Question

The difference between bond orders of CO and NO$^\oplus$ is $\frac{x}{2}$ where x = _________. (Round off to the Nearest Integer)

Hint:

For diatomic molecules with 14–18 electrons, bond orders decrease by 0.5 with each additional electron: $14e⁻ → 3,\;
15e⁻ → 2.5,\;
16e⁻ → 2,\;
17e⁻ → 1.5,\;
18e⁻ → 1.$

Answer 1

Correct Answer: 1

According to Molecular Orbital Theory, the bond order is given by: \[ \text{Bond Order} = \frac{1}{2}(N_b - N_a) \] where $N_b$ and $N_a$ are the numbers of bonding and antibonding electrons. Bond order of CO Carbon monoxide has a total of: \[ 6 + 8 = 14 \text{ electrons} \] CO is isoelectronic with $N_2$ and has bond order: \[ BO(\text{CO}) = 3 \] Bond order of NO Nitric oxide has: \[ 7 + 8 = 15 \text{ electrons} \] The extra electron enters an antibonding $\pi^*$ orbital. \[ BO(\text{NO}) = \frac{1}{2}(10 - 5) = 2.5 \] Difference in bond orders \[ |BO(\text{CO}) - BO(\text{NO})| = |3 - 2.5| = 0.5 \] Given that: \[ \frac{x}{2} = 0.5 \] \[ \boxed{x = 1} \]