In ${CH}_{3} {CCl}_{3} (I), {CHCl}_{3} (II) and {CH}_{3} Cl (III)$ ln the normal tetrahedral bond angle is maintained. Also given cos ${70.5}^{\circ} = \frac{1}{3}$ Therefore dipole moments of the given compounds are: (given due to -I effect of Cl, the bond moment of H-C bond directed towards the H in ${CHCI}_{3}$ )

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$I = 1.9 D, II = 1.7 D, III = 1.7 D$

$I = 1.9 D, II = 1.9 D, III = 1.7 D$

$I = 1.9 D, II = 1.1 D, III = 1.9 D$

$I = 1.9 D, II = 1.7 D, III = 1.9 D$

answer is D.

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Detailed Solution

In ${CCl}_{4}, μ = 0$ so consider

$μ_{C - Cl} of 3 Cl = μ_{C - Cl}$ of 1 Cl. Similarly, in case of

Now,

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