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?

(N/A) Electronic configuration of carbon atom:
$_6C: 1s^2, 2s^2, 2p^2$
In the excited state,the orbital picture of carbon is:
$1s^2, 2s^1, 2p_x^1, 2p_y^1, 2p_z^1$
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 to undergo $dsp^2$ hybridization. Hence,the structure of $CH_4$ cannot be square planar.
Moreover,with a bond angle of $90^{\circ}$ in square planar,the stability of $CH_4$ will be very low because of the repulsion existing between the bond pairs. Hence,$VSEPR$ theory also supports a tetrahedral structure for $CH_4$.