To determine which species among CN+, CN–, NO, and CN has the highest bond order, we need to calculate the bond order for each species. The bond order helps us understand the stability of the bond and is calculated using the molecular orbital theory.
- First, let's calculate the bond order, which is given by the formula: \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} .
- For CN:
- Electronic configuration: CN consists of a total of 13 valence electrons (C has 6, N has 7).
- Molecular orbitals are filled according to Hund’s rule and the energy order for diatomic species with 14 or fewer electrons: \sigma(1s)^2, \sigma^*(1s)^2, \sigma(2s)^2, \sigma^*(2s)^2, \sigma(2p_z)^2, \pi(2p_x)^2 = \pi(2p_y)^2, \text{and } \pi^*(2p_x, 2p_y) .
- Bond Order: \frac{8 - 3}{2} = \frac{5}{2} = 2.5 .
- For CN+:
- Total valence electrons = 12 (one less than CN).
- Bond Order: \frac{8 - 2}{2} = 3 .
- For CN–:
- Total valence electrons = 14 (one more than CN).
- Bond Order: \frac{8 - 2}{2} = 3 .
- For NO:
- Total valence electrons = 15 (N has 7, O has 8).
- Electron configuration alters as \pi^*(2p_x, 2p_y)^1 is filled.
- Bond Order: \frac{8 - 3}{2} = 2.5 .
- From these calculations, CN– and CN+ have the bond order of 3, being the highest.
Thus, by convention, due to the simpler electronic structure in CN-, it is often considered more stable and prevalent. Hence, the species CN- is recognized for having the highest stability among these options.