Chapter 7, Problem 1SAQ

Determine the hybridization about 0 in CH₃OH.

Interpretation Introduction

Interpretation:

Find the hybridization of Oxygen (O) in ${\text{CH}}_{\text{3}}\text{OH}$.

Concept introduction:

For molecule with two or more than two atoms, the formation of hybrid orbital takes place by the mixing of atomic orbitals of atoms possessing nearly same energy in free state. The mixing of orbitals takes place by the overlapping of orbitals of each atom in the molecule, which result in the formation of bonds. The formed hybrid orbitals possess energies that are different from the energy of atomic orbitals of parent atoms. The hybrid orbitals which possess less energy compared to the atomic orbitals of parent atoms result in formation of sigma molecular orbitals whereas hybrid orbitals which possess more energy compared to the atomic orbitals of parent atoms result in formation of pi molecular orbitals.

Answer to Problem 1SAQ

Option (d) is the correct answer. Because, the Oxygen (O) hybridization of ${\text{CH}}_{\text{3}}\text{OH}$ is ${\text{sp}}^3$.

Explanation of Solution

Reason for the correct option:

The electronic configuration of O:

  ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}$

The oxygen can form two bonds with this electronic configuration.

In ${\text{CH}}_{\text{3}}\text{OH}$, this $\text{O}$ forms two single bonds one with $\text{C}$ atom and one with $\text{H}$. ${\text{2p}}_{\text{y}}$ and ${\text{2p}}_{\text{z}}$ are used in this type of the bonding thus, they have $\text{1}$ electron each. $\text{H}$ and $\text{C}$ each contribute $\text{1}$ electron to fill this two half of the p-orbitals of Oxygen($\text{O}$). So the remaining two pairs of electrons are present in $\text{O}$ in the remaining orbitals that is ${\text{2p}}_{\text{x}}{}^{\text{2}}$ and ${\text{2s}}^2$.

The hybridization of $\text{O}$ is ${\text{sp}}^3$ because it forms $\text{2}$ bonds and has $\text{2}$ lone pairs. Its $\text{2s}$ and $\text{2p}$ orbitals are involved in the bonding.