Give the formula and describe the structure of a noble gas species which is isostructural with:
Question: Give the formula and describe the structureof a noble gas species which is isostructural with: (i) $\mathrm{ICl}_{4}^{-}$ (ii) $\mathrm{IBr}_{2}^{-}$ (iii) $\mathrm{BrO}_{3}^{-}$ Solution: (i) $\mathrm{XeF}_{4}$ is isoelectronic with $\mathrm{ICl}_{4}^{-}$and has square planar geometry. (ii)$\mathrm{XeF}_{2}$ is isoelectronic to $\mathrm{IBr}_{2}^{-}$and has a linear structure. (iii) $\mathrm{XeO}_{3}$ is isostructural to $\mathrm{BrO}_{3}^{-}$and has a pyramidal molecular structure....
Read More →A park is in the form of a rectangle 120 m × 100 m.
Question: A park is in the form of a rectangle 120 m 100 m. At the centre of the park there is a circular lawn, The area of park excluding lawn is 8700 m2. Find the radius of the circular lawn. (Use = 22/7). Solution: Let the radius of circular lawn ber. Then, Area of circular lawn $=\pi r^{2}$ It is given that Area of park excluding lawn $=$ Area of rectangle-Area of circular lawn $8700=120 \times 100-\pi r^{2}$ $\pi r^{2}=12000-8700$ $\frac{22}{7} r^{2}=3300$ $r^{2}=\frac{3300 \times 7}{22}$ $...
Read More →Give the formula and describe the structure of a noble gas species which is isostructural with:
Question: Give the formula and describe the structureof a noble gas species which is isostructural with: (i) $\mathrm{ICl}_{4}^{-}$ (ii) $\mathrm{IBr}_{2}^{-}$ (iii) $\mathrm{BrO}_{3}^{-}$ Solution: (i) $\mathrm{XeF}_{4}$ is isoelectronic with $\mathrm{ICl}_{4}^{-}$and has square planar geometry. (ii) XeF2is isoelectronic toand has a linear structure. (iii) XeO3is isostructural toand has a pyramidal molecular structure....
Read More →Which one of the following does not exist?
Question: Which one of the following does not exist? (i)XeOF4(ii)NeF2 (iii)XeF2(iv)XeF6 Solution: NeF2does not exist....
Read More →Arrange the following in the order of property indicated for each set:
Question: Arrange the following in the order of property indicated for each set: (i)F2, Cl2, Br2, I2- increasing bond dissociation enthalpy. (ii)HF, HCl, HBr, HI - increasing acid strength. (iii)NH3, PH3, AsH3, SbH3, BiH3 increasing base strength. Solution: (i)Bond dissociation energy usually decreases on moving down a group as the atomic size increases. However, the bond dissociation energy of F2 is lower than that of Cl2 and Br2. This is due to the small atomic size of fluorine. Thus, the incr...
Read More →Solve the following
Question: How are XeO3and XeOF4prepared? Solution: (i)XeO3can be prepared in two ways as shown. $6 \mathrm{XeF}_{4}+12 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{Xe}+2 \mathrm{XeO}_{3}+24 \mathrm{HF}+3 \mathrm{O}_{2}$ $\mathrm{XeF}_{6}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{XeO}_{3}+6 \mathrm{HF}$ (ii)XeOF4can be prepared using XeF6. $\mathrm{XeF}_{6}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{XeOF}_{4}+2 \mathrm{HF}$...
Read More →With what neutral molecule is
Question: With what neutral molecule is ClOisoelectronic? Is that molecule a Lewis base? Solution: ClOis isoelectronic to ClF. Also, both species contain 26 electrons in all as shown. Total electrons ClO= 17 + 8 + 1 = 26 In ClF = 17 + 9 = 26 ClF acts like a Lewis base as it accepts electrons from F to form ClF3....
Read More →If a square is inscribed in a circle,
Question: If a square is inscribed in a circle, find the ratio of the areas of the circle and the square. Solution: Let ABCD be the square inscribed in a circle of radiusr. Here, OA = OB =r. $\therefore O A^{2}+O B^{2}=A B^{2}$ $\Rightarrow r^{2}+r^{2}=\mathrm{AB}^{2}$ $\Rightarrow 2 r^{2}=\mathrm{AB}^{2}$ Now, area of square $\mathrm{ABCD}=\mathrm{AB}^{2}=2 r^{2}$ Area of circle $=\pi r^{2}$ Now we will find the ratio of area of the circle and the square. $\frac{\text { Area of circle }}{\text ...
Read More →How are xenon fluorides
Question: How are xenon fluorides XeF2, XeF4and XeF6obtained? Solution: XeF2, XeF4,and XeF6are obtained by a direct reaction between Xe and F2. The condition under which the reaction is carried out determines the product....
Read More →Write balanced equations for the following:
Question: Write balanced equations for the following: (i)NaCl is heated with sulphuric acid in the presence of MnO2. (ii)Chlorine gas is passed into a solution of NaI in water. Solution: (i) $4 \mathrm{NaCl}+\mathrm{MnO}_{2}+4 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{MnCl}_{2}+4 \mathrm{NaHSO}_{4}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{Cl}_{2}$ (ii) $\mathrm{Cl}_{2}+\mathrm{NaI} \longrightarrow 2 \mathrm{NaCl}+\mathrm{I}_{2}$...
Read More →What are the oxidation states of phosphorus in the following:
Question: What are the oxidation states of phosphorus in the following: (i)H3PO3 (ii)PCl3 (iii)Ca3P2 (iv)Na3PO4 (v)POF3? Solution: Let the oxidation state of $p$ be $x$. (i)H3PO3 $3+x+3(-2)=0$ $3+x-6=0$ $x-3=0$ $x=+3$ (ii)PCl3 $x+3(-1)=0$ $x-3=0$ $x=+3$ (iii)Ca3P2 $3(+2)+2(x)=0$ $6+2 x=0$ $2 x=-6$ $x=-3$ (iv)Na3PO4 $3(+1)+x+4(-2)=0$ $3+x-8=0$ $x-5=0$ $x=+5$ (v)POF3 $x+(-2)+3(-1)=0$ $x-5=0$ $x=+5$...
Read More →What inspired N. Bartlett for carrying out reaction between
Question: What inspired N. Bartlett for carrying out reaction between Xe and PtF6? Solution: Neil Bartlett initially carried out a reaction between oxygen and $\mathrm{PtF}_{6}$. This resulted in the formation of a red compound, $\mathrm{O}_{2}^{+}\left[\mathrm{PtF}_{6}\right]^{-}$. Later, he realized that the first ionization energy of oxygen ( $1175 \mathrm{~kJ} / \mathrm{mol}$ ) and $\mathrm{Xe}(1170 \mathrm{~kJ} / \mathrm{mol})$ is almost the same. Thus, he tried to prepare a compound with $...
Read More →How can you prepare
Question: How can you prepare Cl2from HCl and HCl from Cl2? Write reactions only. Solution: (i)Cl2can be prepared from HCl by Deacons process. (ii)HCl can be prepared from Cl2on treating it with water....
Read More →Write the reactions of
Question: Write the reactions of F2and Cl2with water. Solution: (i) (ii)...
Read More →Why are halogens coloured?
Question: Why are halogens coloured? Solution: Almost all halogens are coloured. This is because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour....
Read More →Write two uses of
Question: Write two uses of ClO2. Solution: Uses of ClO2: (i)It is used for purifying water. (ii)It is used as a bleaching agent....
Read More →In ∆ABC, D and E are the midpoints of AB and AC, respectively.
Question: In ∆ABC,DandEare the midpoints ofABandAC,respectively. Find the ratio of the areas of ∆ADEand ∆ABC. Solution: It is given that D and E are midpoints of AB and AC.Applying midpoint theorem, we can conclude thatDE∥BC.Hence, by B.P.T., we get: $\frac{A D}{A B}=\frac{A E}{A C}$ Also, $\angle A=\angle A$ Applying SAS similarity theorem, we can conclude that $\triangle A D E \sim \triangle A B C$. Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of t...
Read More →Explain why inspite of nearly the same electronegativity,
Question: Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not. Solution: Both chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume....
Read More →A field is in the form of a circle. A fence is to be erected around the field.
Question: A field is in the form of a circle. A fence is to be erected around the field. The cost fencing would be Rs. 2640 at the rate of Rs. 12 per metre. The, the field is to be thoroughly ploughed at the cost of Re. 0.50 per m2. What is the amount required to plough the field? [Take = 22/7]. Solution: We know that the circumferenceCof a circle of radiusris $C=2 \pi r$ It is given that cost of fencing around the circular field would be Rs.2640 at the rate of Rs.12 per meter. So, $2 \pi r \tim...
Read More →Explain why fluorine forms only one oxoacid,
Question: Explain why fluorine forms only one oxoacid, HOF. Solution: Fluorine forms only one oxoacid i.e., HOF because of its high electronegativity and small size....
Read More →Why are halogens strong oxidising agents?
Question: Why are halogens strong oxidising agents? Solution: The general electronic configuration of halogens is np5, where n = 2-6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron. Hence, they act as strong oxidizing agents....
Read More →In the given figure, DE ∥ BC and DE : BC = 3 : 5.
Question: In the given figure,DE∥BCandDE:BC= 3 : 5. Calculate the ratio of the areas of ∆ADEand the trapeziumBCED. Solution: It is given thatDE∥ BC. $\therefore \angle A D E=\angle A B C$ (Corresponding angles) $\angle A E D=\angle A C B$ (Corresponding angles) Applying AA similarity theorem, we can conclude that $\triangle A D E \sim \triangle A B C$. $\therefore \frac{a r(\Delta A B C)}{a r(\Delta A D E)}=\frac{B C^{2}}{D E^{2}}$ Subtracting 1 from both sides, we get: $\frac{\operatorname{ar}(...
Read More →Why are halogens strong oxidising agents?
Question: Why are halogens strong oxidising agents? Solution: The general electronic configuration of halogens is np5, where n = 2-6. Thus, halogens need only one more electron to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron. Hence, they act as strong oxidizing agents....
Read More →Solve the following
Question: How is SO2an air pollutant? Solution: Sulphur dioxide causes harm to the environment in many ways: 1.It combines with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble. 2.Even in very low concentrations, SO2causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness. 3.It is extremely harmful to pla...
Read More →The area of a circle inscribed in an equilateral
Question: The area of a circle inscribed in an equilateral triangle is $154 \mathrm{~cm}^{2}$. Find the perimeter of the triangle. [Use $\pi=22 / 7$ and $\sqrt{3}=1.73$ ] Solution: It is given that the areaAof circle inscribed in an equilateral triangle is 154 cm2. We know that the areaAof circle inscribed in an equilateral triangle is $A=\pi r^{2}$ Now, we will find the value ofr. Substituting the value of area, $154=3.14 \times r^{2}$ $r^{2}=\frac{154}{3.14}$ $r^{2}=49$ $r=\sqrt{49}$ $r=7 \mat...
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