Question

The hybridization of the chlorine atom in ${\mathrm{CIF}}_3$ molecule is:

• A. $sp$
• B. $s{p}^2$
• C. $s{p}^3$
• D. $s{p}^{3}d$

Answer 1

The correct option is D

$s{p}^{3}d$

Explanation for correct option

(D)$s{p}^{3}d$

Step 1: Hybridization:

• “Hybridization is defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape.”

Step 2: Formula used for calculation:

• Hybridization=$\frac{1}{2}\left(V+M-C+A\right)$
• Where $V$ is the number of valence electrons, $M$ is the monovalent atom attached to the central atom, $C$ is the positive charge and $A$ represents the negative charge.

Step 3: Calculation for Hybridisation of ${\mathrm{CIF}}_3$

• The number of valence electrons in $\text{Cl}$ is $7$ and there are $3$ monovalent atoms.
• Hybridization=$\frac{1}{2}\left(7+3\right)=5$
• The hybridization number is $5$, thus the hybridization of ${\mathrm{CIF}}_3$ is $s{p}^{3}d$.
• Three bond pair and two lone pair is present in ${\mathrm{CIF}}_3$.
• So, the shape of ${\mathrm{CIF}}_3$ is T-shape.

Explanation for incorrect options

The hybridization of the chlorine atom in ${\mathrm{CIF}}_3$ molecule is $s{p}^{3}d$. Hence, the options (A), (B), and (C) are incorrect.

Hence, option (D) is correct. The hybridization of the chlorine atom in ${\mathrm{CIF}}_3$ molecule is $s{p}^{3}d$.