Question

Match the species in List-I with the number of lone pairs given in List-II.

Answer 1

$Xe{F}_2$:Total valence electrons (TVE)$=8+2$

$=10$

$5$ pairs in total

$2$bp and hence $3$lp

$\Rightarrow Xe{F}_2$

electrons -16 e(8\times 2)=6 e\rightarrow \quad 3\quad lone\quad pairs$

2)$Xe{F}_4=8+7\left(4\right)=36e-32e(8\times 4)=4e\to 2lonepairs$

3)$Xe{F}_6=8+7\left(6\right)=50e-48e(8\times 6)=2e\to 1lonepair$

4)${Sf}_6=6+7\left(6\right)=48e-48e(8\times 6)=0e\to Nolonepairs$