Question

Question: Nitrous oxide (N₂O) has these possible Louis structure:

Given the following bond lengths,

$$\begin{gathered} \mathbf{N}\mathbf{-}\mathbf{N}\mathbf{167}\mathbf{pm}\mathbf{N}\mathbf{=}\mathbf{O}\mathbf{115}\mathbf{pm} \\ \mathbf{N}\mathbf{=}\mathbf{N}\mathbf{120}\mathbf{pm}\mathbf{N}\mathbf{-}\mathbf{O}\mathbf{147}\mathbf{pm} \\ \mathbf{N}\mathbf{\equiv }\mathbf{N}\mathbf{110}\mathbf{pm} \end{gathered}$$

Rationalize the observations that the N-N bond length in N₂O isand that the N-O bond length is 119 Pm. Assign formal charges to the resonance structures on the basis of formal charges? Is this consistent with observation?

Short answer

Answer

The nitrogen-nitrogen bond length (112 pm) is in between a double bond (120 pm) and triple bond (115 pm). This can be explained by both resonance and formal charge.

Step-by-step solution

Draw the resonance structures of N2O.

Step 2:

We can easily eliminate$N-N\equiv O$because of formal charge on oxygen and also because there is no $N\equiv O$or N- N bond.

This explains why the N-N bond is intermediate between a double and triple bond whereas, N-O bond is found to be between a single and double bond.