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

Explanation

For a square planar geometry, the hybridisation required is `dsp^2`. Carbon atom has configuration `1s^2 2s^2 2p_(x)^(1) 2p_(y)^(1) 2p_(z)^(1)` in the excited state. It does not contain any d-orbital and therefore cannot undergo `dsp^2` hybridisation. Hence, square planar shape is not possible for `CH_4`. In `CH_4`, carbon atom undergoes `sp^3` hybridisation which makes `CH_4` molecule tetrahedral in shape.