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Consider the Lewis structure for glycine, the simplest amino acid: (a) What are the approximate bond angles about each of the two carbon atoms, and what are the hybridizations of the orbitals on each of them? (b) What are the hybridizations of the orbitals on the two oxygens and the nitrogen atom, and what are the approximate bond angles at the nitrogen? (c) What is the total number of \(\sigma\) bonds in the entire molecule, and what is the total number of \(\pi\) bonds?

Short Answer

Expert verified
The first carbon atom (C1) is sp3 hybridized with bond angles of approximately 109.5°. The second carbon atom (C2) is sp2 hybridized with bond angles of approximately 120°. The nitrogen atom is sp3 hybridized with bond angles of approximately 109.5°. The first oxygen atom (O1) is sp2 hybridized, while the second oxygen atom (O2) is sp3 hybridized. There are a total of 7 σ bonds and 1 π bond in the glycine molecule.

Step by step solution

01

Carbon atom 1 (C1) hybridization

The first carbon atom (C1) is connected to a nitrogen atom, two hydrogen atoms, and the second carbon atom (C2). Since it has four electron domains, it is sp3 hybridized.
02

Carbon atom 1 (C1) bond angles

The bond angles in a perfect tetrahedron (sp3 hybridization) are approximately 109.5°, so the bond angles around C1 should be around 109.5°.
03

Carbon atom 2 (C2) hybridization

The second carbon atom (C2) is connected to the first carbon atom (C1), a double-bonded oxygen atom, and an oxygen atom with a hydrogen atom. It has three electron domains and is sp2 hybridized.
04

Carbon atom 2 (C2) bond angles

The bond angles in a trigonal planar (sp2 hybridization) are approximately 120°, so the bond angles around C2 should be around 120°. (b) Hybridizations on the two oxygen atoms and the nitrogen atom, and bond angles at the nitrogen atom.
05

Nitrogen (N) hybridization

The nitrogen atom is connected to two hydrogen atoms and the first carbon atom (C1). As there are three electron domains, the nitrogen atom is sp3 hybridized.
06

Nitrogen (N) bond angles

The bond angles in a perfect tetrahedron (sp3 hybridization) are approximately 109.5°, so the bond angles around the nitrogen atom should also be around 109.5°.
07

Oxygen 1 (O1) hybridization

The first oxygen atom (O1) is connected to the second carbon atom (C2) through a double bond. It has two electron domains and is sp2 hybridized.
08

Oxygen 2 (O2) hybridization

The second oxygen atom (O2) is connected to the second carbon atom (C2) and hydrogen atom. It has three electron domains and is sp3 hybridized. (c) Total number of σ bonds and π bonds in the entire glycine molecule.
09

Number of σ bonds

Counting the single bonds in the molecule: N-H, N-C, C-H x2, C-C, C-O, and O-H, we have a total of 7 σ bonds.
10

Number of π bonds

Counting the π bonds in the molecule, the only one is the double bond between C2 and O1. Thus, there is 1 π bond.

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Key Concepts

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

Bond Angles
In chemistry, bond angles are the angles between adjacent bonds emanating from the same atom. These angles are crucial as they define the 3D geometry of a molecule, influencing its reactivity and interaction with other molecules. Generally, bond angles depend on the hybridization of the atom and the number of electron domains around it.

- **For sp3 hybridization**, like in the case of the first carbon atom (C1) in glycine, bond angles are about 109.5°. This is the angle you'd find in a perfect tetrahedral shape, where a central atom is bonded to four other atoms with no lone pairs to disrupt the symmetry.
- **For sp2 hybridization**, which is seen in the second carbon atom (C2) and one of the oxygens in glycine, bond angles are around 120°. This configuration arises from a trigonal planar structure where three bonds need to share 360° equally around the central atom. The nitrogen atom in glycine also showcases sp3 hybridization, resulting in bond angles of about 109.5°. This similar tetrahedral structure can form due to the atom being bound to other atoms and sometimes lone pairs.
Hybridization of Orbitals
Hybridization is a concept that describes the mixing of atomic orbitals into new hybrid orbitals of equivalent energy. This mixing better accounts for the geometry and bonding properties of molecules.

- **sp3 hybridization** involves the mixing of one s orbital and three p orbitals, resulting in four hybridized orbitals. Each orbital forms one bond, seen in the tetrahedral arrangement around carbon atom 1 (C1) and nitrogen in glycine.
- **sp2 hybridization** involves the mixing of one s orbital with two p orbitals, leading to three planar hybrid orbitals. Carbon atom 2 (C2) and one oxygen atom ( ext{O1 ext{) feature this hybridization, causing bond angles around 120°. Hybridization explains the bond angles and molecular shapes, allowing predictions about molecular interactions. As evidenced in molecules like glycine, different atoms within a molecule can have distinct hybridization states, adapting to their bonding environments.
Sigma and Pi Bonds
In the context of covalent bonding, sigma ( σ) and pi ( π) bonds describe the types of covalent bonds formed between atoms. These bonds have unique properties and different roles in molecular structure.

- **Sigma bonds ( σ)** are formed by the head-on overlap of orbitals. They are the strongest type of covalent bond and allow for free rotation around the bond axis. In glycine, every single bond corresponds to a sigma bond, tallying up to seven σ bonds in the molecule.
- **Pi bonds ( π)** arise from the side-on overlap of p orbitals. These bonds are less rigid and usually come in addition to a sigma bond. Glycine has one π bond, found in the double bond between carbon atom 2 (C2) and one of the oxygen atoms, giving it unique reactivity and properties. Understanding sigma and pi bonds helps comprehend molecular stability and reactions, as they define how molecules can interact or transform in chemical reactions.
Electron Domains
Electron domains refer to the regions around a central atom where electrons are likely to be found, including bonded atoms and lone pairs. Identifying these domains is critical for predicting molecular shapes and bond angles using the VSEPR (Valence Shell Electron Pair Repulsion) theory.

- A lone pair or atom bonded to the central atom counts as one electron domain. For instance, in glycine, the nitrogen atom has three electron domains - two for the hydrogen atoms and one for the lone pair. This total gives a tetrahedral-like arrangement, leading to approximately 109.5° bond angles.
- Similarly, carbon atom 1 (C1), with its four bonding partners, has four electron domains, fitting an sp3 hybrid model, whereas carbon atom 2 (C2) holds three electron domains corresponding to its sp2 hybrid structure.
Electron domains are a simplistic yet effective way to predict the geometry of a molecule, significantly impacting its physical and chemical properties. They provide a foundational understanding needed to explain molecular shapes and behavior in different environments.

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Most popular questions from this chapter

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(0.96 \mathrm{~A}\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ}\). The dipole moment of the water molecule is \(1.85 \mathrm{D}\). (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? In what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxygen?

What is the hybridization of the central atom in (a) \(\mathrm{SiCl}_{4}\) (b) \(\mathrm{HCN},(\mathrm{c}) \mathrm{SO}_{3}\), (d) \(\mathrm{ICl}_{2}^{-}\), (e) \(\mathrm{BrF}_{4}\) ?

Many compounds of the transition-metal elements contain direct bonds between metal atoms. We will assume that the \(z\) -axis is defined as the metal-metal bond axis. (a) Which of the \(3 d\) orbitals (Figure 6.24) can be used to make a \(\sigma\) bond between metal atoms? (b) Sketch the \(\sigma_{3 d}\) bonding and \(\sigma_{3 d}^{*}\) antibonding MOs. (c) With reference to the "Closer Look" box on the phases of orbitals, explain why a node is generated in the \(\sigma_{3 d}^{*}\) MO. (d) Sketch the energy-level diagram for the \(\mathrm{Sc}_{2}\) molecule, assuming that only the \(3 d\) orbital from part (a) is important. (e) What is the bond order in \(\mathrm{Sc}_{2} ?\)

(a) What are the relationships among bond order, bond length, and bond energy? (b) According to molecular orbital theory, would either \(\mathrm{Be}_{2}\) or \(\mathrm{Be}_{2}{ }^{7}\) be expected to exist? Explain.

What are the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom? (a) three bonding domains and no nonbonding domains, (b) three bonding domains and one nonbonding domain, (c) two bonding domains and two nonbonding domains.

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