Question

part 2 (1 point)
the average single and double n - o bond lengths are 136 pm and 122 pm, respectively. assuming these two structures were to make equal contributions to the observed bonding in hno₂, predict the n - o bond length.

part 3 (1 point)
do you think these two structures contribute equally to the observed molecular structure of hno₂?
choose one:
yes
no
no way to tell

Explanation:

Step1: Calculate average bond - length

Since the two structures make equal contributions, we use the formula for the weighted - average. The formula for the average of two values \(x_1\) and \(x_2\) with equal weights is \(\frac{x_1 + x_2}{2}\). Here, \(x_1=136\) pm and \(x_2 = 122\) pm. So, the average \(=\frac{136+122}{2}\).
\(\frac{136 + 122}{2}=\frac{258}{2}=129\) pm

Step2: Analyze resonance contribution

The two resonance structures of \(HNO_2\) do not contribute equally. The more stable resonance structure contributes more to the actual structure. The structure with the double - bond has less formal charge separation and is more stable. So, the answer is no.

Answer:

Part 2: 129
Part 3: B. No