Question

The hybridization of atomic orbitals of nitrogen in $N{O}_2^{+}$, $N{O}_3^{-}$ and $N{H}_4^{+}$ are:

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

Answer 1

Answer: (B)

Here's a nice short trick.

1) Count and add all the electrons in valence shell

2) Divide it by $8$

3) On dividing by $8$ you will get a remainder and quotient, add quotient $+$ (remainder / $2$)

4) Now get the hybridization corresponding to the number what you got

If $2$ its $sp$,if $3$ its $s{p}^2$, if $4$ its $s{p}^3$, if $5$ its $s{p}^{3}d$ and so on.

Ex: $N{O}^{2+}$

-> Total electrons in valence shell : $5+6\times 2-1=16$.

->$16/8=2$. Quotient = $2$ and Remainder =$0$

-> $2+0/2=2$ i.e $sp$

$N{H}^{4+}$

-> Total electrons in valence shell : $5+7\times 4-1=32$.

-> $32/8=4$. Quotient = $4$ and Remainder = $0$

-> $4+0/2=4$ i.e $s{p}^3$

$N{O}^{3-}$

-> Total electrons in valence shell : $5+6\times 3+1=24$.

-> $24/8=3$. Quotient = $3$ and Remainder = $0$

-> $3+0/2=3$ i.e $s{p}^2$

Hence option B is correct.