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

`C=(Z=6)rArr 1s^(2),2s^(1),2p_(x)^(1),2p_(y)^(1)`.
Hence, it undergoes `sp^(3)` hydbidisation which give tetrahedral shape `(TH)`. For square planar, the hybridisation should be `dsp^(2)`, which is not possible for C-atom in `CH_(4)`, as it does not have `d-`orbitals.