Question
The hybridisation of atomic orbitals of nitrogen in $$NO_2^ + ,NO_3^ - $$ and $$NH_4^ + $$ are
A.
$$sp,s{p^3}$$ and $$s{p^2}$$ respectively
B.
$$sp,s{p^2}$$ and $$s{p^3}$$ respectively
C.
$$s{p^2},sp$$ and $$s{p^3}$$ respectively
D.
$$s{p^2},s{p^3}$$ and $$sp$$ respectively
Answer :
$$sp,s{p^2}$$ and $$s{p^3}$$ respectively
Solution :
For $$NO_2^ + \,:\,H = \frac{1}{2}\left( {5 + 0 + 0 - 1} \right) = 2;$$
∴ $$sp$$ hybridisation
For $$NO_3^ - :H = \frac{1}{2}\left[ {5 + 0 + 1 - 0} \right] = 3;$$
∴ $$s{p^2}$$ hybridisation
For $$NH_4^ + :H = \frac{1}{2}\left[ {5 + 4 + 0 - 1} \right] = 4;$$
∴ $$s{p^3}$$ hybridisation