Question:

Identify the sets of species having the same bond order: 
(i) \( {F}_2, {O}_2^{2-} \) 
(ii) CO, NO\(^{+}\)
(iii) \(N_2\), \(O_2\)
(iv) \(H_2\), \(B_2\) 
The correct option is:

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When analyzing molecular species for bond orders, consider the total number of bonding and antibonding electrons and how these affect molecular stability and reactivity.
Updated On: Mar 19, 2025
  • ii only
  • i, ii, iv only
  • ii, iv only
  • i, ii, iii only
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The Correct Option is B

Solution and Explanation

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).
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