Question

Match the following:
$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{a.}& POC{l}_3& 1.& \text{Oxyacid formed during hydrolysis}\\ & & & \text{undergoes a tautomeric change}\\ \text{b.}& SO{F}_2& 2.& \text{Oxidation state of the central atom does}\\ & & & \text{not change during hydrolysis}\\ \text{c.}& XeO{F}_4& 3.& \text{Complete as well as partial hydrolysis is}\\ & & & \text{possible}\\ \text{d.}& H_2{S}_2{O}_8& 4.& \text{Hydrolysed product reacts with glass}\\ & & 5.& \text{Hybridization of the central atom in the}\\ & & & \text{final product remains the same as the}\\ & & & \text{substrate on hydrolysis}\end{array}$

• A. a - 2,5
  b - 1,2,4,5
  c - 2,3,4
  d - 2,3,5
• B. a - 2,3,4
  b - 2,3,5
  c - 2,5
  d - 1,2,4,5
• C. a - 1,2,4,5
  b - 2,5
  c - 2,3,5
  d - 2,3,4
• D. a - 2,3,5
  b - 2,3,4
  c - 1,2,4,5
  d - 2,5

Answer 1

The correct option is A a - 2,5

b - 1,2,4,5
c - 2,3,4
d - 2,3,5
(a) $POC{l}_3+3H_{2}O\to H_{3}P{O}_4+3HCl$
The oxidation state and the hybridisation of $P$ is the same in $POC{l}_3$ and $H_{3}P{O}_4$ which is +5 and $s{p}^3$ respectively.

(b) $SO{F}_2+H_{2}O\to S{O}_2+HF$
The oxidation state of $S$ is same in $SO{F}_2$ and $S{O}_2$ which is +4 but the hydridisation in $SO{F}_2$ is $s{p}^3$ and that of $S{O}_2$ is $s{p}^2$.
Hydrolysis product $HF$ reacts with glass $Si{O}_2$ to give $Si{F}_4$

(c) Partial hydrolysis:
$XeO{F}_4+H_{2}O\to Xe{O}_2{F}_2+2HF$
complete hydrolysis:
$Xe{O}_2{F}_2+H_{2}O\to Xe{O}_3+2HF$
Oxidation state of $Xe$ remain unchanged which is +6
during partial hydrolysis hybridisation of Xe is unchanged which is $s{p}^{3}d$
Hydrolysis product $HF$ reacts with glass $Si{O}_2$ to give $Si{F}_4$

(d)$H_2{S}_2{O}_8+H_{2}O\to H_2{S}_2{O}_5+H_{2}S{O}_4$
Hybridisation of $S$ remain unchanged which is $s{p}^3$
Oxidation state of sulphur remain unchanged which is +6