Q.Species having same bond order are : (Note: more than one of the given options may be correct.)

The key idea is Bond Order, which quantifies the number of chemical bonds between two atoms. It is calculated using molecular orbital theory as half the difference between the number of electrons in bonding molecular orbitals ($N_b$) and antibonding molecular orbitals ($N_a$).

For diatomic species, bond order can be quickly determined by their total number of electrons:

1. Calculate total electrons for each species:

   • (A) N$_2$: $2 \times 7 = 14$ electrons
   • (B) N$_2^-$: $2 \times 7 + 1 = 15$ electrons
   • (C) F$_2^+$: $2 \times 9 - 1 = 17$ electrons
   • (D) O$_2^-$: $2 \times 8 + 1 = 17$ electrons

2. Determine bond order for each species: …

In chemistry, bond order is a fundamental concept that describes the number of chemical bonds between a pair of atoms. It's a direct indicator of bond strength and stability: a higher bond order generally means a stronger, shorter, and more stable bond. We determine bond order using Molecular Orbital (MO) theory, which describes how atomic orbitals combine to form molecular orbitals.

The key idea is that electrons occupy these molecular orbitals, some of which are bonding (stabilizing the molecule) and some are anti-bonding (destabilizing the molecule). The net effect of these electrons determines the bond order.

A crucial aspect of MO theory for diatomic molecules is the energy ordering of the molecular orbitals, which changes depending on the total number of electrons in the molecule. This change is due to the extent of s-p mixing.

Let's calculate the bond order for each given species:

1. N$_2$

   • Total number of electrons = $7 (\text{from N}) + 7 (\text{from N}) = 14$ electrons.
   • Since it has 14 electrons, we use the MO energy order for $\le 14$ electrons.
   • MO configuration: \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2
   • Number of bonding electrons ($N_b$) = $2 (\sigma_{1s}) + 2 (\sigma_{2s}) + 4 (\pi_{2p}) + 2 (\sigma_{2p_z}) = 10$
   • Number of anti-bonding electrons ($N_a$) = $2 (\sigma^*_{1s}) + 2 (\sigma^*_{2s}) = 4$
   • Bond Order = $\frac{1}{2} (10 - 4) = \frac{1}{2} (6) = \mathbf{3}$

2. N$_2^-$

   • Total number of electrons = $7 (\text{from N}) + 7 (\text{from N}) + 1 (\text{for negative charge}) = 15$ electrons.
   • Even though it has 15 electrons, the s-p mixing effect is still significant for nitrogen-based molecules. Therefore, we still use the MO energy order for $\le 14$ electrons.
   • MO configuration: \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi^*_{2p_x}^1
   • Number of bonding electrons ($N_b$) = $2 (\sigma_{1s}) + 2 (\sigma_{2s}) + 4 (\pi_{2p}) + 2 (\sigma_{2p_z}) = 10$
   • Number of anti-bonding electrons ($N_a$) = $2 (\sigma^*_{1s}) + 2 (\sigma^*_{2s}) + 1 (\pi^*_{2p_x}) = 5$
   • Bond Order = $\frac{1}{2} (10 - 5) = \frac{1}{2} (5) = \mathbf{2.5}$

3. F$_2^+$

   • Total number of electrons = $9 (\text{from F}) + 9 (\text{from F}) - 1 (\text{for positive charge}) = 17$ electrons. …