Question

The $'d'$ orbital involved in the hybridisation in the $PCl_{5}$ molecule is

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

Answer 1

Answer: (B)

$3d_{z^{2}}$

Answer 2

In $PCl_{5},$
No. of hybrid orbitals = no. of a bonds + lone pair
$=5+0=5$
Thus, hybridisation is $sp^{3}d$.
The $d$-orbital involved in the
$sp^{3}d$ hybridisation is $3d_{z^{2}}.$