Question

According to Molecular orbital theory, which of the following is correct with respect to bond order?

• A. Bond order of $\text{N}_2^+$ and $\text{O}_2^+$ is less than $\text{O}_2$
• B. Bond order of $\text{N}_2^+$ and $\text{O}_2^+$ is more than $\text{N}_2$
• C. Bond order of $\text{N}_2^+$ is less than $\text{N}_2$ while that of $\text{O}_2^+$ is more than $\text{O}_2$
• D. Bond order of $\text{N}_2^+$ is less than $\text{O}_2$ while that of $\text{O}_2^+$ is more than $\text{O}_2$

Hint:

Remember this straightforward shortcut for diatomic MO configurations:
• Removing an electron from Nitrogen targets a bonding orbital, lowering its bond order from $3$ down to $2.5$.
• Removing an electron from Oxygen targets an antibonding orbital, raising its bond order from $2$ up to $2.5$!

Answer 1

The Correct Option is C

Concept: According to Molecular Orbital (MO) Theory, the bond order of a diatomic molecule is calculated from its electronic configuration using the expression: \[ \text{Bond Order} = \frac{N_b - N_a}{2} \] where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons. Removing an electron from a bonding orbital decreases the bond order, while removing an electron from an antibonding orbital increases it.

Step 1: Calculate the bond order for Nitrogen species ($\text{N}_2$ and $\text{N}_2^+$).

• $\text{N}_2$ has 14 electrons, filling orbitals up to $\sigma_{2p_z}^2$. It has $10$ bonding and $4$ antibonding electrons: $\text{B.O.} = \frac{10 - 4}{2} = 3$.
• $\text{N}_2^+$ has 13 electrons. The electron is removed from the bonding $\sigma_{2p_z}$ orbital, reducing $N_b$ to 9: $\text{B.O.} = \frac{9 - 4}{2} = 2.5$.
• Thus, $\text{Bond Order of N}_2^+ < \text{Bond Order of N}_2$.

Step 2: Calculate the bond order for Oxygen species ($\text{O}_2$ and $\text{O}_2^+$).

• $\text{O}_2$ has 16 electrons, with two electrons occupying the antibonding $\pi_{2p}^*$ orbitals ($N_b = 10, N_a = 6$): $\text{B.O.} = \frac{10 - 6}{2} = 2$.
• $\text{O}_2^+$ has 15 electrons. The electron is removed from an antibonding $\pi_{2p}^*$ orbital, reducing $N_a$ to 5: $\text{B.O.} = \frac{10 - 5}{2} = 2.5$.
• Thus, $\text{Bond Order of O}_2^+ > \text{Bond Order of O}_2$. Combining these two results matches statement (C).