Question

Hydrolysis of $Xe{{F}_{6}}$ in a strongly alkaline solution is an auto-redox process.
If true enter 1, else enter 0.

Answer 1

Auto-redox reactions are those in which the same atom undergoes oxidation and reduction at the same time. This is also called a disproportionation reaction.

Complete step by step solution:
A reaction in which the same species undergoes oxidation, as well as reduction, is called a disproportionation reaction. For such redox reactions to occur, the reacting species must contain an element which has at least three oxidation states. Such reactions are called auto-redox reactions.
The reaction of hydrolysis of $Xe{{F}_{6}}$ in strong alkaline solution is given below:
$2Xe{{F}_{6}}+16O{{H}^{-}}\to 8{{H}_{2}}O+12{{F}^{-}}+XeO_{6}^{4-}+Xe+{{O}_{2}}$
There is one reactant in which xenon is $Xe{{F}_{6}}$.
The oxidation state of xenon in $Xe{{F}_{6}}$ is:
The oxidation state of fluorine is -1. So,
$x\text{ + 6(-1) = 0}$
$x=+6$
The oxidation state of xenon in $Xe{{F}_{6}}$ is +6.
In the product side there are 2 molecules in which the xenon atom is present i.e., $Xe$ and $XeO_{6}^{4-}$
The oxidation state of xenon in $XeO_{6}^{4-}$. The oxidation state of oxygen is -2. The overall charge of the molecule is -4. So,
$x\text{ + 6(-2) = -4}$
$x=-4+12$
$x=+8$
The oxidation state of xenon in $XeO_{6}^{4-}$ is +8.
And the oxidation state of Xe is 0 because xenon is in its elementary state.
So, the $Xe{{F}_{6}}$ undergoes oxidation to form $XeO_{6}^{4-}$ because the oxidation number increases from +6 to +8 and the same $Xe{{F}_{6}}$ undergoes reduction because the oxidation number decreases from +6 to 0. Hence, this reaction is an auto-redox reaction or disproportionation reaction.

Hence, the statement “Hydrolysis of $Xe{{F}_{6}}$ in a strongly alkaline solution is an auto-redox process” is true.

Note: The hydrolysis of $Xe{{F}_{6}}$ in strongly alkaline solution is an auto-redox reaction but the hydrolysis of $Xe{{F}_{6}}$ is not a redox reaction because the oxidation number remains the same.
$Xe{{F}_{6}}+3{{H}_{2}}O\to Xe{{O}_{3}}+6HF$