Question

The hybridization and geometry of ${\mathrm{BrF}}_3$ molecules are:

• A. $s{p}^{3}d$ and T shaped
• B. $s{p}^2{d}^2$ and tetragonal
• C. $s{p}^{3}dandbent$
• D. None of these

Answer 1

The correct option is A

$s{p}^{3}d$ and T shaped

Explanation for correct option

(A)$s{p}^{3}d$ and T shaped

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 ${\mathrm{BrF}}_3$

• The number of valence electrons in $\text{Br}$ 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 ${\mathrm{BrF}}_3$ is $s{p}^{3}d$.
• Three bond pair and two lone pair is present in ${\mathrm{BrF}}_3$.
• So, the shape of ${\mathrm{BrF}}_3$ is T-shape.

Explanation for incorrect option

The hybridization and geometry of ${\mathrm{BrF}}_3$ molecules are$s{p}^{3}d$ and T shaped. Hence, the options (B), (C), and (D) are incorrect.

Hence, option (A) is correct. The hybridization and geometry of ${\mathrm{BrF}}_3$ molecules are$s{p}^{3}d$ and T shaped.