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?

Short Answer

Expert verified
The second resonance structure (N≡N-O) is the most plausible among all the resonance structures provided for Nitrous oxide (N2O) as it has the closest bond lengths to the observed values and consistent formal charges. The other resonance structures can be eliminated on the basis of formal charges and bond length consistency.

Step by step solution

01

Calculate Average Bond Lengths for N2O

To begin with, we will calculate the expected N-N and N-O bond lengths in N2O using the bond length data provided in the Lewis structures. Let's just focus on the bond lengths for now: - 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 In the first Lewis structure, we have a double bond between the two Nitrogen atoms and a single bond with an unshared electron on the Oxygen atom. While in the second Lewis structure, we have a triple bond between the two Nitrogen atoms and a single bond between N and O. Lastly, in the third Lewis structure, we have a single bond between the Nitrogen atoms and a double bond between N and O.
02

Calculate Bond Lengths in Resonance Structures

Now that we have defined the average bond lengths for N-N and N-O bonds, we will try to find the expected N-N and N-O bond lengths for each resonance structure using the following method: \[\text{Expected Bond Length} = \frac{\text{(Single Bond Length + Double Bond Length)}{2}\] For the first resonance structure (N=N=O.), N-N and N-O bond lengths can be calculated as: N-N bond length: \[\frac{(167 + 120)}{2} = 143.5 \mathrm{pm}\] N-O bond length: \[147 \mathrm{pm}\] (single bond) For the second resonance structure (N≡N-O), N-N and N-O bond lengths can be calculated as: N-N bond length: \[110 \mathrm{pm}\] (triple bond) N-O bond length: \[147 \mathrm{pm}\] (single bond) For the third resonance structure (N-N≡O), N-N and N-O bond lengths can be calculated as: N-N bond length: \[167 \mathrm{pm}\] (single bond) N-O bond length: \[\frac{(147 + 115)}{2} = 131 \mathrm{pm}\]
03

Compare the Calculated Bond Lengths and Assign Formal Charges

Next, we will compare the calculated bond lengths from each resonance structure to see which is consistent with the observed bond lengths given (N-N: 112 pm, N-O: 119 pm). For the first resonance structure (N=N=O.): N-N bond length: \(143.5 \mathrm{pm}\) vs Observed: \(112 \mathrm{pm}\) N-O bond length: \(147 \mathrm{pm}\) vs Observed: \(119 \mathrm{pm}\) For the second resonance structure (N≡N-O): N-N bond length: \(110 \mathrm{pm}\) vs Observed: \(112 \mathrm{pm}\) N-O bond length: \(147 \mathrm{pm}\) vs Observed: \(119 \mathrm{pm}\) For the third resonance structure (N-N≡O): N-N bond length: \(167 \mathrm{pm}\) vs Observed: \(112 \mathrm{pm}\) N-O bond length: \(131 \mathrm{pm}\) vs Observed: \(119 \mathrm{pm}\) Now we'll assign formal charges to each resonance structure: First resonance structure (N=N=O.): Formal charges: N: +1, N: 0, O: -1 Second resonance structure (N≡N-O): Formal charges: N: 0, N: +1, O: -1 Third resonance structure (N-N≡O): Formal charges: N: -1, N: 0, O: +1
04

Analyze and Conclude

Between the calculated bond lengths and the formal charges of each resonance structure, we can see that the second resonance structure (N≡N-O) has the closest bond lengths to the observed values. Further, the formal charges of each atom in this Lewis structure are consistent with their overall bonding patterns. The other resonance structures are less consistent with the observed bond lengths and have more polar atoms leading to instability. Thus, we can eliminate the first and third resonance structures on the basis of formal charges and bond length consistency. Therefore, the second resonance structure (N≡N-O) is the most plausible among all the resonance structures provided for Nitrous oxide (N2O).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Predict the type of bond (ionic, covalent, or polar covalent) one would expect to form between the following pairs of elements. a. \(\mathrm{Rb}\) and \(\mathrm{Cl}\) b. \(S\) and \(S\) c. \(C\) and \(F\) d. Ba and S e. \(\mathrm{N}\) and \(\mathrm{P}\) f. \(B\) and \(H\)

Given the following information: Energy of sublimation of \(\mathrm{Li}(s)=166 \mathrm{kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{HCl}=427 \mathrm{kJ} / \mathrm{mol}\) Ionization energy of \(\mathrm{Li}(g)=520 . \mathrm{kJ} / \mathrm{mol}\) Electron affinity of \(\mathrm{Cl}(g)=-349 \mathrm{kJ} / \mathrm{mol}\) Lattice energy of \(\mathrm{LiCl}(s)=-829 \mathrm{kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{H}_{2}=432 \mathrm{kJ} / \mathrm{mol}\) Calculate the net change in energy for the following reaction: $$ 2 \mathrm{Li}(s)+2 \mathrm{HCl}(g) \longrightarrow 2 \mathrm{LiCl}(s)+\mathrm{H}_{2}(g) $$

Name the compounds in parts a-d and write the formulas for the compounds in parts e-h. a. \(\mathrm{Hg}_{2} \mathrm{O}\) b. \(\operatorname{FeBr}_{3}\) c. CoS d. \(\mathrm{TiCl}_{4}\) e. \(\operatorname{tin}(\text { II })\) nitride f. cobalt(III) iodide g. mercury(II) oxide h. chromium(VI) sulfide

Lewis structures can be used to understand why some molecules react in certain ways. Write the Lewis structures for the reactants and products in the reactions described below. a. Nitrogen dioxide dimerizes to produce dinitrogen tetroxide. b. Boron trihydride accepts a pair of electrons from ammonia, forming \(\mathrm{BH}_{3} \mathrm{NH}_{3}\) Give a possible explanation for why these two reactions occur.

Without using Fig. \(3-4,\) predict the order of increasing electronegativity in each of the following groups of elements. a. \(C, N, O\) b. \(\mathbf{s}, \mathbf{S e}, \mathbf{C l}\) \(\mathbf{c}_{*} \mathrm{Si}, \mathrm{Ge}, \mathrm{Sn}\) d. \(\mathrm{TI}, \mathrm{S}, \mathrm{Ge}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free