The hybridisations of atomic orbitals of nitrogen in `NO_2, NO_3^(-) and N_4^(-)` respectively are

Options

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

Answer

C

Explanation

Hybridisation state = Number of `sigma` bond + number of lone pair or
Hybridisation state `rarr` from steric number rule
Hybridisation state = `1/2` (V.E + MA - C+ a)
For `NO_2^(+)=1/2(5+0-1)`
`=2rarrsp`
`NO_3^(-)=1/2 [5+0+1]`
`=3rarrsp^2`
`NH_4^(+)=1/2 [5+4-1]`
`= 4 rarr sp^3`