Question

The sum of the number of lone pairs of electrons on each central atom in the following species is
$[TeB{r}_6{]}^{2-},[Br{F}_2{]}^{+},SN{F}_3$ and $[Xe{F}_3{]}^{-}$
(Atomic numbers: $N=7,F=9,S=16,Br=35,Te=52,Xe=54)$

Answer 1

Answer: 6

Number of electron pairs around the central atom $=\frac{V+M\pm C}{2}$
$\begin{array}{cc}\text{Compounds}& \text{No. of lone pairs on central atom}\\ {\left[TeB{r}_6\right]}^{2-}& \frac{\left(6+6+2\right)}{2}-6=1\\ {\left[Br{F}_2\right]}^{+}& \frac{\left(7+2-1\right)}{2}-2=2\\ SN{F}_3& \frac{6-1+3}{2}-4=0\\ {\left[Xe{F}_3\right]}^{-}& \frac{\left(8+3+1\right)}{2}-3=3\end{array}$
$\therefore$ Sum of number of lone pairs $=1+2+0+3=6$