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HINT: Bond order is inversely proportional to bond length. Higher the bond order, shorter the bond length. Explanation: To determine the species with the minimum bond length, we need to calculate the bond order for each species using Molecular Orbital Theory (MOT). The bond order is given by the formula: Bond Order = $\frac{1}{2} (\text{N}_b - \text{N}_a)$, where $\text{N}_b$ is the number of electrons in bonding molecular orbitals and $\text{N}_a$ is the number of electrons in antibonding molecular orbitals. 1. For $\text{O}_2$ (16 electrons): Electronic configuration: $(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2pz})^2 (\pi_{2px})^2 (\pi_{2py})^2 (\pi_{2px}^*)^1 (\pi_{2py}^*)^1$ $\text{N}_b = 10$, $\text{N}_a = 6$ Bond Order = $\frac{1}{2} (10 - 6) = 2.0$ 2. For $\text{O}_2^+$ (15 electrons): Electronic configuration: $(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2pz})^2 (\pi_{2px})^2 (\pi_{2py})^2 (\pi_{2px}^*)^1$ $\text{N}_b = 10$, $\text{N}_a = 5$ Bond Order = $\frac{1}{2} (10 - 5) = 2.5$ 3. For $\text{O}_2^-$ (17 electrons): Electronic configuration: $(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2pz})^2 (\pi_{2px})^2 (\pi_{2py})^2 (\pi_{2px}^*)^2 (\pi_{2py}^*)^1$ $\text{N}_b = 10$, $\text{N}_a = 7$ Bond Order = $\frac{1}{2} (10 - 7) = 1.5$ 4. For $\text{O}_2^{2-}$ (18 electrons): Electronic configuration: $(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2pz})^2 (\pi_{2px})^2 (\pi_{2py})^2 (\pi_{2px}^*)^2 (\pi_{2py}^*)^2$ $\text{N}_b = 10$, $\text{N}_a = 8$ Bond Order = $\frac{1}{2} (10 - 8) = 1.0$ Comparing the bond orders: $\text{O}_2^+ (2.5) > \text{O}_2 (2.0) > \text{O}_2^- (1.5) > \text{O}_2^{2-} (1.0)$ Since bond length is inversely proportional to bond order, the species with the highest bond order will have the minimum bond length. Therefore, $\text{O}_2^+$ has the minimum bond length.