Consider the following species: CN^{+},CN^{-} | vedclass

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(B) The bond order of a diatomic species can be calculated using the formula: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of b

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$CN^{-}$

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(B) The bond order of a diatomic species can be calculated using the formula: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.For species with total electrons up to $14$,the configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$.$1$. $CN^{+}$ ($12$ electrons): $	ext{Bond Order} = \frac{8-4}{2} = 2.0$$2$. $CN$ ($13$ electrons): $	ext{Bond Order} = \frac{9-4}{2} = 2.5$$3$. $CN^{-}$ ($14$ electrons): $	ext{Bond Order} = \frac{10-4}{2} = 3.0$$4$. $NO$ ($15$ electrons): $	ext{Bond Order} = \frac{10-5}{2} = 2.5$Comparing the values,$CN^{-}$ has the highest bond order of $3.0$.

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

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