Question

In which pair or pairs is the stronger bond found in the first species?
I. O\(_2^{2-}\), O\(_2\)
II. N\(_2\), N\(_2^+\)
III. NO\(^+\), NO\(^-\)

• A. I only
• B. II only
• C. I and II only
• D. II and III only

Hint:

Bond order increases with a stronger bond. A higher bond order corresponds to a stronger bond, as seen in N\(_2\) compared to N\(_2^+\), and NO\(^+\) compared to NO\(^-\).

Answer 1

The Correct Option is D

To answer this question, we need to consider the bond order and the electronic configurations of the species involved. Bond order is a measure of the strength of a bond, and it is determined by the number of bonding electrons minus the number of anti-bonding electrons. 
1. Pair I: O\(_2^{2-}\), O\(_2\): - O\(_2\) (oxygen molecule) has 16 electrons. The bond order is calculated as: \[ {Bond order} = \frac{(8 { bonding electrons}) - (4 { anti-bonding electrons})}{2} = 2 \] - O\(_2^{2-}\) (oxide ion) has 18 electrons. The bond order is: \[ {Bond order} = \frac{(8 { bonding electrons}) - (6 { anti-bonding electrons})}{2} = 1 \] Therefore, O\(_2\) has a stronger bond than O\(_2^{2-}\), meaning the first species has the weaker bond. 
2. Pair II: N\(_2\), N\(_2^+\): - N\(_2\) has 10 electrons. The bond order is: \[ {Bond order} = \frac{(6 { bonding electrons}) - (2 { anti-bonding electrons})}{2} = 3 \] - N\(_2^+\) has 9 electrons. The bond order is: \[ {Bond order} = \frac{(6 { bonding electrons}) - (3 { anti-bonding electrons})}{2} = 2.5 \] Therefore, N\(_2\) has a stronger bond than N\(_2^+\), meaning the first species has the stronger bond. 
3. Pair III: NO\(^+\), NO\(^-\): - NO\(^+\) has 10 electrons. The bond order is: \[ {Bond order} = \frac{(6 { bonding electrons}) - (2 { anti-bonding electrons})}{2} = 2 \] - NO\(^-\) has 11 electrons. The bond order is: \[ {Bond order} = \frac{(6 { bonding electrons}) - (3 { anti-bonding electrons})}{2} = 1.5 \] Therefore, NO\(^+\) has a stronger bond than NO\(^-\), meaning the first species has the stronger bond. 
Conclusion: The stronger bonds are found in the first species in pairs II and III. Thus, the correct answer is \( \boxed{{D}} \).