Question

The type of hybrid orbital of nitrogen in NO2^+,,NO3^-,,Nh4+ respectively are expected to be

Answer 1

Answer:

The number of orbitals involved in hybridization can be determined by the application of formula:

H = 1/2[V+M-C+A]

Here H = number of orbitals involved in hybridization

V= valence electrons of central atom

M= number of monovalent atoms linked with central atom

C = charge on the cation

A = charge on the anion

$NO_{2} ^{+}:H=\frac{1}{2} [5+0-1+0]=2 or sp$

$NO_{3} ^{-} :H=\frac{1}{2} [5+0-0+1]=3  or sp^{2}$

$NH_{4} ^{+} :H=\frac{1}{2} [5+4-1+0]=4  or sp^{3}$