Question

The sum of the number of lone pairs of electrons on each central atom in the following species is
$[TeBr_{6}]^{2-}, [BrF_{2}]^{+}, SNF_{3}$ and $[XeF_{3}]^{-}$
(Atomic numbers: $N=7, F=9, S=16, Br=35, Te=52, Xe=54)$

Answer 1

Number of electron pairs around the central atom $=\frac{V +M \pm C}{2}$
$\begin{matrix}{\text{Compounds}}&{\text{No. of lone pairs on central atom}} \\ \left[TeBr_{6}\right]^{2-}&\frac{\left(6+6+2\right)}{2}-6=1\\ \left[BrF_{2}\right]^{+}&\frac{\left(7+2-1\right)}{2}-2=2\\ SNF_{3}&\frac{6-1+3}{2}-4=0\\ \left[XeF_{3}\right]^{-}&\frac{\left(8+3+1\right)}{2}-3=3\end{matrix}$
$\therefore $ Sum of number of lone pairs $=1+2+0+3=6$