1.

Identify the correct option :-

Answer»

`NO_(3)^(-) gt NH_(3) gt NH_(2)^(-)` (order of bond angle)
`(CH_(3))_(3)` B is a planar MOLECULE
In `NH_(4)Cl`, .N. ATOM is in `sp^(3)d` Hybridisation
`BF_(3) lt BCl_(3) lt B Br_(3) lt BI_(2)` (Order of bond angle)

Solution :(i) `underset(sp^(2))(NO_(3)^(-)) gt underset("1 lp")underset(sp^(3))(NH_(3)) gt underset("2 lp")underset(sp^(3))(NH_(2)^(-))` (B.A. order)
(ii) In `(CH_(3))_(3)` B hyb. of carbon is `sp^(3)` (non planar)
(III) `NH_(4)Cl to NH_(4)^(+)Cl^(-)` in `NH_(4)^(+)` hyb. of N is `sp^(3)`
(iv) In `s_(8)` molecule TOTAL 16 lp or 32 non bonded `e^(-)`


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