### UV-Visible Spectrum Analysis:**Task:** Arrange the following compounds in order of increasing wavelength for the HOMO to LUMO transition in the UV-Visible spectrum.**Compounds:**1. (a) Bicyclic with one double bond in each ring.2. (b) Bicyclic with two conjugated double bonds.3. (c) Bicyclic with one double bond in each ring, adjacent to each other.4. (d) Bicyclic with three double bonds in total.**Order:**- **Shortest to Longest Wavelength:**1. (a) Pyridinium ion structure: benzene ring bonded to a nitrogen with a positive charge.2. (b) Benzene ring bonded to an ether group.3. (c) Benzene ring bonded to an amine group.4. (d) Benzene ring bonded to a methyl amine group with a positive charge on nitrogen.**Diagram Explanation:**The diagram shows four bicyclic compounds labeled (a), (b), (c), and (d) to be arranged in order of increasing wavelength based on their electronic transitions. Below it, another set of structures (a), (b), (c), and (d) are to be organized in terms of increasing wavelength as described.---### IR Spectrum Analysis:**Task:** Arrange the following bonds in order of increasing stretching frequency in the IR spectrum.**Bonds:**- Illustrated as part of a molecule with the following functional groups: - (a) The O-H bond in a carboxylic acid. - (b) The C=O bond in a carbonyl group. - (c) The C-O bond in an ether group. - (d) The O-H bond in an alcohol.**Order:**- **Lowest to Highest Frequency (cm⁻¹):**1. (a) O-H bond (carboxylic acid). 2. (b) C=O bond (carbonyl).3. (c) C-O bond (ether).4. (d) O-H bond (alcohol).**Diagram Explanation:**The diagram illustrates a molecule where the indicated bonds are marked for analysis of their stretching frequencies. Each bond is labeled (a), (b), (c), and (d), representing different functional groups that are organized in increasing order of their IR stretching frequencies.

Higher the conjugation in Molecules more will wavelength as there would be less energy Gap.

Arrange the following in order of increasing wavelength for the HOMO to LUMO transition in the UV- Visible spectrum. (a) (b) (c) (d) shortest wavelength longest wavelength (a) (b) (c) (d) shortest wavelength longest wavelength Arrange the following in order of increasing stretching frequency of the indicated bondsin the IR spectrum. H C- -CH2-0–H (a) (b) (c) (d) lowest frequency(cm') highest frequency (cm')

Arrange the following in order of increasing wavelength for the HOMO to LUMO transition in the UV- Visible spectrum. (a) (b) (c) (d) shortest wavelength longest wavelength (a) (b) (c) (d) shortest wavelength longest wavelength Arrange the following in order of increasing stretching frequency of the indicated bondsin the IR spectrum. H C- -CH2-0–H (a) (b) (c) (d) lowest frequency(cm') highest frequency (cm')

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Electronic Transitions and Spectroscopy

The term “electronic” connotes electron, and the term “transition” implies transformation. In a molecule, the electrons move from a lower to a higher energy state due to excitation. The two energy states, the ground state and the excited state are the lowest and the highest energy states, respectively. An energy change is observed with this transition, which depicts the various data related to the molecule.

Photoelectron Spectroscopy

Photoelectron spectroscopy (PES) is a part of experimental chemistry. It is a technique used in laboratories that involves projecting intense beams of radiation on a sample element. In response, the element ejects electrons for which the relative energies are measured.

Question

### UV-Visible Spectrum Analysis:

**Task:** Arrange the following compounds in order of increasing wavelength for the HOMO to LUMO transition in the UV-Visible spectrum.

**Compounds:**
1. (a) Bicyclic with one double bond in each ring.
2. (b) Bicyclic with two conjugated double bonds.
3. (c) Bicyclic with one double bond in each ring, adjacent to each other.
4. (d) Bicyclic with three double bonds in total.

**Order:**

- **Shortest to Longest Wavelength:**

1. (a) Pyridinium ion structure: benzene ring bonded to a nitrogen with a positive charge.
2. (b) Benzene ring bonded to an ether group.
3. (c) Benzene ring bonded to an amine group.
4. (d) Benzene ring bonded to a methyl amine group with a positive charge on nitrogen.

**Diagram Explanation:**

The diagram shows four bicyclic compounds labeled (a), (b), (c), and (d) to be arranged in order of increasing wavelength based on their electronic transitions. Below it, another set of structures (a), (b), (c), and (d) are to be organized in terms of increasing wavelength as described.

---

### IR Spectrum Analysis:

**Task:** Arrange the following bonds in order of increasing stretching frequency in the IR spectrum.

**Bonds:**

- Illustrated as part of a molecule with the following functional groups:
- (a) The O-H bond in a carboxylic acid.
- (b) The C=O bond in a carbonyl group.
- (c) The C-O bond in an ether group.
- (d) The O-H bond in an alcohol.

**Order:**

- **Lowest to Highest Frequency (cm⁻¹):**

1. (a) O-H bond (carboxylic acid).
2. (b) C=O bond (carbonyl).
3. (c) C-O bond (ether).
4. (d) O-H bond (alcohol).

**Diagram Explanation:**

The diagram illustrates a molecule where the indicated bonds are marked for analysis of their stretching frequencies. Each bond is labeled (a), (b), (c), and (d), representing different functional groups that are organized in increasing order of their IR stretching frequencies.

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