Question

Among the following species, identify the pair having same bond order \( CN^{-} \), \( O_{2}^{-} \), \( NO^{+} \), \( CN^{+} \):

• A. \( CN^{-} \) and \( O_{2}^{-} \)
• B. \( O_{2}^{-} \) and \( NO^{+} \)
• C. \( CN^{-} \) and \( NO^{+} \)
• D. \( CN^{-} \) and \( CN^{+} \)
• E. \( NO^{+} \) and \( CN^{+} \)

Hint:

Memorize the "14-electron rule": Any diatomic species with 14 electrons (like \(N_2, CO, CN^-, NO^+\)) has a bond order of 3. This is a common shortcut for competitive exams.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Bond order is related to the number of electrons in a molecule or ion. Species that are isoelectronic (have the same number of electrons) typically have the same bond order, provided they follow similar Molecular Orbital (MO) configurations.

Step 2: Key Formula or Approach
For diatomic species with 14 electrons (like \(N_2\)), the bond order is 3. For every electron added or removed from this 14-electron "peak," the bond order generally decreases by 0.5.

Step 3: Detailed Explanation
1. Count electrons for each species: - \( CN^{-} \): \( 6 (C) + 7 (N) + 1 (\text{charge}) = 14 \) electrons. - \( O_{2}^{-} \): \( 8 + 8 + 1 = 17 \) electrons. - \( NO^{+} \): \( 7 (N) + 8 (O) - 1 (\text{charge}) = 14 \) electrons. - \( CN^{+} \): \( 6 + 7 - 1 = 12 \) electrons.
2. Identify isoelectronic pair: Both \( CN^{-} \) and \( NO^{+} \) have 14 electrons. 3. Calculate Bond Orders: - 14 electrons (\( CN^{-}, NO^{+} \)): Bond Order = 3. - 17 electrons (\( O_{2}^{-} \)): Bond Order = 1.5. - 12 electrons (\( CN^{+} \)): Bond Order = 2.

Step 4: Final Answer
The pair with the same bond order is \( CN^{-} \) and \( NO^{+} \).