Page No 208: - Chapter 7 - P - Block Elements - Exercise Solutions - Ncert Solutions class 12 - Chemistry

Share this to Friend on WhatsApp

Question 7.32:

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.

Answer:

(i) 

(ii) 

Question 7.33:

How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained?

Answer:

XeF₂, XeF_4, and XeF₆ are obtained by a direct reaction between Xe and F2. The condition under which the reaction is carried out determines the product.

Question 7.34:

With what neutral molecule is ClO⁻ isoelectronic? Is that molecule a Lewis base?

Answer:

ClO⁻ is 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.

Question 7.35:

How are XeO₃ and XeOF₄ prepared?

Answer:

(i) XeO₃ can be prepared in two ways as shown.

(ii) XeOF₄ can be prepared using XeF₆.

Question 7.36:

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.

Answer:

(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 increasing order for bond dissociation energy among halogens is as follows:

I₂ < F₂ < Br₂ < Cl₂

(ii) HF < HCl < HBr < HI

The bond dissociation energy of H-X molecules where X = F, Cl, Br, I, decreases with an increase in the atomic size. Since H-I bond is the weakest, HI is the strongest acid.

(iii) BiH₃ ≤ SbH₃ < AsH₃ < PH₃ < NH₃

On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom decreases. Thus, the basic strength decreases.

Question 7.37:

Which one of the following does not exist?

(i) XeOF₄ (ii) NeF₂

(iii) XeF₂ (iv) XeF₆

Answer:

NeF₂ does not exist.

Question 7.38:

Give the formula and describe the structure of a noble gas species which is isostructural with:

(i) 

(ii) 

(iii) 

Answer:

(i)

XeF₄ is isoelectronic with   and has square planar geometry.

(ii)

XeF₂ is isoelectronic to   and has a linear structure.

(iii)

XeO₃ is isostructural to  and has a pyramidal molecular structure.

Question 7.39:

Why do noble gases have comparatively large atomic sizes?

Answer:

Noble gases do not form molecules. In case of noble gases, the atomic radii corresponds to van der Waal’s radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, van der Waal’s radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period.

Question 7.40:

List the uses of Neon and argon gases.

Answer:

Uses of neon gas:

(i) It is mixed with helium to protect electrical equipments from high voltage.

(ii) It is filled in discharge tubes with characteristic colours.

(iii) It is used in beacon lights.

Uses of Argon gas:

(i) Argon along with nitrogen is used in gas-filled electric lamps. This is because Ar is more inert than N.

(ii) It is usually used to provide an inert temperature in a high metallurgical process.

(iii) It is also used in laboratories to handle air-sensitive substances.