Question

Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar?

Answer 1

Electronic configuration of carbon atom:
$_6C$: $1s^ 2 2s^ 2 2p^ 2$
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 $d-orbit$ alsto undergo $dsp ^2$ hybridization. Hence, the structure of $CH_4$ cannot be square planar.
Moreover, with a bond angle of $90\degree$ in square planar, the stability of $CH_4$ will be very less because of the repulsion existing between the bond pairs.

Hence, VSEPR theory also supports a tetrahedral structure for $CH_4$.