Question

The type of hybridisation of $P$ atom in $PC{l}_5,PC{l}_4^{+}$ and $PC{l}_6^{-}$ is : (respectively)

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

Answer 1

The correct option is B $s{p}^{3}d,s{p}^3,s{p}^2{d}^2$

In $PC{l}_5$ No, of $e^{-}$ pairs around central atom $P=5$
No of lone pairs $=0$ and hybridisation - $s{p}^{3}d$ Shape $=$ trigonal bipyramidal. In $PC{l}_4^{+}-number$. of $e^{-}$ pairs $=4$, lone pairs $=0$, hybridisation $=s{p}^3$ and $shape-tetrahedral$.
In $PC{l}_6^{-}-$ number of $e^{-}$ pairs $=6$, lone pairs $=0,hybridisation=s{p}^3{d}^2$ and shape is octahedral.