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Chapter 9: Problem 58

The nitrogen atoms in \(N_{2}\) participate in multiple bonding, whereas those in hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), do not. (a) Draw Lewis structures for both molecules. (b) What is the hybridization of the nitrogen atoms in each molecule? (c) Which molecule has the stronger \(\mathrm{N}-\mathrm{N}\) bond?

Short Answer

Expert verified
(a) Lewis structures: N鈧� has a triple bond between the nitrogen atoms (N鈮) each with one lone pair of electrons, while N鈧侶鈧� has single bonds between nitrogen atoms (N-N) with each nitrogen bonded to two hydrogens and having one lone pair of electrons (H-N-N-H, with hydrogens single-bonded on the sides of nitrogen atoms). (b) Hybridization: For N鈧�, each nitrogen atom has an sp hybridization due to two electron groups (one triple bond and one lone pair). In N鈧侶鈧�, each nitrogen atom has sp鲁 hybridization with four electron groups (one single bond with another nitrogen, two single bonds with hydrogens, and one lone pair). (c) The N-N bond in N鈧� is stronger due to the triple bond between the nitrogen atoms, while in N鈧侶鈧�, the N-N bond is a single bond.

Step by step solution

01

Drawing Lewis Structures

We will begin with drawing the Lewis structures for both molecules. We need to remember the valence electrons for each atom and distribute them accordingly. 1) N鈧�: Nitrogen has 5 valence electrons. In the N鈧� molecule we have two nitrogen atoms, hence a total of 10 valence electrons. These electrons can be distributed as follows: N鈮 Here, we have a triple bond between the two nitrogen atoms and each nitrogen atom has one lone pair of electrons, making a total of 10 valence electrons. 2) N鈧侶鈧� (Hydrazine): In hydrazine, we have two nitrogen atoms and four hydrogen atoms. Hydrogen has 1 valence electron each, giving a total of 4 valence electrons for 4 hydrogen atoms. Nitrogen has 5 valence electrons, so two nitrogen atoms together have 10 valence electrons. Therefore, N鈧侶鈧� has a total of 14 valence electrons. These electrons can be distributed as follows: H | N-N | | H H In this structure, the nitrogen atoms are single-bonded to one another and each nitrogen atom has two single bonds with two hydrogen atoms. Each nitrogen also has one lone pair of electrons, adding up to the total of 14 valence electrons.
02

Determine Hybridization of Nitrogen Atoms

To determine the hybridization, we will count the number of electron groups around each nitrogen atom (where "electron group" means either a single, double or triple bond, or a lone pair) and use the result to identify the hybridization in each case. 1) N鈧�: Each nitrogen atom has one triple bond (1 electron group) and one lone pair (1 electron group). Therefore, there are 2 electron groups around each nitrogen atom. This configuration corresponds to sp hybridization. 2) N鈧侶鈧� (Hydrazine): Each nitrogen atom has one single bond with its neighbor nitrogen atom (1 electron group), two single bonds with two hydrogen atoms (2 electron groups), and one lone pair (1 electron group). Therefore, there are 4 electron groups around each nitrogen atom. This configuration corresponds to sp鲁 hybridization.
03

Compare N-N bond strength

To compare the strength of the N-N bonds in both molecules, we can use the information about their bonding: 1) N鈧�: The N-N bond in N鈧� is a triple bond, which consists of 1 sigma bond and 2 pi bonds. 2) N鈧侶鈧� (Hydrazine): In hydrazine, the N-N bond is a single sigma bond. Triple bonds are stronger than single bonds. Thus, the N-N bond in the N鈧� molecule is stronger than the N-N bond in the N鈧侶鈧� (hydrazine) molecule.

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Key Concepts

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

Lewis Structures
Lewis structures are a useful way to visualize the distribution of valence electrons in molecules. They provide insight into how atoms bond and the shape of a molecule. Let's examine the Lewis structures for nitrogen (\(N_2\)) and hydrazine (\(N_2H_4\)).

- For \(N_2\), each nitrogen atom has 5 valence electrons, making a total of 10 for the molecule. The Lewis structure shows a triple bond between the nitrogen atoms, which means there are three shared pairs of electrons. Each nitrogen also has one lone pair of electrons. This structure completes each nitrogen atom's valence shell, achieving the stable configuration of 8 electrons.

- In hydrazine (\(N_2H_4\)), there are two nitrogen atoms and four hydrogen atoms. Hydrogens contribute 4 valence electrons (1 each), and nitrogen offers 10. Therefore, hydrazine has 14 electrons available for bonding. The Lewis structure depicts each nitrogen single bonded to another nitrogen atom and also bonded to two hydrogen atoms, with lone pairs remaining on each nitrogen. This setup results in a single bond between the two nitrogen atoms, differing significantly from the triple bond structure in \(N_2\).
Hybridization
The concept of hybridization helps us understand the bonding and geometry around central atoms in a molecule. It involves the mixing of atomic orbitals to form new hybrid orbitals which can then bond with other atoms.

- In \(N_2\), hybridization of the nitrogen atoms occurs as \(sp\). Each nitrogen atom in \(N_2\) forms one triple bond and has a lone pair, totaling two electron groups. The \(sp\) hybridization is a result of mixing one s orbital with one p orbital, forming two identical \(sp\) hybrid orbitals, ideal for linear geometry, explaining the strong linear triple bonds.

- In contrast, hydrazine (\(N_2H_4\)) displays \(sp^3\) hybridization for its nitrogen atoms. Each nitrogen forms single bonds with one another, two hydrogen atoms, and has one lone pair, totaling four electron groups. This \(sp^3\) hybridization results from mixing one s orbital with three p orbitals, giving four equivalent orbitals that adapt to a tetrahedral shape, which matches the three-dimensional structure of the molecule.
Bond Strength
Bond strength refers to how strongly atoms are held together. It is crucial in determining a molecule's stability and reactivity.

- In a nitrogen molecule (\(N_2\)), the \(N-N\) bond consists of a triple bond. This bond type comprises one sigma bond and two pi bonds. Triple bonds are the strongest of covalent bonds because more electrons are shared between the atoms, creating a stronger attraction.

- On the other hand, hydrazine (\(N_2H_4\)) contains a single \(N-N\) bond, which incorporates just one sigma bond with no pi components. Single bonds are typically weaker than multiple bonds because they involve fewer shared electrons. Consequently, the \(N-N\) bond in \(N_2\) is significantly stronger than in \(N_2H_4\).

Understanding these differences in bond types and strengths sheds light on why some molecules are more stable or reactive than others. The \(N_2\) molecule, with its robust triple bond, is much more resilient than hydrazine, which has weaker single bonds.

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Most popular questions from this chapter

The energy-level diagram in Figure 9.36 shows that the sideways overlap of a pair of porbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding \(\pi\) orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the \(\pi_{2 p}\) to the $\pi_{2 p}^{\star}$ molecular orbital. (a) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the HOMO in ethylene? (b) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the LUMO in ethylene? (c) Is the C-Cbond in ethylene stronger or weaker in the excited state than in the ground state? Why? (d) Is the \(C-C\) bond in ethylene easier to twist in the ground state or in the excited state?

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2 p}^{*}\) orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\)-axis is vertical in the plane of the paper and the \(x\)-axis horizontal. Write " \(\mathrm{M}^{"}\) at the origin to denote a metal atom. (b) Now, on the \(x\) axis to the right of \(M\), draw the Lewis structure of a CO molecule, with the carbon nearest the \(M\). The CO bond axis should be on the \(x\)-axis. (c) Draw the \(\mathrm{CO} \pi_{2 p}^{*}\) orbital, with phases (see the "Closer Look" box on phases) in the plane of the paper. Two lobes should be pointing toward M. (d) Now draw the \(d_{x y}\) orbital of \(\mathrm{M}\), with phases. Can you see how they will overlap with the \(\pi_{2}^{*}\) orbital of \(\mathrm{CO}\) ? (e) What kind of bond is being made with the orbitals between \(M\) and \(\mathrm{C}_{,} \sigma\) or \(\pi\) ? (f) Predict what will happen to the strength of the CO bond in a metal\(\mathrm{CO}\) complex compared to \(\mathrm{CO}\) alone.

(a) What is meant by the term orbital overlap? (b) Describe what a chemical bond is in terms of electron density between two atoms.

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) $$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a \(\sigma\) bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two \(s\) orbitals combine to form a \(\pi\) bond? Explain.
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