A $\text{ Xe}{{\text{F}}_{\text{6}}}\xrightarrow{\text{Complete hydrolysis}}\text{P+other product}\xrightarrow{\text{O}{{\text{H}}^{\text{-}}}\text{/}{{\text{H}}_{\text{2}}}\text{O}}\text{Q}\xrightarrow{\text{Slow disproportionation reaction in O}{{\text{H}}^{\text{-}}}\text{/}{{\text{H}}_{\text{2}}}\text{O}}\text{Products }$
Under ambient conditions, the total number of gases released as products in final step of reaction scheme below is:
(A) 0
(B) 1
(C) 2
(D) 3

Hint: $\text{ }Xe{{F}_{6}}\text{ }$ is a powerful fluorinating agent. It is readily hydrolyzed. Hydrolysis leads to formation of xenon oxides. Disproportionation reaction is a reaction in which the same ion or molecule undergoes oxidation and reduction occurs simultaneously. Xenon has 0, $\text{ }+2\text{ }$, $\text{ }+4\text{ }$, $\text{ }+6\text{ }$ and $\text{ }+8\text{ }$ as oxidation states.

Complete Solution:
-Xenon has eight electrons in its valence shell.
-six electrons form six bonds with fluorine and two electrons remain as lone pairs. It has an octahedral structure. It has $\text{ }s{{p}^{3}}{{d}^{3}}\text{ }$ hybridization.

-As $\text{ }Xe{{F}_{6}}\text{ }$ is a powerful fluorinating agent. In $\text{ }Xe{{F}_{6}}\text{ }$, oxidation state of fluorine is $\text{ }-1\text{ }$ and oxidation state of $\text{ }Xe\text{ }$ is $\text{ }+6\text{ }$.
Hydrolysis of $\text{ }Xe{{F}_{6}}\text{ }$ process xenon trioxide and hydrogen fluoride.

-Xenon trioxide reacts with hydroxyl ion to produce $\text{ }HXe{{O}_{4}}^{-}\text{ }$
$\text{ Xe}{{\text{O}}_{\text{3}}}\xrightarrow{\text{O}{{\text{H}}^{\text{-}}}\text{/}{{\text{H}}_{\text{2}}}\text{O}}\text{HXe}{{\text{O}}_{\text{4}}}^{\text{-}}\text{ }$

- The $\text{ }HXe{{O}_{4}}^{-}\text{ }$ undergoes a disproportionation reaction in which xenon undergoes oxidation and reduction simultaneously.
$\text{ }HXe{{O}_{4}}^{-}\to Xe{{O}_{6}}^{4-}+Xe+{{H}_{2}}O+{{O}_{2}}\text{ }$

Oxidation number of xenon is $\text{ }HXe{{O}_{4}}^{-}\text{ }$ is $\text{ }+6\text{ }$ and it undergoes reduction to form $\text{ }Xe\text{ }$ which has oxidation number zero and it undergoes oxidation simultaneously to form $\text{ }Xe{{O}_{6}}^{4-}\text{ }$ in which xenon has oxidation number $\text{ }+8\text{ }$. As xenon and oxygen are produced in gaseous state,

Note: $\text{ }Xe{{F}_{6}}\text{ }$ has seven electron pairs, 6 bonding pairs and 1 lone pair. It undergoes complete hydrolysis to produce xenon trioxide and hydrogen fluoride. It undergoes partial hydrolysis to form oxyfluorides and hydrogen fluoride. $\text{ }Xe{{O}_{3}}\text{ }$ is highly explosive and is a powerful oxidizing agent.