The bond order of a species can be determined using the molecular orbital theory. The bond order is given by:
\[
\text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}
\]
Step 1: Bond Order Calculation for Each Species
i. \( F_2, O_2^{2-} \)
For \( F_2 \): The electron configuration is \( (1\sigma_g^2)(1\sigma_u^2)(2\sigma_g^2)(2\sigma_u^2)(2\pi_u^4)(2\pi_g^4) \). The bond order is 1.
For \( O_2^{2-} \): The electron configuration is \( (1\sigma_g^2)(1\sigma_u^2)(2\sigma_g^2)(2\sigma_u^2)(2\pi_u^4)(2\pi_g^4)(2\sigma_g^2)(2\sigma_u^2) \). The bond order is 1.
Thus, \( F_2 \) and \( O_2^{2-} \) have the same bond order.
ii. \( CO, NO^+ \)
For \( CO \): The electron configuration is \( (1\sigma_g^2)(1\sigma_u^2)(2\sigma_g^2)(2\sigma_u^2)(2\pi_u^4)(2\pi_g^2) \). The bond order is 3.
For \( NO^+ \): The electron configuration is \( (1\sigma_g^2)(1\sigma_u^2)(2\sigma_g^2)(2\sigma_u^2)(2\pi_u^4)(2\pi_g^1) \). The bond order is 3.
Thus, \( CO \) and \( NO^+ \) have the same bond order.
iii. \( N_2, O_2 \)
For \( N_2 \): The electron configuration is \( (1\sigma_g^2)(1\sigma_u^2)(2\sigma_g^2)(2\sigma_u^2)(2\pi_u^4)(2\pi_g^4) \). The bond order is 3.
For \( O_2 \): The electron configuration is \( (1\sigma_g^2)(1\sigma_u^2)(2\sigma_g^2)(2\sigma_u^2)(2\pi_u^4)(2\pi_g^4) \). The bond order is 2.
Thus, \( N_2 \) and \( O_2 \) have different bond orders, so this set does not match.
iv. \( H_2, B_2 \)
For \( H_2 \): The electron configuration is \( (1\sigma_g^2) \). The bond order is 1.
For \( B_2 \): The electron configuration is \( (1\sigma_g^2)(1\sigma_u^2)(2\sigma_g^2)(2\sigma_u^2) \). The bond order is 1.
Thus, \( H_2 \) and \( B_2 \) have the same bond order.
Step 2: Conclusion
The correct sets of species with the same bond order are \( i, ii, iii \) only. Therefore, the correct answer is Option (2).