The 'd' orbital involved in the hybridisation in the PCl5 molecule is

Question

The 'd' orbital involved in the hybridisation in the PCl5 molecule is (A)  3dx2-y2  (B)  3dz2  (C)  3dxy  (D)  4dx2-y2

Tardigrade Admin · Accepted Answer

In PCl5, No. of hybrid orbitals = no. of a bonds + lone pair =5+0=5 Thus, hybridisation is sp3d. The d-orbital involved in the sp3d hybridisation is 3dz2.

Q. The $^{'} d^{'}$ orbital involved in the hybridisation in the $PC l_{5}$ molecule is

$3 d_{x^{2} - y^{2}}$

$3 d_{z^{2}}$

$3 d_{x y}$

$4 d_{x^{2} - y^{2}}$

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

Q. The ′d′ orbital involved in the hybridisation in the PCl5​ molecule is

3dx2−y2​

3dz2​

3dxy​

4dx2−y2​