Question

In the structure of $ClF_{ 3 }$, the number of lone pairs of electrons of central atoms Cl is:
A. Four
B. One
C. Three
D. Two

Answer 1

Hint: To answer this question first we need to know the hybridization of this compound and then we can easily find the lone pairs present in this compound.

Complete step by step answer:
You can write the formula for the number of electrons as,
Number of electrons = $\dfrac { 1 }{ 2 } [V+N-C+A]$
where,
V = number of valence electrons present in the central atom
N = number of monovalent atoms bonded to the central atom
C = charge of cation
A = charge of the anion
Now we have to determine the hybridization of the ClF₃ molecule.
Number of electrons = $\dfrac { 1 }{ 2 } [7+3]$ = 5
The number of electrons is 5 that means the hybridization will be $sp^{ 3 }d$ and we can say electronic geometry of the molecule will be trigonal bipyramidal.

Here Cl is attached to three F atoms, so,
Bond pair electrons = 3
Lone pair electrons = 5 - 3 = 2

Hence, the number of lone pairs of electrons present on the central atom Chlorine is two.

Therefore, we can conclude that the correct answer to this question is option D.

Note: We should know that $ClF_{ 3 }$ molecular geometry is said to be a T-shaped. It has such shape because of the presence of these two lone pairs at equatorial positions and there are greater repulsions.
As we have already discussed, the electron geometry of this compound is trigonal bipyramidal with a 175$^{ o }$ F-Cl-F bond angle.