Question
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 centre. Explain why $\mathrm{CH}_4$ is not square planar?
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Answer
Electronic configuration of carbon atom:
${ }_6 \mathrm{C}: 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2$
In the excited state, the orbital picture of carbon can be represented as:
Hence, carbon atom undergoes $s p^3$ hybridization in $\mathrm{CH}_4$ molecule and takes a tetrahedral shape.
For a square planar shape, the hybridization of the central atom has to be $d s p^2$. However, an atom of carbon does not have $d$-orbitalsto undergo $d s p^2$ hybridization. Hence, the structure of $\mathrm{CH}_4$ cannot be square planar.
Moreover, with a bond angle of $90^{\circ}$ in square planar, the stability of $\mathrm{CH}_4$ will be very less because of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral structure for $\mathrm{CH}_4$.
${ }_6 \mathrm{C}: 1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2$
In the excited state, the orbital picture of carbon can be represented as:
Hence, carbon atom undergoes $s p^3$ hybridization in $\mathrm{CH}_4$ molecule and takes a tetrahedral shape.
For a square planar shape, the hybridization of the central atom has to be $d s p^2$. However, an atom of carbon does not have $d$-orbitalsto undergo $d s p^2$ hybridization. Hence, the structure of $\mathrm{CH}_4$ cannot be square planar.
Moreover, with a bond angle of $90^{\circ}$ in square planar, the stability of $\mathrm{CH}_4$ will be very less because of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral structure for $\mathrm{CH}_4$.
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