pBlock ElementsII 2 Question 10
10. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is
(2014 Adv.)
(a) 0
(b) 1
(c) 2
(d) 3
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Solution:
- PLAN This problem can be solved by using concept involved in chemical properties of xenon oxide and xenon fluoride.
$\mathrm{XeF}{6}$ on complete hydrolysis produces $\mathrm{XeO}{3}$.
$\mathrm{XeO}{3}$ on reaction with $\mathrm{OH}^{-}$produces $\mathrm{HXeO}{4}^{-}$which on further treatment with $\mathrm{OH}^{-}$undergo slow disproportionation reaction and produces $\mathrm{XeO}{6}^{4-}$ along with $\mathrm{Xe}(g), \mathrm{H}{2} \mathrm{O}(l)$ and $\mathrm{O}_{2}(g)$ as a by-product.
Oxidation half-cell in basic aqueous solution
$$ \mathrm{HXeO}{4}^{-}+5 \mathrm{OH}^{-} \longrightarrow \mathrm{XeO}{6}^{4-}+3 \mathrm{H}_{2} \mathrm{O}+2 e^{-} $$
Reduction half-cell in basic aqueous solution
$$ \mathrm{HXeO}{4}^{-}+3 \mathrm{H}{2} \mathrm{O}+6 e^{-} \longrightarrow \mathrm{Xe}+7 \mathrm{OH}^{-} $$
Balanced overall disproportionation reaction is
$$ 4 \mathrm{HXeO}{4}^{-}+8 \mathrm{OH}^{-} \longrightarrow \underbrace{3 \mathrm{XeO}{6}^{4-}+\mathrm{Xe}}{2 \text { products }}+6 \mathrm{H}{2} \mathrm{O} $$
Complete sequence of reaction can be shown as
$\mathrm{XeF}{6}+3 \mathrm{H}{2} \mathrm{O} \longrightarrow \mathrm{XeO}{3}+3 \mathrm{H}{2} \mathrm{~F}_{2}$
Thus, (c) is the correct answer.
