Question

Increasing order of bond order of oxygen and its ions is

• A. \( O_2^{+} < O_2^{2-} < O_2^{-} < O_2 \)
• B. \( O_2^{-} < O_2^{2-} < O_2^{+} < O_2 \)
• C. \( O_2^{+} < O_2 < O_2^{-} < O_2^{2-} \)
• D. \( O_2^{2-} < O_2^{-} < O_2^{+} < O_2 \)

Hint:

In molecular orbital theory, the bond order helps determine the strength and stability of a molecule. For ions, bond order increases with electron removal and decreases with electron addition.

Answer 1

The Correct Option is D

The bond order for different oxygen species can be determined by the Molecular Orbital Theory (MOT). According to MOT: 1. Oxygen molecule (O$_2$): The molecular orbital configuration for O$_2$ is: \[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^1 \pi^*_{2p_y}^1 \] The bond order is calculated as: \[ \text{Bond Order} = \frac{(8 - 4)}{2} = 2 \] 2. Oxygen ion O$_2^+$: The removal of an electron from O$_2$ leads to: \[ \text{Bond Order} = \frac{(8 - 3)}{2} = 2.5 \] 3. Oxygen ion O$_2^{-}$: The addition of an electron results in: \[ \text{Bond Order} = \frac{(8 - 5)}{2} = 1.5 \] 4. Oxygen ion O$_2^{2-}$: The addition of two electrons results in: \[ \text{Bond Order} = \frac{(8 - 6)}{2} = 1 \] Thus, the increasing order of bond order is: \[ O_2^{2-} < O_2^{-} < O_2^{+} < O_2 \]