JEE Main 2022 — Chemistry Inorganic Chemistry

Number of lone pairs on the central atom=Number of hybrid orbitals of the central atom$-$−Number of sigma bonds of the central atom.

Number of hybrid orbitals or

Hybridisation for a molecule is given by $=\frac{1}{2}(V+H-C+A)$=21​(V+H−C+A)

Where V= Number of valance electrons in central atom.

H = Number of surrounding monovalent atoms.

C= Cationic charge

A = Anionic charge.

$${\mathrm{XeF}}_{6}:$$XeF6​:

Number of hybrid orbitals are 7.

Number of sigma bonds formed by the central atom are 6

Number of lone pairs on the central atom=7-6=1 lone pair

$${\mathrm{XeOF}}_{4}:\frac{8+4}{2}=6$$XeOF4​:28+4​=6

Number of hybrid orbitals are 6

Number of sigma bonds formed by the central atom are 5

Number of lone pairs on the central atom=6$-$−5=1

${\mathrm{XeO}}_{3}$XeO3​:

Number of hybrid orbitals=4

Number of sigma bonds formed by the central atom are 3

Number of lone pairs on the central atom=4$-$−3=1

Hence, the total number of lone pairs on the central atom of ${\mathrm{XeO}}_{3},{\mathrm{XeOF}}_{4}\mathrm{and}{\mathrm{XeF}}_{6}$XeO3​,XeOF4​andXeF6​ are 3.