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 ${CHCI}_{3}$ )

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

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

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

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

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

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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,

JEE

NEET

Foundation JEE

Foundation NEET

CBSE