The correct sequence of dipole moments among the chlorides of methane is

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$C H C l_{3} < C H_{2} C l_{2} > C H_{3} C l > C C l_{4}$

$C H_{2} C l_{2} > C H_{3} C l > C H C l_{3} > C C l_{4}$

$C H_{3} C l > C H_{2} C l_{2} > C H C l_{3} > C C l_{4}$

$C H_{2} C l_{2} > C H C l_{3} > C H_{3} C l > C C l_{4}$

answer is C.

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Detailed Solution

$\begin{array}{l} μ_{C C l_{4}} = 0 \\ μ_{\vec{C - C l}} = 1.5 D μ_{\vec{H - C}} = 0.4 D \end{array}$

The resultant dipole moment of three $C - D$ bonds = 0.4 D.

Net dipole moment = 1.9 D

The resultant of two C - H bonds

$\begin{array}{l} = \sqrt{{(0.4)}^{2} + {(0.4)}^{2} + 2 \times {(0.4)}^{2} \times \cos {109.5}^{0}} \\ = \sqrt{0.16 + 16 - 0.2873} = 0.18 D \\ T h e r e s u l \tan t o f t w o C - C l b o n d s \\ = \sqrt{{(1.5)}^{2} + {(1.5)}^{2} + 2 \times {(1.5)}^{2} \times \cos {109.5}^{0}} \\ = \sqrt{2.25 + 2.25 + 2 \times 2.25 \times \cos {109.5}^{0}} \\ = \sqrt{4.5 - 4.04} = \sqrt{0.46} = 0.678 \end{array}$

$\begin{array}{l} ∴ μ_{c a l} = 0.858 D b u t \exp e r i m e n t a l l y f o r C H_{2} C l_{2}, μ = 1.59 D . \\ T h e r e s u l \tan t o f t h r e e C - C l b o n d s = 1.5 D \\ E x p e r i m e n t a l d i p o l e m o m e n t o f C H C l_{3} = 1.15 D \end{array}$