# NEET Chemistry The p-Block Elements (XII) Questions Solved

BOARD

(a) Write the balanced chemical equations for obtaining $Xe{O}_{3}$ and $XeO{F}_{4}$ from $Xe{F}_{6}.$

(b) Account for the following :

(i) ${H}_{2}S$ is less acidic than ${H}_{2}Te$.

(ii)${H}_{3}P{O}_{2}$ has reducing nature.

(iii) $S{O}_{2}$ is an air pollutant.

(a) $Xe{F}_{6}+3{H}_{2}O\to Xe{O}_{3}+6HF$

$Xe{F}_{6}+{H}_{2}O\to XeO{F}_{4}+2HF$

(b) (i) In the group with increase in size of the element and increased bond distance, the bond dissociation energy decreases, therefore H-S bond dissociation energy is higher than H-Te and hence H-S bond breaks less easily than H-Te bond and H2S is a weaker acid than ${H}_{2}Te$.

(ii) Because it contains two P-H bonds and thus reduces $AgN{O}_{3}$ to metallic silver

$4AgN{O}_{3}+{H}_{3}P{O}_{2}+2{H}_{2}O\to 4Ag↓+{H}_{3}P{O}_{4}+4HN{O}_{3}$

(iii) $S{O}_{2}$ is a pungent and irritating gas. It acts as an air pollutant due to the following reasons :

1) It causes throat and eye irritation as it is absorbed readily by respiratory tract.

2) It combines with moisture forming sulphurous acid. It is then converted into ${H}_{2}S{O}_{4}$. Both these acids cause acid rain and destroy the marble, corrode metals, deteriorate fabrics, paper, leather etc.

Difficulty Level:

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