Question

The hybridization of carbon in diamond, graphite, and acetylene is respectively:

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

Answer 1

The correct option is C

$s{p}^3,s{p}^2,sp$

Explanation for correct option

(C)$s{p}^3,s{p}^2,sp$

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.”

Hybridization in diamond

• Each carbon atom in a diamond undergoes $s{p}^3$ hybridisation and is bonded to four other carbon atoms in a tetrahedral fashion.
• The structure expands throughout space, forming a hard three-dimensional carbon atom network.
• Throughout the lattice, directional covalent bonds are present.

Hybridization in graphite

• In a hexagonal ring, the hybridization of each carbon atom is $s{p}^2$.
• $\mathrm{\pi }$ bond is formed by the fourth electron. The electrons are delocalized over the whole sheet.
• Graphite is extremely slippery and soft because it cleaves easily between layers.

Hybridization in acetylene

• The structure of acetylene is $\mathrm{CH}\equiv \mathrm{CH}$.
• In acetylene, one carbon combines with other carbon atoms with three bonds having two pi bonds, and one sigma bond.
• The hybridization of carbon in acetylene is $sp$.

Explanation for incorrect option

The hybridization of carbon in diamond, graphite, and acetylene is respectively $s{p}^3,s{p}^2,sp$. Hence, the options (A), (B), and (D) are incorrect.

Hence, option(C) is correct. The hybridization of carbon in diamond, graphite, and acetylene is respectively $s{p}^3,s{p}^2,sp$.