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Chapter 11: Problem 37

Consider the molecules \(\mathrm{NO}^{+}\) and \(\mathrm{N}_{2}^{+}\) and use molecular orbital theory to answer the following: (a) Write the molecular orbital configuration of each ion (ignore the 1 s electrons). (b) Predict the bond order of each ion. (c) Which of these ions is paramagnetic? Which is diamagnetic? (d) Which of these ions do you think has the greater bond length? Explain.

Short Answer

Expert verified
\(\mathrm{NO}^{+}\) has a bond order of 3 and is diamagnetic, while \(\mathrm{N}_{2}^{+}\) has a bond order of 3.5 and is paramagnetic. Therefore, \(\mathrm{N}_{2}^{+}\) has a shorter bond length.

Step by step solution

01

Molecular Orbital Configuration

We first need to determine the electron configuration. Nitrogen atom has 7 electrons, Oxygen atom has 8. \(\mathrm{NO}^{+}\) loses one electron, so it has 14 electrons. The molecular orbital configuration for 14 electrons is: \(\sigma_{1s}^{2}\), \(\sigma_{2s}^{2}\), \(\sigma_{2p}^{2}\), \(\pi_{2p}^{4}\), \(\sigma_{2p}^{2}\), \(\pi_{2p}^{2}\). \(\mathrm{N}_{2}^{+}\) has 13 electrons. The molecular orbital configuration for 13 electrons is: \(\sigma_{1s}^{2}\), \(\sigma_{2s}^{2}\), \(\sigma_{2p}^{2}\), \(\pi_{2p}^{4}\), \(\sigma_{2p}^{2}\), \(\pi_{2p}^{1}\).
02

Predict the Bond Order

Next, we have to calculate the bond order. The formula for bond order is \(\frac{1}{2}(n_{b} - n_{a})\), where \(n_{b}\) is the number of electrons in bonding orbitals and \(n_{a}\) is the number of electrons in antibonding orbitals. For \(\mathrm{NO}^{+}\), it has 10 bonding electrons and 4 antibonding electrons, giving a bond order of 3. For \(\mathrm{N}_{2}^{+}\), it has 10 bonding electrons and 3 antibonding electrons, hence bond order of 3.5.
03

Check for Paramagnetism or Diamagnetism

A molecule is paramagnetic if it has unpaired electrons, and diamagnetic if all electrons are paired. In \(\mathrm{NO}^{+}\), all electrons are paired, so it is diamagnetic. In \(\mathrm{N}_{2}^{+}\), there is one unpaired electron, hence it is paramagnetic.
04

Determine the Bond Length

Lastly, when comparing bond lengths, it’s useful to remember that, all else being equal, a greater bond order generally indicates a shorter bond length. Therefore, \(\mathrm{N}_{2}^{+}\), with a bond order of 3.5, likely has a shorter bond length than \(\mathrm{NO}^{+}\) with a bond order of 3.

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Most popular questions from this chapter

Methyl nitrate, \(\mathrm{CH}_{3} \mathrm{NO}_{3}\), is used as a rocket propellant. The skeletal structure of the molecule is \(\mathrm{CH}_{3} \mathrm{ONO}_{2}\). The N and three O atoms all lie in the same plane, but the \(\mathrm{CH}_{3}\) group is not in the same plane as the \(\mathrm{NO}_{3}\) group. The bond angle \(\mathrm{C}-\mathrm{O}-\mathrm{N}\) is \(105^{\circ},\) and the bond angle \(\mathrm{O}-\mathrm{N}-\mathrm{O}\) is \(125^{\circ} .\) One nitrogen-to-oxygen bond length is \(136 \mathrm{pm},\) and the other two are \(126 \mathrm{pm}\) (a) Draw a sketch of the molecule showing its geometric shape. (b) Label all the bonds in the molecule as \(\sigma\) or \(\pi\), and indicate the probable orbital overlaps involved. (c) Explain why all three nitrogen-to-oxygen bond lengths are not the same.

A solar cell that is \(15 \%\) efficient in converting solar to electric energy produces an energy flow of \(1.00 \mathrm{kW} / \mathrm{m}^{2}\) when exposed to full sunlight. (a) If the cell has an area of \(40.0 \mathrm{cm}^{2}\), what is the power output of the cell, in watts? (b) If the power calculated in part (a) is produced at \(0.45 \mathrm{V},\) how much current does the cell deliver?

Show that both the valence-bond method and molecular orbital theory provide an explanation for the existence of the covalent molecule \(\mathrm{Na}_{2}\) in the gaseous state. Would you predict \(\mathrm{Na}_{2}\) by the Lewis theory?

For the following pairs of molecular orbitals, indicate the one you expect to have the lower energy, and state the reason for your choice. (a) \(\sigma_{1 s}\) or \(\sigma_{1 s}^{*} ;\) (b) \(\sigma_{2 s}\) or \(\sigma_{2 p}\) (c) \(\sigma_{1 s}^{*}\) or \(\sigma_{2 s} ;\) (d) \(\sigma_{2 p}\) or \(\sigma_{2 p}^{*}\)

The structure of the molecule allene, \(\mathrm{CH}_{2} \mathrm{CCH}_{2}\), is shown here. Propose hybridization schemes for the \(C\) atoms in this molecule.
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