Question

The bond order of N\(_2^+\) ion is ................. (Round off to one decimal place)

Hint:

Bond order can be determined using the molecular orbital theory. For positive ions, the bond order will be higher than the neutral molecule due to the loss of electrons from the antibonding orbitals.

Answer 1

Correct Answer: 2.5

Step 1: Understanding the bond order formula.
The bond order of a molecule or ion is given by the formula: \[ \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \] Step 2: Molecular orbital configuration of N\(_2^+\).
For the N\(_2\) molecule, the electron configuration in the molecular orbitals is: \[ ( \sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^0 (\pi^*_{2p_y})^0 \] In N\(_2^+\), one electron is removed from the antibonding \(\pi^*\) orbital, making the configuration: \[ ( \sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^0 (\pi^*_{2p_y})^1 \] Step 3: Calculating the bond order.
Number of bonding electrons = 2 (from \(\sigma_{1s}\)) + 2 (from \(\sigma_{2s}\)) + 4 (from \(\pi_{2p_x}\) and \(\pi_{2p_y}\)) = 8
Number of antibonding electrons = 2 (from \(\sigma^*_{1s}\)) + 2 (from \(\sigma^*_{2s}\)) + 1 (from \(\pi^*_{2p_y}\)) = 5
Therefore, the bond order is: \[ \text{Bond Order} = \frac{8 - 5}{2} = 1.5 \] Thus, the bond order of N\(_2^+\) is 2.5.