Question

In the reaction $2PC{l}_5\rightleftharpoons PC{l}_4^{+}+PC{l}_6^{-}$, the change in hybridization is from:

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

Answer 1

The correct option is A $s{p}^{3}d$ to $s{p}^3$and $s{p}^3{d}^2$

$PC{l}_5$ is $s{p}^3$d-hybridized with trigonal bipyramidal. In $PC{l}_4^{+}$ phosphorous is $s{p}^3$ - hybridized.
$PC{l}_6^{-}$ is $s{p}^3{d}^2$ hybridized with octahedral shape.