Question

In a change from $PC{l}_3\to PC{l}_5$. The hybrid state of $p$ change from:

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

Answer 1

The correct option is C $s{p}^3$ to $s{p}^{3}d$

As hybridisation is equal to sum of sigma bonds + number of lone pairs. Hence in $P{Cl}_3$ we have 3 sigma bonds and one lone pair so total 4, hence hybrdisation is $s{p}^3$.

Hybridization of $P{Cl}_5$ is $s{p}^{3}d$

The Hybrid state of $p$ changes from $s{p}^3$ to $s{p}^{3}d$