Question

Assertion (A): Geometry of $IC{l}_3$ is tetrahedral.
Reason (R): Its shape is T-shape , due to the presence of two lone pairs.
A. Both A and R are true and R is the correct explanation of A
B. Both A and R are true but R is not the correct explanation of A
C. A is true R is false
D. A is false but R is true

Answer 1

Hint: To solve this question use the VSEPR theory and hybridization concept. If the molecule is of T- shape then its hybridization should be $s{p}^{3}d$ in this case there will be three bond pairs and two lone pairs. The formula to find the hybridization is,

 $$H={\displaystyle \frac{1}{2}}(V+H-C+A)$$

Complete step by step solution:
From your chemistry lessons you have read about the VSEPR theory which helps in assigning the shapes and geometry of any compound. To find the shape of any compound, the most important thing is to find the hybridization of that compound. We can find the shape of a molecule through Lewis dot structure also but here we are going to use VSEPR theory to solve this question.
Firstly we should know how to calculate the hybridization and the formula is,

$$H={\displaystyle \frac{1}{2}}(V+H-C+A)$$
Where, V= no. of valence electron a central atom have
H= no. of surrounding monovalent atom
C= cationic charge
A= anionic charge
Now let us the cases one by one, firstly we will go through the assertion
(A). Geometry of $IC{l}_3$ is tetrahedral.
Here, to find the geometry firstly we have to find the hybridization
In $IC{l}_3$ the central atom will be iodine and the surrounding atom are chlorine
No. of valence electron in iodine is 7 and no. of chlorine present is 3
Therefore, hybridization of $IC{l}_3$ = ${\displaystyle \frac{1}{2}}(7+3-0+0)={\displaystyle \frac{10}{2}}=5$
So, the no is 5 and it represents $s{p}^{3}d$ hybridization with a real geometry of trigonal bipyramidal.
Thus the assertion is wrong because $IC{l}_3$ is not tetrahedral in geometry.
(2). Reason (R): Its shape is T-shape , due to the presence of two lone pairs.
The geometry of $IC{l}_3$ should be trigonal bipyramidal due to $s{p}^{3}d$ hybridization but the geometry gets distorted due to the presence of lone pairs. In this compound there are 3 bond pairs and two lone pairs. Thus it shows T- shape instead of trigonal bipyramidal.

Thus, the reason is correct.

Thus, the correct option will be (D).

Note: In $s{p}^{3}d$ hybridization the lone pairs are firstly placed to the equatorial and then the axial position to attain the maximum stability and minimum repulsion thus in this case also the lone pairs are placed at equatorial position. The central metal atom has greater electronegativity than surrounding atoms.