Ncert Exercises & Solutions

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 $s p^{3}$ 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-orbitals to 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 $_{4}$ .

4.22 Explain why BeH $_{2}$ molecule has a zero dipole moment although the Be–H bonds are polar.

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 $s p^{3}$ 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-orbitals to 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 $_{4}$ .

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