Question

The bond order and magnetic property of acetylide ion are same as that of

• A. \(O^+_2\)
• B. \(N^+_2\)
• C. \(NO^+\)
• D. \(O^-_2\)

Answer 1

The Correct Option is C

To determine the bond order and magnetic property of the acetylide ion and compare it with the given options, we should first understand each entity's electronic configuration and bond characteristics.

Step-by-Step Solution

1. Acetylide Ion \(C_2^{2-}\):
   • The acetylide ion has 14 electrons in total: each carbon contributes 6 electrons, and the negative charges contribute 2 additional electrons.
   • The molecular orbital configuration is similar to a diatomic molecule:
   • \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2\)
   • Bond order is calculated as:
   • \(\text{Bond order} = \frac{\text{(Number of bonding electrons - Number of antibonding electrons)}}{2} = \frac{8-2}{2} = 3\)
   • This indicates a triple bond.
   • With all electrons paired, the acetylide ion is diamagnetic.
2. \(NO^+\):
   • It has 14 electrons, similar to \(N_2\). It shares the same molecular orbital configuration as \(N_2\).
   • The molecular orbital configuration for \(NO^+\) is:
   • \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2\)
   • Bond order is calculated as:
   • \(\text{Bond order} = \frac{8-2}{2} = 3\)
   • Like the acetylide ion, \(NO^+\) is also diamagnetic due to all paired electrons.
3. Other Options:
   • \(O^+_2\) and \(O^-_2\): Both have different electron counts leading to different bond orders and magnetic properties. Neither is diamagnetic.
   • \(N^+_2\): Although it has a bond order of 2.5, it cannot have the same magnetic property (complete electron pairing) as the acetylide ion.

Conclusion:

The bond order and magnetic property of the acetylide ion are the same as that of \(NO^+\), with both having a bond order of 3 and being diamagnetic due to all electrons being paired. Thus, the correct answer is \(NO^+\).

This understanding is crucial because knowing electron configurations and molecular orbital theory benefits various competitive exams and aids in predicting compound behavior.