The hybridisation of atomic orbitals of nitrogen in `NO_2^+ , NO_3^- ` and `NH_4^(+)` are :

Options

A. `sp^2, sp^3 and sp^2` respectively
B. `sp, sp^2 and sp^3` respectively
C. `sp^2, sp and sp^3` respectively
D. `sp^2, sp^3 and sp` respectively

Answer

B

Explanation

In `NO_2^+` total no. of valence electrons = 5 + 2 x 6-1 =16
Now 16+8 = `2(Q_1) + 0(R_1)` = 2 Thus, type of hybridization is sp. In `NO_3^(-)` total no. of valence electrons =5+ 3 x 6+1=24
Now, 24 + 8 = 3`(Q_1) + 0(R_1)` = 3
Thus type of hybridisation =`sp^2`
In `NH_4^(+)` ions, total no of valence electrons = 5+1 x 4-1 =8
Now 8+2 = 4`(Q_1)+0(R_1)=4`
Thus type of hybridization is `sp^3`
Combining all the results, option (b) is correct.