Question

If bond angle in $CH_{4}$ is $x^{^\circ }y^{'}$ then write your answer as $x$ (Only integer value)

Answer 1

Electronic configuration of carbon
$\text{In ground state :} _{6} \text{C} - 1 \text{s}^{2} \text{,} 2 \text{s}^{2} \text{,} 2 \text{p}_{x}^{1} \text{,} 2 \text{p}_{y}^{1}$
In CH₄ molecule, carbon is sp³ hybridized, so it is tetrahedral in shape. For square planner dsp² hybridization is required which is not possible in carbon due to absence of d-orbitals. Furthermore according to VSEPR theory, the four bonded electron pairs around carbon atom arranged themselves in a regular tetrahedron geometry. For tetrahedral structure, the bond angle is 109°28' while in square planar structure, the bond angle is 90°. Therefore, in tetrahedral structure repulsion between bonded electron pairs is less than that of the square planner.