Question

what is the bond order of the cn molecule? do not use fractions to report your answer.

question 7
1 pts
do you expect cn to be a stable ion?
stable
unstable

Explanation:

Step1: Determine the electron configuration of CN

Carbon has 6 electrons and nitrogen has 7 electrons. So CN has 13 electrons. The molecular orbital configuration of CN is \(\sigma_{1s}^{2}\sigma_{1s}^{*2}\sigma_{2s}^{2}\sigma_{2s}^{*2}\sigma_{2p}^{2}\pi_{2p}^{4}\).

Step2: Calculate the bond order

The formula for bond order is \(BO=\frac{1}{2}(N_{b}-N_{a})\), where \(N_{b}\) is the number of bonding electrons and \(N_{a}\) is the number of antibonding electrons. Here \(N_{b} = 9\) and \(N_{a}=4\). So \(BO=\frac{1}{2}(9 - 4)=2.5\). But we need to round it as per the question's instruction (do not use fractions). So bond order is \(3\) (since \(2.5\) is closer to \(3\) than \(2\) in terms of whole - number approximation for bond order concept in basic understanding).

Step3: Analyze the stability of \(CN^{-}\)

\(CN^{-}\) has 14 electrons. Its molecular orbital configuration is \(\sigma_{1s}^{2}\sigma_{1s}^{*2}\sigma_{2s}^{2}\sigma_{2s}^{*2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p}^{*2}\). Using the bond - order formula \(BO=\frac{1}{2}(N_{b}-N_{a})\), where \(N_{b} = 10\) and \(N_{a}=4\). So \(BO=\frac{1}{2}(10 - 4)=3\). A bond order of \(3\) (a triple bond) indicates a stable ion.

Answer:

For the bond - order of \(CN\): \(3\)
For the stability of \(CN^{-}\): Stable