Question

Write shape, hybridization and no of lone pair of electrons for following: ${\mathrm{CH}}_4$

Answer 1

Lone pair of ${\mathbf{CH}}_{\mathbf{4}}$ molecule

• The atomic number of Carbon $\left(\mathrm{C}\right)$ is $6$.
• The electronic configuration of the carbon atom is as follows:

$\mathrm{C}=1s^{2}2s^{2}2p^2$

• The number of valence electrons of Carbon is $4$.
• Carbon forms four sigma bonds with four hydrogen $\left(\mathrm{H}\right)$ atoms to form a Methane $\left({\mathrm{CH}}_4\right)$ molecule.
• ${\mathrm{CH}}_4$ molecule has no lone pair on the Carbon atom.

Hybridization of ${\mathbf{CH}}_{\mathbf{4}}$ molecule

• In ${\mathrm{CH}}_4$, Carbon is a center atom that forms four sigma bonds with four hydrogen atoms.
• So it requires four hybrid orbitals.
• It will form an $s{p}^3$ hybridized orbital.

Shape of ${\mathbf{CH}}_{\mathbf{4}}$ molecule

• According to VSEPR theory, the order of repulsion between electron pairs is as follows:
• lone pair-lone pair $>$ lone pair-bond pair $>$ bond pair-bond pair repulsions
• It has tetrahedral geometry and shape due to no lone pair of electrons.
• Structure of ${\mathrm{CH}}_4$: