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) According to the $VSEPR$ theory,the geometry of a molecule is determined by the minimization of repulsion between electron pairs in the valence shell of the central atom.
In a square planar geometry,the bond angle would be $90^{\circ}$,which leads to greater repulsion between the bonding electron pairs compared to the tetrahedral geometry where the bond angle is $109^{\circ}28'$.
Therefore,the tetrahedral geometry is more stable as it minimizes the repulsion between the four bonding electron pairs around the central carbon atom.