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

The tetrahedral and square planar structures of `CH_(4)` are shown as follows :

According to VSEPR theory, the shared electron pairs aroung the central atom in a covalent molecule are so arranged in space that the force of repulsion in them is the minimum. Now, in a square planar geometry, the bond angle is `90^(@)` while it is `109^(@)-28` in tetrahedral geometry. This clearly show that the electron repulsion are less in the tetrahedral geometry as compared to the square planat geometry. Thus, methanel cannot be repressented by square planar structure.