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 an indicator of the stability and strength of a chemical bond.
According to Molecular Orbital (MO) theory, bond order is calculated as half the difference between the number of bonding electrons and anti-bonding electrons.
A highly useful shortcut in MO theory is that isoelectronic species (molecules or ions with the exact same total number of electrons) generally possess identical bond orders because their molecular orbital filling patterns are the same.
Step 2: Key Formula or Approach:
1. Calculate the total number of electrons for each given species by adding the atomic numbers of the constituent atoms and adjusting for any charge (add for negative charge, subtract for positive).
2. Use the standard correlation between total electron count and bond order for diatomic molecules:
- 14 electrons $\rightarrow$ Bond Order 3
- 15 or 13 electrons $\rightarrow$ Bond Order 2.5
- 16 or 12 electrons $\rightarrow$ Bond Order 2
- 17 or 11 electrons $\rightarrow$ Bond Order 1.5
Step 3: Detailed Explanation:
Let's determine the total electron count and the corresponding bond order for each species:
1. $CN^-$ (Cyanide ion):
Total electrons = $6$ (from Carbon) $+ 7$ (from Nitrogen) $+ 1$ (for the negative charge) = $14$ electrons.
Species with 14 electrons are isoelectronic with $N_2$ and have a bond order of 3.
2. $O_2^-$ (Superoxide ion):
Total electrons = $8$ (from Oxygen) $+ 8$ (from Oxygen) $+ 1$ (for the negative charge) = $17$ electrons.
Species with 17 electrons have a bond order of 1.5.
3. $NO^+$ (Nitrosonium ion):
Total electrons = $7$ (from Nitrogen) $+ 8$ (from Oxygen) $- 1$ (for the positive charge) = $14$ electrons.
Species with 14 electrons are isoelectronic with $N_2$ and have a bond order of 3.
4. $CN^+$ (Cyanogen cation):
Total electrons = $6$ (from Carbon) $+ 7$ (from Nitrogen) $- 1$ (for the positive charge) = $12$ electrons.
Species with 12 electrons are isoelectronic with $C_2$ and have a bond order of 2.
Comparing the results, we can see that both $CN^-$ and $NO^+$ possess exactly 14 electrons and consequently share the same bond order of 3.
Step 4: Final Answer:
The pair with the same bond order is $CN^-$ and $NO^+$.