To determine which species has the highest bond order, we need to calculate the bond order for each species.$\\$Bond order is calculated using the formula:$\\$Bond order$= \frac{Number \ of \ bonding \ electrons \ - \ Number \ of \ antibonding \ electrons}{2} \\$Let's analyze each species:$\\$1. $$CN$^+ \\$• Total electrons: $6 ($C$) + 7 ($N$) - 1 = 12 \\$• Electron configuration: $( \sigma_{1s})^2 ( \sigma^*_{1s})^2 ( \sigma_{2s})^2 ( \sigma^*_{2s})^2 ( \pi_{2p_x})^2 ( \pi_{2p_y})^2 ( \sigma_{2p_z})^2 \\$• Bond order: $\frac{8 - 4}{2} = 2 \\$2. $$CN$^- \\$• Total electrons: $6 ($C$) + 7 ($N$) + 1 = 14 \\$• Electron configuration: $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\sigma_{2p_z})^2 (\pi^*_{2p_x})^2 \\$• Bond order: $\frac{10 - 4}{2} = 3 \\$3. $$NO$\\$• Total electrons: $7 ($N$) + 8 ($O$) = 15 \\$• Electron configuration: $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\sigma_{2p_z})^2 (\pi^*_{2p_x})^1 \\$• Bond order: $\frac{10 - 5}{2} = 2.5 \\$4. $$CN$\\$• Total electrons: $6 ($C$) + 7 ($N$) = 13 \\$• Electron configuration: $(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\sigma_{2p_z})^1 \\$• Bond order: $\frac{9 - 4}{2} = 2.5 \\$Comparing the bond orders, $$CN$^-$ has the highest bond order of 3.$\\$Therefore, the species with the highest bond order is $$CN$^-$, which corresponds to Option 2.