The sum of number of lone pairs of electrons present on the central atoms of XeO 3 , XeOF 4 and XeF 6 is

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 $= \frac{1}{2}\left(\mathrm{V}+\mathrm{H}-\mathrm{C}+\mathrm{A}\right)$

Where V= Number of valance electrons in central atom.

H = Number of surrounding monovalent atoms.

C=  Cationic charge

A = Anionic charge.

${\mathrm{XeF}}_6:$

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

${\mathrm{XeOF}}_4:\frac{8+4}{2}=6$

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

${\mathrm{XeO}}_3$:

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 ${\mathrm{XeO}}_3,{\mathrm{XeOF}}_4 \mathrm{and} {\mathrm{XeF}}_6$ are 3.