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Apart from tetrahedral geometry, another possible geometry for \mathrm{CH}_4  is square planar with the four H atoms at the corners of the square and the C atom at its center. Explain why \mathrm{CH}_4 is not square planar.

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The electronic configuration of carbon atoms is C: 1 s^2 2 s^2 2 p^2.

Where it has s-orbital, p-orbital only and there is no d-orbital present.

Hence the carbon atom undergoes s p^3 hybridization in methane molecule and takes a tetrahedral shape.

And for a molecule to have a square planar structure it must have d orbital present.

But here the absence of d-orbital, as a result, does not undergo d s p^2 hybridization, the structure of methane cannot be square planar.

Also, the reason that the bond angle in square planar 90^{\circ}  makes the molecule more unstable is because of the repulsion between the bond pairs.

Hence according to VSEPR theory, \mathrm{CH}_4 molecules take a tetrahedral structure.

Posted by

Divya Prakash Singh

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