Question

The bond order of CO is x. The bond order of N\(_2^+\) and C\(_2\) are respectively:

• A. \(\frac{x}{2}, \frac{2x}{3}\)
• B. \(x, \frac{x}{2}\)
• C. \(\frac{2x}{3}, \frac{5x}{6}\)
• D. \(\frac{5x}{6}, \frac{2x}{3}\)

Hint:

Bond order is calculated using: \[ \text{Bond Order} = \frac{N_b-N_a}{2} \] where \(N_b\) and \(N_a\) represent the number of bonding and antibonding electrons respectively.

Answer 1

The Correct Option is D

Step 1: Recall bond order formula.
\[ \text{Bond order} = \frac{N_b - N_a}{2} \] where \(N_b\) = number of bonding electrons, \(N_a\) = antibonding electrons.

Step 2: CO bond order.
Let bond order of CO = x → reference for proportional calculation.

Step 3: N\(_2^+\) bond order.
N\(_2\) neutral bond order = 3 → removing one electron (cation) → bond order increases slightly. Express in terms of x → \(5x/6\).

Step 4: C\(_2\) bond order.
C\(_2\) molecular orbital calculation → bond order = 2 → proportional to CO bond order → \(2x/3\).

Step 5: Compare with options.
Option (4) correctly matches \(\text{N}_2^+ = 5x/6, \text{C}_2 = 2x/3\).

Step 6: Conclusion.
Thus, the bond orders are \(\frac{5x}{6}\) and \(\frac{2x}{3}\).