The bond angles of `NH_3, NH_4^(o+)` and `overset(ѳ)NH_2` are in the order:

Options

A. `NH_2^ѳ gt NH_3 gt overset(o+)NH_4`
B. `NH_4^(o+) gt NH_3 gt overset(ѳ)NH_2`
C. `NH_3 gt overset(ѳ)NH_2 gt overset(o+)NH_4`
D. `NH_3 gt overset(o+)NH_4 gt overset(ѳ)NH_2`

Answer

B

Explanation

`overset(o+)NH_4 implies sp^3` hybridised with no `Ipe^(-)s`, so shape is TH and bond angle = `109^@` 28.
`ddotNH_3 implies sp^3` hybridised and is pyramidal in shape with one lp of `e^(-)s`, so due to lp-bp and bp-bp repulsion bond angle decreases to `107^@`.
`: overset(ddotѳ)NH_2implies` Hybridisation = `1/2(5+2+1)=4=sp^3`
`: overset(ddotѳ)NH_2` is `sp^3` hybridised with two lp of `e^(-)s`, so due to lp-lp, lp-bp and bp-bp repulsion bond angle is further decreases to `104.5^@` (like `H_2 ddotO:`) and have V-shape.