Question

Hybridisation state of $Xe$ in $Xe{F}_2,Xe{F}_4$ and $Xe{F}_6$ respectively are

• A. $s{p}^2,s{p}^{3}d,s{p}^3{d}^2$
• B. $s{p}^{3}d,s{p}^3{d}^2,s{p}^3{d}^3$
• C. $s{p}^3{d}^2,s{p}^{3}d,s{p}^3{d}^3$
• D. $s{p}^2,s{p}^3,s{p}^{3}d$

Answer 1

Answer: (B)

$Xe{F}_2-s{p}^{3}d;$
Total no. of valence electrons $=22$
$\frac{22}{8}=2\left(Q\right)+6\left(R\right),\frac{6}{2}=3\left(Q\right)$
$X=2+3=5$
Hybridisation is $s{p}^{3}d$.
$Xe{F}_4-s{p}^3{d}^2$
Total no. of electrons in outermost shells $=8+28=36$
$\frac{36}{8}=4\left(Q\right)+4\left(R\right),\frac{4}{2}=2\left(Q\right)+0\left(R\right)$
$X=4+2+0=6$
Hybridisation is $s{p}^3{d}^2$.
$Xe{F}_6-s{p}^3{d}^3$
Total no. of valence electrons $=8+42=50$
$\frac{50}{8}=6\left(Q\right)+2\left(R\right),\frac{2}{2}$
$=1\left(Q\right),X=6+1=7$
Hybridisation is $s{p}^3{d}^3.$