Problem 104
Question
Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: $$\therefore N=N=O^{\cdot} \leftrightarrow: N \equiv N-\vec{O}: \longleftrightarrow: N-N \equiv 0$$ Given the following bond lengths, $$\begin{aligned} &\mathrm{N}-\mathrm{N} \quad 167 \mathrm{pm} \quad \mathrm{N}=\mathrm{O} \quad 115 \mathrm{pm}\\\ &\mathrm{N}=\mathrm{N} \quad 120 \mathrm{pm} \quad \mathrm{N}-\mathrm{O} \quad 147 \mathrm{pm}\\\ &\mathrm{N} \equiv \mathrm{N} \quad 110 \mathrm{pm} \end{aligned}$$ rationalize the observations that the \(\mathrm{N}-\mathrm{N}\) bond length in \(\mathrm{N}_{2} \mathrm{O}\) is \(112 \mathrm{pm}\) and that the \(\mathrm{N}-\mathrm{O}\) bond length is \(119 \mathrm{pm}\). Assign formal charges to the resonance structures for \(\mathrm{N}_{2} \mathrm{O}\). Can you eliminate any of the resonance structures on the basis of formal charges? Is this consistent with observation?
Step-by-Step Solution
VerifiedKey Concepts
Resonance Structures
- N=N=O
- N≡N-O
- N-N≡O
Resonance helps explain why molecular properties, like bond lengths, are sometimes averages of single, double, or even triple bonds. None of the resonance structures alone fully describe the actual electron distribution of the molecule, but together, they provide a complete picture.
Bond Length Calculation
- N-N single bond: 167 pm
- N=N double bond: 120 pm
- N≡N triple bond: 110 pm
- N-O single bond: 147 pm
- N=O double bond: 115 pm
Formal Charges
- First structure (N=N=O): N(+1), N(0), O(-1)
- Second structure (N≡N-O): N(0), N(+1), O(-1)
- Third structure (N-N≡O): N(-1), N(0), O(+1)