. $32.9 \mathrm{g}$ of $\mathrm{K}$ combines with $67.1 \mathrm{g}$ of $\mathrm{Br}$

$3.60 \mathrm{g}$ of $\mathrm{K}$ combines with $\frac{67.1}{32.9} \times 3.60=\frac{241.56}{32.9}=7.342 \mathrm{g}$

since we have $6.4 \mathrm{g}$ of bromine, it means $B r_{2}$ is limiting reactant.

Number of moles of $B r_{2}=\frac{6.4 \mathrm{g}}{160}=0.04$ moles

$2 K+B r_{2} \longrightarrow 2 K B r$ mole of $B r_{2}$ reacts with 2 moles of Potassium to form $K B r$

0.04 mole of $B r_{2}$ will react with $2 \times 0.04$ moles of $K$ to form $K B r$

$=0.08$ moles $=8 \times 10^{-2}$ moles of $K$

Q. Calculate the number of atoms in each of the following:

(i) 52 moles of $A r$

(iii) $52 g$ of $H e$

(ii) $52 u$ of $H e$

[NCERT]

Sol. (i) Number of atoms of

$^{\prime} A r^{\prime}=52 \times 6.02 \times 10^{23}=3.132 \times 10^{25}$ atoms.

(ii) 1 atom of He has a mass of 4 u.

Number of atoms of ‘ $^{\prime} H e^{\prime}=\frac{52}{4}=13$ atoms.

(iii) Number of moles of ‘ $\mathrm{He}^{\prime}=\frac{52}{4}=13$ moles

$\therefore$ Number of atoms of $^{\prime} H e^{\prime}=13 \times 6.023 \times 10^{23}=7.826 \times 10^{24}$ atoms.

Q. Chlorophyll, the green colouring material of plants contains 2.68% of magnesium by mass. Calculate the number of magnesium atoms in 5.00 g of this complex.

[NCERT]

Sol. – Mass of magnesium in $5.00 \mathrm{g}$ of complex

$=\frac{2.68}{100} \times 5.00=0.134 g$

No. of moles of $M g=\frac{0.134}{24}$

atoms of $M g=6.023 \times 10^{23} \times \frac{0.134}{24}=3.36 \times 10^{22}$ atoms

Q. In three moles of ethane $\left(C_{2} H_{6}\right)$ calculate the following:

(i) Number of moles of carbon

(ii) Number of moles of hydrogen atoms

(iii) Number of molecules of ethane.

Sol. (i) 1 mole of $C_{2} H_{6}$ contains 2 moles of carbon $\therefore \quad$ Number of moles of carbon in 3 moles of $C_{2} H_{6}=6$

(ii) 1 mole of $C_{2} H_{6}$ contain 6 mole atoms of hydrogen

$\therefore \quad$ Number of moles of hydrogen atoms in 3 mole of $C_{2} H_{6}=3 \times 6=18$

(iii) 1 mole of $C_{2} H_{6}=6.022 \times 10^{23}$ molecules

$\therefore \quad$ Number of molecules in 3 mole of $C_{2} H_{6}=3 \times 6.022 \times 10^{23}=1.807 \times 10^{24}$ molecules

Q. Calculate the mass of sodium acetate $\left(\mathrm{CH}_{3} \mathrm{COONa}\right)$ required to make $500 \mathrm{mL}$ of 0.375 molar aqueous solution. Molar mass of sodium acetate is $82.0245 \mathrm{g} \mathrm{mol}^{-1}$

Sol. $-0.375 M$ aqueous solution means that $1000 \mathrm{mL}$ of the solution contain sodium acetate $=0.375$ mole

$\therefore \quad 500 \quad \mathrm{mL}$ of the solution should contain sodium

acetate $=\frac{0.375}{2}$ mole

Molar mass of sodium acetate $=82.0245 \mathrm{g} \mathrm{mol}^{-1}$

$\therefore\ Mass of sodium acetate acquired$=\frac{0.375}{2}$mole$\times 82.0245 \mathrm{g} \mathrm{mol}^{-1}=15.380 \mathrm{g}$Q. – How much copper can be obtained from$100 \mathrm{g}$of copper sulphate$\left(C u S O_{4}\right) ?$[Atomic mass of$C u=63.5$amu]. [NCERT] Sol. I mole of$\mathrm{CuSO}_{4}$contains 1 mole$(1 \mathrm{gatom})$of$^{c} \mathrm{Cu}^{\prime}\mathrm{CuSO}_{4}=63.5+32+4 \times 16=159.5 \mathrm{gmol}^{-1}$Thus,$\mathrm{Cu}$that can be obtained from$159.5 \mathrm{g}$of$\mathrm{CuSO}_{4}=63.5 \mathrm{g}\therefore \quad \mathrm{Cu}$that can be obtained from$100 \mathrm{g}$of$\mathrm{CuSO}_{4}=\frac{63.5}{159.5} \times 100 \mathrm{g}=39.81 \mathrm{g}$Molar mass of Q. Find out volume of the following at$S . T . P .$(i)$14 \mathrm{g}$of nitrogen (ii)$6.023 \times 10^{22}$molecules of ammonia$\left(N H_{3}\right)$(iii) 0.1 mole of sulphur dioxide$\left(S O_{2}\right)$Sol. (i) Molecular mass of$N_{2}=14 \times 2=2828 g$of$N_{2}$occupy at S.T.P.$=22.4$litres$\therefore 14 g$of$N_{2}$occupy at S.T.P.$=\frac{22.4}{28} \times 14=11.2$litre. (ii)$6.023 \times 10^{23}$molecules of$\mathrm{NH}_{3}$occupy at S.T.P.$=22.4$litres.$\therefore 6.023 \times 10^{22}$molecules of$\mathrm{NH}_{3}$occupy at S.T.P.$=\frac{22.4}{6.023 \times 10^{23}} \times 6.023 \times 10^{22}=2.24 L$(iii) 1 mole of$S O_{2}$occupies at S.T.P.$=22.4$litres$\therefore 0.1 \mathrm{mol}$of$\mathrm{SO}_{2}$occupies at S.T.P.$=\frac{22.4}{1} \times 0.1=2.24 L$Q. Ferric sulphate is a crystalline compound of iron. It is used in water and sewage treatment to help the removal of suspended impurities. Calculate the mass percentage of iron, sulphur and oxygen in the compound. [NCERT] Sol. The formula of the compound is$F e_{2}\left(S O_{4}\right)_{3}$The formula mass$=2 \times 56+3(32+64)=400 u\therefore \quad$Percentage of iron$=\frac{2 \times 56 \times 100}{400}=28 \%$$\begin{array}{l} {\text { Percentage of sulphur }=\frac{3 \times 32 \times 100}{400}=24 \%} \\ {\text { Percentage of oxygen }=\frac{3 \times 64 \times 100}{400}=48 \%} \end{array}$ [esquestion] Calculate the mass percent of different elements present in sodium sulphate$\left(N a_{2} S O_{4}\right)$. #tag# [NCERT] Q. Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxides is$\left.159.8 g \mathrm{mol}^{-1} \text { [ Atomic mass: } \mathrm{Fe}=55.85, O=16.00 \mathrm{amu}\right]$[NCERT] Sol. Hence, molecular formula is same as empirical formula, viz..$\mathrm{Fe}_{2} \mathrm{O}_{3}$Q. Zinc and hydrochloric acid react according to the reaction:$Z n(s)+2 H C l(a q) \longrightarrow Z n C l_{2}(a q)+H_{2}(g)$If 0.30 mol$Z n$are added to hydrochloric acid containing 0.52 mol$\mathrm{HCl},$how many moles of$\mathrm{H}_{2}$are produced? Which is the limiting reactant? [NCERT] Sol.$Z n(s)+2 H C l(a q) \longrightarrow Z n C l_{2}(a q)+H_{2}(g)$1 mole of$Z n$reacts with 2 moles of$H C l$0.3 mole of$Z n$will react with$2 \times 0.30=0.6$moles of$\mathrm{HCl}$But we have only 0.52 moles of$H C l,$therefore,$H C l$is limiting reactant. 2 moles of$\mathrm{HCl}$react with 1 mole of$\mathrm{Zn}$to form 1 mole of$\mathrm{H}_{2}$. 0.52 moles of$\mathrm{HCl}$react with$\frac{0.52}{2}$moles of$\mathrm{Zn}$to form$\frac{0.52}{2}=0.26$moles of$H_{2}$Q. In a reaction$A+B_{2} \longrightarrow A B_{2}$identify the limiting reagent if any in the following reaction mixtures: (i) 300 atoms of$A+200$molecule of$B$(ii)$2 \mathrm{mol}$of$A+3 \mathrm{mol}$of$B$. (iii) 100 atoms of$A+100$molecules of$B$(iv) 5 mol of$A+2.5$mol of$B$(v)$2.5 \mathrm{mol}$of$A+5 \mathrm{mol}$of$B$[NCERT] Sol. According to the equation, 1 mole of Areacts with 1 mole of$B_{2}$and 1 atom of Areacts with 1 molecule of$B_{2}$(i)$\quad B$is limiting reagent because 200 molecules of$B_{2}$will react with 200 atoms of$A$and 100 atoms will be left in excess. (ii)$\quad A$(iii) Both will react completely because it is stoichiometric mixture. No limiting reagent. (iv)$\quad B$(v)$A$Q. Reaction of$2 B r^{-}(a q)+C l_{2}(a q) \longrightarrow 2 C l^{-}(a q)+B r_{2}(a q)$is used for commercial preparation of bromine from its salts. Suppose we have$50.0 \mathrm{mL}$of 0.060 M solution of$\mathrm{NaBr},$what volume of$0.050 M$solution of$C l_{2}$is needed to react completely with the$B r^{-} ? \quad$[NCERT] Sol. Q. A solution is prepared by dissolving$18.25 \mathrm{g}$of$\mathrm{NaOH}$in distilled water to give$200 \mathrm{cm}^{3}$of solution. Calculate the molarity of$\mathrm{NaOH}$in the solution.$\quad$[Atomic weight of$N a=23, O=16, H=1 u]$[NCERT] Sol.$M=\frac{W_{B}}{M_{B}} \times \frac{1000}{\text { Vol. of solution in } \mathrm{cm}^{3}}=\frac{18.25}{40} \times \frac{1000}{200}=2.28 \mathrm{mol} L^{-1}$Q. Chlorophyll contains$2.68 \%$of$M g$by weight. Calculate the number of$M g$atoms in$2.0 \mathrm{g}$of Chlorophyll. [ Atomic weight of$M g=24 u]$[NCERT] Sol. Amount of$M g$present in chlorophyll$=$Mass of chlorophyll$\times \%$of$M g=2 g \times \frac{2.68}{100}=0.0536 \mathrm{g}$of$\mathrm{Mg}24 g$of$M g$contains$6.023 \times 10^{23}$atoms$0.0536 g$of$M g$contains$\frac{6.023 \times 10^{23}}{24} \times 0.0536=1.345 \times 10^{21}$atoms Q. A sample of$\mathrm{NaNO}_{3}$weighing$0.83 \mathrm{g}$is placed in a$50 \mathrm{ml}$flask. What is molarity of solution? [Atomic weight of$N a=23, N=14, O=16 u] \quad \quad$[NCERT] Sol. Molecular weight of$\mathrm{NaNO}_{3}=23+14+3 \times 16=23+14+48=85 g \mathrm{mol}^{-1}M=\frac{W_{B}}{M_{B}} \times \frac{1000}{\text { Volume of solution in } m L}=\frac{0.83}{85} \times \frac{1000}{50}=\frac{830}{4250}=\frac{83}{425}=0.1952 M$Q. In a reaction vessel,$0.184 g$of$\mathrm{NaOH}$is required to be added for completing the reaction. How many milli litre of$0.150 M$NaOH solution should be added for this requirement? [NCERT] Sol.$M=\frac{W_{B}}{M_{B}} \times \frac{1000}{\text { Volume of solution }}0.15=\frac{0.184}{40} \times \frac{1000}{\text { Volume of solution }}$Volume of$\mathrm{NaOH}$solution$=\frac{184}{40 \times 0.15}=\frac{184}{6}=30.66 \mathrm{mL}$Q. How many grams of dinitrogen gas are required to completely react with 0.3 gm dihydrogen gas to yield ammonia ? Also calculate the amount of ammonia formed. [NCERT] Sol. Q. If the density of methanol is$0.793 \mathrm{kg} L^{-1},$what is its volume needed for making$2.5 L$of its$0.25 M$solution$?$[NCERT] Sol. Amount of methanol required$m(g)=\frac{\text { Molarity } \times \mathrm{M} . \text { Mass } \times V_{i n} m L}{1000}=\frac{0.25 \times 32 \times 2500}{1000}=20 g$since, density$=\frac{\text { Mass }}{\text { Volume }}\therefore$Volume of$\mathrm{CH}_{3} \mathrm{OH}$required$=\frac{\text { Mass }}{\text { density }}=\frac{20}{0.793}=25.22 \mathrm{mL}$Q. (i) A sample of$N a O H$weighing$38 g$is dissolved in water and the solution is made to$50.0 \mathrm{mL}$in a volumetric flask. What is the molarity of the resulting solution? (ii) How many moles of$\mathrm{NaOH}$are contained in$27 \mathrm{mL}$of$0.15 M N a O H ?$[NCERT] Sol. (i) Moles of$\mathrm{NaOH}$in$50.0 \mathrm{mL}$solution$=\frac{38}{40}\therefore$moles of$\mathrm{NaOH}$in$1.0 \mathrm{L}$solution $=\frac{38}{40} \times \frac{1000}{50}=19 M$ (ii)$1.0 L$of solution contains moles$=0.1527.0 \mathrm{mL}$of solution contains moles$=\frac{0.15}{1000} \times 27=4.05 \times 10^{-3} \mathrm{moles}$Q. A sample of drinking water was found to be severely contaminated with chloroform$\left(\mathrm{CHCl}_{3}\right)$supposed to be carcinogen. The level of contamination has 15 ppm (by mass) (i) Express this in percent by mass (ii) Determine molality of chloroform in water sample. [NCERT] Sol. (i) 15 ppm means 15 parts by mass of$\mathrm{CHCl}_{3}$in$10^{6}$parts by mass of water Hence, percent by mass $=\frac{15}{10^{6}} \times 100=1.5 \times 10^{-3} \%$ (ii) Moles of chloroform$=\frac{15}{119.5}=0.1255$Hence, molality$(m)$of$\mathrm{CHCl}_{3}=\frac{0.1255}{10^{6}} \times 10^{3}=1.25 \times 10^{-4} \mathrm{m}$Q.$500 \mathrm{mL}$of$0.25 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$solution is added to an aqueous solution of$15.00 \mathrm{g}$of$\mathrm{BaCl}_{2}$resulting in the formation of the white precipitate of insoluble$B a S O_{4} .$How many moles and how many grams of$B a S O_{4}$are formed?$\quad$[At. Wt. of$B a=137 u, S=3 u, O=16 u]$[NCERT] Sol.$B a C l_{2}(a q)+N a_{2} S O_{4}(a q) \longrightarrow B a S O_{4}(s)+2 N a C l(a q)$Number of moles of$N a_{2} \mathrm{SO}_{4}=\left(\frac{500 m L}{1000 m L}\right) \times 0.25 m o l=\frac{0.25}{2}=0.125 \mathrm{mol} \mathrm{Na}_{2} \mathrm{SO}_{4}$Molar mass of$B a C l_{2}=208.2 \mathrm{g}$Number of moles of$B a C l_{2}=\frac{15}{208.2}=0.072 \mathrm{mol}$of$\mathrm{BaCl}_{2}$0.072 moles of$B a C l_{2}$reacts with 0.072 moles of$B a C l_{2}$to form 0.072 mole of$B a S O_{4}$Mass of$B a S O_{4}$formed$=0.072 \times 233.4=16.80 g$of$B a S O_{4}$Q. (i) A sample of$N a O H$weighing 0.38 is dissolved in water and the solution is made to$50.0 \mathrm{cm}^{3}$in a volumetric flask. What is the molarity of the solution$\quad$(ii) State and explain law of multiple proportion. [NCERT] Sol. (i)$\quad M=\frac{W_{B}}{M_{B}} \times \frac{1000}{\text { Volume of solution in } m L}$$=\frac{0.38}{40} \times \frac{1000}{50}=0.19 \mathrm{mol} L^{-1}$ (ii) Law of multiple proportion states whenever two elements reacts to form two or more compounds, the ratio between different weights of one of the elements which combine with fixed weight of another is always simple, e.g., Nitrogen reacts with oxygen to form$\mathrm{NO}$and$\mathrm{NO}_{2} \cdot$In$\mathrm{NO}, 14 \mathrm{g}$of$N$reacts with$16 \mathrm{g}$of$\mathrm{O}$In$\mathrm{NO}_{2} 14 \mathrm{g}$of$N$reacts with$32 \mathrm{g}$of$\mathrm{O} .$The ratio between weights of oxygen which combine with fixed weight of nitrogen is$16: 32,$i.e., 1: 2 Q. Express each of the following in S.I. units : (i) 93 million miles (this is the distance between the earth and the sun). (ii) 5 feet 2 inches (this is the average height of an Indian female). (iii) 100 miles per hour (this is the typical speed of Rajdhani Express). (v)$46^{\circ} \mathrm{C}$(this is the peak summer temperature in Delhi). (vi) 150 pounds (this is the average weight of an Indian male). [NCERT] Sol. (i) The S.I. unit of distance is metre ($m$) $\begin{array}{l} {1 \text { mile }=1.60 \text { kilometre }=1.60 \times 1000 \mathrm{m}} \\ {\text { Unit factor }=\frac{1.60 \times 1000(\mathrm{m})}{1 \mathrm{mile}}=\frac{1.6 \times 10^{3}(\mathrm{m})}{1 \mathrm{mile}}} \end{array}$ 93 millionmiles$=\frac{93 \times 10^{6}(\text { miles }) \times 1.6 \times 10^{3}(\mathrm{m})}{1 \text { mile }}\therefore=93 \times 1.6 \times 10^{9} \mathrm{m}=148.8 \times 10^{9} \mathrm{m}=1.49 \times 10^{11} \mathrm{m}$(ii) 5 feet 2 inches$=62$inches $\begin{array}{l} {1 \text { inch }=2.54 \times 10^{-2} \mathrm{m}} \\ {\text { Unit factor }=\frac{2.54 \times 10^{-2}(\mathrm{m})}{1 \mathrm{inch}}} \\ {62 \text { inches }=\frac{62(\text { inches }) \times 2.54 \times 10^{-2} \mathrm{m}}{1 \mathrm{inch}}} \\ {=62 \times 2.54 \times 10^{-2} \mathrm{m}} \\ {=157.48 \times 10^{-2} \mathrm{m}=1.57 \mathrm{m}} \end{array}$ (iii) 1 mile$=1.60 \mathrm{km}=1.60 \times 10^{3} \mathrm{m}$$\begin{array}{l} {\text { Unit factor }=\frac{1.60 \times 10^{3} \mathrm{m}}{1 \mathrm{mile}}} \\ {1 \mathrm{hr}=60 \times 60 \mathrm{s}=3.6 \times 10^{3} \mathrm{s}} \\ {\text { Unit factor }=\frac{3.6 \times 10^{3} \mathrm{s}}{1 \mathrm{hr}}} \end{array}$$\therefore \quad$Speed$=\frac{100 \text { miles }}{h r}=\frac{100 \text { miles }}{h r} \times \frac{1.60 \times 10^{3} \mathrm{m}}{1 \mathrm{mile}} \times \frac{1 \mathrm{hr}}{3.6 \times 10^{3} \mathrm{s}}=44 \mathrm{ms}^{-1}$(iv)$1 \dot{A}=10^{-10} \mathrm{m}$$\text { Unit factor }=\frac{10^{-10} \mathrm{m}}{1(\hat{A})}$$\therefore \quad 0.74 \mathrm{A}=\frac{0.74 \hat{A} \times 10^{-10}(m)}{1(\hat{A})}$$=0.74 \times 10^{-10} \mathrm{m} \text { or }=7.4 \times 10^{-11} \mathrm{m}$ (v)$\quad 0^{\circ} \mathrm{C}=273.15 \mathrm{K}\quad 46^{\circ} \mathrm{C}=273.15 \mathrm{K}+46 \mathrm{K}=319.15 \mathrm{K}$(vi) 1 pound$=454 \times 10^{-3} \mathrm{kg}$$\text { Unit factor }=\frac{454 \times 10^{-3}(k g)}{1(\text { pound })}$$\begin{aligned} \therefore \quad 150 \text { pound }=\frac{150(\text { pound }) \times 454 \times 10^{-3}(k g)}{1(\text { pound })} & \\=& 150 \times 454 \times 10^{-3} \mathrm{kg}=68.1 \mathrm{kg} \end{aligned}$Q. Using the unit conversion factors, express the given measurements in the designated units. (i)$25 L$to$m^{3}$(ii)$25 g L^{-1}$to$m g d L^{-1}$(iii)$1.54 \mathrm{mms}^{-1}$to$\mathrm{pm} \mu \mathrm{s}^{-1}$(iv)$2.66 g \mathrm{cm}^{-3}$to$\mu g \mu m^{-3}$(v)$4.22 \mathrm{Lh}^{-2}$to$\mathrm{mLs}^{-1}$[NCERT] Sol. (i) Conversion factor$=\frac{\left(10^{-3} \mathrm{m}^{3}\right)}{(1 . L)}$$25 L=(25 L) \times \frac{\left(10^{-3} m^{3}\right)}{(1 L)}=2.5 \times 10^{-2} m^{3}$ (ii) Conversion factor=$=\frac{(1000 m g)}{(1 g)} \times \frac{(10 d L)}{(1 L)}$$\begin{array}{l} {25 g L^{-1}=(25 g) \times \frac{\left(10^{3} m g\right)}{(1 g)} \times\left(1 L^{-1}\right) \times \frac{\left(10 d L^{-1}\right)}{\left(1 L^{-1}\right)}} \\ {=2.5 \times 10^{3} \mathrm{mg} d L^{-1}} \end{array}$ (iii) Conversion factor$=\frac{\left(10^{9} p m\right)}{(1 m m)} \times \frac{\left(10^{-6} \mu s\right)}{(1 s)}$$\begin{array}{l} {1.54 \mathrm{mm} \mathrm{s}^{-1}=(1.54 \mathrm{mm}) \times \frac{\left(10^{9} \mathrm{pm}\right)}{(1 \mathrm{mm})}\left(1 \mathrm{s}^{-1}\right) \times \frac{\left(10^{-6} \mu \mathrm{s}^{-1}\right)}{\left(1 \mathrm{s}^{-1}\right)}} \\ {=1.54 \times 10^{3} \mathrm{pm} / \mathrm{s}^{-1}} \end{array}$ (iv) Conversion factor$=\frac{\left(10^{6} \mu g\right)}{(1 g)} \times \frac{\left(10^{4} \mu m\right)}{(1 \mathrm{cm})}2.66 g \mathrm{cm}^{-3}=(2.66 g) \times \frac{\left(10^{6} \mu g\right)}{(1 g)} \times\left(\mathrm{cm}^{-3}\right) \frac{\left(10^{4} \mu m\right)^{-3}}{\left(1 \mathrm{cm}^{-3}\right)}=2.66 \times\left(10^{6} \mu g\right) \times\left(10^{-12} \mu m^{-3}\right)$$=2.66 \times 10^{-6} \mu g \mu m^{-3}$ (v) Conversion factor$=\frac{10^{3} m L}{(1 L)} \times \frac{(60 \times 60 \mathrm{s})}{1 h}\quad 4.2 \mathrm{Lh}^{-2}=(4.2 \mathrm{L}) \times \frac{\left(10^{3} \mathrm{mL}\right)}{(1 \mathrm{L})} \times\left(h^{-2}\right) \times \frac{(60 \times 60 \mathrm{s})^{-2}}{\left(h^{-2}\right)}=3.24 \times 10^{-4} \mathrm{mL} s^{-2}$Q. The following data were obtained when dinitrogen and dioxygen react together to form different compounds : [NCERT] Sol. (a) Fixing the mass of dinitrogen as 28 g, masses of dioxygen combined will be 32, 64, 32 and 80 g in the given four oxides. These are in the ratio 1 : 2 : 1 : 5 which is a simple number ratio. Hence, the given data obey the law of multiple proportion. Q. Explain ‘Avogadro’s Number’ and ‘Mole’. What is their importance ? Avogadro’s number may be defined as the number of atoms present in one gram atom of the element or the number of molecules present in one gram molecule of the substance. The amount of the substance containing Avogadro’s number of atoms or molecules is called a mole (or simply written as mol). Thus, a mole is a chemist’s unit of counting particles such as atoms, molecules, ions, electrons, protons etc., which represents a value of$6.022 \times 10^{23}$A mole of hydrogen atoms means$6.022 \times 10^{23}$atoms of hydrogen whereas a mole of hydrogen molecules means$6.022 \times 10^{23}$molecules of hydrogen or$2 \times 6.022 \times 10^{23}$atoms of hydrogen (because each hydrogen molecule contains 2 atoms of hydrogen). Similarly, a mole of oxygen molecules means$6.022 \times 10^{23}$molecules of oxygen or$2 \times 6.022 \times 10^{23}$atoms of oxygen. A mole of sand particles means$6.022 \times 10^{23}$sand particles. A mole of electrons means$6.022 \times 10^{23}$electrons and so on. Sol. Q. Calculate number of moles in each of the following : (i)$11 \mathrm{g}$of$\mathrm{CO}_{2}$(ii)$3.01 \times 10^{22}$molecules of$\mathrm{CO}_{2}$(iii) 1.12 litre of$\mathrm{CO}_{2}$at$\mathrm{S.T.P.}$Sol. (i) Molecular mass of$C O_{2}$$=(12+2 \times 16)=44$$44 \mathrm{g}$of$\mathrm{CO}_{2}=1 \mathrm{mole}$of$\mathrm{CO}_{2}11 \mathrm{g}$of$\mathrm{CO}_{2}=\frac{1}{44} \times 11=0.25 \mathrm{mol}$(ii)$6.02 \times 10^{23}$molecules of$\mathrm{CO}_{2}=1$mole of$C O_{2}3.01 \times 10^{22}$molecules of$\mathrm{CO}_{2}=\frac{1}{6.02 \times 10^{23}} \times 3.01 \times 10^{22}=0.05 \mathrm{mol}$(iii) 22.4 litres of$\mathrm{CO}_{2}$at S.T.P.$=1$mole of$C O_{2}$1.12 litres of$\mathrm{CO}_{2}$at S.T.P.$=\frac{1}{22.4} \times 1.12=0.05 \mathrm{mol}$Q. Butyric acid contains C, H, O elements. A 4.24 mg sample of butyric acid is completely burnt in oxygen. It gives 8.45 mg of carbon dioxide and 3.46 mg of water. What is the mass percentage of each element ? Determine the empirical and molecular formula of butyric acid if molecular mass of butyric acid is determined to be 88 u. [NCERT] Sol. Mass of carbon present in$8.45 \mathrm{mg}$of$\mathrm{CO}_{2}=\frac{8.45 \times 12}{44} m g=2.30 m g$Percentage of carbon$=\frac{2.30 \times 100}{4.24}=54.24 \%$Mass of hydrogen in$3.46 \mathrm{mg}$of$\mathrm{H}_{2} \mathrm{O}=\frac{3.46 \times 2}{18} m g=0.384 m g$Percentage of hydrogen$=\frac{0.384 \times 100}{4.24}=9.05 \%$Percentage of oxygen$=100-54.24-9.05=36.71 \%$Q. A welding fuel contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g of carbon dioxide and 0.690 g of water and no other products. 10.0 L of this welding gas at S.T.P. is found to weigh 11.6 g. Calculate : (i) empirical formula, (ii) molar mass, and (iii) molecular formula of welding gas. [NCERT] Sol. (i) Calculation of empirical formula: Mass of carbon in 3.38 g carbon dioxide$=\frac{12 \times 3.38}{44}=0.922 g$Mass of hydrogen in 0.690 g of water$=\frac{2 \times 0.690}{18}=0.077 g$The ratio by mass of$C$and$H$in the sample$=0.922: 0.077$The mole ratio of$C$and$H$in the sample$=\frac{0.922}{12}: \frac{0.077}{1}=0.077: 0.077=1: 1$Thus empirical formula of the gas is$C H$(ii) Calculation of molar mass:$10.0 \mathrm{L}$of gas at S.T.P. weighs$=11.6 \mathrm{g}22.4 \mathrm{L}$of gas at S.T.P. weighs$=\frac{11.6 \times 22.4}{10}=25.98 g$Thus, molar mass of gas$=25.98 \mathrm{g} \mathrm{mol}^{-1}$(iii) Calculation of molecular formula:$n=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{25.98}{13}=1.998 \approx 2$Thus, molecular formula is$(C H)_{2}=C_{2} H_{2}$Q. Naphthalene (moth balls) contains$93.71 \%$carbon and$6.29 \%$hydrogen. If its molar mass is$128 g \mathrm{mol}^{-1},$calculate its molecular formula. [NCERT] Sol. From the data given it is clear that if we have$100.0 \mathrm{g}$of naphthalene, there are$93.71 \mathrm{g}$of$\mathrm{C}$and$6.29 \mathrm{g}$of$\mathrm{H.Number}$of moles of$\mathrm{C}$and H in$100.0 \mathrm{g}$sample of naphthalene are$93.71 g C\left(\frac{1 \mathrm{mol} C}{12.0 \mathrm{g} C}\right)=7.80 \mathrm{mol} \mathrm{C}6.29 g H\left(\frac{1 \mathrm{mol} H}{1.00 g H}\right)=6.29 \mathrm{mol} \mathrm{H}$Mole ratio$=\left(\frac{7.80 \mathrm{mol} \mathrm{C}}{6.29 \mathrm{mol} \mathrm{H}}\right)=\left(\frac{1.24 \mathrm{mol} \mathrm{C}}{1.00 \mathrm{mol} \mathrm{H}}\right)$Changing the decimal fraction into whole number ratio of$C$and$H,$we know$1.24 \cong \frac{5}{4}$Therefore, the ratio of$C$to$H$is Mole ratio$=\frac{5 / 4 \mathrm{mol} \mathrm{C}}{1 \mathrm{mol} \mathrm{H}}=\frac{5 \mathrm{mol} \mathrm{C}}{4 \mathrm{mol} \mathrm{H}}$The ratio of carbon to hydrogen atoms in naphthalene is 5: 4 giving the empirical formula,$C_{5} H_{4} .$The ratio of molecular mass to the empirical mass gives the factor with which the empirical formula is multiplied to obtain the molecular formula.$\frac{128 g / m o l \text { of napthalene }}{64.0 g / m o l \text { of } C_{5} H_{4}}=2.00$Thus, molecular formula of naphthalene is given by empirical formula multiplied by$2,$i.e.$\left(C_{5} H_{4}\right)_{2}$or$C_{10} H_{8}$Q. Calculate the amount of lime$C a(O H)_{2}$required to remove the hardness of$60,000$litres of well water containing$16.2 \mathrm{g}$of calcium bicarbonate per hundred litres. [ Atomic masses$C a=40, \quad C=12, \quad O=16, H=1]$Sol. (i) Calculation of the weight of calcium bicarbonate present. Wt. of calcium bicarbonate present in 100 litres of well water$=16.2 \mathrm{g}$Wt. of calcium bicarbonate present in$60,000$litres of $\text { well water }=\frac{16.2}{100} \times 60000 g=9720 g$ (ii) Calculation of the quantity of lime required. The equation involved is: Q. A solution of glucose in water is labelled as$10 \%(\mathrm{w} / \mathrm{w}) .$The density of the solution is$1.20 g \mathrm{mL}^{-1} .$Calculate (i) Molality (ii) Molarity, and (iii) Mole fraction of each component in solution. Sol.$10 \%$(w/w) solution of glucose means that$10 g$of glucose is present in$100 g$of solution or in$90 g$of water. (i) Calculation of molality $\begin{array}{l} {\text { Mass of glucose }=10 g} \\ {\text { Moles of glucose }=\frac{10}{180}=0.0556} \end{array}$ (Molar mass of glucose$=180$) Mass of water$=90 g$Molality$=\frac{\text { Moles of glucose }}{\text { Mass of water }} \times 1000$$=\frac{0.0556}{90} \times 1000=0.618 m$ (ii) Calculation of molarity Moles of glucose$=0.0556$Volume of solution$=\frac{\text { Mass }}{\text { Density }}=\frac{100}{1.20}=83.3 \mathrm{mL}$Molarity$=\frac{\text { Moles of glucose }}{\text { Vol. of solution }} \times 1000=\frac{0.0556}{83.3} \times 1000=0.667 M$(iii) Calculation of mole fraction of components $\begin{array}{l} {\text { Moles of glucose }=0.0556} \\ {\text { Moles of water }=\frac{90}{18}=5.0} \\ {\text { Total moles }=5.0+0.0556=5.0556} \end{array}$ Mole fraction of glucose$=\frac{0.0556}{5.0556}=0.011$Mole fraction of water$=\frac{5.0}{5.0556}=0.989$Q.$100 \mathrm{mL}$of$0.2 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$is mixed with$200 \mathrm{mL}$of$0.05 M N H_{4} O H .$Calculate the amount of$A l(O H)_{3}$precipitated and the molarity and normality of the resulting solution in terms of each ion. Sol. Carbon Family | Question Bank for Class 11 Chemistry Get Carbon Family important questions for Boards exams. Download or View the Important Question bank for Class 11 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 11 chemistry chapter wise CBSE. Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, 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. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. 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. Is carbon dioxide poisonous or not ? Sol. Carbon dioxide is not poisonous. Q. What is dry ice ? Why is it so called ? Sol. Solid carbon dioxide gas is called dry ice, solid carbon dioxide can sublime to the gaseous state without passing through the liquid state and therefore, it is called dry ice. Q. What is state of hybridisation of carbon in (i)$\mathrm{CO}_{3}^{2-}$(ii) diamond (iii) graphite? [NCERT] Sol. The hybridisation of carbon in (i)$\mathrm{CO}_{3}^{2-}$is$s p^{2}$(ii) in diamond is$s p^{3}$and in (iii) graphite is$\mathrm{sp}^{2}$Q. Name the isotopes of carbon used for radio carbon dating and for international standard for atomic mass. Sol. Isotope of carbon used for radio carbon dating is 14 C. Isotope of carbon used for international standard for atomic mass is$12 \mathrm{C}$. Q. What is allotropy ? Sol. Allotropy. “The existence of an element in two or more different forms which have different physical properties but same chemical properties, is called allotropy and the forms are called allotropes.” The allotropic forms of carbon are: (i) Crystallined forms – Diamond, graphite. (ii) Non-Crystallined forms – Charcoal, lamp-black. Q. Explain the differences in the properties of diamond and graphite on the basis of their structures. [NCERT] Sol. Diamond is very hard crystalline solid because each carbon is linked to four other carbon atoms tetrahedrally by using$s p^{3}$hybrid orbitals giving rise to a rigid three dimensional network of carbon atoms. It does not conduct electricity because it does not have free electrons. Graphite has structure in which each carbon is bonded to three other carbon atoms to form a hexagonal sheet. Each ‘CI atoms forms three sigma bonds by means of$s p^{2}$hybrid orbitals. The unhybridised$p$-orbitals of each carbon atoms having electrons overlaps with$p$-orbitals of other carbon atoms to form$\pi$-bonds. The hexagonal layers are held together by weak Vander Wall’s force of attraction, that is why it is soft and slippery and used as lubricant. It is conductor of electricity because -electrons within layer are free to move and this accounts for electrical conductivity. Q. What is the oxidation state of carbon in each of the followingcompounds? ( i )$\mathrm{CO}$(ii)$\mathrm{HCN}$(iii)$\mathrm{H}_{2} \mathrm{CO}_{3}$(iv)$\mathrm{CaC}_{2}$Sol. (i) 2 (ii) 2 (iii) 4 (iv) –1 Q. Why carbon forms covalent compounds whereas lead forms ionic compounds ? Sol. – Carbon cannot lose electrons to form$C^{4+}$because the ionization energy required is very high. It cannot gain electrons to form$C^{4-}$because it is energetically not favourable. Hence$C$forms covalent compounds. Down the group, the$I . E .$decreases.$P b$being the last element has so low, I.E. that it can lose electrons to form ionic compounds. Q. What happens when : (i) Quick lime is heated with coke? (ii) Carbon monoxide reacts with$C l_{2} ?$(iii) Plants absorb$\mathrm{CO}_{2} ?$Sol. (i) Calcium carbide and Carbon monoxide are formed. $\mathrm{CaO}+3 \mathrm{C} \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaC}_{2}+\mathrm{CO}$ (ii) Phosgene gas is formed. $\mathrm{CO}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2}$ [Phosgene gas (Carbonyl Chloride)] (iii) They form glucose and starch. Q. What are carbides? How are they classified? Give one example of each type of carbides. Sol. Compounds of electropositive elements with carbon are called carbides. There are three types of Carbides: (i) Ionic Carbide : Most electropositive metals form ionic carbide, e.g., calcium carbide. (ii) Covalent Carbide : Less electropositive metals and nonmetals form covalent carbides, e.g., silicon carbide, boron carbides. (iii) Interstitial Carbides : Those carbides which are formed by transition metals in which carbon atoms occupy interstitial voids,$e . g ., W C(\text { Tungsten Carbide })$Q. Explain the following: (i) Silanes are few in number while alkanes are large in number. (ii) Conc.$H N O_{3}$can be stored in aluminium container but it cannot be stored in zinc container. (iii) Burning magnesium continues to burn while burning sulphur gets extinguished when dropped in gas containing$\mathrm{NO}$(iv) Aqueous solution of carbonates and bicarbonates of alkali metals is alkaline. Sol. (i) Carbon has the maximum tendency for catenation due to stronger$C-C \quad\left(353.5 \mathrm{kJ} \mathrm{mol}^{-1}\right)$and hence it forms large number of alkanes. silicon, on the other hand, bonds has much lesser tendency for catenation due to weak$S i-S i \quad\left(200 k J \mathrm{mol}^{-1}\right)$and hence it forms only few silanes (ii) In presence of conc.$H N O_{3}, A l$becomes passive, due to a thin protective layer of its oxide$\left(A l_{2} O_{3}\right)$which is formed on its surface which prevents the further action between the metal and the acid. Therefore$A l$containers can be used for storing conc. However, it cannot be stored in zinc vessels because zinc reacts with$\mathrm{HNO}_{3}$. $\mathrm{Zn}+4 \mathrm{HNO}_{3} \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$ (iii) Heat evolved during burning of$M g$is enough to decompose$N O$to$\mathrm{N}_{2}$and$\mathrm{O}_{2} .$The$\mathrm{O}_{2}$thus, produced support combustion of$M g .$In contrast, the heat produced during burning of$S$is not quite sufficient to decompose$N O .$As a result, sulphur stops burning. (iv) The alkaline nature of aqueous solution is due to the hydrolysis of these salts.$\underset{\text { Bicarbonates }}{H C O_{3}^{-}}+H_{2} O \rightleftharpoons H_{2} C O_{3}+\overline{O H}$$\underset{\text { Cattonates }}{C O_{3}^{2-}}+H_{2} O \rightleftharpoons H C O_{3}^{-}+\overline{O H}$ Q. Explain the following: (i)$\quad \mathrm{SiO}_{2}$is solid but$\mathrm{CO}_{2}$is a gas at room temperature. (ii)$\quad \mathrm{SiCl}_{4}$can be easily hydrolysed but$\mathrm{CCl}_{4}$does not hydrolyse. Sol. (i) Carbon is able to form$p \pi-p \pi$bond with$O$atom and constitute a stable non-polar molecule$O=C=O .$Due to weak interparticle forces its boiling point is low and it is a gas at room temperature. Si$,$on the other hand is not able to form bond with$O$atoms because of its relatively large size. In order to complete it octet sillicon is linked to four$O$atoms around it by sigma bonds and thus it constitutes network structure which is responsible for its solid state. (ii) In$\mathrm{SiCl}_{4},$the silicon atom has empty 3 dorbitals in its valence shell which can accept an electron pair from$H_{2} O$molecule resulting the co-ordinate bond. The co-ordinated molecule loses a molecule of$H C l$and as a result, a chlorine atom gets replaced by an$O H$group as shown. In the same way, all the four$C l$atoms are replaced by$O H$group and the molecule gets completely hydrolyse. Carbon atom, on the other hand does not have vacant$d$-orbitals in 2nd level. Hence, it cannot accept electron pair form$H_{2} O$molecule and therefore,$\mathrm{CCl}_{4}$is not hydrolysed. p block Elements Class 11 Important Questions with Answers Get Boron Family (P-Block elements) important questions with answers for Class 11 and other exams. View the Important Question bank for Class 11 & 12 Chemistry syllabus. These important questions and answers will play significant role in clearing concepts of Chemistry. This question bank is designed keeping NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 & 12 subjects. Learn the important concepts of p block elements with these important question and you can also access the notes of each subject from class 11 & 12 subjects. Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, 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. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. 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. Write the electronic configuration of p-block elements. Sol. Electronic configuration of$\mathrm{p}$-block elements is$n s^{2} n p^{1-6}$Q. Name the element having highest melting point. Sol. The element having highest melting point is carbon (diamond). Q. The element having highest melting point is carbon (diamond). Sol. Amorphous and Crystalline. Q. What is tincal ? Give its chemical formula. Sol. Tincal is an ore of boron. Its formula is$\mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{10} \cdot 10 \mathrm{H}_{2} \mathrm{O}$Q. Why does boron not form$B^{3+}$ions? Sol. Boron has very small size and has very high sum of three ionisation enthalpies$\left(I E+I E_{2}+I E_{3}\right)$therefore it cannot lose its three electrons to form$B^{3+}$ions. Q. Write the balanced equation for the preparation of elemental boron by reduction with dihydrogen. Sol.$2 \mathrm{BCl}_{3}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \frac{\mathrm{sun}}{_{1270 \mathrm{K}}} 2 \mathrm{B}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{g})$Q. Why is$B F_{3}$weaker Lewis acid than$B C l_{3} ?$Sol.$B F_{3}$is weaker lewis acid than$B C l_{3}$because of more effecttive back bonding in case of$F$due to smaller size than$C l$Q. Why does boric acid act as Lewis acid ? Sol. It is because in boric acid, boron does not have its octet complete. It accepts$O H^{-}$from water. Q. Aluminium forms$\left[A I F_{6}\right]^{3-}$but boron does not form$\left[B F_{6}\right]^{3-}$Explain why? Sol. Boron does not have vacant d-orbitals as second shell is the outer most shell. Q. By means of a balanced chemical equation show how$B(O H)_{3}$behaves as an acid in water.$\quad$[NCERT] Sol.$B(O H)_{3}$is a weak monobasic acid. It acts as a lewis acid by accepting electrons from a hydroxy ion in water.$B(O H)_{3}+2 H O H \longrightarrow\left[B(O H)_{4}\right]^{-}+H_{3} O^{+}$Q. How does sodium hydride react with diborane ? Sol. Sodium borohydride is formed$2 \mathrm{NaH}+\mathrm{B}_{2} \mathrm{H}_{6} \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{NaBH}_{4}$Q. What type of glass is obtained when borax is added ? Sol. Pyrex glass is a glass which is heat resistant. It can with stand high temperature. Q. Describe briefly how element boron can be prepared. [NCERT] Sol. Boron can be obtained by reduction of Boron oxide with electropositive metal like Mg.$B_{2} O_{3}(s)+3 M g(g) \longrightarrow 2 B(s)+3 M g O(s)$Q. What is the oxidation state of boron in boric acid ? What is the basicity of this acid ? Sol. Boric acid is$H_{3} B O_{3}$Oxidation state of boron$=+3$Boric acid is tribasic. Q. Boron is unable to form$B F_{6}^{3-}$ion. Explain. Sol. Boron does not have vacant$d$-orbitals in its valence shell. Therefore, it cannot expand its octet. Therefore, maximum covalency of boron cannot exceed and it does not form$B F_{6}^{3-}$ion. Q. Why do boron halides form “addition compounds” with amines ? Sol. Boron halides are electron deficient, therefore, they can form addition compounds with electron rich amines$e . g .\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} \longrightarrow \mathrm{BF}_{3}$Q. Describe what happens when boric acid is heated ? Sol. When boric acid is heated at$370 K .$metaboric acid$\left(H B O_{2}\right)$is formed. On heating strongly$B_{2} O_{3}$is formed. (i)$\mathrm{H}_{3} \mathrm{BO}_{3} \stackrel{370 \mathrm{K}}{\longrightarrow} \mathrm{HBO}_{2}+\mathrm{H}_{2} \mathrm{O}$(ii)$2 H B O_{2} \stackrel{\text { heat }}{\longrightarrow} B_{2} O_{3}+H_{2} O$Q. What happens when a borax solution is acidified ? Write a balanced equation of the reaction. Sol.$\mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{7} \cdot 10 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{HCl}+5 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaCl}+4 H_{2} B Q_{3}+12 H_{3} Q$Q. Complete and balance the following chemical equation : (i)$B F_{3}+L i H \stackrel{450 K}{\longrightarrow}$(ii)$\mathrm{B}_{2} \mathrm{H}_{6}+\mathrm{NaH} \longrightarrow$Sol. Q. If$B-C l$bond has a dipole moment, why does$B C l_{3}$have zero dipole moment? [NCERT] Sol.$=B-C l$bond has dipole moment due to difference in electronegativity of$B$and chlorine atom.$B C l_{3}$on the contrary has a zero dipole moment because dipole moment is a vector quantity. The net resultant of the three bonds is zero due to its symmetrical planer triangular structure. Q. Describe the shape of$B F_{3}$and$B H_{4}^{-}$. Assign hybridisation of boron in these species.$\quad$[NCERT] Sol. – In$B F_{3}$the hybridisation of boron is$s p^{2}$and hence the molecule has a planar triangular shape. In$B H_{4}^{-}$the hybridisation of boron is$s p^{3}$and hence it has a tetrahedral shape. Q. What are electron deficient compound? Are$B C l_{3}$and$S i C l_{4}$electron deficient species? Explain. Sol. Electron deficient compounds are those which have incomplete octet.$B C l_{3}$is an electron deficient compound as it has sextet of electrons. On the contrary$\mathrm{SiCl}_{4}$is not an electron deficient compound. However, due to the presence of vacant$d$-orbitals in its valence shell it can extend its co-ordination number and form species like$S i F_{3}^{-}, S i F_{6}^{2-},$etc. Q. Discuss the pattern of variation in oxidation state from$B$to Tl in boron family. [NCERT] Sol. – The boron family exhibits two oxidation states of$+3$and$+1$in their compounds on moving down the group from boron to thallium, the stability of lower oxidation state increases$i . e .,+1$oxidation state becomes more prominent due to ” inert pair effect? Q. How can you explain higher stability of$B C l_{3}$as compared to$T l C l_{3} ? \quad$[NCERT] Sol.$B C l_{3}$has more stability than$T I C l_{3} .$It is because boron due to its small size can exhibit only one oxidation state of$+3$in its compounds. On the contrary. Thallium exhibits two oxidation states of$+3$and$+1$in its compounds. The lower oxidation state of$+1$is more stable because of inert pair effect. To add further the increase in the magnitude of nuclear charge is not compensated by the increase in size as a result of which the$n s$-electrons do not participate in bonding. Q. How is lithium aluminium hydride (LiAlH_{4} ) ~ p r e p a r e d ? ~ W h a t ~ is its important use? [NCERT] Sol. It is obtained by reacting$A I C I_{3}$with$L i H$in the presence of dry ether Uses : It is a very good reducing agent used in many organic synthesis. Q. Write balanced equation to show what happens when aq$\mathrm{NaOH}$is added dropwise to aq$\mathrm{GaCl}_{3} .$The precipitate formed initially dissolves on further addition of$N a O H$solution. [NCERT] Sol.$\mathrm{GaCl}_{3}+\mathrm{NaOH} \longrightarrow \mathrm{Ga}(\mathrm{OH})_{3}+3 \mathrm{NaCl}\mathrm{Ga}(\mathrm{OH})_{3}+\mathrm{NaOH} \longrightarrow \underset{\text { Souble Complex }}{\mathrm{NaGaO}_{2}}+2 \mathrm{H}_{2} \mathrm{O}$Q. Write the uses of boron compounds. Sol. Uses of boron compounds : (i) The main use of borax and boric acid is in the preparation of heat resistant borosilicate glass such as pyrex glass. (ii) Borax is used as flux for soldering of metals and procelain enamels. (iii) Dilute aqueous solution of boric acid is used as mild antiseptic. (iv) Boron rods are used as control rods in nuclear reactors. Q. How does boron react with (i) metals like$C r$(ii)$N a O H$(iii) acids like$H_{2} S O_{4}$(conc.)? Sol. Q. Write balanced equations for the reaction of elemental boron with elemental chlorine, oxygen and nitrogen at a high temperature. [NCERT] Sol. (i)$\quad 2 B+3 C l_{2} \stackrel{\text { heat }}{\longrightarrow} 2 B C l_{3}$(ii)$\quad 4 B+3 O_{2} \stackrel{\text { heat }}{\longrightarrow} 2 B_{2} O_{3}$(iii)$\quad 2 B+N_{2} \stackrel{\text { heat }}{\longrightarrow} 2 B N$Q. Suggest reason why$B-F$bond length in$B F_{3}(130 p m)$and$B F_{4}^{-}(143 p m)$differ$?$[NCERT] Sol.$B-F$bond length in$B F_{3}$is 130 pm because boron in is hybridised and has a planar triangular structure. As a result of deficiency of electrons on the boron atom the excessive electron density on the fluorine atoms (due to its small size and strong interelectronic repulsions) is pushed on to the boron atom$i . e .$back donation or back bonding occurs. This introduces a partially double bond character shortening the bond. However, in the boron atom is hybridised with bond length of$143 \mathrm{pm}$Q. Why$B B r_{3}$is a stronger Lewis acid as compared to$B F_{3}$though fluorine is more electronegative than chlorine? Sol. Both$B F_{3}$and$B B r_{3}$are electron deficient molecules due to presence of only six electrons around boron atom. Therefore, these molecules act as Lewis acids due to their tendency to accept a pair of electrons. In , the 2 p-orbital of adjoining$F$atom containing lone pair of electrons form effective bonding with empty$2 \mathrm{p}$-orbital of boron atom. As a result, the lone pair of$F$is donated to$B$atom and hence the electron-deficiency ofboron decreases. In contrast in, the size of$B r$atom is much bigger than the$F$atom, hence donation of lone pair of electrons of$B r$to$B$does not occur to any significant extent. As a result, the electron-deficiency of$B$is much higher in$\mathrm{BBr}_{3}$than that in$\mathrm{BF}_{3},$and hence$\mathrm{BBr}_{3}$is a stronger Lewis acid than$\mathrm{BF}_{3}$Q.$A I F_{3}$is insoluble in anhydrous$H F$but dissolves on addition of$\mathrm{NaF.}$Aluminium trifluoride precipitates when gaseous$B F_{3}$is bubbled through it. Give reasons.$\quad$[NCERT] Sol. Q. Give chemical equations to show the reaction between (i) Borax with ethanol in the presence of conc.$H_{2} \mathrm{SO}_{4}$(ii) Diborane and chlorine (iii) Diborane and caustic potash (iv) Magnesium boride and phosphoric acid (v) Boron trifluoride and hydrofluoric acid. Sol. Q. What is the structure of boric acid and what is its basicity? How can we get boric acid from, (i) colemanite (ii) borax? or What is the action of heat on boric acid ? Sol. – Boric acid is$H_{3} B O_{3} .$It can also be represented by$B(O H)_{3} .$It has layer type structure in which$B O_{3}$units are linked to one another through$H$-bonds. The$H$-atom constitutes covalent bond with one unit and$H$-bond with other unit. Boric acid is a weak monobasic acid, it does not act as proton donor but as Lewis acid.$\mathrm{B}(\mathrm{OH})_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{B}(\mathrm{OH})_{4}^{-}+\mathrm{H}^{+}$Preparation (i) By the action of$\mathrm{SO}_{2}$on boiling suspension of colemanite in water. $\mathrm{Ca}_{2} \mathrm{B}_{6} \mathrm{O}_{11}+2 \mathrm{SO}_{2}+9 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{CaSO}_{3}+6 \mathrm{H}_{3} \mathrm{BO}_{3}$ (ii) By acidifying aqueous solution of borax with mineral acids. $\mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{7}+2 \mathrm{HCl}+5 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaCl}+4 \mathrm{H}_{3} \mathrm{BO}_{3}$ Effect of heat. Heating orthoboric acid above$370 K$leads to partial removal of water to yield metaboric acid,$\left(\mathrm{HBO}_{2}\right) ;$further heating yields. States of matter Class 11 Questions and Answers- Important for Exams Get states of matter class 11 important questions and answers for various exams.View the Important Question bank for Class 11 Chemistry and other subjects too. These important questions will play significant role in clearing concepts of Chemistry. These questions and answers are designed, keeping States of matter Class 11 syllabus as per NCERT in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 11 chemistry. Learn the concepts of States of matter and other topics of class 11 with these questions and answers. Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, 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. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. 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 triple point of a substance ? Sol. It is the point at which solid, liquid and vapour of a substance are in equilibrium with each other. For example, triple point of water is the temperature at which ice, water and its vapour coexist. Q.$H C l$is gas while$H F$is liquid at room temperature, why? Sol. HF molecules are associated with intermolecular$H-$bonding, therefore, it is liquid whereas$H C l$is gas because less Vander Waal’s forces of attraction. Q. What is boiling point of water at (i) higher altitudes, (ii) in pressure cooker? Sol.$(i)<100^{\circ} C(\text { ii })>100^{\circ} C$Q. Why is air denser at lower level than at higher altitudes ? Sol. Heavier air will come down and lighter air goes up. Air at lower level is denser since it is compressed by mass of air above it. Q. What is dry ice ? Why is it so called ? Sol. Solid$\mathrm{CO}_{2}$It is because solid$\mathrm{CO}_{2}$is directly converted into gaseous state (sublimes) and does not change into liquid, so it is called dry ice. Q. A rubber balloon permeable to hydrogen in all its isotopic forms is filled with deuterium$\left(D_{2}\right)$and then placed in a box containing pure hydrogen. Will the balloon expand or contract or remains as it is? Sol. Balloon will expand because rate of diffusion of$H_{2}$is greater than that of$D_{2}$Q. What type of graph would you get when$P V$is plotted against$P$at constant temperature? Sol. A straight line parallel to pressure axis. Q. Name and state the law governing the expansion of gases when they are heated or cooled at constant pressure. Sol. Charle’s Law. Q. At a certain altitude, the density of air is$1 / 10$th of the density of the earth’s atmosphere and temperature is$-10^{\circ} \mathrm{C} .$What is the pressure at that altitude? Assume that air behaves like an ideal gas, has uniform composition and is at S.T.T. at the earth’s surface. Sol.$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$or$P_{2}=\frac{P_{1} V_{1} T_{2}}{T_{1} V_{2}}$But$d \propto \frac{1}{V} .$Hence$P_{2}=\frac{P_{1} T_{2}}{T_{1}}\left(\frac{d_{2}}{d_{1}}\right)=\frac{760 \times 263}{273}\left(\frac{1}{10}\right)=73.2 \mathrm{mm}$Q. How much time it would take to distribute one Avogadro’s number of wheat grains, if$10^{10}$grains are distributed each second? [NCERT] Sol. Number of years$=\frac{6.023 \times 10^{23}}{10^{10} \times 365 \times 24 \times 60 \times 60}=1,908,00$years. Q. A chamber of constant volume contains hydrogen gas. When the chamber is immersed in a bath of melting ice$\left(0^{\circ} \mathrm{C}\right),$the pressure of the gas is$1.07 \times 10^{2} \mathrm{kPa} .$What pressure will be indicated when the chamber is brought to$100^{\circ} \mathrm{C} ?$[NCERT] Sol. Q. What would be the S.I. unit for the quantity$p V^{2} T^{2} / n ?$[NCERT] Sol.$\frac{p V^{2} T^{2}}{n}=\frac{\left(N m^{-2}\right)\left(m^{3}\right)^{2}(K)^{2}}{m o l}=N m^{4} K^{2} m o l^{-1}$Q. What is the value of universal gas constant ? What is its value in SI units ? Sol.$R=8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$Q. Explain the significance of the vander Waal’s parameters ? Sol.$^{c} a^{\prime}$is a measure of the magnitude of the intermolecular forces of attraction while$b$is a measure of the effective size of the gas molecules. Q. Give most common application of Dalton’s law. Sol. The air pressure decreases with increases in altitude. That is why jet aeroplane flying at high altitude need pressurization of the cabin so that partial pressure of oxygen is sufficient for breathing. Q.$N_{2} O$and$C O_{2}$have the same rate of diffusion under same conditions of temperature and pressure. Why? Sol. Both have same molar mass$\left(=44 g m o l^{-1}\right)$. According to Graham’s law of diffusion, rates of diffusion of different gases are inversely proportional to the square root of their molar masses under same conditions of temperature and pressure. Q. At what temperature will oxygen molecules have the same K.E. as ozone molecules at$30^{\circ} \mathrm{C} ?$Sol. At$30^{\circ} \mathrm{C},$kinetic energy depends only on absolute temperature and not on the identity of a gas. Q. Which two postulates of the kinetic molecular theory are only approximations when applied to real gases? Sol. (i) Inter molecular forces between molecules are negligible. (ii) Molecules of a gas have negligible volumes. Q. Account for the following properties of gases on the basis of kinetic molecular theory, (i) High compressibility (ii) Gases occupy whole of the volume available to them. Sol. (i) High compressibility is due to large empty spaces between the molecules. (ii) Due to absence of attractive forces between molecules, the molecules of gases can easily separate from one another. Q. Critical temperature of carbon dioxide and$C H_{4}$are$31.1^{\circ} \mathrm{C}$and$-81.9^{\circ} \mathrm{C}$respectively. Which of these has stronger intermolecular forces and why?$\quad$[NCERT] Sol. Higher the critical temperature, more easily the gas can be liquefied i.e. greater are the intermolecular forces of attraction. Therefore,$\mathrm{CO}_{2}$has stronger intermolecular forces than$\mathrm{CH}_{4}$Q. (i) What will be the pressure exerted by a mixture of$3.2 \mathrm{g}$of methane and$4.4 \mathrm{g}$of carbon dioxide contained in a$9 d m^{3}$flask at$27^{\circ} \mathrm{C} ?$(ii) Give two example of Covalent solids. OR (i) Calculate the total number of electrons present in 1.4$g$of nitrogen gas. (ii) Which of the two gases, ammonia and hydrogen chloride, will diffuse faster and by what factor? (iii) Why urea has sharp melting point but glass does not? Sol.$(\mathrm{i}) \mathrm{pCH}_{4}=\frac{n R T}{\mathrm{V}}=\frac{3.2}{16} \times \frac{0.0821 \mathrm{L} \operatorname{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 300 \mathrm{K}}{9 \mathrm{L}}p C H_{4}=\frac{0.2 \times 24.63}{9}=\frac{4.926}{9}=0.547 \mathrm{atm}p C H_{4}=\frac{0.2 \times 24.63}{9}=\frac{4.926}{9}=0.547 \mathrm{atm}$(ii) Boron and silicon are example of Covalent solids. Or (i)$28 g$of Nitrogen gas contains$2 \times 7 \times 6.023 \times 10^{23}$electrons.$1+1\begin{array}{llll}{1.4} & {g} & {\text { of }} & {\text { Nitrogen }} & {\text { gas }} & {\text { contains }}\end{array}\frac{2 \times 7 \times 6.023 \times 10^{23}}{28} \times 1.4=\frac{2 \times 7 \times 6.023 \times 10^{23}}{20}=\frac{84.32 \times 10^{23}}{20}=4.2161 \times 10^{23}$electrons (ii)$N H_{3}$will diffuse faster$\frac{r_{N H_{3}}}{r_{H C l}}=\sqrt{\frac{36.5}{17}}=\sqrt{2.14}=1.46$times faster. (iii) Urea is a crystalline solid therefore it has sharp melting point whereas glass does not have sharp melting point because it is amorphous, i.e., does not have regular three dimensional structure. Q. What will be the minimum pressure required to compress$500 d m^{3}$of air at 1 bar to$200 d m^{3}$at$30^{\circ} \mathrm{C} ? \quad

Sol.

Q. A manometer is connected to a gas containing bulb. The open arm reads 43.7 cm whereas the arm connected to the bulb reads 15.6 cm. If the barometric pressure is 743 mm mercury, what is the pressure of the bulb reads 15.6 cm. If the barometric pressure is 743 mm mercury, what is the pressure of the gas in bar ?

[NCERT]

Sol. Pressure of gas $=$ Atmospheric pressure $+$ Difference between levels of $\mathrm{Hg}$

$=743 \mathrm{mm}+(43.7 \mathrm{cm}-15.6 \mathrm{cm})=74.3 \mathrm{cm}+28.1 \mathrm{cm}$

$P=102.4 \mathrm{cm} P$ in bar $=\frac{102.4}{76}=1.347 \mathrm{bar}$

Q. Calculate the number of moles of hydrogen $\left(H_{2}\right)$ present in a $500 \mathrm{cm}^{3}$ sample of hydrogen gas at a pressure of $760 \mathrm{mm}$ Hg and $27^{\circ} \mathrm{C}$.

[NCERT]

Sol. $P V=n R T, \quad P=1$ atm, $\quad V=500 \mathrm{cm}^{3}, \quad n=?$

$R=82.1 \mathrm{atm} \mathrm{cm}^{3} \mathrm{K}^{-1} \mathrm{mol}^{-1}, T=300 \mathrm{K}$

$n=\frac{1 a t m, \times 500 \mathrm{cm}^{3}}{\left(82.1 \mathrm{atm} \mathrm{cm}^{3} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right) \times(300 \mathrm{K})}=0.02 \mathrm{mol}$

Q. $34.05 \mathrm{ml}$ of phosphorus weight $0.0625 \mathrm{g}$ at $546^{\circ} \mathrm{Cand} 1$

bar pressure. What is the molar mass of phosphorus?

Or

In terms of Charle’s law explain why $-273^{\circ} \mathrm{C}$ is the lowest possible temperature.

[NCERT]

Sol.

$p V=n R T$

1 bar $\times \frac{34.05}{1000} L=\frac{0.0625}{M} \times 0.083 \times 819 K$

$M=\frac{0.0625 \times 0.083 \times 819 K \times 1000}{34.05}=\frac{4248.5625}{34.05}$

$M=124.77 \mathrm{g} \mathrm{mol}^{-1}$

Or

Charle’s plotted the volume against temperature in $^{\circ} C$. These plots when extraplotted intersect the temperature axis at the same point $-273^{\circ} \mathrm{C} .$ He concluded that all gases at this temperature could have zero volume and below this temperature volume would be negative. It shows $-273^{\circ} \mathrm{C}$ is lowest temperature attainable.

Q. A vessel of $120 \mathrm{mL}$ capacity contains a certain amount of gas at $35^{\circ} \mathrm{C}$ and 1.2 bar pressure. The gas is transferred to another vessel of volume $180 \mathrm{mL}$ at $35^{\circ} \mathrm{C} .$ What would be its pressure?

[NCERT]

Sol. Since temperature and amount of gas remains constant, therefore, Boyle’s law is applicable.

Q. A balloon is filled with hydrogen at room temperature. It will burst if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies 2.27 L volume, upto what volume can the balloon be expanded ?

[NCERT]

Sol. According to Boyle’s law, at constant temperature,

Since balloon bursts at 0.2 bar pressure, the volume of the balloon should be less than 11.35 L.

Q. A student forgot to add the reaction mixture to the round bottomed flask at $27^{\circ} \mathrm{C}$ but instead, he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer, he found the temperature of the flask was $477^{\circ} \mathrm{C} .$ What fraction of air would have been expelled out?

[NCERT]

Sol. Suppose volume of vessel $=V c m^{3}$

i.e., volume of air in the flask at $27^{\circ} \mathrm{C}=\mathrm{Vcm}^{3}$

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}, \quad$ i.e., $\quad \frac{V}{300}=\frac{V_{2}}{750} \quad$ or $\quad V_{2}=2.5 \mathrm{V}$

$\therefore$ Volume expelled $=2.5 \mathrm{V}-\mathrm{V}=1.5 \mathrm{V}$

Fraction of air expelled $=\frac{1.5 \mathrm{V}}{2.5 \mathrm{V}}=\frac{3}{5}$

Q. What is the difference in pressure between the top and bottom of a vessel $76 \mathrm{cm}$ deep at $27^{\circ} \mathrm{C}$ when filled with (i) water (ii) mercury? Density of water at $27^{\circ} \mathrm{C}$ is $0.990 \mathrm{g} \mathrm{cm}^{-3}$ and that of mercury is $13.60 \mathrm{g} \mathrm{cm}^{-3}$

Sol. Pressure $=$ height $\times$ density $\times g$

Case (i). Pressure

$=76 \mathrm{cm} \times 0.99 \mathrm{g} / \mathrm{cm}^{3} \times 981 \mathrm{cm} / \mathrm{s}^{2}$

$=7.38 \times 10^{4}$ dynes $\mathrm{cm}^{-2}$

$=0.073$ atm $\left(1 \mathrm{atm}=1.013 \times 10^{6} \text { dynes } \mathrm{cm}^{-2}\right)$

Case (ii). Pressure

$=76 \mathrm{cm} \times 13.6 \mathrm{g} / \mathrm{cm}^{3} \times 981 \mathrm{cm} / \mathrm{s}^{2}$

$=1.013 \times 10^{6}$ dynes $\mathrm{cm}^{-2}=1 \mathrm{atm}$

Q. An iron cylinder contains helium at a pressure of $250 \mathrm{kPa}$ at $300 K .$ The cylinder can withstand a pressure of

$1 \times 10^{6}$ pa. Theroom in which cylinder is placed catches fire. Predict whether the cylinder will blow up before it melts or not. (M.P. of the cylinder $=1800 K$ )

Sol.

Q. On a ship sailing in a pacific ocean where temperature is $23.4^{\circ} \mathrm{C},$ a balloon is filled with $2 \mathrm{L}$ air. What will be the volume of the balloon when the ship reaches Indian ocean, where temperature is $26.1^{\circ} \mathrm{C} ? \quad$

[NCERT]

Sol. According to Charles’ law

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

$V_{1}=2 L$

$V_{2}=?$

$T_{1}=273+23.4=296.4 \mathrm{K} \quad T_{2}=273+26.1=299.1$

$\therefore V_{2}=\frac{V_{1} T_{2}}{T_{1}}=\frac{2 L \times 299.1 K}{296.4 K}=2.018 L$

Q. What is the increase in volume when the temperature of 800 $m L$ of air increases from $27^{\circ} \mathrm{C}$ to $47^{\circ} \mathrm{Cunder}$ constant pressure of $1 \mathrm{bar} ?$

[NCERT]

Sol. Since the amount of gas and the pressure remains constant, Charles’ law is applicable. i.e.

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

$V_{1}=800 \mathrm{mL}$

$V_{2}=?$

$T_{1}=273+27=300 K ; \quad T_{2}=273+47=320 K$

$\frac{800 m L}{300 K}=\frac{V_{2}}{320 K}$

or $\quad V_{2}=\frac{(800 \mathrm{mL})}{(300 \mathrm{K})} \times(320 \mathrm{K})=853.3 \mathrm{mL}$

Increase in volume of air $=853.3-800=53.3 \mathrm{mL}$

Q. A cylinder containing cooking gas can withstand a pressure of 14.9 atmospheres. The pressure gauge of the cylinder indicates 12 atmosphere at $27^{\circ} \mathrm{C} .$ Due to sudden fire in the building, temperature starts rising. At what temperature will the cylinder explode?

Sol. since gas is confined in a cylinder, its volume will remain constan

Q. What will be the pressure exerted by a mixture when $0.5 L$ of $H_{2}$ at 0.8 bar and $2.0 L$ of oxygen at 0.7 bar are introduced in 1 container at $27^{\circ} \mathrm{C}$.

[NCERT]

Sol.

Q. Calculate the temperature of 4.0 moles of a gas occupying $6 d m^{3}$ at 3.32 bar $\left(R=0.083 \text { bar } d m^{3} K^{-1} \mathrm{mol}^{-1}\right)$

[NCERT]

Sol. According to ideal gas equation:

$p V=n R T$

$p=3.32$ bar, $V=5 d m^{3}, n=4.0 \mathrm{mol}$

$R=0.083$ bar $d m^{3} K^{-1} \mathrm{mol}^{-1}$

$T=\frac{p V}{n R}=\frac{3.32 \text { bar } \times 5 d m^{3}}{4.0 \mathrm{mol} \times 0.083 \text { bar } d m^{3} K^{-1} \mathrm{mol}^{-1}}=50 \mathrm{K}$

Q. When $2 g$ of gas $A$ is introduced into an evacuated flask kept at $25^{\circ} \mathrm{C},$ the pressure is found to be 1 atm. If $3 \mathrm{g}$ of another gas $B$ is then added to the same flask the total pressure becomes 1.5 atm at the same temperature. Assuming ideal behaviour of gases, calculate the ratio of molecular weights $M_{A}: M_{B}$

Sol. According to ideal gas equation,

Q. An evacuated glass vessel weighs $50.0 \mathrm{g}$ when empty, $148.0 \mathrm{g}$ when filled with a liquid of density $0.98 \mathrm{g} \mathrm{m}^{-1}$ and $50.5 \mathrm{g}$ when filled with an ideal gas at $760 \mathrm{mm} \mathrm{Hg}$ at $300 \mathrm{K}$. Determine the molecular weight of the gas.b

Sol. Weight of liquid $=148-50=98 g$

Density of liquid $=0.98 g m l^{-1}$

$\therefore$ Volume of liquid $=\frac{98}{0.98}=100 \mathrm{ml}$

Weight of gas $=50.5-50.0=0.5 g$

Volume $=\frac{100}{1000} L, P=1$ atm, $T=300 K$

$p V=n R T$ or $p V=\frac{W}{M} R T$

$\frac{1 \times 100}{1000}=\frac{0.5}{M} \times 0.082 \times 300$

$\therefore M=123.15 g \mathrm{mol}^{-1}$

Q. At $0^{\circ} \mathrm{C},$ the density of a gaseous oxide at 2 bar pressure is same as that of nitrogen at 5 bar. What is the molecular mass of the gaseous oxide? $\quad$

[NCERT]

Sol. Calculation of density of $N_{2}$ at 5 bar and $0^{\circ} \mathrm{C}$

$d=\frac{P M}{R T}=\frac{5(\text { bar }) \times 28\left(g m o l^{-1}\right)}{0.0831\left(L \text { bar } K^{-1} \mathrm{mol}^{-1}\right) \times 273.15(K)}$

$=6.168 g L^{-1}$

Calculation of molar mass of gaseous oxide

$M=\frac{d R T}{P}$

$=\frac{6.168\left(g L^{-1}\right) \times 0.0831\left(L \text { bar } K^{-1} \mathrm{mol}^{-1}\right) \times 273.15(K)}{2}$

$=70.0 \mathrm{g} \mathrm{mol}^{-1}$

Q. Density of a gas is found to be $5.46 g / d m^{3}$ at $27^{\circ} \mathrm{C}$ and 2 bar pressure. What will be its density at S.T.P.?

[NCERT]

Sol.

Q. What will be pressure exerted by a mixture of $3.2 \mathrm{g}$ of methane and $4.4 \mathrm{g}$ of carbon dioxide contained in a $9 \mathrm{dm}^{3}$ flask at $27^{\circ} \mathrm{C} ?$

Sol. $p=\frac{n}{V} R T=\frac{w}{M} \frac{R T}{V}$

$p_{C H_{4}}=\left(\frac{3.2}{16} m o l\right) \frac{0.0821 d m^{3} a t m K^{-1} m o l^{-1} \times 300 K}{9 d m^{3}}$

$=0.55 \mathrm{atm}$

$p_{\mathrm{CO}_{2}}=\left(\frac{4.4}{44} \mathrm{mol}\right) \frac{0.0821 \mathrm{dm}^{3} \mathrm{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 300 \mathrm{K}}{9 \mathrm{dm}^{3}}$

$=0.27 \mathrm{atm}$

$p_{\text {Total }}=0.55+0.27=0.82 \mathrm{atm}$

Q. Calculate the total pressure in a mixture of $8 g$ of $O_{2}$ and $4 g$

of $H_{2}$ confined in a vessel of $d m^{3}$ at $27^{\circ} C, R=0.083$ bar $d m^{3} K^{-1} \mathrm{mol}^{-1} .$ Atomic mass of $O=16 u, H=1 u$

[NCERT]

Sol. $p V=n R T$

$p \times 1=\frac{8}{32} \times 0.083 \times 300=\frac{24.9}{4}=6.225$ bar

$p \times 1=\frac{4}{2} \times 0.083 \times 300=49.80 \mathrm{bar}$

Total Pressure $=6.225+49.80=56.025$ bar

Q. For 10 minutes each, at $27^{\circ} \mathrm{C}$, from two identical holes, nitrogen and an unknown gas are leaked into common vessel of 3 L capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen. What is the molar mass of the unknown gas ?

[NCERT]

Sol.

Q. Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius $10 \mathrm{m},$ mass $100 \mathrm{kg}$ is filled with helium at 1.66 bar at $27^{\circ} \mathrm{C}$. (Density of air $=1.2$ $\left.k g m^{-3} \text { and } R=0.083 \text { bar } d m^{3} K^{-1} \mathrm{mol}^{-1}\right)$

[NCERT]

Sol. Pay load is the difference between the mass of displaced air and the mass of the balloon.

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

Radius of balloon, $r=10 \mathrm{m}$

$V=\frac{4}{3} \times 3.14 \times(10)^{3}=4186.7 \mathrm{m}^{3}$

Mass of displaced air $=4186.7 \mathrm{m}^{3} \times 1.2 \mathrm{kg} \mathrm{m}^{-3}$

$=5024.04 \mathrm{kg}$

Moles of gas present $=\frac{p V}{R T}$

$=\frac{1.66 \times 4186.7 \times 10^{3}}{0.083 \times 300}=279.11 \times 10^{3} \mathrm{mol}$

Mass of helium present $=279.11 \times 10^{3} \times 4$

$=1116.44 \times 10^{3} g=1116.4 \mathrm{kg}$

Mass of filled balloon $=100+1116.4=1216.4 \mathrm{kg}$ Pay load $=$ mass of displaced air – Mass of balloon

$=5024.4-1216.44=3807.6 \mathrm{kg}$

Q. A balloon of diameter $20 \mathrm{m}$ weights $100 \mathrm{kg} .$ Calculate the pay load if it is filled with $H e$ at 1.1 atm and $27^{\circ} \mathrm{C}$. Density of air is $1.2 \mathrm{kg} \mathrm{m}^{-3}\left(\mathrm{R}=0.082 \mathrm{dm}^{3} \mathrm{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right)$

Sol. Volume of balloon,

$V=\frac{4}{3} \pi r^{3} r=10 m$

$\therefore V=\frac{4}{3} \times 3.14 \times(10)^{3}=4187 \mathrm{m}^{3}$

Mass of displaced air $=4187 \mathrm{m}^{3} \times 1.2 \mathrm{kg} \mathrm{m}^{-3}$

$=5024.44 \mathrm{kg}$

Moles of gas present, $n=\frac{P V}{R T}=\frac{1 \times 4187 \times 10^{3}}{0,082 \times 298}$

$=171.3 \times 10^{3} \mathrm{mol}$

Mass of He present $=171.3 \times 10^{3} \times 4$

$=685.3 \times 10^{3} g$

$=685.3 \mathrm{kg}$

Mass of filled balloon $=100+685.3=785.3 \mathrm{kg}$

Pay load $=$ Mass of displaced air – Mass of balloon $=5024.4-785.3=4239.1 \mathrm{kg}$

Q. Using vander Waal’s equation, calculate the constant ” $a$ ‘ when two moles of a gas confined in a $4 L$ flask exerts a pressure of 11.0 atmospheres at a temperature of $300 K .$ The value of $b$ is $0.05 L \mathrm{mol}^{-1}$

Sol. $\left(p+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n R T$

$n=2 \mathrm{mol}, V=4 L, p=11 \mathrm{atm}$

$T=300 K, b=0.05 L m o l^{-1}$

Substituting the values

$\left(11+\frac{a \times 4}{16}\right)(4-2 \times 0.05)=2 \times 0.082 \times 300$

$\frac{176+4 a}{16} \times 3.9=49.2$

$(176+4 a) 3.9=49.2 \times 16$

$15.6 a=787.2-686.4$

$a=6.4616 \mathrm{atm} \mathrm{L}^{2} \mathrm{mol}^{-2}$

Q. $34.05 \mathrm{mL}$ of phosphorus vapour weigh $0.0625 \mathrm{g}$ at $546^{\circ} \mathrm{C}$ and 1 bar pressure. What is the molar mass of phosphorus?

[NCERT]

Sol. Mole of phosphorus vapour $(n)=\frac{P V}{R T}$

$=\frac{1(\text {bar}) \times 34.05 \times 10^{-3}(L)}{0.0831\left(\text {bar } L K^{-1} \mathrm{mol}^{-1}\right) \times 819.15(K)}=5.0 \times 10^{-4}$

Let molar mass of phosphorus be $M g m o l^{-1}$

$\therefore$ Mole of phosphorus vapour $=\frac{0.0625}{M}$

Now, $\frac{0.0625}{M}=5.0 \times 10^{-4}$ or $M=\frac{0.0625}{5.0 \times 10^{-4}}$

$=125 g \mathrm{mol}^{-1}$

Q. The drain cleaner, Drainex contain small bits of aluminium which react with caustic soda to produce hydrogen. What volume of hydrogen at $20^{\circ} \mathrm{C}$ and 1 bar will be released when $0.15 g$ of aluminium reacts?

[NCERT]

Sol. The chemical reaction taking place is

Q. For a real gas obeying vander Waal’s equation, a graph is plotted between $P V_{m}(y \text { -axis })$ and $P(x \text { -axis })$ where $V_{m}$ is the molar volume. Find $y$ -intercept of the graph.

Sol. For a real gas, the plot of $P V_{m}$ vs $P$ can be of the type $A$ or $B$ but at the point of intercept, $P=0$ and at any low pressure, vander Waal’s equation reduce to ideal gas equation.

$\mathrm{PV}=\mathrm{nRT}$ or $\mathrm{PV}_{\mathrm{m}}=\mathrm{RT}$

Hence, $y$ -intercept of graph will be $=R T$

Q. (i) State Graham’s law of diffusion. Arrange the gases $\mathrm{CO}_{2}, \mathrm{SO}_{2}, \mathrm{NO}_{2}$ in order of increasing rates of diffusion.

(ii) What is the volume of $0.300 \mathrm{mol}$ of ideal gas at $60^{\circ} \mathrm{C}$ and 0.821 atm pressure?

Sol. (i) For statement of the Graham’s law. Molar mass of:

$\mathrm{CO}_{2}=44 \mathrm{u} ; \mathrm{SO}_{2}=64 \mathrm{u}$ and $\mathrm{NO}_{2}=46 \mathrm{u}$

As $r_{d i f f} \propto \frac{1}{\sqrt{M}},$ therefore, larger the molar mass lesser

will be the rate of diffusion under similar condition. Thus, increasing order of rates of diffusion is

$r_{S O_{2}}<r_{N O_{2}}<r_{C O_{2}}$

(ii) $V=\frac{n R T}{P}=\frac{0.30(\text { mol }) \times 0.0821\left(L \text { atm } K^{-1} \mathrm{mol}^{-1}\right) \times 333(\mathrm{K})}{0.821(\mathrm{atm})}$

$=9.99 L \approx 10.8 L$

Q. Through two ends of a glass tube of length $200 \mathrm{cm} \mathrm{HCl}$ and $N H_{3}$ gases are allowed to enter. At what distance ammonium chloride will first appear?

[NCERT]

Sol. $\frac{r_{N H_{3}}}{r_{H C l}}=\frac{l_{1}}{200-l_{1}}=\sqrt{\frac{M_{H C l}}{M_{N H_{3}}}}=\sqrt{\frac{36.5}{17}}$

$\frac{l_{1}}{200-l_{1}}=\sqrt{2.147}=1.465$

$l_{1}=293-1.465 l_{1} ; 2.465 l_{1}=293$

$l_{1}=\frac{293}{2.465}=118.88 \mathrm{cm}$

$200-l_{1}=200-118.88=81.12 \mathrm{cm}$ from $\mathrm{HCl}$ end.

Q. Equal volumes of two gases A and B diffuse through a porous pot in 20 and 10 seconds respectively. If the molar mass of A be 80, find the molar mass of B.

[NCERT]

Sol. $\frac{r_{A}}{r_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$ or $\frac{t_{B}}{t_{A}}=\sqrt{\frac{M_{B}}{M_{A}}}$

$t_{A}=20$ sec, $t_{B}=10$ sec, $M_{A}=80, M_{B}=?$

$\frac{10}{20}=\sqrt{\frac{M_{B}}{80}}$

$M_{B}=\frac{80}{4}=20 g \mathrm{mol}^{-1}$

Q. A mixture of hydrogen and oxygen at one bar pressure contains 20% by weight of hydrogen. Calculate the partial pressure of hydrogen.

[NCERT]

Sol. As the mixture $H_{2}$ and $O_{2}$ contains $20 \%$ by weight of

hydrogen, therefore, if $H_{2}=20 g,$ then $O_{2}=80 g$

$n_{H_{2}}=\frac{20}{2}=10$ moles, $n_{O_{2}}=\frac{80}{32}=2.5$ moles

$p_{H_{2}}=\frac{n_{H_{2}}}{n_{H_{2}}+n_{O_{2}}} \times P_{\text {total }}=\frac{10}{10+2.5} \times 1$ bar $=0.8$ bar.

Q. Calculate the total pressure in a mixture of $8 g$ of oxygen and $4 g$ of hydrogen confined in a vessel of $1 d m^{3}$ at $27^{\circ} \mathrm{C}$ $\left(R=0.083 \text { bar } d m^{3} K^{-1} \mathrm{mol}^{-1}\right)$

[NCERT]

Sol. Partial pressure of oxygen gas,

$p=\frac{n R T}{V}$

$n=\frac{8}{32}$ mol, $\quad \mathrm{V}=1 \mathrm{dm}^{3}, T=300 \mathrm{K}$

$p_{\left(O_{2}\right)}=\frac{8 \times 0.083 \times 300}{32 \times 1}=6.225$ bar

Partial pressure of hydrogen gas

$p=\frac{n R T}{V}$

$n=\frac{4}{2}=2 m o l$

$p\left(H_{2}\right)=\frac{2 \times 0.083 \times 300}{1}=49.8 \mathrm{bar}$

Total pressure $=p_{\left(O_{2}\right)}+p_{\left(H_{2}\right)}$

$=6.225+49.8=56.025$ bar

Q. A vessel of $1.00 \mathrm{dm}^{3}$ capacity contains $8.00 \mathrm{g}$ of oxygen and $4.00 \mathrm{g}$ of hydrogen at $27^{\circ} \mathrm{C} .$ Calculate the partial pressure of each gas and also the total pressure in the container $\left(R=0.083 \text { bar } d m^{3} K^{-1} \mathrm{mol}^{-1}\right)$

[NCERT]

Sol. Let the partial pressure of hydrogen be $P_{H_{2}}$ and the partial

pressure of oxygen be $P_{O_{2}}$

The number of mole of hydrogen $\left(n_{1}\right)=\frac{4}{2}=2$ mole

The number of mole of oxygen $\left(n_{2}\right)=\frac{8}{32}=0.25$ mole

Now, applying ideal gas equation for each gas $p_{H_{2}} \times V=n_{1} R T$

$p_{H_{2}}=\frac{n_{1} R T}{V}=\frac{2 \times 0.083 \times 300}{1}=49.8$ bar

Similarly, $p_{O_{2}} V=n_{2} R T$

$p_{O_{2}}=\frac{n_{2} R T}{V}=\frac{0.25 \times 0.083 \times 300}{1}=6.225 \mathrm{bar}$

Total pressure of gaseous mixture

$=p_{H_{2}}+p_{O_{2}}=49.8+6.225=56.025$ bar.

Q. Calculate (i) Average kinetic energy of $32 g$ of methane molecules at $27^{\circ} \mathrm{C} R=8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$

(ii) Root mean square speed and (ii) Most probable speed of methane molecule at $27^{\circ} \mathrm{C}$

Sol. (i) Average kinetic energy is given as $E_{k}=\frac{3 n R T}{2}$

Here, $n=\frac{32}{16}=2 ; R=8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1} ; T=300 \mathrm{K}$

$\therefore E_{k}=\frac{3 \times 2 \times 8.314 \times 300}{2}=7482.6 \mathrm{J}$

(ii) Root mean square speed is given as

$u_{r m s}=\sqrt{\frac{3 R T}{M}}$

Here use $R=8.314 \times 10^{7}$ ergs $K^{-1} \mathrm{mol}^{-1}$ to get speed in

$\mathrm{cm} \mathrm{s}^{-1}$

$u_{m s}=\sqrt{\frac{3 \times 8.314 \times 10^{7} \times 300}{16}}=68385.85 \mathrm{cm} s^{-1}=68.38 \mathrm{ms}^{-1}$

(iii) Most probable speed

$(\alpha)=\frac{u_{r m s}}{1.224}=\frac{683.8}{1.224}=558.7 \mathrm{ms}^{-1}$

Q. Calculate root mean square speed $(r m s)$ of ozone which is kept in a closed vessel at $20^{\circ} \mathrm{C}$ and a pressure of $82 \mathrm{mm}$ of $H g$

Sol. Volume occupied by $1 \mathrm{mol}$ of $\mathrm{O}_{3}$ at $20^{\circ} \mathrm{C}$ and $82 \mathrm{mm}$ pressure is calculated by applying general gas equation,

$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

$\therefore \quad V_{2}=\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}=\frac{76 \times 22400 \times 293}{273 \times 82}=22281.92 \mathrm{cm}^{3}$

Now, $u=\sqrt{\frac{3 P V}{M}}$

Here, we use $P$ in dyne/cm$^{2}, P=82 \times 13.6 \times 981$ dyne/cm$^{2}$

$\therefore u=\sqrt{\frac{3 \times 82 \times 13.6 \times 981 \times 22281.92}{48}}=3.90 \times 10^{4} \mathrm{cm} \mathrm{s}^{-1}$

Q. (i) Explain Boyle’s law with the help of Kinetic theory of gases.

(ii) An open beaker at $27^{\circ} \mathrm{C}$ is heated to $477^{\circ} \mathrm{C} .$ What fraction of air would have been expelled out?

[NCERT]

Sol. (i) The kinetic theory of gases assumes that pressure of gas is due to collision of gas molecules with the walls of the container. The more will be frequency of collision, more will be pressure. The reduction in volume of gas increases no. of molecules per unit volume to which pressure is directly proportional. Therefore, the volume of the gas is reduced if pressure is increased or pressure is inversely proportional to volume.

$\frac{1}{2} m u^{2}=\frac{3}{2} k T$

$p=\frac{1}{3} \frac{N}{V} m u^{2}$ or $p=\frac{2}{3} \frac{N}{V} \times \frac{1}{2} m u^{2}=\frac{2}{3} \frac{N}{V} \times \frac{3}{2} k T$

$\Rightarrow p V=N k T \Rightarrow P \alpha \frac{1}{V}$

It can be seen that at a constant temperature for a fixed number of gas molecules, the pressure is inversely proportional to volume.

(ii) $\frac{T_{1}}{T_{2}}=\frac{n_{1}}{n_{2}} \quad \begin{array}{cc}{T_{1}=27^{\circ} C+273} & {=300 K} \\ {T_{2}=477+273=} & {750 K}\end{array}$

$\frac{300 K}{750 K}=\frac{n_{1}}{n_{2}}, \quad \frac{n_{1}}{n_{2}}=\frac{2}{5} b v$

fraction of air escaped $=1-\frac{2}{5}=\frac{3}{5}$

Q. At which temperature average velocity of oxygen molecules is equal to the rms velocity at $27^{\circ} \mathrm{C} ?$

Sol. Average velocity $=\sqrt{\frac{8 R T}{\pi M}}$

Root mean square velocity

$=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 R \times 300 K}{M}}$

For equal values, $\sqrt{\frac{8 R T}{\pi M}}=\sqrt{\frac{3 R \times 300}{M}}$

or $\quad \frac{8 R T}{\pi M}=\frac{3 R \times 300}{M}$ or $\frac{8 T}{\pi}=900$

or $\quad T=353.57 K=80.57^{\circ} \mathrm{C}$

Q. At a constant temperature, a gas occupies a volume of 200 mL at a pressure of 0.720 bar. It is subjected to an external pressure of 0.900 bar. What is the resulting volume of the gas ?

[NCERT]

Sol. Boyle’s law is applicable as the amount and temperature are unaltered

$p_{1} V_{1}=p_{2} V_{2}$

or $p_{1} / p_{2}=V_{2} / V_{1}$ ‘Substituting the values

0.720 bar/ 0.900 bar $=V_{2} / 200 \mathrm{mL}$

$V_{2}=\frac{720}{900} \times 200 m L=160 m L$

Boyle’s law is manifested in the working of many devices used in daily life such as cycle pump, aneroid barometer and tyre pressure gauge etc.

Q. Calculate the number of nitrogen molecules present in 2.8 g of nitrogen gas.

[NCERT]

Sol. Number of moles of nitrogen

$=2.8 g / 28 g \mathrm{mol}^{-1}=0.1 \mathrm{mol}$

Number of nitrogen molecules $=0.1 \mathrm{mol} \times 6.022 \times 10^{23}$

$m o l^{-1}=6.022 \times 10^{22}$

Q. If the density of a gas at the sea level at $0^{\circ} \mathrm{C}$ is $1.29 \mathrm{kg} \mathrm{m}^{-3}$, what is its molar mass? (Assume that pressure is equal to 1 bar)

[NCERT]

Sol. $p V_{m}=R T$ or $p M / d=R T$ or $M=d R T / p$

$=\frac{1.29 \mathrm{kg} m^{-3} \times 8.314 \mathrm{NmK}^{-1} \mathrm{mol}^{-1} \times 273.15 \mathrm{K}}{1.0 \times 10^{5} \mathrm{Nm}^{-2}(\mathrm{or} P a)}$

$=\frac{1.29 \times 8.314 \times 273.15 k g m o l^{-1}}{1 \times 10^{5}}$

$=0.0293 \mathrm{kg} \mathrm{mol}^{-1}$ or molar mass is $29.3 \mathrm{g} \mathrm{mol}^{-1}$

Q. Which of the two gases, ammonia and hydrogen chloride, will diffuse faster and by what factor ?

[NCERT]

Sol. $r_{N H_{3}} / r_{H C 1}=\left(M_{H C 1} / M_{N H_{3}}\right)^{1 / 2}$

$=(36.5 / 17)^{1 / 2}=1.46$ or $r_{N H_{3}}=1.46 r_{H C 1}$

Thus ammonia will diffuse 1.46 times faster than hydrogen chloride gas.

Q. A 2.5 flask contains 0.25 mol each of sulphur dioxide, and nitrogen gas at $27^{\circ} \mathrm{C}$. Calculate the partial pressure exerted by each gas and also the total pressure.

[NCERT]

Sol. Partial pressure of sulphur dioxide,

$p_{S O_{2}}=n R T / V$

$=\frac{0.25 \mathrm{mol} \times 8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \times 300 \mathrm{K}}{2.5 \times 10^{-3} \mathrm{m}^{3}}$

$=2.49 \times 10^{5} \mathrm{Nm}^{-2}=2.49 \times 10^{5} \mathrm{Pa}$

Similarly $p_{N_{2}}=2.49 \times 10^{5} \mathrm{Pa}$

Following Dalton’s law $p_{\text {Total }}=p_{N_{2}}+p_{S O_{2}}$

$=2.49 \times 10^{5} P a+2.49 \times 10^{5} P a=4.98 \times 10^{5} P a$

Q. An open vessel at $27^{\circ} \mathrm{C}$ is heated until $\frac{3}{5}$ parts of the air in it has been expelled. Assuming that the volume of the vessel remains constant, find the temperature to which the vessel has been heated.

Sol. As the vessel is open, pressure and volume remain constant. Thus, if $n_{1}$ moles are present at $T_{1}$ and $n_{2}$ moles are present at $T_{2},$ we can write

$P V=n_{1} R T_{1} ; P V=n_{2} R T_{2}$

Hence, $n_{1} R T_{1}=n_{2} R T_{2}$ or $n_{1} T_{1}=n_{2} T_{2}$

or, $\quad \frac{n_{1}}{n_{2}}=\frac{T_{2}}{T_{1}}$

Suppose the no. of moles of air originally present $=n$

After heating, no. of moles of air expelled $=\frac{3}{5} n$

$\therefore$ No. of moles left after heating $=n-\frac{3}{5} n=\frac{2}{5} n$

Thus, $n_{1}=n, T_{1}=300 K ; n_{2}=\frac{2}{5} n, T_{2}=?$

$\frac{n}{\frac{2}{5} n}=\frac{T_{2}}{300}$ or $\frac{5}{2}=\frac{T_{2}}{300}$ or $, T_{2}=750 \mathrm{K}$

Alternatively, suppose the volume of the vessel $=V$

i.e. Volume of air initially at $27^{\circ} C=V$

Volume of air expelled $=\frac{3}{5} V$

$\therefore$ Volume of air left at $27^{\circ} \mathrm{C}=\frac{2}{5} \mathrm{V}$

However, on heating to $T^{\circ} K,$ it would become $=V$ As pressure remains constant, (vessel being open),

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$ i.e. $\frac{2 / 5 \mathrm{V}}{300 \mathrm{K}}=\frac{\mathrm{V}}{T_{2}}$ or $T_{2}=750 \mathrm{K}$

Q. A glass bulb contains $2.24 L$ of $H_{2}$ and $1.12 L$ of $D_{2}$ at

S.T.P. It is connected to fully evacuated bulb by a stopcock with a small opening. The stopcock is opened for sometime and then closed. The first bulb now contains $0.10 \mathrm{g}$ of $D_{2} .$ Calculate the percentage composition by weight of the gases in the second bulb.

Sol. Weight of $2.24 \mathrm{L}$ of $\mathrm{H}_{2}$ at $\mathrm{S} . T . P .=0.2 \mathrm{g}$

(Mol. mass of $\left.H_{2}=2\right)=0.1 \mathrm{mol}$

Weight of $1.12 \mathrm{L}$ of $D_{2}$ at $S . T . P .=0.2 \mathrm{g}$

(Mol. mass of $\left.D_{2}=4\right)=0.05 \mathrm{mol}$

As number of moles of two gases are different but V and T are same, therefore, their partial pressures will be different, i.e., in the ratio of their number of moles. Thus,

$\frac{P_{H_{2}}}{P_{D_{2}}}=\frac{n_{H_{2}}}{n_{D_{2}}}=\frac{0.1}{0.5}=2$

Now, $D_{2}$ present in the first bulb $=0.1 g$ (Given)

$D_{2}$ diffused into the second bulb

$=0.2-0.1=0.1 g=0.56 L$ at $S . T . P .$

Now, $\frac{r_{H_{2}}}{r_{D_{2}}}=\frac{P_{H_{2}}}{P_{D_{2}}} \times \sqrt{\frac{M_{D_{2}}}{M_{H_{2}}}}$

Or, $\frac{v_{H_{2}}}{t} \times \frac{t}{v_{D_{2}}}=\frac{P_{H_{2}}}{P_{D_{2}}} \times \sqrt{\frac{M_{D_{2}}}{M_{H_{2}}}}$

$\frac{v_{H_{2}}}{t} \times \frac{t}{0.56 L}=2 \times \sqrt{\frac{4}{2}}$

or $\quad v_{H_{2}}=1.584 L=0.14 g$ of $H_{2}$

$\therefore$ Weight of the gases in 2 nd bulb

$=0.10 g\left(D_{2}\right)+0.14 g\left(H_{2}\right)=0.24 g$

Hence, in the 2 nd bulb,

$\%$ of $D_{2}$ by weight $=\frac{0.10}{0.24} \times 100=41.67 \%$

$\%$ of $H_{2}$ by weight $=100-41.67=58.33 \%$

Q. An $L P G$ (Liquefied Petroleum Gas) cylinder weights 14.8 kg when empty. When full, it weighs 29.0 kg and shows a pressure of 2.5 atm. In the course of use at $27^{\circ} \mathrm{C},$ the mass of the full cylinder is reduced to 23.2 kg. Find out the volume of the conditions, and the final pressure inside the cylinder. Assume $L P G$ to be $n$ -butane with normal boiling point of $0^{\circ} \mathrm{C}$

Sol. Weight of $L P G$ originally present $=29.0-14.8$ $=14.2 \mathrm{kg}$

Pressure $=2.5$ atm Weight of LPG present after use $=23.2-14.8$ $=8.4 \mathrm{kg}$

since volume of the cylinder is constant, applying

since volume of the cylinder is constant, applying

$p V=n R T$

$\frac{p_{1}}{p_{2}}=\frac{n_{1}}{n_{2}}=\frac{W_{1} / M}{W_{2} / M}=\frac{W_{1}}{W_{2}}$

$\frac{2.5}{p_{2}}=\frac{14.2}{8.4}$

$p_{2}=\frac{2.5 \times 8.4}{14.2}=1.48 \mathrm{atm}$

or Weight of used gas $=14.2-8.4=5.8 \mathrm{kg}$

Moles of gas $=\frac{5.8 \times 10^{3}}{58}=100$ moles

Normal conditions

$p=1$ atm;

$T=273+27=300 K$

Volume of 100 moles of $L P G$ at 1 atm and $300 \mathrm{K}$

$V=\frac{n R T}{p}=\frac{100 \times 0.082 \times 300}{1}=2460$ litre

$V=2.460 \mathrm{m}^{3}$

Q. The compressibility factor for one mole of a vander Waals gas at $0^{\circ} \mathrm{C}$ and 100 atm pressure is found to be $0.5 .$ Assume that the volume of gas molecule is negligible, calculate yander

Sol. Compressibility factor

$Z=\frac{p V}{R T} ; \quad 0.5=\frac{100 \times V}{0.082 \times 273}$

$\therefore V=\frac{0.5 \times 0.082 \times 273}{100}=0.1119 L$

If volume of molecules is negligible i.e. $b$ is negligible vander Waals’ equation:

$\left(p+\frac{a}{V^{2}}\right)(\mathrm{V})=R T$

or $p V=R T-a / V$ or $a=R T V-P V^{2}$

$=(0.082 \times 273 \times 0.1119)-(100 \times 0.1119 \times 0.1119)$

$=1.253 \mathrm{atm} L^{2} \mathrm{mol}^{-2}$

Ionic Equilibrium | Question Bank for Class 11 Chemistry

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Q. How can we predict whether a precipitate will be formed or not on mixing two solutions ?

Sol. A precipitate will be formed if ionic product > solubility product.

Q. Define Lewis acid and Lewis base.

Sol. Lewis acid is a species which is electron deficient or positively charged Lewis base is a species which is electron rich, i.e., has lone pair of electron or is negatively charged.

Q. Define Bronsted acid and Bronsted base.

Sol. Bronsted acid can donate $H^{+}$ where as bronsted base can accept.

Q. ‘All Lewis bases are also Brosted base’. Is it true ? If yes, why ?

Sol. Yes, it is true, It is because Lewis bases are – velycharged or electron rich. They are bronsted bases also because they can accept $H^{+}$ easily.

Q. $p K_{a}$ values of acids $A, B, C, D$ are $1.5,3.5,2.0$ and $5.0 .$ Which of them is strongest acid?

Sol. Acid $^{\prime} A^{\prime}$ with $\mathrm{pKa}=1.5$ is strongest acid, lower the value of $P K_{a}$ stronger will be the acid.

Q. Predict if the solution of the following salts are neutral, acidic or basic: $N a C l, K B r, N a C N, N H_{4} N O_{3}, N a N O_{2}$ and $K F$

Sol. Neutral: $\mathrm{NaCl}, \mathrm{KBr} \quad$ Acidic : $\mathrm{NH}_{4} \mathrm{NO}_{3}$

Basic : $\mathrm{NaCN}, \mathrm{NaNO}_{2}$ and $\mathrm{KF}$

Q. What is the meant by conjugate acid-base pair ? Find the conjugate acid/base for the following species :

Sol. An acid-base which differ by a proton is called conjugate acid base pair.

$\mathrm{NO}_{2}^{-}, \mathrm{HCN}, \mathrm{ClO}_{4}^{-}, \mathrm{HF}, \mathrm{H}_{2} \mathrm{O}(\mathrm{acid})$ or $\mathrm{O}^{2-}$ (base)

$\mathrm{HCO}_{3}^{-}$ and $\mathrm{HS}^{-}$

Q. Which of the following are Lewis acids?

$H_{2} O, B F_{3}, H^{+}, N H_{4}^{+}$

Sol. $p H=-\log \left[H^{+}\right]=-\log \left(3.8 \times 10^{-3}\right)$

$=-\log 3.8+3=3-0.5798=2.4202=2.42$

Q. The $p H$ of a sample of vinegar is $3.76 .$ Calculate the concentration of hydrogen ion in it.

Sol. $p H=-\log \left[H^{+}\right]$ or $\log \left[H^{+}\right]=-p H=-3.76=\overline{4} .24$

$\therefore\left[H^{+}\right]=$ Antilog $\overline{4} 24=1.738 \times 10^{-4} M=1.74 \times 10^{4} M$

Q. Which of the following combination would result in the formation of a buffer solution?

(i) $\mathrm{NH}_{4} \mathrm{Cl}+\mathrm{NH}_{3}$

(ii) $\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{HCl}$

(iii) $\mathrm{CH}_{3} \mathrm{COONa}+\mathrm{CH}_{3} \mathrm{COOH}$

(iv) $\mathrm{NH}_{3}+\mathrm{HCl}$ in the molar ratio of 2: 1

(v) $H C l+N a O H$

Sol. (i), ( iii), (iv)

Q. A $0.02 \mathrm{M}$ solution of pyridinium hydrochloride has $p H=3.44 .$ Calculate the ionization constant of pyridine.

Sol.

Q. The $p H$ of $0.005 M$ codeine $\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NQ}_{3}\right)$ solution is 9.95 Calculate the ionization constant and $p K_{b}$

Sol. $\mathrm{Cod}+H_{2} \mathrm{O} \rightleftharpoons \mathrm{CodH}^{+}+\mathrm{OH}^{-}$

$p H=9.95 \therefore p O H=14-9.95=4.05$

i.e., $-\log \left[\mathrm{OH}^{-}\right]=4.05$

or $\log \left[O H^{-}\right]=-4.05=5.95$ or $\left[O H^{-}\right]=8.913 \times 10^{-5}$

$K_{b}=\frac{\left[\mathrm{Cod} H^{+}\right][\mathrm{OH}]}{[\mathrm{Cod}]}=\frac{[\mathrm{OH}]^{2}}{[\mathrm{Cod}}=\frac{\left(8.91 \times 10^{5}\right)^{2}}{5 \times 10^{-3}}=1.588 \times 10^{-6}$

$p K_{b}=-\log \left(1.588 \times 10^{-6}\right)=6-0.20=5.8$

Q. The ionization constant of HCOOH is $1.8 \times 10^{-4} .$ Around what $p H$ will its mixture with sodium formate gives buffer solution of higher capacity. Calculate the ratio of sodium formate \& formic acid in a buffer of $p H=4.25$

Sol. $p K_{a}=-\log K_{a}=-\log 1.8 \times 10^{-4}=3.74$

Buffer capacity is maximum near $p K_{a}$ of the acid $i . e .,$ at $p H=3.74 .$ In order to obtain a buffer solution of

$p H=4.25$ we have

$\log \frac{[\text { salt }]}{[\text { acid }]}=p H-p K_{a}=4.25-3.74=0.51$

$\frac{[\text { salt }]}{[\text { acid }]}=$ Antilog $0.51=3.24$

It means concentration of salt should be 3.24 times that of acid.

Q. The ionization constant of dimethylamine is $5.4 \times 10^{-4}$ Calculate the degree of ionization in its $0.02 M$ solution. What percentage of dimethylamine is ionized if the solution is also $0.1 M N a O H ?$

Sol.

Q. The $p H$ of $0.1 M$ solution of cyanic acid (HCNO) is $2.34 .$ Calculate the ionization constant of the acid and its degree of ionization in the solution.

Sol. HCNO ionizes as:

$H C N O=H^{+}+C N O^{-}$

The dissociation constant

$K_{a}=\frac{\left[H^{+}\right][C N O]}{[H C N O]} \Rightarrow\left[H^{+}\right]=\left[C N O^{-}\right]$

Now, $p H=2.34$

$-\log \left[H^{+}\right]=2.34 \Rightarrow \log \left[H^{+}\right]=-2.34=3.36$

$\left(H^{+}\right)=$ Antilog $(3.36)=4.57 \times 10^{-3} \mathrm{M}$

$\therefore\left[H^{+}\right]=\left[C N O^{-}\right]=4.57 \times 10^{-3} M$ and $[H C N O]=0.1 M$

$K_{a}=\frac{\left(4.57 \times 10^{-3}\right) \times\left(4.57 \times 10^{-3}\right)}{0.1}=2.09 \times 10^{-4}$

Degree of dissociation $=\alpha=\sqrt{\frac{K_{a}}{c}}$

$=\sqrt{\frac{2.09 \times 10^{-4}}{0.1}}=4.57 \times 10^{-2} \mathrm{or}$

Q. The $p H$ of $0.004 M$ hydrazine solution is $9.7 .$ Calculate its ionization constant $K_{b}$ and $p K_{b}$

Sol.

Q. Calculate the pH of the mixture formed by the addition of 5, 9, 9.5, 9.9, 9.95, 10, 10.05 and 10.1 mL of 0.5M KOH solution to 100 mL of 0.05 HBr solution. What will be the most suitable indicator for this titration ?

Sol. $5 m L$ of $0.5 M K O H=2.5$ millimoles

$100 \mathrm{mL}$ of $0.05 \mathrm{M} \mathrm{HBr}=5$ milli moles

$105 \mathrm{mL}$ of solution contains $5-2.5=2.5$ milli moles $=2.5 \times 10^{-3}$ moles of acid

$1000 \mathrm{mL}$ of contains $=\frac{2.5 \times 10^{-3}}{105} \times 1000$

$=2.38 \times 10^{-2} \mathrm{mol} L^{-1}$

$p H=-\log 2.38 \times 10^{-2}=-\log 2.38-\log 10^{-2}$

$=-0.38+2.000=1.62$

Phenolpthalein and Thymolpt halein are suitable indicators.

Q. Benzoic acid is monobasic and when 1.22 g of its pure sample after dissolution in distilled water is titrated against a base it uses 50 mL of 0.2 M NaOH. Calculate the molar mass of benzoic acid.

Sol. $50 \mathrm{mL}$ of $0.2 \mathrm{MNaOH}=50 \times 0.2=10 \mathrm{millimoles}$ of $\mathrm{NaOH}$

10 millimoles of $N a O H$ will neutralize 10 millimoles of Benzoic acid.

10 millimoles of benzoic acid $=1.22 \mathrm{g}$

1 mole $=1000$ millimoles $=\frac{1.22}{10} \times 1000=122 \mathrm{g} \mathrm{mol}^{-1}$

Molecular weight $=122 \mathrm{g} \mathrm{mol}^{-1}$

Q. Equal volumes of $0.002 M$ solutions of sodium iodate and copper chlorate are mixed together. Will it lead to precipitation of copper iodate? For copper iodate $K_{s p}=7.4 \times 10^{-8}$

Sol. When equal volume of $\mathrm{NalO}_{3} \& \mathrm{CuClO}_{3}$ are mixed, the concentration of each of the solution will become half.

$\left[\mathrm{IO}_{3}^{-}\right]=\frac{1}{2} \times 0.002=0.001=10^{-3}$

$\left[\mathrm{Cu}^{2+}\right]=\frac{1}{2} \times 0.002=0.001=10^{-3}$

$\mathrm{Cu}\left(\mathrm{IO}_{3}\right)_{2} \mathrm{H}$ tyth $\quad \mathrm{Cu}^{2+}+2 \mathrm{IO}_{3}^{-}$

I.P (Ionic product) $=\left[C u^{2+}\right]\left[I O_{3}^{-}\right]^{2}=\left[10^{-3}\right]\left[10^{-3}\right]^{2}=10^{-9}$

since $I . P .\left(1 \times 10^{-9}\right)$ is less than $K_{s p}\left(7.4 \times 10^{-8}\right),$ therefore No precipitation will take place.

Q. What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, therefore, is no precipitation of iron sulphide ? For iron sulphide

$K_{s p}=6.3 \times 10^{-8}$

Sol. $K_{s p}=6.3 \times 10^{-8}$

FeS $\mathrm{HHH} \quad \mathrm{Fe}_{‘^{\prime} \mathrm{S}}^{+2}+\mathrm{S}^{-2}$

$K_{s p}=6.3 \times 10^{-18}$

$K_{s p}=s \times s=s^{2}=6.3 \times 10^{-18}$

$s=2.51 \times 10^{-9}$

since volume of solution will become double on mixing equal volume of each solution.

$\therefore$ concentration will become half on mixing max. concentration of each ion $=2 \times S$

$=2 \times 2.51 \times 10^{-9}=5.02 \times 10^{-9} \mathrm{m}$

Q. The ionization constant of phenol is $1.0 \times 10^{-10} .$ What is the concentration of phenate ion in $0.05 M$ solution of phenol? What will be its degree of ionization if the solution is also $0.01 M$ sodium phenate?

Sol.

Q. The ionization constants of $H F, H C O O H$ and $H C N$ at

$298 K$ are $6.8 \times 10^{4}, 1.8 \times 10^{4}$ and $4.8 \times 10^{9}$ respectively. Calculate the ionization constants of the corresponding conjugate base.

Sol. The relation between ionization constant of an acid and that of its conjugate base is

$K_{a} \times K_{b}=K_{w}$

$\therefore \quad K_{b}=\frac{K_{w}}{K_{a}}$

The conjugate base of $H F$ is $F^{-}$

$K_{b\left(F^{-}\right)}=\frac{K_{w}}{K_{a}(H F)}=\frac{1 \times 10^{-14}}{6.8 \times 10^{-4}}=1.47 \times 10^{-11}$

The conjugate base of $H C O O H$ is $H C O O^{-}$

$K_{b}\left(H C O O^{-}\right)=\frac{K_{w}}{K_{a(H C O O H)}}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-4}}=5.56 \times 10^{-11}$

The conjugate base of $H C N$ is $C N^{-}$

$K_{b\left(C N^{-}\right)}=\frac{K_{w}}{K_{a(H C N)}}=\frac{1 \times 10^{-14}}{4.8 \times 10^{-9}}=2.08 \times 10^{-6}$

Q. The ionization constant of $H F$ is $3.2 \times 10^{-4} .$ Calculate the degree of dissociation of $H F$ in its $0.02 M$ solution. Calculate the concentration of all species present $\left(H_{3} O^{4}, F^{-} \text {andHH } \text { in the solution and its } p H\right.$

Sol. If $C$ is the initial concentration and $\alpha$ is the degree of ionization of $H F$ then

Q. The $p H$ of a $0.01 M$ solution of an organic acid is 4.15 Calculate

(i) the concentration of the anion

(ii) ionization constant of the acid

(iii) $\quad p K_{a}$

Sol. $p H=4.15$

$-\log \left[H_{3} O^{+}\right]=4.15$

or $\log \left[H_{3} O^{+}\right]=-4.15=\overline{5} .85$

$\therefore \log \left[H_{3} O^{+}\right]=$ antilog $(\overline{5} .85)=7.08 \times 10^{-5}$

$\mathrm{Now},\left[H_{3} \mathrm{O}^{+}\right]=\left[A^{-}\right]=7.08 \times 10^{-5} \mathrm{M}$

(i) Concentration of anion $=7.08 \times 10^{-5} \mathrm{M}$

(ii) Ionization constant of acid

$H A(a q)+H_{2} O(l) ! H_{3} O^{+}(a q)+A^{-}(a q)$

$K_{a}=\frac{\left[H_{3} O^{+}\right]\left[A^{-}\right]}{[H A]}$

Concentration of undissociated acid $=0.01-0.000078$

$=0.009922$

$\therefore K_{a}=\frac{\left(7.08 \times 10^{-5}\right) \times\left(7.08 \times 10^{-5}\right)}{0.009922}=5.05 \times 10^{-7}$

(iii) $p K_{a}=-\log K_{a}=-\left(5.05 \times 10^{-7}\right)=6.30$

Q. Calculate the degree of ionization and $p H$ of $0.05 M$ ammonia solution. The ionization constant of ammonia $\left(K_{b}\right)$ is $1.77 \times 10^{-5} .$ Also calculate the ionic constant of the conjugate acid of ammonia.

Sol.

Q. The ionization constant of benzoic acid is $6.46 \times 10^{-5}$ and $K_{s p}$ for silver benzoate is $2.5 \times 10^{-13} .$ How many times is silver benzoate more soluble in a buffer of $p H$ 3.19 compared to its solubility in pure water?

Sol.

Silver benzoate is 3.16 times more souble in buffer solution of $p H=3.19$ than in pure water.

Q. The indicators will be suitable for the following acid $-$ base titration? (i) $\mathrm{HCOOH}$ against $\mathrm{NaOH}$ (ii) $\mathrm{HBr}$ against $K O H$ (iii) $N H_{4} O H$ against $H N O_{3} .$

Sol. (i) Phenolpthalein (ii) Phenolpthalein (iii) Methyl orange

Q. (i) Write the conjugate acids for the following Bronsted

$\text { bases: } \mathrm{NH}_{2}^{-}, \mathrm{NH}_{3} \text { and } \mathrm{HCOO}^{-}$

(ii) Write the conjugate bases for the following

$\text { Bronsted acids: } H F, H_{2} S O_{4}, H C O_{3}^{-}$

Sol. (1) Conjugate acids

(a) $\mathrm{NH}_{2}^{-}: \mathrm{NH}_{3}$

(b) $N H_{3}: \mathrm{NH}_{4}^{+}$

(c) $\mathrm{HCOO}^{-}: \mathrm{HCOOH}$

(ii) Conjugate bases

$\begin{array}{l} {H F: F^{-}} \\ {H_{2} S O_{4}: H S O_{4}^{-}} \\ {H C O_{3}^{-}: C O_{3}^{2-}} \end{array}$

Q. The species $H_{2} \mathrm{O}, \mathrm{HCO}_{3}^{-}, \mathrm{HSO}_{4}^{-}$ and $\mathrm{NH}_{3}$ can act both as Bronsted acids and bases. Give the corresponding conjugate acid and base.

Sol.

Q. Classify the following species into Lewis acids and Lewis bases and show how these act as such:

$\begin{array}{llll}{\text { (i) } H O^{-}} & {\text {(ii) } F^{-}} & {\text {(iii) } H^{+}} & {\text {(iv) } B C l_{3}}\end{array}$

Sol. (i) $\mathrm{HO}^{-}$ is a Lewis base because it can donate a pair of electrons.

(ii) $F^{-}$ acts as a Lewis base because it can donate a pair of electron.

(iii) $H^{+}$ is a Lewis acid because it can accept a lone pair of electrons from base like $\mathrm{OH}^{-}$ and ions. a

(iv) acts as a Lewis acid because it is electron deficient and can accept a lone pair of electrons.

Q. What is meant by the conjugate acid base pair? Find the conjugate acid/base for the following compounds and ions: $H N O_{2}, C N^{-}, H C l O_{4}, F^{-}, O H^{-}, C O_{3}^{2-}$ and $S^{2-}$

Sol. Acid can donate a proton to form base, which is called conjugate base of that acid. Base can accept a proton to form acid, which is called conjugate acid of that base, e.g.

Q. Assuming complete dissociation calculate the $p H$ of the following solution:

$0.003 M H C l,(\text { ii }) 0.005 M N a O H,$ (iii) $0.002 M H B r$

(iv) $0.002 M K O H$

Sol.

Q. If $0.56 g K O H$ is dissolved in water to give $200 \mathrm{mL}$ of solution at $298 \mathrm{K}$ Calculate the concentration of potassium, hydrogen and hydroxyl ions. What is its $p H ?$

Sol.

Q. The solubility of $\operatorname{Sr}(\mathrm{OH})_{2}$ at $298 \mathrm{K}$ is $19.23 \mathrm{g} / \mathrm{L}$ of solution. Calculate the concentration of strontium and hydroxyl ions and the $p H$ of the solution.

Sol.

Q. Calculate the $p H$ of $0.05 M$ sodium acetate solution if the $p K_{a}$ of acetic acid is 4.74

Sol.

Q. A solution gives the following colour with different indicators methyl orange-yellow, methyl red-yellow, bromothymol blue-orange. What is approximate pH of the solution? Explain giving reason.

Sol. The colour in methyl orange indicates that $p H>4.5$ The colour in methyl red indicates that $p H>6.0$ The colour in bromothymol blue indicates that $p H<6.3$ Therefore, $p H$ of the solution is between 6.0 to 6.3

Q. The colour in methyl orange indicates that $p H>4.5$ The colour in methyl red indicates that $p H>6.0$ The colour in bromothymol blue indicates that $p H<6.3$ Therefore, $p H$ of the solution is between 6.0 to 6.3

Sol.

Q. The ionization constant of nitrous acid is $4.5 \times 10^{-4}$ Calculate the $p H$ of $0.04 M$ sodium nitrite solution and also its degree of hydrolysis.

Sol. Sodium nitrite is a salt of strong base and weak acid. Its degree of hydrolysis, h is given by the relation :

Q. The ionization constant of the $\mathrm{CH}_{3} \mathrm{COOH}$ is

$1.74 \times 10^{-5} .$ Calculate the degree of the dissociation of acetic acid in its $0.05 \mathrm{M}$ solution. Calculate the concentration of the $\mathrm{CH}_{3} \mathrm{COO}^{-}$ and its $p H$

Sol.

Q. $p H$ of $0.01 M$ of an organic acid is $4.15 .$ Calculate the concentration of the anion and $K_{a} \&$ its $p K_{a}$

Sol.

Q. How much volume of $0.1 \mathrm{MCH}_{3} \mathrm{COOH}[\mathrm{HAc}]$ should be added to $50 \mathrm{mL}$ of $0.2 \mathrm{M}$ NaAc (sodium acetate) solution if we want to prepare a buffer solution of $p H=4.91$

Sol.

Q. How much of $0.3 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}$ should be mixed with $30 \mathrm{ml}$ of $0.2 M N H_{4} C l$ to give buffer solution of $p H 8.65$ to 10 $\left(p K_{b} \text { of } N H_{4}^{\pm}=4.75\right)$

Sol. $p K_{a}$ for $N H_{4}^{+}=14-4.75=9.25 \quad\left[\because p K_{b}=4.75\right]$

$p H=p K_{a}+\log \frac{N H_{4} O H}{\left[N H_{4}^{+}\right]}$

$8.65=9.25+\log \frac{\left[N H_{4} O H\right]}{\left[N H_{4}^{+}\right]}$

$\log \frac{\left[N H_{4} O H\right]}{\left[N H_{4} C l\right]}=-0.60+1-1=\overline{1} .40$

$\frac{\left[N H_{4} O H\right]}{\left[N H_{4} C l\right]}=$ antilog $\overline{1} .40=0.25$

$\frac{\left[N H_{4} O H\right]}{\left[N H_{4} C l\right]}=\frac{0.3 \times V / 1000}{0.2 \times 30 / 1000}=0.25$

$V=5.00 \mathrm{mL}$

for $p H=10$

$10=9.25+\log \frac{\left[N H_{4} O H\right]}{\left[N H_{4} C l\right]}=0.75=\log \frac{\left[N H_{4} O H\right]}{\left[N H_{4} C l\right]}$

$\log \frac{\left[N H_{4} O H\right]}{\left[N H_{4} C l\right]}=0.75 ; \frac{\left[N H_{4} O H\right]}{\left[N H_{4} C l\right]}=5.8$

$\frac{0.3 \times V / 1000}{0.2 \times 30 / 1000}=5.8, V=20 \times 5.8=116 \mathrm{mL}$

Q. The ionization constant of acetic acid is $1.74 \times 10^{-5}$ Calculate the degree of dissociation of acetic acid in its $0.05 M$ solution. Calculate the concentration of acetate ion in the solution and its $p H ?$

Sol.

$p H=-\log \left(9.33 \times 10^{-4}\right)=4-0.9699=4-0.97=3.03$

Q. $100 \mathrm{mL}$ of $\frac{N}{10} \mathrm{NaOH}$ solution is mixed with $100 \mathrm{mL}$ of $\frac{N}{5} H C l$ solution and the whole solution is made up to 1 litre. What is the $p H$ of resulting solution?

Sol. Number of meq of $\mathrm{NaOH}=100 \times \frac{1}{10}=10 \mathrm{meq}$

Number of meq of $H C l=100 \times \frac{1}{5}=20$ meq No. of meq of $H C l$ in $200 \mathrm{mL}$ solution on mixing $=10 \mathrm{meq}$ $200 m L$ of solution contains $10 \mathrm{meq}$ $=10 \times 10^{-3} e q=10 \times 10^{-3} \mathrm{moles}$ of $\mathrm{HCl}$

$=\frac{10^{-2}}{200} \times 1000=5 \times 10^{-2} \mathrm{mol} L^{-1}$ of $H C l$

$p H=-\log \left[H_{3} O^{+}\right]=-\log 5 \times 10^{-2}=-\log 5-\log 10^{-2}$

$=-0.6990+2.000=1.3010$

Q. Ionic product of water at $310 K$ is $2.7 \times 10^{-14} .$ What is the neutral $p H$ of water at this temperature?

Sol.

$K_{w}=2.7 \times 10^{-14}$

$K_{w}=\left[H_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=2.7 \times 10^{-14}$

$\left[H_{3} O^{+}\right]=\left[O H^{-}\right] ;\left[H_{3} O^{+}\right]^{2}=2.7 \times 10^{-14}$

$\left[H_{3} O^{+}\right]=\sqrt{2.7} \times 10^{-7} ;\left[H_{3} O^{+}\right]=1.65 \times 10^{-7}$

$\left[H_{3} O^{+}\right]=1.65 \times 10^{-7}$

$p H=-\log 1.65 \times 10^{-7}=-\log 1.65-\log 10^{-7}$

$=-0.217+7.000=6.78$

Q. $K_{b}$ for $N H_{4} O H$ is $1.77 \times 10^{-5}$ at $298 K .$ calculate the hydrolysis constant of $N H_{4} C l$ and $p H$ of $0.04 M N H_{4} C l$ solution.

Sol.

Q. The degree of ionization of a $0.1 M$ bromo acetic acid solution is $0.132 .$ Calculate the $p H$ of solution and the $p K_{a}$ of bromo acetic acid.

Sol.

Degree of dissociation, $\alpha=0.132$

Concentration of $H_{3} \mathrm{O}^{+}$ ion

$\left[H_{3} O^{+}\right]=c, \alpha=0.1 \times 0.132=0.0132 M$

$p H=-\log \left[H_{3} O^{+}\right]$

$=-\log (0.0132)=-(-1.88)=1.88$

The dissociation constant,

$K_{a}=\frac{c \alpha^{2}}{1-\alpha}=\frac{0.1 \times(0.132)^{2}}{1-0.132}=2.01 \times 10^{-3}$

$p K_{a}=-\log K_{a}=-\log \left(2.01 \times 10^{-3}\right)=-0.303+3=2.7$

Q. The $p K_{b}$ of acetic acid and $p K_{a}$ of $N H_{4} O H$ are 4.76

and 4.75 respectively. Calculate the $K_{h}$ of $\mathrm{CH}_{3} \mathrm{COONH}_{4}$ at $298 K$ also the degree of the hydrolysis and $p H$ of its (a) $0.01 \mathrm{Mand}$ (b) $0.04 \mathrm{M}$

Sol.

Q. Calculate the degree of hydrolysis and $p H$ of $0.02 M$ $N H_{4} C N$ solution at $298 K . K_{b}$ for $N H_{4} O H=1.77 \times 10^{5} K_{a}$

for $H C N=4.99 \times 10^{-10}$

Sol.

Q. (i) Define ionic equilibrium giving an example.

(ii) For the reaction $\left[A g(C N)_{2}\right]^{-}$ E $\mathbb{E} \mathbb{Z} \quad A g^{+}+2 C N^{-}$

$K=4.0 \times 10^{-19}$ at $25^{\circ} \mathrm{C} .$ Calculate the silver ion concentration in a solution which was originally $0.10 M$ in $K C N$ and $0.03 M$ in $A g N O_{3}$

Sol.

Q. What is the $p H$ of $0.001 M$ aniline solution? The ionization constant of aniline is $4.27 \times 10^{-10} .$ Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Sol.

Q. Calculate the solubility of $A g C l, F e(O H)_{3}, A g_{2} S O_{4}$ and

$H g_{2} B r_{2} \quad$ from $\quad$ their solubility products. $K_{s p}(A g C l)=1.8 \times 10^{-10}, \quad K_{s p}\left[F e(O H)_{3}\right]=1.0 \times 10^{-38}$

$K_{s p}\left[A g_{2} S O_{4}\right]=1.4 \times 10^{-5}, \quad K_{s p}\left(H g_{2} B r_{2}\right)=5.6 \times 10^{-23}$

Also calculate the solubilities of salts in $g L^{-1}$ and molarities of the ions.

Sol.

Q. Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at $298 \mathrm{K.~Given~} K_{s p}$ values:

$A g_{2} C r O_{4}=1.1 \times 10^{-12}, B a C r O_{4}=1.2 \times 10^{-10}, F e(O H)_{3}$

$=1.0 \times 10^{-38}$

$P b C l_{2}=1.6 \times 10^{-5}, H g_{2} I_{2}=4.5 \times 10^{-29}$

Determine also the molarities of individual ions.

Sol.

Q. (i) State the formula of conjugate base of each of the following acids:

(a) $H_{3} \mathrm{O}^{-}$

(b) $\mathrm{HSO}_{4}^{-}$

(c) $H_{3} P O_{4}$

(d) $\mathrm{CH}_{3} \mathrm{N}^{+} \mathrm{H}_{3}$

(ii) State the formula of conjugate acid of the following

(a) $\mathrm{OH}^{-}$ (b) $\mathrm{CO}_{3}^{2-}$ (c) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$ (d) $\mathrm{HPO}_{4}^{2-}$

(iii) Arranging the following in increasing order of basic IT ;

$\left.F^{-}, B r^{-}, C l^{-}, I^{-} .\right]$

Sol. (i) $(\mathrm{a}) \mathrm{H}_{2} \mathrm{O},(\mathrm{b}) \mathrm{SO}_{4}^{2-},(\mathrm{c}) \mathrm{H}_{2} \mathrm{PO}_{4}^{-},(\mathrm{d}) \mathrm{CH}_{3} \mathrm{NH}_{2}$

(ii) $(\mathrm{a}) \mathrm{H}_{2} \mathrm{O},(\mathrm{b}) \mathrm{HCO}_{3}^{-},(\mathrm{c})\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2},(\mathrm{d})$

(iii) $F^{-}>C l^{-}>B r^{-}>I^{-}$

Q. The first ionization constant of $H_{2} S$ is $9.1 \times 10^{-8} .$ Calculate

the concentration of $H S^{-}$ ion is its $0.1 M$ solution. How will this concentration be effected if the solution is $0.1 M H C l$ also. If the second dissociation constant of

$H_{2} S$ is $1.2 \times 10^{-13}$, calculate the concentration of $S^{-2}$ ion under these conditions.

Sol.

Q. (i) Calculate $p H$ of an aqueous solution of $1.0 \mathrm{M}$ ammonium formate assuming complete dissociation. $\left(p K_{a} \text { of } H C O O H=3.8 \text { and } p K_{b} \text { of } N H_{3}=4.8\right)$

(ii) Saccharine $\left(K_{a}=2 \times 10^{-12}\right)$ is a weak acid represented by formula Hsac. $4 \times 10^{-4} \mathrm{mol}$ of saccharine is dissolved in $200 \mathrm{cm}^{3}$ water $p H=3 .$ Assuming no change in volume, calculate the concentration of $S a c^{-}$ ion in resulting solution.

Sol.

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Q. What is the relationship between the standard EMF of the cell and the equilibrium constant of the cell reaction at 298 K ?

Sol. $E_{c a l l}^{o}=\frac{0.0591}{n} \log K_{c}$

Q. Why is it not possible to measure the voltage of an isolated reduction half reaction or the potential of a single electrode?

[H.S.B. 1995, H.S.B. 2000, P.S.B. 2000]

Sol. The reduction half reaction cannot take place alone.

Q. State the factors that influence the value of cell potential

of the following cell: $M g(s)\left|M g^{2+}(a q) \| A g^{+}(a q)\right| A g(s)$

[A.I.S.B. 2003]

Sol. Concentration of $M g^{2+}$ and $A g^{+}$ ions in the solution and temperature.

Q. How is an electrochemical cell represented by cell notation?

[D.S.B. 1994 Supp.]

Sol. The cell notation of an electro-chemical cell is:

Anode | Anode electrolyte || Cathode electrolyte | Cathode

Q. After sometime, the voltage of an electrochemical cell becomes zero. Comment on the statement

[Pb. Board 2002]

Sol. The statement is correct because the reduction potentials (E’) of the two electrodes become equal at that stage. What is over voltage?

Q. What is over voltage?

Sol. It is the additional voltage required to cause electrolysis.

Q. Which allotropic form of carbon is used for making electrodes?

[H.P. Board 2002]

Sol. Graphite is used for making electrodes as it is a good conductor of electricity.

Q. Write Nernst equation for the following cell reaction:

$Z n(s)\left|Z n^{2+}(a q) \| \quad C u^{2+}(a q)\right| C u(s)$

[Pb. Board 2006]

Sol. $E_{c e l l}=E_{c l l}^{o}-\frac{0.0591}{2} \log \frac{\left[Z n^{2+}(a q)\right]}{\left[C u^{2+}(a q)\right]}$

Q. Can a solution of $\mathrm{ZnSO}_{4}$ be stored in a vessel made up of copper?

Sol. Yes, the solution can be safely stored since copper is placed above zinc in the electrochemical series and is a weaker reducing agent. It cannot lose electrons to $Z n^{2+}(a q)$ ions.

Q. What are the units of molar conductivity?

[A.I.S.B. 1994, H.S.B. 2001]

Sol. $\mathrm{Ohm}^{-1} \mathrm{cm}^{2} \mathrm{mol}^{-1}$ or $\mathrm{S} \mathrm{cm}^{2} \mathrm{mol}^{-1}$

Q. What is the relationship between specific conductance and equivalent conductance?

[P.S.B. 1997, 2000]

Sol. $\Lambda_{\mathrm{eq}}=k \times V$ where $V=$ volume in $\mathrm{cm}^{3}$ containing $1 \mathrm{g} \mathrm{eq}-$

of the substance.

Q. Why is not possible to determine $\wedge_{m}^{\infty}$ for weak electrolytes by extrapolation?

[Haryana Board 2001]

Sol. Because $\wedge_{m}^{\infty}$ for weak electrolytes does not increase linearly with dilution as in case of strong electrolytes.

Q. How is $\wedge_{m}^{\infty}$ of weak electrolytes calculated according to Kohlrausch’s law?

[Haryana Board 2005]

Sol. $\wedge_{m}^{\infty}$ of weak electrolytes is calculated in terms of strong electrolytes. For example,

$\wedge_{m}^{\infty} C H_{3} C O O H=\wedge_{m}^{\infty} C H_{3} C O O N a+\wedge_{m}^{\infty} H C l-\wedge_{m}^{\infty} N a C l$

Q. Which is evolved at cathode when an aqueous solution of NaClis electrolysed?

[D.S.B. 2004 Supp.]

Sol. Hydrogen gas is evolved at cathode.

Q. What is the function of platinised platinum in the standard hydrogen electrode?

[D.S.B. 1996]

Sol. It acts as inert electrode responsible for the electrical conductance in the standard hydrogen electrode.

Q. Which is the basic difference in a primary and secondary cell?

[A.I.S.B. 1997]

Sol. A primary cell cannot be recharged while it is possible to recharge a secondary cell.

Q. Write an expression to relate molar conductivity of an electrolyte to its degree of dissociation.

[Delhi Board 1997, A.I.S.B. 1999]

Sol. $\alpha=\frac{\wedge_{m}^{c}}{\wedge_{m}^{\infty}}$

Q. What will happen to the value of emf of the cell

Zn $|$ Zn $^{2+}(0.1 M) \| C u^{2+}(0.1 M) | C u,$ if the concentration of the electrolyte in the anode compartment is increased

[Delhi Board 2001]

Sol. EMF of the cell increases

Q. What would happen if the protective tin coating over on iron bucket is broken in some places

[Delhi Board 2002]

Sol. If the protective tin coating over an iron bucket is broken at some place then iron corrodes faster than it does in the absence of Sn as oxidation potential of Fe is higher than that of Sn.

Q. Which solution will allow greater conductance of electricity, $1 M N a C l$ at $293 \mathrm{Kor} 1 \mathrm{MNaCl}$ at $323 \mathrm{K}$

[Delhi Board 2003]

Sol. $1 M N a C l$ at $323 K$ as the ionic mobilities increase with increase in temperature.

Q. Give one point of distinction between emf and potential difference

[Haryana Board 1997]

Sol. emf is the difference in the electrode potentials when no current flows through the cell while potential difference is the difference in the electrode potentials when current flows through the cell.

Q. Define the following terms: Specific and equivalent conductance

[H.S.B. Sample Paper 1991, D.S.B. 1992]

Sol. Specific conductance is the conductance of 1 cm3 of the solution whereas equivalent conductance is the conductance of a solution containing one gram equivalent of the electrolyte provided the distance between the electrodes is 1 cm and area of the electrodes is so large that the whole solution is contained between them.

Q. On passing one faraday of charge, on gram mole of the substance is deposited on the cathode. Comment

[Haryana Board 2000]

Sol. The statement is wrong. One gram equivalent of the substance is deposited on passing one Faraday or 96500 C of charge.

Q. Rusting of iron is quicker in saline water than in ordinary water. Give reason

[AISB 2002; CBSE PMT (Mains) 2005]

Sol. In saline water, the presence of $N a^{+}$ and $C l^{-}$ ions increases the conductance of the solution in contact with the metal surface. This accelerates the formation of $F e^{2+}$ ion and hence that of rust, $F e_{2} O_{3} . x H_{2} O$

Saline medium has extra salts such as sodium chloride dissolved in water. This means that it has a greater concentration of electrolyte than ordinary medium. The ions present will favour the formation of more electrochemical cells and will thus promote rusting or corrosion.

Q. What is standard hydrogen electrode? Give the reaction that occurs at this electrode when it acts as positive electrode in an electrochemical cell

[Delhi Board 1996]

Sol. Standard Hydrogen electrode: It is a glass tube broaden at the base. It is provided with a side tube, through which $H_{2}$ gas at one atmosphere is passed. A platinum wire is fused in the tube. A platinum black strip is attached to the lower end of the platinum wire. It is immersed in $1 M H_{3} O^{+}$ ion solution.

Q. What is fuel cell? Give the electrode reactions of hydrogen oxygen fuel cells

[A.I.S.B. 1996,2001,2006; Delhi Board 2002]

Sol. Fuel cells : These are the galvanic cells in which the reactants are fed continuously into the cell and simultaneously the products are continuously removed. A most popular fuel cell is Hydrogen-oxygen fuel cell.

The reactions taking place in hydrogen – oxygen fuel cell:

At cathode : $O_{2}+4 H^{+}+4 e^{-} \longrightarrow 2 H_{2} O$

At anode : $2 H_{2} \longrightarrow 4 H^{+}+4 e^{-}$

Net reaction $: 2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}$

Q. With the help of a graph explain why it is not easy to

determine $\lambda_{m}^{\infty}$ for a weak electrolyte by extrapolating the concentration $-$ molar conductance curve as for strong electrolytes

[Delhi Board 1998]

Sol. Extrapolation of concentration molar concentration curve $\rightarrow$ The curve in case of weak electrolytes is shown below:

Obviously $\wedge_{m}^{\infty}$ means molar conductance at infinite dilution. This can only be obtained by extrapolating the curve. The point at $y$ -axis where it meets of extrapolation is the value of It is apparent from the graph that in case of weak electrolytes the curve does not meet $y$ -axis on extrapolation, hence the value of can not be determined to the contrary in case of strong electrolytes the curve does meet the y-axis on extrapolation.

Q. Write the cell reactions which occur in lead storage battery (i) When the battery is in use and (ii) when the battery is on charging

[Delhi Board 2004]

Sol. (i) The cell reactions during the use (discharge) of lead storage battery are

At anode $: P b(s)+S O_{4}^{2-} \longrightarrow P b S O_{4}(s)+2 e^{-}$

At cathode:

$P b O_{2}+4 H^{+}+S O_{4}^{2-}+2 e^{-} \longrightarrow P b S O_{4}(s)+2 H_{2} O(l)$

Net reaction

$P b+P b O_{2}+4 H^{+}+2 S O_{4}^{2-} \longrightarrow 2 P b S O_{4}(s)+2 H_{2} O(l)$

(ii) During charging of the battery, the above reactions are reversed

Q. Explain why electrolysis of aqueous solution of $N a C l$ gives $H_{2}$ at cathode and $C l_{2}$ anode. Write overall reaction

[Delhi Board 2004,2006]

Sol. Sodium chloride and water ionize as follows

$N a C l(a q .) \longrightarrow N a^{+}(a q .)+C l^{-}(a q .)$

$H_{2} O(l) \rightarrow H^{+}(a q .)+O H^{-}(a q .)$

At cathode: Both $N a^{+}$ and $H^{+}$ ions are present near the cathode. But the discharge potential of is lower than that

of ions. So ions are discharged in preference to $N a^{+}$ ions

$H^{+}+e^{-} \longrightarrow H ; H+H \longrightarrow H_{2}(g)$

Thus, $\mathrm{H}_{2}$ gas is liberated at the cathode and $\mathrm{Na}^{+}$ ions remain in the solution

At the anode: Both $C l^{-}$ and $O H^{-}$ ions are present near

the anode. As the discharge potential of $C l^{-}$ ions is lower

than that of $O H^{-}$ ions, so $C l^{-}$ ions are discharged in

preference to $\mathrm{OH}^{-}$ ions $C l^{-}-e^{-} \longrightarrow C l^{\bullet} ; \quad C l^{\bullet}+C l^{\bullet} \longrightarrow C l_{2}$

Thus $\mathrm{Cl}_{2}$ is liberated at anode and $\mathrm{OH}^{-}$ ions remain in the solution. The overall reaction is

$N a C l(a q .)+H_{2} O(l) \longrightarrow N a^{+}(a q .)+O H^{-}(a q .)+1 / 2 H_{2}(g)$

$+1 / 2 C l_{2}(g)$

Q. What is corrosion? Describe the electrochemical phenomenon of rusting of Iron

[Delhi Board 2005]

Sol. Corrosion is the process of slowly eating away of the metal due to attack of the atmospheric gases on the surface of the metal resulting into the formation of compounds. Such as oxides, sulphides, carbonates, etc. The corrosion of iron is called rusting. According to theory of rusting, impure iron surface behaves as a small electrochemical cell in the presence of According to theory of rusting, impure iron surface behaves as a small electrochemical cell in the presence of

water containing dissolved oxygen or $\mathrm{CO}_{2} .$ The pure iron acts as anode and impure surface as cathode At anode: Iron atoms undergo oxidation spontaneously forming $F e^{2+}$ ions

$F e \longrightarrow F e^{2+}(a q .)+2 e^{-} \quad E_{o x i}^{o}=-0.44 V$

$F e^{2+}$ ions move into solution and electrons into cathodic area where they are picked up by ions of the solution. At cathode:

$2 H^{+}+1 / 2 O_{2}+2 e^{-} \longrightarrow H_{2} O E_{\mathrm{red}}^{\circ}=1.23 \mathrm{V}$

$\mathrm{H}^{+}$ ions are produced by secondary reactions either from $\mathrm{H}_{2} \mathrm{O}$ or from $\mathrm{H}_{2} \mathrm{CO}_{3}\left(\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\right)$

$H_{2} O \longrightarrow H^{+}+O H^{-}$

$H_{2} C O_{3} \longrightarrow H^{+}+H C O_{3}^{-}$

The overall reaction of the corrosion cell may be represent

$F e(s)+2 H^{+}(a q)+1 / 2 O_{2} \longrightarrow F e^{2+}(a q)+H_{2} O ; E_{c e l l}^{o}=1.67 V$

The Fe $\mathrm{Fe}^{2+}$ ions more through water and come at the surface where these are further oxidised into $F e^{3+}$ ions by atmospheric oxygen to form hydrate ferric oxide known as rust, $F e_{2} O_{3} . . H_{2} O$

\begin{aligned} 2 \mathrm{Fe}^{2+}+1 / 2 \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} & \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}+4 \mathrm{H}^{+} \\ \mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{xH}_{2} \mathrm{O} \longrightarrow & \mathrm{Fe}_{2} \mathrm{O}_{3} \cdot \mathrm{xH}_{2} \mathrm{O} \end{aligned}

Q. The values of $\Lambda_{m}^{\Theta}$ for $N H_{4} C l, N a O H$ and $N a C l$ at infinite are respectively $129.8,248.1$ and 126.4 ohm $^{-1} \mathrm{cm}^{2} \mathrm{mol}^{-1}$ Calculate $\wedge_{m}^{o}$ of $N H_{4} O H$

[MP Board 2005]

Sol.

Q. Calculate the standard free energy change for the reaction occurring in the cell

$Z n(s)\left|Z n^{2+}(1 M) \| C u^{2+}(1 M)\right| C u(s)$

$\left[\text { Given } E_{Z n^{2+} / Z n}^{o}=-0.76 V, E_{C u^{2+} / C u}^{o}\right.$

$\left.=0.34 V ; F=96500 \mathrm{Cmol}^{-1}\right]$

How is it related to equilibrium constant for the reaction

[Delhi Board 1998]