Question

Consider the following species :
CN⁺, CN^–, NO and CN
Which one of these will have the highest bond order ?

• A. NO
• B. CN⁺
• C. CN^-
• D. CN

Answer 1

The Correct Option is C

To determine which species has the highest bond order, we need to evaluate the molecular orbital electron configurations for each species. Bond order is defined as the difference between the number of bonding electrons and antibonding electrons divided by two:

\text{Bond Order} = \frac{\text{Number of Bonding Electrons - Number of Antibonding Electrons}}{2}

Let's look at the electron configurations of the given species:

1. CN⁺: This species has a total of 12 electrons (from C and N, 6 each, minus 1 due to the positive charge). Its electronic configuration is similar to that of N₂, which has a bond order of 3.
2. CN^-: This species contains a total of 14 electrons (from C and N, 6 each, plus 2 from the negative charge). The bond order calculation is similar to O₂, which results in a bond order of 3.5.
3. CN: This neutral species has 13 electrons. Its electron configuration leads to a bond order of about 2.5, similar to the CO molecule.
4. NO: The NO molecule has 15 electrons. It typically has a bond order of 2.5.

Comparing the bond orders:

• CN⁺ has a bond order of 3.
• CN^- has a bond order of 3.5.
• CN has a bond order of 2.5.
• NO has a bond order of 2.5.

Thus, among the given species, CN^- has the highest bond order of 3.5.

Therefore, the correct answer is CN^-.