Question

The diatomic molecule(s) that has/have bond order of one is/are

• A. B₂
• B. \(N^{2-}_2\)
• C. \(Li_2\)
• D. \(O^{2-}_2\)

Answer 1

The Correct Option is A, C, D

Problem 

Identify the diatomic molecule(s) that have a **bond order of 1**.

Working

Bond order formula: $$\text{Bond order} = \frac{N_b - N_a}{2}$$ where $N_b =$ number of electrons in bonding orbitals $N_a =$ number of electrons in antibonding orbitals.

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1. B₂

Total electrons = 10. MO configuration (for Z < 8): $$\sigma_{2s}^2 \sigma_{2s}^*{}^2 (\pi_{2p_x}^1 = \pi_{2p_y}^1)$$ Bond order: $$\text{BO} = \frac{6 - 4}{2} = 1$$ ✔ B₂ has bond order **1**.

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2. N₂^{2−}

N₂ has 14 electrons → N₂²⁻ has 16 electrons. N₂ MO filling gives: $$\text{BO} = 3 - 1 = 2$$ Bond order = **2**, not 1. ✘ Not correct.

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3. Li₂

Total electrons = 6. MO configuration: $$\sigma_{1s}^2 \sigma_{1s}^*{}^2 \sigma_{2s}^2$$ Bond order: $$\text{BO} = \frac{2 - 0}{2} = 1$$ ✔ Li₂ has bond order **1**.

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4. O₂^{2−}

O₂ has 16 electrons → O₂²⁻ has 18 electrons. Bond order of O₂ = 2 O₂⁻ = 1.5 O₂²⁻ = 1 ✔ O₂²⁻ has bond order **1**.

Answer

Correct molecules with bond order 1: B₂, Li₂, O₂²⁻
✔ Corresponding to options 1, 3 and 4.