Question

The species having $s{p}^3$ hybridisation of atomic orbitals of nitrogen is/are $N{O}_2^{+},N{O}_3^{-}andN{H}_4^{+},N{F}_3$ are

• A. $N{H}_4^{+}$ only
• B. $N{O}_2^{+}$ only
• C. $N{O}_3^{-}$only
• D. $N{H}_4^{+}$ and $N{F}_3$

Answer 1

The correct option is D $N{H}_4^{+}$ and $N{F}_3$

$\text{Steric Number}\left(x\right)=\frac{1}{2}\left(M+V-C+A\right)$.
where, M = number of monovalent surrounding atoms
V = number of valence electrons on the central atom
C = charge (with sign)
A=charge on anion
For $N{O}_2^{+}:H=\frac{1}{2}(5+0+0-1)=2$
$\therefore sp$ hybridisation

For $N{O}_3^{-}:H=\frac{1}{2}(5+0+1-0)=3$
$\therefore s{p}^2$ hybridisation

For $N{H}_4^{+}:H=\frac{1}{2}(5+4+0-1)=4$
$\therefore s{p}^3$ hybridisation

$N{F}_3$
$\text{Steric Number (x)}=\frac{1}{2}(5+3-0+0)$.
$\text{Steric Number (x)}=4$.
Hybridisation =$s{p}^3$