JEE Chemistry 2025 Question with Solution

Given: The species are $SO_2$, $NO_2^-$ and $N_3^-$.

Find: The hybridization of the central atom in each species.

Use the steric number concept:

$$\text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs}$$

For resonance structures, the hybridization of the central atom remains the same across all significant contributing structures.

For $SO_2$:

• Central atom is sulfur.
• It has two bonds with oxygen atoms and one lone pair on sulfur.
• So there are three regions of electron density around the central atom.

$$\text{Steric Number} = 2 + 1 = 3$$

Hence, the hybridization is $sp^2$.

For $NO_2^-$:

• Central atom is nitrogen.
• It has two bonds with oxygen atoms and one lone pair on nitrogen.
• Therefore, there are three regions of electron density around nitrogen.

$$\text{Steric Number} = 2 + 1 = 3$$

Hence, the hybridization is $sp^2$.

For $N_3^-$:

• The azide ion is linear and the central nitrogen is bonded to two terminal nitrogen atoms.
• The central nitrogen has two regions of electron density and no lone pair in the key resonance description used for hybridization.

$$\text{Steric Number} = 2 + 0 = 2$$

Hence, the hybridization is $sp$.

Therefore, the hybridizations are $sp^2$, $sp^2$ and $sp$. The correct option is A.

Given: Determine the hybridization of the central atom in $SO_2$, $NO_2^-$ and $N_3^-$.

Find: The correct sequence of hybridizations.

1. For $SO_2$:

• Central atom: sulfur.
• Number of bond pairs: $2$.
• Number of lone pairs: $1$.

$$\text{Steric Number} = 2 + 1 = 3$$

So the hybridization is $sp^2$.

2. For $NO_2^-$:

• Central atom: nitrogen.
• Resonance structures are $[O=N-O^-]$ and $[O^-N=O]$.
• Number of bond pairs around nitrogen: $2$.
• Number of lone pairs on nitrogen: $1$.

$$\text{Steric Number} = 2 + 1 = 3$$

So the hybridization is $sp^2$.

3. For $N_3^-$:

• Central atom: middle nitrogen.
• Resonance structures show the central nitrogen bonded to two terminal nitrogens.
• Number of bond pairs around the central nitrogen: $2$.
• Number of lone pairs on the central nitrogen: $0$.

$$\text{Steric Number} = 2 + 0 = 2$$

So the hybridization is $sp$.

4. Combining all three results gives:

$$sp^2,\ sp^2,\ sp$$

This matches option (1).

Therefore, the correct option is A.