Whether the vibration that formaldehyde undergoes is a symmetric stretch, asymmetric stretch, an in-plane bend, or an out-of-plane is to be classified. Concept introduction: The atoms of a molecule can undergo different types of regular oscillations in which interatomic distances and bond angles vary rhythmically. There are two types of vibrational motions that are stretching and bending. In stretching, the distance between two atoms in a chemical bond grows longer and shorter. If the distance between two atoms in a chemical bond is equal during the period of vibrations, it is symmetric stretching, and if the distance between two atoms in a chemical bond is unequal during the period of vibrations, it is asymmetric stretching. Whereas in bending vibration, an angle (either a bond angle or a dihedral angle) becomes larger or shorter. If all of the atoms that initially lie in the same plane remain in the same plane during the period of vibration, it is in-plane vibration, and if all of the atoms that initially lie in the same plane not remain in the same plane during the period of vibration, it is out-of-plane vibration

Question

Want to see the full answer?

Answer 1

Question

Chapter 15, Problem 15.14P

Interpretation Introduction

Interpretation:

Whether the vibration that formaldehyde undergoes is a symmetric stretch, asymmetric stretch, an in-plane bend, or an out-of-plane is to be classified.

Concept introduction:

The atoms of a molecule can undergo different types of regular oscillations in which interatomic distances and bond angles vary rhythmically. There are two types of vibrational motions that are stretching and bending. In stretching, the distance between two atoms in a chemical bond grows longer and shorter. If the distance between two atoms in a chemical bond is equal during the period of vibrations, it is symmetric stretching, and if the distance between two atoms in a chemical bond is unequal during the period of vibrations, it is asymmetric stretching. Whereas in bending vibration, an angle (either a bond angle or a dihedral angle) becomes larger or shorter. If all of the atoms that initially lie in the same plane remain in the same plane during the period of vibration, it is in-plane vibration, and if all of the atoms that initially lie in the same plane not remain in the same plane during the period of vibration, it is out-of-plane vibration

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 2

Question

Chapter 15, Problem 15.14P

Interpretation Introduction

Interpretation:

Whether the vibration that formaldehyde undergoes is a symmetric stretch, asymmetric stretch, an in-plane bend, or an out-of-plane is to be classified.

Concept introduction:

The atoms of a molecule can undergo different types of regular oscillations in which interatomic distances and bond angles vary rhythmically. There are two types of vibrational motions that are stretching and bending. In stretching, the distance between two atoms in a chemical bond grows longer and shorter. If the distance between two atoms in a chemical bond is equal during the period of vibrations, it is symmetric stretching, and if the distance between two atoms in a chemical bond is unequal during the period of vibrations, it is asymmetric stretching. Whereas in bending vibration, an angle (either a bond angle or a dihedral angle) becomes larger or shorter. If all of the atoms that initially lie in the same plane remain in the same plane during the period of vibration, it is in-plane vibration, and if all of the atoms that initially lie in the same plane not remain in the same plane during the period of vibration, it is out-of-plane vibration

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 15, Problem 15.14P

Interpretation Introduction

Interpretation:

Whether the vibration that formaldehyde undergoes is a symmetric stretch, asymmetric stretch, an in-plane bend, or an out-of-plane is to be classified.

Concept introduction:

The atoms of a molecule can undergo different types of regular oscillations in which interatomic distances and bond angles vary rhythmically. There are two types of vibrational motions that are stretching and bending. In stretching, the distance between two atoms in a chemical bond grows longer and shorter. If the distance between two atoms in a chemical bond is equal during the period of vibrations, it is symmetric stretching, and if the distance between two atoms in a chemical bond is unequal during the period of vibrations, it is asymmetric stretching. Whereas in bending vibration, an angle (either a bond angle or a dihedral angle) becomes larger or shorter. If all of the atoms that initially lie in the same plane remain in the same plane during the period of vibration, it is in-plane vibration, and if all of the atoms that initially lie in the same plane not remain in the same plane during the period of vibration, it is out-of-plane vibration

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 15, Problem 15.14P

Interpretation Introduction

Interpretation:

Whether the vibration that formaldehyde undergoes is a symmetric stretch, asymmetric stretch, an in-plane bend, or an out-of-plane is to be classified.

Concept introduction:

The atoms of a molecule can undergo different types of regular oscillations in which interatomic distances and bond angles vary rhythmically. There are two types of vibrational motions that are stretching and bending. In stretching, the distance between two atoms in a chemical bond grows longer and shorter. If the distance between two atoms in a chemical bond is equal during the period of vibrations, it is symmetric stretching, and if the distance between two atoms in a chemical bond is unequal during the period of vibrations, it is asymmetric stretching. Whereas in bending vibration, an angle (either a bond angle or a dihedral angle) becomes larger or shorter. If all of the atoms that initially lie in the same plane remain in the same plane during the period of vibration, it is in-plane vibration, and if all of the atoms that initially lie in the same plane not remain in the same plane during the period of vibration, it is out-of-plane vibration

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 15, Problem 15.14P

Interpretation Introduction

Interpretation:

Whether the vibration that formaldehyde undergoes is a symmetric stretch, asymmetric stretch, an in-plane bend, or an out-of-plane is to be classified.

Concept introduction:

The atoms of a molecule can undergo different types of regular oscillations in which interatomic distances and bond angles vary rhythmically. There are two types of vibrational motions that are stretching and bending. In stretching, the distance between two atoms in a chemical bond grows longer and shorter. If the distance between two atoms in a chemical bond is equal during the period of vibrations, it is symmetric stretching, and if the distance between two atoms in a chemical bond is unequal during the period of vibrations, it is asymmetric stretching. Whereas in bending vibration, an angle (either a bond angle or a dihedral angle) becomes larger or shorter. If all of the atoms that initially lie in the same plane remain in the same plane during the period of vibration, it is in-plane vibration, and if all of the atoms that initially lie in the same plane not remain in the same plane during the period of vibration, it is out-of-plane vibration

Answer 6

Chapter 15, Problem 15.14P

Interpretation Introduction

Interpretation:

Whether the vibration that formaldehyde undergoes is a symmetric stretch, asymmetric stretch, an in-plane bend, or an out-of-plane is to be classified.

Concept introduction:

The atoms of a molecule can undergo different types of regular oscillations in which interatomic distances and bond angles vary rhythmically. There are two types of vibrational motions that are stretching and bending. In stretching, the distance between two atoms in a chemical bond grows longer and shorter. If the distance between two atoms in a chemical bond is equal during the period of vibrations, it is symmetric stretching, and if the distance between two atoms in a chemical bond is unequal during the period of vibrations, it is asymmetric stretching. Whereas in bending vibration, an angle (either a bond angle or a dihedral angle) becomes larger or shorter. If all of the atoms that initially lie in the same plane remain in the same plane during the period of vibration, it is in-plane vibration, and if all of the atoms that initially lie in the same plane not remain in the same plane during the period of vibration, it is out-of-plane vibration

Answer 7

Chapter 15, Problem 15.14P

Interpretation Introduction

Interpretation:

Whether the vibration that formaldehyde undergoes is a symmetric stretch, asymmetric stretch, an in-plane bend, or an out-of-plane is to be classified.

Concept introduction:

The atoms of a molecule can undergo different types of regular oscillations in which interatomic distances and bond angles vary rhythmically. There are two types of vibrational motions that are stretching and bending. In stretching, the distance between two atoms in a chemical bond grows longer and shorter. If the distance between two atoms in a chemical bond is equal during the period of vibrations, it is symmetric stretching, and if the distance between two atoms in a chemical bond is unequal during the period of vibrations, it is asymmetric stretching. Whereas in bending vibration, an angle (either a bond angle or a dihedral angle) becomes larger or shorter. If all of the atoms that initially lie in the same plane remain in the same plane during the period of vibration, it is in-plane vibration, and if all of the atoms that initially lie in the same plane not remain in the same plane during the period of vibration, it is out-of-plane vibration

Answer 8

Expert Solution & Answer

Answer 9

Expert Solution & Answer

Answer 10

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 11

Ch. 15 - Prob. 15.1P Ch. 15 - Prob. 15.2P Ch. 15 - Prob. 15.3P Ch. 15 - Prob. 15.4P Ch. 15 - Prob. 15.5P Ch. 15 - Prob. 15.6P Ch. 15 - Prob. 15.7P Ch. 15 - Prob. 15.8P Ch. 15 - Prob. 15.9P Ch. 15 - Prob. 15.10P

Want to see the full answer?

Chapter 15 Solutions