Question

Which one of the following compounds involve in disproportionation reaction with water?

• A. $Xe{F}_2$
• B. $Xe{F}_4$
• C. $Xe{F}_6$
• D. $Xe{O}_2{F}_2$

Answer 1

The correct option is B $Xe{F}_4$

$XeF4$ reacts rapidly with water by undergoing disproportionation.

$6Xe{F}_4+12H_{2}O⟶2Xe{O}_3+4Xe+3O_2+24HF$

$A$ $B$ $C$

Oxidation state of $Xe$ in $6Xe{F}_4$ $\left(A\right)$

$6x-24=0$

$x=4$

Oxidation state of $Xe$ in $2Xe{O}_3$ $\left(B\right)$

$2x-12=0$

$x=6$

Oxidation state of $Xe$ in $C=0$

Disproportionation reaction is that reaction in which a substance is simultaneously oxidized and reduced, giving two different products.

In the reaction of $Xe{F}_4$ with water the oxidation state changes from $+4\left(A\right)$ to $+6\left(B\right)$ and $0\left(C\right)$

Hence, the correct option is B.