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:

• A. ii only
• B. i, ii, iv only
• C. ii, iv only
• D. i, ii, iii only

Hint:

When analyzing molecular species for bond orders, consider the total number of bonding and antibonding electrons and how these affect molecular stability and reactivity.

Answer 1

The Correct Option is B

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).

Answer 2

To solve the problem, we need to identify which sets of species have the same bond order based on molecular orbital theory.

1. Understanding Bond Order:
Bond order is the number of bonds between two atoms in a molecule and is calculated using the formula: \[ \text{Bond Order} = \frac{1}{2} \left( \text{(Number of bonding electrons)} - \text{(Number of anti-bonding electrons)} \right) \] Higher bond orders indicate stronger bonds between the atoms.

2. Analyzing Each Species:

F₂ and O₂²⁻ (Option i): These species are both diatomic molecules with similar bonding. Their bond orders are both 1 (single bond), so they have the same bond order.

CO and NO⁺ (Option ii): Both these molecules have a triple bond between the atoms, resulting in a bond order of 3. So, they have the same bond order.

N₂ and O₂ (Option iii): N₂ has a bond order of 3 (triple bond), while O₂ has a bond order of 2 (double bond). These molecules do not have the same bond order.

H₂ and B₂ (Option iv): H₂ has a bond order of 1, and B₂ has a bond order of 1 as well (both have a single bond). So, they have the same bond order.

3. Final Answer:
The species with the same bond order are:
F₂ and O₂²⁻ (Option i)
CO and NO⁺ (Option ii)
H₂ and B₂ (Option iv)

Hence, the correct sets of species with the same bond order are i, ii, iv.

Final Answer:
The correct option is (B) i, ii, iv only.