Question

If bond angle in $C{H}_4$ is $x^{{}^{\circ }}{y}^{{}^{\prime }}$ then write your answer as $x$ (Only integer value)

Answer 1

Answer: 109

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.