The correct order of increasing bond order among the given is:

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Accepted Answer

1. **Formula:** Bond Order (B.O.) = $\frac{1}{2} (N_b - N_a)$.

2. **Analyze He₂⁺:**
   - Total electrons: $2 + 2 - 1 = 3$.
   - Configuration: $\sigma_{1s}^2 \sigma_{1s}^{*1}$.
   - B.O. = $\frac{1}{2} (2 - 1) = 0.5$.

3. **Analyze O₂⁻ (Superoxide ion):**
   - Total electrons: $8 + 8 + 1 = 17$.
   - Configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) (\pi_{2p_x}^{*2} = \pi_{2p_y}^{*1})$.
   - $N_b = 10, N_a = 7$.
   - B.O. = $\frac{1}{2} (10 - 7) = 1.5$.

4. **Analyze NO:**
   - Total electrons: $7 + 8 = 15$.
   - Similar to $O_2^+$. Configuration ends in $\pi^*$ having 1 electron.
   - $N_b = 10, N_a = 5$.
   - B.O. = $\frac{1}{2} (10 - 5) = 2.5$.

5. **Analyze C₂²⁻ (Acetylide ion):**
   - Total electrons: $6 + 6 + 2 = 14$. Isoelectronic with $N_2$.
   - Configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} (\pi_{2p_x}^2 = \pi_{2p_y}^2) \sigma_{2p_z}^2$.
   - $N_b = 10, N_a = 4$.
   - B.O. = $\frac{1}{2} (10 - 4) = 3.0$.

6. **Conclusion:** Increasing order is $0.5 < 1.5 < 2.5 < 3.0$, which corresponds to $He_2^+ < O_2^- < NO < C_2^{2-}$.

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