Question

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

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

Answer 1

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

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 $s{p}^3$ and so on.

For $N{O}_2^{+}$

Total electrons in valence shell : 5 + 6 × 2 − 1 = 16.

$\to$ 16/8 = 2. Quotient = 2 and Remainder = 0

$\to$ 2+0/2 = 2 i.e. $sp$

For $N{O}_3^{+}$

Total electrons in valence shell : 5 + 6 × 3 + 1 = 24.

$\to$ 24/8 = 3. Quotient = 3 and Remainder = 0

$\to$ 3+0/2 = 3 i.e. $s{p}^2$

For $N{H}_4^{+}$

Total electrons in valence shell : 5 + 7 × 4 − 1 = 32.

$\to$ 32/8 = 4. Quotient = 4 and Remainder = 0

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

Hence, option A is correct .