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Apart from tetrahedral geometry, another...

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 ?

Text Solution

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`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.
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