Question

In $ICl_4^ \ominus $ the shape is square planar. The number of bond pair-lone repulsion at $90^\circ $ are:
(A). $6$
(B). $8$
(C). $12$
(D). $4$

Answer 1

Hint:
We know that $ICl_4^ \ominus $ is $A{B_4}{E_2}$ type molecule and it shows:
Electronic Structure: $A{B_4}{E_2}$
Electronic Geometry: Octahedral
Hybridization of central atom is $s{p^3}{d^2}$

Complete step by step answer:
There are two lone pairs of electrons which are perpendicular to the square plane. In the square plane there are $8$ lone pair electrons.
Thus, repulsion at $90^\circ = 8$

Therefore, the correct option is (B) $8$.

Additional Information:
In the lewis structure of $ICl_4^ \ominus $ there are total $36$ valence electrons.
Since Iodine $\left( I \right)$ is below period $3$ on the periodic table, it can have more than $8$ electrons. In the lewis structure of $ICl_4^ \ominus $ the iodine atom has $12$ valence electrons.
$3 - D$ structure.

Important Points:
In $ICl_4^ \ominus $ lewis structure, Iodine $\left( I \right)$ has the least electronegativity and goes in the center of lewis structure.
The $ICl_4^ \ominus $ lewis structure you’ll need to put more than eight electrons on the iodine atom.
In the lewis structure for $ICl_4^ \ominus $, there are a total of $36$ electrons.

Note: Note that you should put $ICl_4^ \ominus $ lewis structure in brackets with $ - 1$ charge outside to show that it is an ion with negative one charge.