Question

(a) Give reasons :
(i) $H_{3}P{O}_3$ undergoes disproportionation reaction but $H_{3}P{O}_4$ does not.

(ii) When $C{l}_2$ reacts with excess of $F_2,Cl{F}_3$ is formed and not $FC{l}_3.$

(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.

(b) Draw the structures of the following :
(i) $Xe{F}_4$

(ii) $HCl{O}_3$

Answer 1

(a) (i) In $H_{3}P{O}_3$ O.S. of P is +3. So this O.S. can increase due to dispropornation. But, in $H_{3}P{O}_4,$ O.S. of P is +5, which is the highest O.S. of P.

(ii) F cannot expand its covalency as F is a small sized atom, it cannot pick three large sized Cl atoms around it.

(iii) Oxygen has multiple bonding (O = O) whereas Sulphur shows catenation. Due to $p\pi$-$p\pi$ bonding in oxygen, it is in gas form and sulphur forms single bonds, it is in solid form.

$\left(b\right)$
(i)
(ii)