Q.11 Hybridizations of nitrogen in NO2+, NO3−, NH4+ respectively are
Nitrogen Hybridization in NO₂⁺, NO₃⁻, NH₄⁺: sp, sp², sp³
Complete analysis of steric number and molecular geometry
✅ (C) sp, sp², sp³
NO₂⁺ (linear), NO₃⁻ (trigonal planar), NH₄⁺ (tetrahedral)
NO₂⁺ (linear), NO₃⁻ (trigonal planar), NH₄⁺ (tetrahedral)
Steric Number = Hybridization Rule
Hybridization = ½(Valence e⁻ + Monovalent atoms ± charge)
2 → sp (linear) | 3 → sp² (trigonal planar) | 4 → sp³ (tetrahedral)1. NO₂⁺: sp Hybridization
🧪 NO₂⁺ (Nitronium ion)
Lewis Structure: O=N=O⁺ (linear)
Nitrogen: 2 σ bonds, 0 lone pairs
Steric #: 2 → sp
Geometry: Linear (180°)
2. NO₃⁻: sp² Hybridization
🧪 NO₃⁻ (Nitrate ion)
Lewis Structure: 3 resonance N-O bonds
Nitrogen: 3 σ bonds, 0 lone pairs
Steric #: 3 → sp²
Geometry: Trigonal planar (120°)
3. NH₄⁺: sp³ Hybridization
🧪 NH₄⁺ (Ammonium ion)
Lewis Structure: 4 N-H bonds
Nitrogen: 4 σ bonds, 0 lone pairs
Steric #: 4 → sp³
Geometry: Tetrahedral (109.5°)
Complete Hybridization Summary
| Species | σ Bonds | Steric # | Hybridization | Geometry |
|---|---|---|---|---|
| NO₂⁺ | 2 | 2 | sp | Linear |
| NO₃⁻ | 3 | 3 | sp² | Trigonal planar |
| NH₄⁺ | 4 | 4 | sp³ | Tetrahedral |
Options Breakdown
✅ (C) sp, sp², sp³
Perfect match: NO₂⁺(sp), NO₃⁻(sp²), NH₄⁺(sp³)
❌ (A) sp, sp², sp³
NO₃⁻ can’t be sp (needs 3 orbitals)
❌ (B) sp, sp³, sp²
NH₄⁺ never sp² (needs 4 orbitals)
❌ (D) sp³, sp², sp
NO₂⁺ never sp³ (only 2 bonds)
⚡ Bond Count = Hybridization
2 bonds → sp | 3 bonds → sp² | 4 bonds → sp³No lone pairs in these ions → Count sigma bonds only!
Memorize These Classics
- sp (linear): NO₂⁺, CO₂, BeCl₂
- sp² (trigonal): NO₃⁻, BF₃, SO₃, C₂H₄
- sp³ (tetrahedral): NH₄⁺, CH₄, H₂O, NH₃
