Molecular Orbital Theory Properties of Diatomic Species

To determine the correctness of the statements, we analyze each species using Molecular Orbital (MO) Theory.

Analysis of Option (A):

• Species: $\text{C}_2^{2-}$
• Total number of electrons: $2 \times 6 + 2 = 14\text{ electrons}$
• MO Electronic Configuration (for species with $\le 14$ electrons): $\sigma 1s^2 \; \sigma^* 1s^2 \; \sigma 2s^2 \; \sigma^* 2s^2 \; (\pi 2p_x^2 = \pi 2p_y^2) \; \sigma 2p_z^2$
• Magnetic Nature: Since all the electrons are fully paired in the molecular orbitals, $\text{C}_2^{2-}$ does not contain any unpaired electrons.
• Therefore, $\text{C}_2^{2-}$ is diamagnetic.

Thus, Option (A) is correct.

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Analysis of Option (B):

• Species: $\text{O}_2$ and $\text{O}_2^{2+}$

• For $\text{O}_2$ (16 electrons): $\sigma 1s^2 \; \sigma^* 1s^2 \; \sigma 2s^2 \; \sigma^* 2s^2 \; \sigma 2p_z^2 \; (\pi 2p_x^2 = \pi 2p_y^2) \; (\pi^* 2p_x^1 = \pi^* 2p_y^1)$ $\text{Bond Order of } \text{O}_2 = \frac{N_b - N_a}{2} = \frac{10 - 6}{2} = 2$

• For $\text{O}_2^{2+}$ (14 electrons): Two electrons are removed from the anti-bonding orbitals ($\pi^* 2p_x$ and $\pi^* 2p_y$). $\text{Bond Order of } \text{O}_2^{2+} = \frac{N_b - N_a}{2} = \frac{10 - 4}{2} = 3$

Since Bond Length is inversely proportional to Bond Order ($\text{Bond Length} \propto \frac{1}{\text{Bond Order}}$): $\text{Bond Order of } \text{O}_2^{2+} (3) > \text{Bond Order of } \text{O}_2 (2) \implies \text{Bond Length of } \text{O}_2^{2+} < \text{Bond Length of } \text{O}_2$

Therefore, $\text{O}_2^{2+}$ has a shorter bond length than $\text{O}_2$.

Thus, Option (B) is incorrect.

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Analysis of Option (C):

• Species: $\text{N}_2^+$ and $\text{N}_2^-$

• For $\text{N}_2^+$ (13 electrons): $\text{Bonding electrons } (N_b) = 9, \quad \text{Anti-bonding electrons } (N_a) = 4$ $\text{Bond Order} = \frac{9 - 4}{2} = 2.5$

• For $\text{N}_2^-$ (15 electrons): $\text{Bonding electrons } (N_b) = 10, \quad \text{Anti-bonding electrons } (N_a) = 5$ $\text{Bond Order} = \frac{10 - 5}{2} = 2.5$

Both $\text{N}_2^+$ and $\text{N}_2^-$ have an identical bond order of $2.5$.

Thus, Option (C) is correct.

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Analysis of Option (D):

• Species: $\text{He}_2^+$
• Total number of electrons: $2 + 2 - 1 = 3\text{ electrons}$
• MO Electronic Configuration: $\sigma 1s^2 \; \sigma^* 1s^1$
• Bond Order: $\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{2 - 1}{2} = 0.5$

Since the bond order is greater than zero ($0.5$), $\text{He}_2^+$ is a stable bound species and its energy is lower than the combined energy of its separated constituents ($\text{He} + \text{He}^+$). Hence, its energy is not equivalent to two isolated $\text{He}$ atoms.

Thus, Option (D) is incorrect.

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Conclusion:

The correct statements are (A) and (C).