Dipole moment

CH₃Cl > CH₂Cl₂ > CHCl₃ > CCl₄
μ of CCl₄ = 0

CCl₄ was not in question :P

Can U xplain why CH₂Cl₂<CH₃Cl
and so on....

I have also read the same... and whoever i have asked has replied the same...

But i think CCl₃H>CCl₂H₂>Ch₃Cl>CCl₄

dipole moment of the hydrocarbon depends on geometry i.e.
μ is inversly proportion to θ , so it iz.
expermental value of μ r as follows:-
CHCl₃ = 1.02 D ,
CH₂Cl₂ = 1.55D ,
CH₃Cl = 1.93 D.

That experimental values....

Ok consider CH₂Cl₂...

μ b/w C-Cl be x

Here two x are at angle of 110° so there vector sum > x (in CH₃Cl)

i am confused...

as i think , explanation is ,Cl is a bulky group , hence as no. of Cl increases , der will steric hindrance increases and the angle increases , and den due to its geometry, the polarity of the C - Cl bond tends to cancele.
and as i already mention'd , μ ≈ 1/θ,so as angle increases (due 2 s.h.), μ decreases..

H -> electropositive than C
Cl -> electronegative than C

In CH₃Cl, three H make C more electronegative which is pulled by Cl. So net dipole is large.

In CH₂Cl₂, shared bet. 2 Cl atoms..

In CHCl₃, 3 Cl atoms are present, which distrubutes more and hence less compared to CH₃Cl and CH₂Cl₂ net dipole moment.

I think this is a reason... please correct me if I am wrong..

draw the tetrahedral diag da. u should get it.