Question

what is the bond order of no? do not use fractions to report your answer.
question 10 1 pts
do you expect no to be a stable ion?
stable
unstable

Explanation:

Step1: Determine the number of bonding and antibonding electrons

The molecular orbital configuration of \(NO^-\) is similar to that of \(O_2\). The valence electrons of \(N\) is \(5\) and of \(O\) is \(6\), and since it is \(NO^-\), there is an extra electron. So total valence electrons \(=5 + 6+1=12\).
The molecular orbital configuration for \(NO^-\) (assuming \(O_2\) - like configuration for simplicity in this context) is \(\sigma_{2s}^2\sigma_{2s}^{*2}\sigma_{2p}^2\pi_{2p}^4\pi_{2p}^{*2}\)
The number of bonding electrons \(N_b = 8\) and the number of antibonding electrons \(N_a=4\)

Step2: Calculate the bond order

The formula for bond order \(BO=\frac{N_b - N_a}{2}\)
Substitute \(N_b = 8\) and \(N_a = 4\) into the formula: \(BO=\frac{8 - 4}{2}=2\)

Step3: Determine stability based on bond order

A bond order of \(2\) indicates a relatively strong bond. Generally, a non - zero bond order implies some degree of stability.

Answer:

Bond order: \(2\)
Stable