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

Explanation

Electronic configuration of carbon atom: C : `sigma1s^(2)2s^(2)2p^(2)`
In the excited state, the orbital picture of carbon can be represented as:

Hence. carbon atom undergoes `sp^(3)` hybridisation in `CH_(4)` molecule and takes tetrahedral shape.

ror a square planar shape, the hybridisation of the central atom has to be `dsp^(3)`. However, an atom of carbon does not have d-orbitals to undergo `dsp^(3)` hybridisation. Hence, the structure of `CH_(4)` is tetrahedral.