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

Electronic configuration of carbon atom:
`""_(6)V:1s^(2)2s^(2)2p^(2)`
In the excited state, the orbital picture of carbon can be represented as :

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

For a square planar shape, the hybridization of the central atom has to be `dsp^(2)`. However, an atom of carbon does not have d-orbitals undergo `dsp^(2)` hybridization. Hence, the sturcture of `CH_(4)` cannot be square planar.
Moreover, with a bond angle of `90^(@)` in square planar, the stability of `CH_(4)` will be very less because of the repulsion existing between the bond pairs. Hence , VSEPR therory also supports a tetrahedral structure for `CH_(4)`.