Question

What is the hybridization of \[{\text{Xe}}\] in cationic part of solid \[{\text{Xe}}{{\text{F}}_{\text{6}}}\] ?
A. \[s{p^3}{d^3}\]
B. \[s{p^3}d\]
C. \[s{p^3}{d^2}\]
D. \[s{p^3}\]

Answer 1

Hint:First write the chemical formula of the cation present. Then, find out the steric number of the central xenon atom in the cation. Then from the steric number, find out the type of hybridisation.

Complete answer:
The solid solid \[{\text{Xe}}{{\text{F}}_{\text{6}}}\] is present as \[{\left[ {{\text{Xe}}{{\text{F}}_{\text{5}}}} \right]^ + }{\left[ {{\text{Xe}}{{\text{F}}_{\text{7}}}} \right]^ - }\].
The cationic part is \[{\left[ {{\text{Xe}}{{\text{F}}_{\text{5}}}} \right]^ + }\] and the anionic part is \[{\left[ {{\text{Xe}}{{\text{F}}_{\text{7}}}} \right]^ - }\].
Now consider the cationic part \[{\left[ {{\text{Xe}}{{\text{F}}_{\text{5}}}} \right]^ + }\].
The atomic number of xenon is 54 and its electronic configuration is \[\left[ {{\text{Kr}}} \right]4{d^{10}}5{s^2}5{p^6}\].
In the outermost shell (the valence shell), a xenon atom has 8 electrons out of which one electron is lost to form the cation. 7 electrons are remaining.
Xenon atoms form five bonds with five fluorine atoms by sharing its five electrons. A Xenon atom has five bond pairs of electrons and one lone pair of electrons. The steric number of xenon in the cation \[{\left[ {{\text{Xe}}{{\text{F}}_{\text{5}}}} \right]^ + }\] is 6.
The type of hybridization associated with the steric number of 6 is \[s{p^3}{d^2}\].
The electron pair geometry is octahedral and the ionic geometry is distorted square pyramidal.
Thus, the hybridization of \[{\text{Xe}}\] in cationic part of solid \[{\text{Xe}}{{\text{F}}_{\text{6}}}\] is \[s{p^3}{d^2}\].

Hence, the correct option is the option C.

Note:
The steric number is the total number of bond pairs and lone pairs of electrons present around the central atom in the molecule / ion. Thus, if 5 bond pairs and 1 lone pair of electrons are present, the steric number is \[5 + 1 = 6\] .