Question

Identify the molecule which undergoes ${\mathrm{sp}}^2$ hybridization.

• A. ${\mathrm{BH}}_3$
• B. ${\mathrm{BeH}}_2$
• C. ${\mathrm{BeF}}_2$
• D. ${\mathrm{BeCl}}_2$

Answer 1

The correct option is A

${\mathrm{BH}}_3$

Explanation for Correct answer:

Option (A) ${\mathrm{BH}}_3$

VSEPR Theory:

• The shape of the molecule is determined by the total number of electron pairs(bonding and nonbonding) around the central atom and the orientation of these electron pairs in the space around the central atom.
• In order to minimize the repulsive forces between them, electron pairs around the central atom, tend to stay as far away from each other as possible.
• Electron pairs around the molecule's central atom can be shared or can be lone pairs. The 'shared pairs' of electrons are also called bond pairs of electrons. The presence of lone pair of electrons on the central atom causes some distortions in the expected regular shape of the molecules.
• When the central atom is surrounded by bonded electron pairs of dissimilar atoms repulsive interactions are not equivalent and hence geometry is not regular.
• When the central atom is surrounded by bonded electron pairs as well as lone pairs which are not involved in bonding, repulsive interactions are not equivalent, and hence molecular geometry will be irregular.
• The strength of repulsions between different electron pairs follows the order, lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.
• According to this theory, the hybridization of molecules can easily find out by using the formula $\mathrm{H}=\frac{1}{2}(\mathrm{v}+\mathrm{m}-\mathrm{c}+\mathrm{a})$, where hybridization is denoted by $\mathrm{H}$, $\mathrm{V}$is the number of electrons in the central atom, $\mathrm{m}$ stands for monovalent atom linked with the central atom, $\mathrm{c}$ is the charge of cation and $\mathrm{a}$ is the charge of anion.

Hybridization of ${\mathrm{BH}}_3$

• The process of Hybridization is the intermixing of atomic orbitals that occurs to form new orbitals of equivalent energy.
• In hybridization not only half-filled orbitals can participate, in certain cases, but even the filled orbitals of the valence shell also participate in hybridization.
• In ${\mathrm{BH}}_3$, the atomic number of Boron$\left(\mathrm{B}\right)$ is 5, the electronic configuration is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^1$
• So, the valence electron of Boron$\left(\mathrm{B}\right)$ atom is $\mathrm{v}=3$, the number of monovalent atoms linked with the central atom $\mathrm{m}=3$ and ${\mathrm{BH}}_3$is a neutral molecule so $\mathrm{c}=0$ and $\mathrm{a}=0$.
• Therefore

$$\begin{gathered} \mathrm{H}=\frac{1}{2}(\mathrm{v}+\mathrm{m}-\mathrm{c}+\mathrm{a}) \\ =\frac{1}{2}(3+3-0+0) \\ =\frac{1}{2}\times 6 \\ =3 \end{gathered}$$

• That is 3 hybridized orbitals are present, out of which 3 are bond pair $(\mathrm{B}-\mathrm{H})$, so ${\mathrm{BH}}_3$ is. ${\mathrm{sp}}^2$ hybridized.
• Hence, Option A ${\mathrm{BH}}_3$ is correct.

Explanation for incorrect answers:

Option (B) ${\mathrm{BeH}}_2$

• In ${\mathrm{BeH}}_2$, the atomic number of Beryllium$\left(\mathrm{Be}\right)$ is 4, the electronic configuration is $1{\mathrm{s}}^{2}2{\mathrm{s}}^2$
• So, the valence electron of Beryllium$\left(\mathrm{Be}\right)$ atom is $\mathrm{v}=2$, the number of monovalent atoms linked with the central atom $\mathrm{m}=2$ and ${\mathrm{BeH}}_2$is a neutral molecule so $\mathrm{c}=0$ and $\mathrm{a}=0$.
• Therefore

$$\begin{gathered} \mathrm{H}=\frac{1}{2}(\mathrm{v}+\mathrm{m}-\mathrm{c}+\mathrm{a}) \\ =\frac{1}{2}(2+2-0+0) \\ =\frac{1}{2}\times 4 \\ =2 \end{gathered}$$

• That is 2 hybridized orbitals are present, so ${\mathrm{BeH}}_2$ is. $\mathrm{sp}$ hybridized.
• Hence, Option B ${\mathrm{BeH}}_2$ is incorrect.

Option (C) ${\mathrm{BeF}}_2$

• In ${\mathrm{BeF}}_2$, the atomic number of Beryllium$\left(\mathrm{Be}\right)$ is 4, the electronic configuration is $1{\mathrm{s}}^{2}2{\mathrm{s}}^2$
• So, the valence electron of Beryllium$\left(\mathrm{Be}\right)$ atom is $\mathrm{v}=2$, the number of monovalent atoms linked with the central atom $\mathrm{m}=2$ and ${\mathrm{BeF}}_2$is a neutral molecule so $\mathrm{c}=0$ and $\mathrm{a}=0$.
• Therefore

$$\begin{gathered} \mathrm{H}=\frac{1}{2}(\mathrm{v}+\mathrm{m}-\mathrm{c}+\mathrm{a}) \\ =\frac{1}{2}(2+2-0+0) \\ =\frac{1}{2}\times 4 \\ =2 \end{gathered}$$

• That is 2 hybridized orbitals are present, so ${\mathrm{BeF}}_2$ is. $\mathrm{sp}$ hybridized.
• Hence, Option C ${\mathrm{BeF}}_2$ is incorrect.

Option (D) ${\mathrm{BeCl}}_2$

• In ${\mathrm{BeCl}}_2$, the atomic number of Beryllium$\left(\mathrm{Be}\right)$ is 4, the electronic configuration is $1{\mathrm{s}}^{2}2{\mathrm{s}}^2$
• So, the valence electron of Beryllium$\left(\mathrm{Be}\right)$ atom is $\mathrm{v}=2$, the number of monovalent atoms linked with the central atom $\mathrm{m}=2$ and ${\mathrm{BeH}}_2$is a neutral molecule so $\mathrm{c}=0$ and $\mathrm{a}=0$.
• Therefore

$$\begin{gathered} \mathrm{H}=\frac{1}{2}(\mathrm{v}+\mathrm{m}-\mathrm{c}+\mathrm{a}) \\ =\frac{1}{2}(2+2-0+0) \\ =\frac{1}{2}\times 4 \\ =2 \end{gathered}$$

• That is 2 hybridized orbitals are present, so ${\mathrm{BeCl}}_2$ is. $\mathrm{sp}$ hybridized.
• Hence, Option D ${\mathrm{BeCl}}_2$ is incorrect.

Therefore, option (A) ${\mathrm{BH}}_3$ is the correct answer.