Question

$\text{Xe}{{\text{F}}_{\text{6}}}$ on complete hydrolysis gives:
(A) Xe
(B) $\text{Xe}{{\text{O}}_{\text{2}}}$
(C) $\text{Xe}{{\text{O}}_{\text{3}}}$
(D) $\text{Xe}{{\text{O}}_{\text{4}}}\text{F}$

Answer 1

Hint: Start by writing the electronic configuration of Xenon. Determine in xenon hexafluoride which has hybridization. Accordingly write the reaction of complete hydrolysis and get the answer.


Complete step by step solution:
- Xenon is a noble gas element which belongs to the fifth period.
- Let’s write the electronic configuration of Xenon.
\[Xe=\left[ Kr \right]4{{d}^{10}}5{{s}^{2}}5{{p}^{6}}\]
- In xenon, there are eight electrons in its valence shell. It also has an empty 5d orbital. There are six fluorine atoms in xenon hexafluoride, six covalent bonds are formed. Xenon undergoes $s{{p}^{3}}{{d}^{3}}$ hybridization to form seven $s{{p}^{3}}{{d}^{3}}$ hybrid orbitals. Six $s{{p}^{3}}{{d}^{3}}$ orbitals have one electron each and one $s{{p}^{3}}{{d}^{3}}$ contain a lone pair of electrons. The six unpaired electrons get paired with six electrons from six fluorine atoms to form xenon hexafluoride, $\text{Xe}{{\text{F}}_{\text{6}}}$.
- Due to the presence of one lone pair of electrons, the geometry is distorted octahedral.
- Now, when xenon hexafluoride undergoes complete hydrolysis, it forms xenon trioxide, $\text{Xe}{{\text{O}}_{\text{3}}}$ and hydrofluoric acid, HF.
\[Xe{{F}_{6}}+3{{H}_{2}}O\to Xe{{O}_{3}}+6HF\]
- In xenon trioxide, xenon is $s{{p}^{3}}$ hybridized. There are four $s{{p}^{3}}$ orbitals out of which three $s{{p}^{3}}$ orbitals contain one electron each and one $s{{p}^{3}}$ orbital will contain one lone pair of electrons. Three $s{{p}^{3}}$ orbitals will form a sigma bond with three oxygen atoms. There will be three 5d orbitals having remaining three unpaired electrons of xenon. These three 5d orbitals will laterally overlap with three 2p orbitals of three oxygen atoms to form three pi-bonds. The resulting structure is $\text{Xe}{{\text{O}}_{\text{3}}}$, which will have pyramidal geometry due to the presence of one lone pair of electrons.
- Therefore, $\text{Xe}{{\text{F}}_{\text{6}}}$ on complete hydrolysis gives $\text{Xe}{{\text{O}}_{\text{3}}}$.

- The correct answer is option (C).


Note: Remember that xenon has a stable configuration in its monoatomic state. Xenon hexafluoride, $\text{Xe}{{\text{F}}_{\text{6}}}$ on complete hydrolysis will give xenon trioxide, $\text{Xe}{{\text{O}}_{\text{3}}}$ which has a pyramidal geometry.