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Chapter 9: Problem 82

(a) The nitric oxide molecule, NO, readily loses one electron to form the NO \(^{+}\) ion. Which of the following is the best explanation of why this happens: (i) Oxygen is more electronegative than nitrogen, (ii) The highest energy electron in NO lies in a \(\pi_{2 p}^{*}\) molecular orbital, or (iii) The \(\pi_{2 p}^{*}\) MO in NO is completely filled. (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in NO, NO^, and NO', and describe the magnetic properties of each. (c) With what neutral homonuclear diatomic molecules are the NO \(^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

Short Answer

Expert verified
The best explanation for NO losing an electron to form NO鈦� is that the highest energy electron in NO lies in a 蟺鈧俻* molecular orbital. The order of N-O bond strengths in NO, NO鈦�, and NO鈦� is NO鈦� > NO > NO鈦�, and their magnetic properties are NO (paramagnetic), NO鈦� (diamagnetic), and NO鈦� (paramagnetic). The neutral homonuclear diatomic molecules isoelectronic with NO鈦� and NO鈦� are N鈧� and O鈧�, respectively.

Step by step solution

01

Evaluate the explanations for NO losing an electron

Let's assess each explanation from the problem statement: (i) Oxygen is more electronegative than nitrogen: Though it's true that oxygen is more electronegative than nitrogen, the statement alone doesn't fully explain why NO readily loses an electron. Electronegativity plays a role in determining the distribution of the electrons in the molecule but doesn't directly explain the ionization process. (ii) The highest energy electron in NO lies in a 蟺鈧俻* molecular orbital: This statement suggests that the last electron fills an antibonding orbital (indicated by the *), which means the electron is more readily lost to form the ion NO鈦�, with the loss of the electron resulting in an overall strengthening of the bond. (iii) The 蟺鈧俻* MO in NO is completely filled: This statement is incorrect. If the 蟺鈧俻* MO in NO were completely filled, losing an electron would destabilize the molecule. Instead, the highest energy molecular orbital is partially filled. Given these assessments, explanation (ii) is the best reason for why NO readily loses an electron.
02

Predict N-O bond strengths and magnetic properties

To predict the bond strengths in NO, NO鈦�, and NO鈦�, we use molecular orbital theory. The order of bond strengths will be determined by the number of bonding electrons minus the number of antibonding electrons. For magnetic properties, we'll consider whether the molecules have unpaired electrons. NO: Nitrogen has 7 electrons, and oxygen has 8 electrons, so NO has 15 electrons. Using the molecular orbitals, we can determine the electron configuration. NO has 1 unpaired electron, making it paramagnetic. The bond order is (10-5)/2 = 2.5. NO鈦�: When NO loses an electron, there will be 14 electrons in total. NO鈦� will have no unpaired electrons, which means it's diamagnetic. The bond order is (10-4)/2 = 3, so it has a stronger bond than NO. NO鈦�: When NO gains an electron, there will be 16 electrons in total. NO鈦� will have 2 unpaired electrons, making it paramagnetic. The bond order is (10-6)/2 = 2, so it has a weaker bond than NO. Therefore, the order of bond strengths is NO鈦� > NO > NO鈦�, and the magnetic properties are NO (paramagnetic), NO鈦� (diamagnetic), and NO鈦� (paramagnetic).
03

Identify isoelectronic neutral homonuclear diatomic molecules

We now need to find the neutral homonuclear diatomic molecules that have the same number of electrons as NO鈦� and NO鈦�. NO鈦� has 14 electrons. NO鈦� has 16 electrons. For the same number of electrons, we need to identify the neutral homonuclear diatomic molecules that have 14 or 16 electrons. Using the periodic table, we can identify these molecules: 14 electrons: Nitrogen has 7 electrons, so N鈧� has 14 electrons in total. 16 electrons: Oxygen has 8 electrons, so O鈧� has 16 electrons in total. Thus, the NO鈦� ion is isoelectronic with the N鈧� molecule, and the NO鈦� ion is isoelectronic with the O鈧� molecule.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nitric Oxide
Nitric oxide, often abbreviated as NO, is a diatomic molecule composed of one nitrogen and one oxygen atom. It is quite interesting in the realm of chemistry due to its ability to easily lose or gain an electron, affecting its properties significantly. In its neutral state, NO has a total of 15 electrons, making its electron arrangement such that it has one unpaired electron, resulting in paramagnetic properties. This unpaired electron resides in the
  • Highest energy orbital known as the \(\pi_{2p}^{*}\pi_{2p}^{*}\) molecular orbital, which is partially filled.

Due to the nature of this orbital, NO can readily lose an electron to form the NO\(^{+}\) ion, where the electron is removed from the antibonding orbital, thereby strengthening the bond. Its electronic configuration and molecular orbital arrangement make NO unique among small diatomic molecules.
Bond Strength
The bond strength in molecules like NO, NO\(^{+}\), and NO\(^{-}\) is profoundly impacted by their electronic configurations, as explained by molecular orbital theory. Bond strength is directly related to bond order, which can be calculated using the formula:\[ \text{Bond Order} = \frac{(\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons})}{2} \]Let's explore the bond strengths in:
  • NO: The bond order for NO is 2.5, calculated as the difference between the 10 bonding and 5 antibonding electrons, divided by 2. This bond order reflects a relatively strong bond.
  • NO\(^{+}\): With 14 total electrons, losing one antibonding electron results in a bond order of 3 (10 bonding minus 4 antibonding), indicating a stronger bond compared to neutral NO.
  • NO\(^{-}\): Gaining an electron results in a total of 16 electrons, with the bond order being 2 (10 bonding minus 6 antibonding), which indicates a weaker bond compared to both NO and NO\(^{+}\).
Each molecule's bond strength is a balance of these bonding and antibonding interactions, demonstrating how electron removal or addition can significantly influence molecular stability.
Magnetic Properties
The magnetic properties of a molecule are determined by the presence or absence of unpaired electrons in its molecular orbitals. Molecules with unpaired electrons are paramagnetic, while those with all electrons paired are diamagnetic. Let's examine the magnetic characteristics of NO, NO\(^{+}\), and NO\(^{-}\).
  • NO: With 15 electrons, it possesses one unpaired electron in the \(\pi_{2p}^{*}\) molecular orbital, making it paramagnetic.
  • NO\(^{+}\): When NO loses an electron, the resulting 14-electron NO\(^{+}\) species has all its electrons paired, characterizing it as diamagnetic.
  • NO\(^{-}\): With 16 electrons, two unpaired electrons are present, maintaining its paramagnetic nature.
The magnetic properties provide insight into the electronic structures, helping to visualize electron interactions in various molecular states.
Isoelectronic Molecules
Isoelectronic molecules or ions possess the same number of electrons, although they may differ in atomic composition or charge. Identifying isoelectronic species is essential for comparing molecular properties, as they often display similar behaviors. Let's correlate NO\(^{+}\) and NO\(^{-}\) with their isoelectronic neutral homonuclear diatomic counterparts.
  • NO\(^{+} \) : This ion has 14 electrons. A well-known neutral homonuclear diatomic molecule with 14 electrons is nitrogen gas (N\(_2\)), which also has 14 electrons (7 from each nitrogen atom).
  • NO\(^{-}\) : This ion contains 16 electrons. The neutral diatomic molecule oxygen gas (O\(_2\)) is isoelectronic with NO\(^{-}\), having 16 electrons (8 from each oxygen atom).
By understanding and identifying these isoelectronic relationships, one can make predictions about other physical and chemical properties based on the established characteristics of well-studied molecules.

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Most popular questions from this chapter

Name the proper three-dimensional molecular shapes for each of the following molecules or ions, showing lone pairs as needed: \((\mathbf{a}) \mathrm{ClO}_{2}^{-}(\mathbf{b}) \mathrm{SO}_{4}^{2-}(\mathbf{c}) \mathrm{NF}_{3}(\mathbf{d}) \mathrm{CCl}_{2} \mathrm{Br}_{2}(\mathbf{e}) \mathrm{SF}_{4}^{2+}\)

The following is part of a molecular orbital energy-level diagram for MOs constructed from 1 s atomic orbitals. (a) What labels do we use for the two MOs shown? (b) For which of the following molecules or ions could this be the energy-level diagram: $$ \mathrm{H}_{2} \mathrm{He}_{2}, \mathrm{H}_{2}^{+}, \mathrm{He}_{2}^{+}, \mathrm{or} \mathrm{H}_{2}^{-} ? $$ (c) What is the bond order of the molecule or ion? (d) If an electron is added to the system, into which of the MOs will it be added? [Section 9.7\(]\)

Shown here are three pairs of hybrid orbitals, with each set at a characteristic angle. For each pair, determine the type of hybridization, if any, that could lead to hybrid orbitals at the specified angle.

Benzaldehyde, \(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}\), is a fragrant substance responsible for the aroma of almonds. Its Lewis structure is O=Cc1cccc(C=O)c1 (a) What is the hybridization at each of the carbonatoms of the molecule? (b) What is the total number of valence electrons in benzaldehyde? (c) How many of the valence electrons are used to make \(\sigma\) bonds in the molecule? (d) How many valence electrons are used to make \(\pi\) bonds? (e) How many valence electrons remain in nonbonding pairs in the molecule?

The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is \(109.5^{\circ}\), the characteristic angle for tetrahedral molecules.
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