Question

The type of hybridisation and number of lone pair(s) of electrons of Xe in $XeO{F_{4}}\;$ respectively, are:
A.$s{p^3}d\,$and 1
B.$s{p^3}d\,$and 2
C.$s{p^3}{d^2}\,$and 1
D.$s{p^3}{d^2}\,$and 2

Answer 1

To answer this question, you should recall that Xenon is a noble gas and when bonding it will have the tendency to use all the eight electrons in the outermost shell to form bond pairs. We know fluorine is a halogen and can form one sigma bond with the central atom and oxygen can form a double bond with the central atom. Now, use this information to answer the question.

Complete step by step answer:
Hybridization is defined as the concept of mixing two atomic orbitals with the same energy levels to give a degenerated new type of orbitals. This intermixing is based on quantum mechanics. $XeO{F_{4}}\;$has square pyramidal geometry. The central ${\text{Xe}}$atom has one lone pair electrons and five bonding domains where it donates its outermost eight electrons to four sigma bonded fluorine and one double-bonded oxygen. This results in $s{p^3}{d^2}$ hybridisation. The geometry or shape is octahedral and the arrangement of electrons around the central atom in the molecule is square pyramidal. The structure can be drawn as:

Hence, the correct answer to this question is option C.

Note:
Even if you are not able to calculate the hybridisation using the above-mentioned you can find the hybridization $(X)$ using the formula: $\frac{1}{2}(V + H - C + A)$ where
$V$= Number of valence electrons in the central atom
$H$= Number of surrounding monovalent atoms
$C$= Cationic charge
$A$= Anionic charge. The value of X will determine the hybridisation of the molecule. If $X$is 2 then $sp$; is 3 then $s{p^2}$ ; is 4 then $s{p^3}$; is 5 then $s{p^3}d$ ; is 6 then $s{p^3}{d^2}$ ; is 7 then $s{p^3}{d^3}$ hybridization.