Question

$\text{PC}{\text{l}}_5$ has a shape of trigonal bipyramidal whereas $\text{I}{\text{F}}_5$ has the shape of a square pyramid. It is due to
A. presence of an unshared electron pair on I which is oriented so as to minimise the repulsion while in P in $\text{PC}{\text{l}}_5$ has no unshared pair.
B. octet of P is complete while that of I is incomplete.
C. P and I are of different groups.
D. F and Cl have different degrees of repulsion.

Answer 1

Hint: Phosphorus has five electrons in the outermost shell and it can make five bonds with other molecules whereas iodine has 7 electrons and has a tendency to accept a pair of electrons so it forms mainly a single bond.

Complete answer:
-In $\text{PC}{\text{l}}_5$, the electronic configuration of phosphorus is $\text{1}{\text{s}}^2\text{ 2}{\text{s}}^2\text{ 2}{\text{p}}^6\text{ 3}{\text{s}}^2\text{ 3}{\text{p}}^3$ to form five bonds with chlorine electron form s-orbital is excited to the vacant d-orbital due to which phosphorus gains the ability to make 5 bonds.
-Now, it has a total of five unpaired electrons.
-So, the hybridisation of phosphorus is $\text{s}{\text{p}}^3\text{d}$ and the geometry will be trigonal bipyramidal.
-Because it has 5 bond pairs and 0 lone pairs.

-In $\text{I}{\text{F}}_5$, according to the electronic configuration of iodine two electrons from 5p-orbital are excited to move to the 5d-orbital.
-So, now iodine tends to make 5 bonds and along with it have one lone pair.
-The lone pair will cause the repulsion towards the bond pair and increases the bond angles that’s why it is oriented in such a way that it causes less repulsion.
-So, the hybridisation will be $\text{s}{\text{p}}^3{\text{d}}^2$and the geometry will be square pyramidal.
-The structure is:

So, the correct answer is “Option A”.

Note: The lone pair- lone pair repulsion causes maximum repulsion than lone pair-bond pair and lone pair-bond pair repulsion is more than the bond pair-bond pair repulsion.