Question

The bond order in \( N_2^{2-} \) species is:

• A. 2
• B. 2.5
• C. 3
• D. 3.5

Hint:

To find the bond order, always count the number of bonding and antibonding electrons in the molecular orbital diagram and apply the formula.

Answer 1

The Correct Option is A

The bond order in a molecular species can be calculated using the molecular orbital theory. The bond order is given by the formula:
\[ {Bond order} = \frac{1}{2} \left( {Number of bonding electrons} - {Number of antibonding electrons} \right) \] For \( N_2^{2-} \), the electronic configuration of the molecule in its molecular orbitals is as follows:
- The molecular orbital diagram for \( N_2 \) (and ions) places the bonding and antibonding electrons in the \( \sigma_{2s}, \sigma_{2s}^*, \sigma_{2p_z}, \pi_{2p_x} = \pi_{2p_y} \), and \( \pi_{2p_x}^*, \pi_{2p_y}^*, \sigma_{2p_z}^* \) orbitals.
- The \( N_2^{2-} \) ion has 14 electrons, and by filling the molecular orbitals, we get 10 bonding electrons and 4 antibonding electrons.
Thus, the bond order is: \[ {Bond order} = \frac{1}{2} \left( 10 - 4 \right) = 2 \]