How many ions of the following have a bond or | vedclass

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(B) Bond order is 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

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$3$

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(B) Bond order is 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.SpeciesBond Order$N_2^{-}$$2.5$$NO^{-}$$2$$C_2^{-}$$2.5$$N_2^{+}$$2.5$$C_2^{2-}$$3$$CN^{+}$$2$From the table,the ions with a bond order of $2.5$ are $N_2^{-}$,$C_2^{-}$,and $N_2^{+}$.Therefore,there are $3$ such ions.

How many ions of the following have a bond order of $2.5 ?$
$N_2^{-} ; NO^{-} ; C_2^{-} ; N_2^{+} ; C_2^{2-} ; CN^{+}$

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How many ions of the following have a bond order of $2.5 ?$$N_2^{-} ; NO^{-} ; C_2^{-} ; N_2^{+} ; C_2^{2-} ; CN^{+}$