Apart from tetrahedral geometry, another possible geometry for `CH_(4)` is square planar with the four H atoms at the cornerrs of the square and the C atom at its centre. Explain why `CH_(4)` is not square planar?

Explanation

According to VSEPR theory, shared electron pairs around the central atom in a covalent molecule lie apart at the farthest possible distance so as to minimise the electrostatic forces of repulsion between them. For tetrahedral geometry, the bond angle is `109^(@)28'` while for square planar geometry, the bond angleis `90^(@)`. in square planar geometry, repulsions between the bond pairs are greater than that in the tetrahedral geometry. consequently, methane assumes tetrahedral geometry instead of square planar geometry. however, for square planar geometry, required hybridisation is `dsp^(2)` which is not possible for carbon as it has no d-orbitals in the valence shell.