The dipole moments of CCl_{4},CHCl_{3} and CH | vedclass

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Question

(A) $CH_{4}$ is a symmetrical tetrahedral molecule with four identical $C-H$ bonds,so its net dipole moment $\mu_{net} = 0$.$CCl_{4}$ is also a symmet

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$CH_{4} = CCl_{4} < CHCl_{3}$

Vedclass · Answer

(A) $CH_{4}$ is a symmetrical tetrahedral molecule with four identical $C-H$ bonds,so its net dipole moment $\mu_{net} = 0$.$CCl_{4}$ is also a symmetrical tetrahedral molecule with four identical $C-Cl$ bonds,so its net dipole moment $\mu_{net} = 0$.$CHCl_{3}$ is an asymmetrical molecule where the dipole moments of the $C-Cl$ bonds and $C-H$ bond do not cancel each other out,resulting in a non-zero net dipole moment $(\mu_{net} 
eq 0)$.Therefore,the order of dipole moments is $CH_{4} = CCl_{4} < CHCl_{3}$.

The dipole moments of $CCl_{4}$,$CHCl_{3}$ and $CH_{4}$ are in the order:

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