Answer
412.2k+ views
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.
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 $90^\circ = 8$
![seo images](https://www.vedantu.com/question-sets/1544637b-a0f6-4b84-9ffc-79c61fb91fdb656824900765437784.png)
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.
![seo images](https://www.vedantu.com/question-sets/2a1b9f9c-fe87-4b49-a88a-42955f2155bb4323859206290939041.png)
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.
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