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?

#### Solution 1

According to VSEPR theory, if CH_{4}were square planar, the bond angle would be 90°. For tetrahedral structure, the bond angle is 109°28′. Therefore, in square planar structure, repulsion between bond pairs would be more and thus the stability will be less.

#### Solution 2

Electronic configuration of carbon atom:

_{6}C: 1*s*^{2} 2*s*^{2} 2*p*^{2}

In the excited state, the orbital picture of carbon can be represented as:

Hence, carbon atom undergoes *sp*^{3} hybridization in CH_{4} molecule and takes a tetrahedral shape.

For a square planar shape, the hybridization of the central atom has to be *dsp*^{2}. However, an atom of carbon does not have *d*-orbitalsto undergo *dsp*^{2} hybridization. Hence, the structure of CH_{4} cannot be square planar.

Moreover, with a bond angle of 90° in square planar, the stability of CH_{4} will be very less because of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral structure for CH_{4}.