Question

Arrange the following in increasing order of bond order: (A) He\(_2^+\)
(B) O\(_2^-\)
(C) HF
(D) NO\(^-\)

• A. (A), (B), (C), (D)
• B. (C), (D), (A), (B)
• C. (B), (A), (D), (C)
• D. (A), (C), (B), (D)

Hint:

Bond order indicates the strength of the bond: higher bond order means a stronger bond. The bond order formula can help calculate it based on bonding and antibonding electrons.

Answer 1

The Correct Option is A

Step 1: Bond Order Formula.
The bond order in a molecule is given by the formula: \[ \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \]

Step 2: Bond Orders for Each Species.
- (A) He\(_2^+\): For He\(_2^+\), there are 2 electrons in bonding orbitals and 1 electron in antibonding orbitals. So, the bond order is: \[ \text{Bond Order} = \frac{2 - 1}{2} = 0.5 \] - (B) O\(_2^-\): O\(_2^-\) has 10 bonding electrons and 7 antibonding electrons, so the bond order is: \[ \text{Bond Order} = \frac{10 - 7}{2} = 1.5 \] - (C) HF: HF is a simple molecule with a single bond between hydrogen and fluorine. The bond order is 1. - (D) NO\(^-\): NO\(^-\) has 11 bonding electrons and 6 antibonding electrons, so the bond order is: \[ \text{Bond Order} = \frac{11 - 6}{2} = 2.5 \]

Step 3: Conclusion.
The increasing order of bond order is: \[ \text{He}_2^+ (0.5) < \text{O}_2^- (1.5) < \text{HF} (1) < \text{NO}^- (2.5) \] So, the correct order is (A), (B), (C), (D), corresponding to option (1).