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: Revisiting the bond order equation.
Bond order = (Nb - Na) / 2, where Nb denotes bonding electrons and Na antibonding electrons.

Step 2: Defining the CO reference.
Let the bond order of CO be x, serving as the proportional baseline.

Step 3: Determining N₂⁺ bond order.
Neutral N₂ has a bond order of 3; removing one electron to form the cation slightly alters it, expressible as 5x/6 relative to CO.

Step 4: Determining C₂ bond order.
Molecular orbital assessment of C₂ gives a bond order of 2, which proportionally equates to 2x/3.

Step 5: Matching with the provided alternatives.
Option (4) correctly states N₂⁺ = 5x/6 and C₂ = 2x/3.

Step 6: Conclusion.
Thus, the respective bond orders are 5x/6 and 2x/3.