Pblock Elementsii Result 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) $3$
Show Answer
Answer:
Correct Answer: ( c )
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 furtl treatment with $\mathrm{OH}^{-}$undergo slow disproportionation reacti and produces $\mathrm{XeO}_6^{4-}$ along with $\mathrm{Xe}(\mathrm{g}), \mathrm{H}_2 \mathrm{O}(\mathrm{l})$ and $\mathrm{O}_2(\mathrm{~g})$ 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
Thus, (c) is the correct answer.
BETA
