JEE Main: Hybridisation of SO₂, NO₂⁻ & N₃⁻ — Super Easy Concept 💡

 

❓ Question

In $\text{SO}_2$, $\text{NO}_2^-$, and $\text{N}_3^-$, the hybridizations at the central atom are respectively:

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🖼️ Question Image

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✍️ Short Solution

We’ll find hybridization at each central atom using the Steric Number (SN) method:

$$\text{Steric number}=\text{Number of σ-bonds}+\text{Number of lone pairs on central atom}$$

and use:

Steric Number	Hybridization	Example
2	sp	BeCl₂
3	sp²	BF₃
4	sp³	CH₄
5	sp³d	PCl₅
6	sp³d²	SF₆

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🔹 (A) For $\text{SO}_2$

Step 1: Central atom = S
Valence electrons of S = 6

Each O atom forms a bond with S; one double bond and one coordinate bond (resonance).
→ S forms 2 σ-bonds + has 1 lone pair

Steric number = 3 → sp² hybridization

✅ Hybridization of S in SO₂ = sp²

Geometry: Bent / V-shaped (bond angle ≈ 119°)

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🔹 (B) For $\text{NO}_2^-$

Step 1: Central atom = N
Valence electrons of N = 5, plus 1 extra electron (for negative charge) → total 6

N forms 2 σ-bonds with O atoms and has 1 lone pair.

Steric number = 3 → sp² hybridization

✅ Hybridization of N in NO₂⁻ = sp²

Geometry: Bent (≈115°)

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🔹 (C) For $\text{N}_3^-$

Step 1: Linear ion, structure = $^-N = N^+ = N^-$ (resonance possible).

The central N atom forms 2 σ-bonds (one with each terminal N), and no lone pair.

Steric number = 2 → sp hybridization

✅ Hybridization of central N in N₃⁻ = sp

Geometry: Linear (180°)

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🧮 Image Solution

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✅ Final Answer

$$\boxed{{\displaystyle \text{Hybridizations: SO₂ → sp², NO₂⁻ → sp², N₃⁻ → sp}}}$$

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💡 Concept Recap

• Resonance does not change hybridization — it affects bond order, not σ-bonds.

• Steric number is the key shortcut — count σ-bonds + lone pairs on the central atom only.

• Bent geometry arises for SN = 3 (sp²) with one lone pair.