Question

The hybridisation of atomic orbital of Nitrogen in $N{O}_3^{-},N{O}_2^{+},N{H}_4^{+}$ respectively:

• A. $s{p}^2,sp,s{p}^3$
• B. $sp,s{p}^2,s{p}^3$
• C. $s{p}^2,sp,s{p}^2$
• D. $s{p}^3,sp,s{p}^3$

Answer 1

The correct option is A $s{p}^2,sp,s{p}^3$

Steric Number $\left(x\right)$ = $\frac{1}{2}\left(M+V-C+A\right)$.
where, M = number of monovalent surrounding atoms
V = number of valence electrons on the central atom
C = charge (with sign)
A=charge on anion
$N{O}_3^{-}$
$\text{Steric Number (x)}=\frac{1}{2}(0+5-0+1)$.
$\text{Steric Number (x)}=3$.
Hybridisation =$s{p}^2$

$N{O}_2^{+}$
$\text{Steric Number (x)}=\frac{1}{2}(0+5-1+0)$.
$\text{Steric Number (x)}=2$.
Hybridisation =$sp$

$N{H}_4^{+}$
$\text{Steric Number (x)}=\frac{1}{2}(4+5-1+0)$.
$\text{Steric Number (x)}=4$.
Hybridisation =$s{p}^3$