Discussion Forum

1)

How many ions of the following have bond order of 2.5?

  $N_{2}^{-},NO^{-}, C_{2}^{-}, N_{2}^{+}, C_{2}^{2-}, CN^{+}$

Answer:

Option B

Explanation:

 bond order= (Number of bonding electrons- Number of anti-bonding electrons)/ 2

 Bond order of $N_{2}^{-}$ is $\frac{(10-5)}{2}= \frac{5}{2}$=2.5

Bond order of $NO^{-}$ is $\frac{(10-6)}{2}=\frac{4}{2}=2$

Bond order  of $C_{2}^{-}$   is $\frac{1}{2} (9-4)=\frac{5}{2}$=2.5

Bond order of $N_{2}^{+}$ is $\frac{(9-4)}{2}=\frac{5}{2}$=2.5

Bond order of $C_{2}^{2-}$ is $\frac{(10-4)}{2}=\frac{6}{2}=3$

 Bond order of $CN^{+}$ is $\frac{(8-4)}{2}=2$