Explain :
(i) `mu of NH_3 gt NF_3`
(ii) `mu of CH_3 C1 gt CH_3 F gt CH_3 Br gt CH_3 I`
(iii) `mu of CO_2` is zero, nut `mu of SO_2 ne 0`
(iv) Why the lone pair of `overline e^, s` has no effect on the `mu` of `PH_3`. The bond angle in `PH_3` is `92^@`.
(v) `mu of CH_2 C1_2 gt CH C1_3`.

Explanation

`[{:("Resulant"-(mu) of NH_(3),,"Resultant" - (mu) of NF_(3),,),(=4.90 xx 10^(-3)Cm,,= 0. 80 xx 10^(-30) Cm,,):}]`
In `NH_(3)` the net moment of `(N -H)` bonds and the contribution from the lp `e^(-')s` (lone pair `e^(-')`s) are in the same direction and are additive [see (a) given above] The net moment of teh `(N-F)` bond opposes the dipole effect of the `lpe^(-')` in `NH_(3)` and the resultant is less than `mu So, mu` of `Nh_(3) gtNF_(3)`
(b) The electronegatives of halogens decrease from `F` to So `mu` of `HF gt HCI gtHBr gt HI` But `mu` of `CH_(3)F` is smaller than `CH_(3)CI` due to shorter `(C -F)` bond distance, although `EN` of `F` is greater than that of `CI`
(c ) In `CO_(2')C` is sp hybridised and linear The bond moments of `(C -O)` are equal and in opposite directions and cancel each other Hence `mu` is zero
in `SO_(2)` is `sp^(2)` hybridised having one `lp ebar` on `S` atom. The `(O -S-O)` bond angle is nerly `120(S-O)` bond moment does not cancel and shows a net resultant mu
(d) The `92^(@)` bond angle suggests that `P` uses three `P` atomic orbitals in forming bonds with `H` with one `lp e^(-)` in 3s atomic orbital i.e `P` in `PH_(3)` is `sp^(2)` hybridised (unlike `NH_(3)` in which `N` is `sp^(3)` hybridised) Therefore, due to the presence of `lpee^(-')s` in 3s atomic orbital of `P` which is spherical symmetrical the polarity of the molecule is not affected. In order to effect the polarity of the molecule the `e^(-)s` must be in a directional orbital Moreover, `EN` of `P` and `H` are nearly same so `PH_(3)` molecule is almost non-polar
In (I) all bond momnets are in the same direction so they are additive and the net resultant mu is more than (II) In (II) the bond moment of one of the `CI` atoms opposes, the net moment of the other two, so the net resultant mu is less than that of (I)
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