If equatorial plane in PCl 5, molecule is the X - Y plane, the orbital hybridising to produce axial bonds are:

Question

If equatorial plane in PCl 5, molecule is the X - Y plane, the orbital hybridising to produce axial bonds are: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/df562bb19669274ab0954267c5fa1291-.png" /> (A) p z, d z2 (B) p x , d xy  (C) p y , p z  (D) dx y, dx2-y2

Tardigrade Admin · Accepted Answer

Bonds labelled as 1,2 and 3 are lying in a single plane. Bonds labelled 4 and 5 are meant to be formed along Z- axis. So, p z and d z2 hybridise to give two hybridised orbitals along the z - axis.

Q. If equatorial plane in $PC l_{5}$ , molecule is the $X - Y$ plane, the orbital hybridising to produce axial bonds are:

$p_{z}, d_{z^{2}}$

$p_{x}, d_{x y}$

$p_{y}, p_{z}$

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

Bonds labelled as $1, 2$ and $3$ are lying in a single plane. Bonds labelled $4$ and $5$ are meant to be formed along $Z -$ axis. So, $p_{z}$ and $d_{z^{2}}$ hybridise to give two hybridised orbitals along the $z -$ axis.

Q. If equatorial plane in PCl5​, molecule is the X−Y plane, the orbital hybridising to produce axial bonds are:

pz​,dz2​

px​,dxy​

py​,pz​

dxy​,dx2−y2​