Question

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

• A. $CN$
• B. $NO$
• C. $ CN^{+}$
• D. $CN^{-}$

Answer 1

The Correct Option is D

To determine which of the given species has the highest bond order, let us first calculate the bond order for each species. The bond order is a measure of the strength and stability of a bond and is calculated using the Molecular Orbital (MO) theory, according to the formula:

\text{Bond Order} = \frac{1}{2} (\text{Number of bonding electrons} - \text{Number of antibonding electrons})

1. Bond Order of CN^{-}:
   • Electronic configuration for CN^{-}: The addition of an electron results in 14 electrons.
   • Molecular orbital configuration: (1\sigma)^2 (1\sigma^*)^2 (2\sigma)^2 (2\sigma^*)^2 (1\pi)^4 (3\sigma)^2
   • Bond order calculation: \frac{1}{2}(10 - 4) = 3
2. Bond Order of CN:
   • Electronic configuration: CN has 13 electrons.
   • Molecular orbital configuration: (1\sigma)^2 (1\sigma^*)^2 (2\sigma)^2 (2\sigma^*)^2 (1\pi)^4 (3\sigma)^1
   • Bond order calculation: \frac{1}{2}(9 - 4) = 2.5
3. Bond Order of CN^{+}:
   • Electronic configuration: Loss of an electron results in 12 electrons.
   • Molecular orbital configuration: (1\sigma)^2 (1\sigma^*)^2 (2\sigma)^2 (2\sigma^*)^2 (1\pi)^4
   • Bond order calculation: \frac{1}{2}(8 - 4) = 2
4. Bond Order of NO:
   • Electronic configuration: NO has 15 electrons.
   • Molecular orbital configuration: (1\sigma)^2 (1\sigma^*)^2 (2\sigma)^2 (2\sigma^*)^2 (1\pi)^4 (3\sigma)^2 (2\pi_x)^1
   • Bond order calculation: \frac{1}{2}(10 - 5) = 2.5

From the calculations above, the bond orders for CN^{-}, CN, CN^{+}, and NO are 3, 2.5, 2, and 2.5, respectively. Therefore, the species with the highest bond order is CN^{-}, with a bond order of 3.