Concept: Molecular orbital theory (MOT) is used to determine the stability of diatomic molecules and ions. Stability is directly related to the bond order. A higher bond order generally indicates greater stability.

Why (A) is correct:

To determine the stability, we need to calculate the bond order for each species using the molecular orbital configuration.

• N${}_2$: Total electrons = 14. Configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\sigma }_{2p_z}^2$.
  Bond order = $\frac{1}{2}$(Number of electrons in bonding MOs - Number of electrons in antibonding MOs) = $\frac{1}{2}$(10 - 4) = 3.0
• O${}_2$: Total electrons = 16. Configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\pi }_{2p_x}^{\ast 1}{\pi }_{2p_y}^{\ast 1}$.
  Bond order = $\frac{1}{2}$(10 - 6) = 2.0
• O${}_2^{+}$: Total electrons = 15. Configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\pi }_{2p_x}^{\ast 1}$.
  Bond order = $\frac{1}{2}$(10 - 5) = 2.5
• O${}_2^{-}$: Total electrons = 17. Configuration: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p_z}^2{\pi }_{2p_x}^2{\pi }_{2p_y}^2{\pi }_{2p_x}^{\ast 2}{\pi }_{2p_y}^{\ast 1}$.
  Bond order = $\frac{1}{2}$(10 - 7) = 1.5

Since N${}_2$ has the highest bond order (3.0), it is the most stable among the given options.

Option Analysis:

• A) N${}_2$: Bond order = 3.0. This is the highest bond order, indicating maximum stability.
• B) O${}_2$: Bond order = 2.0. Less stable than N${}_2$ and O${}_2^{+}$.
• C) O${}_2^{+}$: Bond order = 2.5. More stable than O${}_2$ and O${}_2^{-}$, but less stable than N${}_2$.
• D) O${}_2^{-}$: Bond order = 1.5. This is the lowest bond order, indicating minimum stability among the options.

Correct Answer: (A)