Question

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

• A. CHCl₃ < CH₂Cl₂> CH₃Cl> CCl₄
• B. CH₂Cl₂ > CH₃Cl > CHCl₃> CCl₄
• C. CH₃Cl> CH₂Cl₂> CHCl₃< CCl₄
• D. CH₂Cl₂ > CHCl₃> CH₃Cl > CCl₄

Answer 1

Answer: (C)

CH₃Cl> CH₂Cl₂> CHCl₃< CCl₄

Answer 2

$\mu _ { \mathrm { CCl } _ { 4 } }$ = 0

= 1.5 D = 0.4 D

The resultant dipole moment of three C–H bonds = 0.4 D.

Net dipole moment = 1.9 D

The resultant of two C–H bonds

= $\sqrt{(0.4)^{2} + (0.4)^{2} + 2 \times (0.4)^{2} \times \cos 109.5º}$

= $\sqrt{0.16 + 0.16 - 0.2873}$ = 0.18 D

The resultant of two C – Cl bonds.

=$\sqrt{(1.5)^{2} + (1.5)^{2} + 2 \times (1.5)^{2} \times \cos 109.5º}$

= $\sqrt{2.25 + 2.25 + 2 \times 2.25 \times \cos 109.5º}$= $\sqrt{4.5 - 4.04}$

= $\sqrt{0.46}$

= 0.678

\ µ_cal = 0.858 D but experimentally for CH₂Cl₂, µ = 1.59 D.

The resultant of three C – Cl bonds = 1.5 D

Experimental dipole moment of CHCl₃ = 1.15 D