Question

The d-orbitial involved in the hybridisation in the $PC{l}_5$ molecule is

• A. $3d_{x^2-y^2}$
• B. $3d_{z^2}$
• C. $3d_{xy}$
• D. $4d_{x^2}$

Answer 1

Answer: (B)

$3d_{z^2}$

In $PC{l}_5,$

Number of hybride orbitals $=$ number of $\sigma -$ bonds $+$ number of lone pairs $=5+0=5$

Thus, hybridisation is $s{p}^{3}d$. The d-orbital involved in the $s{p}^{3}d$ hybridisation is $3d_{z^2}$ because of its proper orientation.