Question

The hybridization of $\mathrm{N}$ in ${{\mathrm{NO}}_2}^{-},{{\mathrm{NO}}_2}^{+},{{\mathrm{NO}}_4}^{-}$are respectively:

• A. ${\mathrm{sp}}^2,\mathrm{sp},{\mathrm{sp}}^3$
• B. ${\mathrm{sp}}^3,\mathrm{sp},{\mathrm{sp}}^3$
• C. $\mathrm{sp},\mathrm{sp},{\mathrm{sp}}^3$
• D. ${\mathrm{sp}}^2,{\mathrm{sp}}^2,{\mathrm{sp}}^3$

Answer 1

The correct option is A

${\mathrm{sp}}^2,\mathrm{sp},{\mathrm{sp}}^3$

The explanation of the correct option is:

(a)

• Explanation: A hybrid orbital is one that is formed by combining two or more atomic orbitals. The resulting orbital is different in shape and energy from the component orbitals. Hybridization is used to explain atomic bonding and model molecular geometry.
• To find the number of hybrid orbitals involved in hybridization, the formula is as follows:

$\mathrm{H}=\frac{1}{2}[\mathrm{V}+\mathrm{M}–\mathrm{C}+\mathrm{A}]$

Where,

H is the number of orbitals involved in hybridization

V is the number of valence electrons on the central atom

M is the number of monovalent atoms attached to the central atom

C is the charge on the cation

A is the charge on the anion

$$\begin{gathered} \mathrm{H}=\frac{1}{2}[\mathrm{V}+\mathrm{M}–\mathrm{C}+\mathrm{A}] \\ \mathrm{H}=\frac{(5+1)}{2}=3={\mathrm{sp}}^2 \\ \mathrm{H}=\frac{(5–1)}{2}=2={\mathrm{sp}}^1 \\ \mathrm{H}=\frac{(5+4–1)}{2}=4={\mathrm{sp}}^3 \end{gathered}$$

The correct option is (a): Hence ${\mathrm{sp}}^2,\mathrm{sp},{\mathrm{sp}}^3$.