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Which of the following species has a bond order equal to 2?

1. NO⁺: (σ1s)², (σ*1s)², (σ2s)², (σ*2s)², (σ2p)², (π2p)⁴ = (π2p)², (π*2p)²
2. CN⁻: (σ1s)², (σ*1s)², (σ2s)², (σ*2s)², (π2p)⁴ = (π2p)², (π*2p)²
3. O₂⁻: (σ1s)², (σ*1s)², (σ2s)², (σ*2s)², (π2p)⁴ = (π2p)², (π*2p)²
4. O₂²⁻: (σ1s)², (σ*1s)², (σ2s)², (σ*2s)², (π2p)⁴ = (π2p)², (π*2p)²

Correct answer: O₂²⁻: (σ1s)², (σ*1s)², (σ2s)², (σ*2s)², (π2p)⁴ = (π2p)², (π*2p)²

Solution

The bond order is calculated as (number of bonding electrons - number of antibonding electrons)/2. For O₂²⁻, the bond order is (8-4)/2 = 2.

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