Problem 12 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

In the infrared spectrum of hydrogen iodide, \(\bar{V}(\mathrm{HI})\) is observed at \(2310 \mathrm{cm}^{-1}\). Predict the value of \(\bar{V}(\mathrm{D} \mathrm{I})\) in the deuterated analogue. (Section 25.4)

Step-by-Step Solution

Answer

The predicted value of \(\bar{V}(\text{DI})\) is approximately 1633 cm\(^{-1}\).

Step 1: Understand the Concept

The frequency of vibration in molecular bonds depends on the reduced mass of the atoms involved and follows the expression \( \bar{V} = \frac{1}{2\pi c} \sqrt{\frac{k}{\mu}} \), where \( \mu \) is the reduced mass, \( k \) is the force constant, and \( c \) is the speed of light. Since the force constant and speed of light are constants, changes in vibration frequency arise due to the change in reduced mass.

Step 2: Calculate the Reduced Mass for H-I

The reduced mass \( \mu \) is given by \( \mu = \frac{m_1 m_2}{m_1 + m_2} \), where \( m_1 \) and \( m_2 \) are the atomic masses. For hydrogen iodide (HI), \( m_1 = 1 \) amu for H and \( m_2 = 127 \) amu for Iodine. Thus, \[ \mu(\mathrm{HI}) = \frac{1 \times 127}{1 + 127} = \frac{127}{128} \approx 0.9922 \text{ amu} \].

Step 3: Calculate the Reduced Mass for D-I

Substitute deuterium (D, atomic mass = 2 amu) in place of hydrogen to form deuterium iodide (DI). The reduced mass \( \mu \) is \( \mu(\mathrm{DI}) = \frac{2 \times 127}{2 + 127} = \frac{254}{129} \approx 1.9682 \text{ amu} \).

Step 4: Apply the Reduced Mass Ratio

The frequency \( \bar{V} \) is inversely proportional to the square root of the reduced mass. Thus, \( \frac{\bar{V}(\mathrm{DI})}{\bar{V}(\mathrm{HI})} = \sqrt{\frac{\mu(\mathrm{HI})}{\mu(\mathrm{DI})}} \). Substitute the values to find \( \frac{2310 \text{ cm}^{-1}}{\bar{V}(\mathrm{DI})} = \sqrt{\frac{0.9922}{1.9682}} \).

Step 5: Solve for \(\bar{V}(\mathrm{DI})\)

Calculate \( \sqrt{\frac{0.9922}{1.9682}} \approx 0.707 \). Then, \( 2310 \times 0.707 = 1633 \text{ cm}^{-1} \) which is \( \bar{V}(\mathrm{DI}) \).

Key Concepts

Vibration Frequency

The concept of vibration frequency in infrared spectroscopy is key to understanding molecular motion. Molecules can be compared to tiny oscillating springs. The vibration frequency of these molecules depends on several factors:

• Reduced mass of the molecule.
• Force constant of the bond.

The vibration frequency, represented by \( \bar{V} \), is calculated with the equation:\[ \bar{V} = \frac{1}{2\pi c} \sqrt{\frac{k}{\mu}} \]where:

• \( k \) is the force constant of the bond reflecting bond strength.
• \( \mu \) is the reduced mass, indicating how the mass is shared between two bonded atoms.
• \( c \) is the speed of light.

Since \( k \) and \( c \) are constants in many scenarios, variations in \( \bar{V} \) typically arise from changes in the reduced mass \( \mu \). This tells us how changing atoms in a bond impacts the vibrational properties.

Reduced Mass

The reduced mass \( \mu \) plays a pivotal role in determining the vibration frequency of a molecular bond. It's a measure of the 'effective' mass when two atoms are bonded together and is calculated using:\[ \mu = \frac{m_1 m_2}{m_1 + m_2} \]Here, \( m_1 \) and \( m_2 \) represent the masses of the two atoms in the molecule.

In practical terms, the reduced mass balances the contribution of each atom in the bond. It simplifies the two-body problem into a one-body problem, making calculations manageable.

The smaller the difference in mass between the two atoms, the higher the reduced mass, and wise versa. A higher reduced mass generally leads to lower vibration frequencies. In bonds, the reduced mass is crucial as it influences how molecules vibrate, which is directly observable in infrared spectroscopy.

Hydrogen Iodide

Hydrogen iodide (HI) is a diatomic molecule composed of hydrogen and iodine. In the context of infrared spectroscopy, HI is used to study the vibration frequencies between these atoms.

In HI, the hydrogen atom, gives a small mass, while iodine, with a larger atomic mass, dominates the reduced mass calculation. Thus, we calculate the reduced mass as:\[ \mu(\mathrm{HI}) = \frac{1 \times 127}{1 + 127} \approx 0.9922 \text{ amu} \]This calculation reveals how the heavier iodine influences the molecule's vibration frequency, which is notably high due to iodine's larger mass role in the system.

Studying HI in IR spectroscopy provides a foundational understanding of how bond vibrations can indicate molecular characteristics, aiding in identifying substances based on their unique vibration frequencies.

Deuterium Iodide

Deuterium iodide (DI) is the deuterated form of hydrogen iodide, where deuterium replaces hydrogen. In DI, deuterium, with a mass of 2 amu (atomic mass units), impacts the reduced mass as:\[ \mu(\mathrm{DI}) = \frac{2 \times 127}{2 + 127} = \frac{254}{129} \approx 1.9682 \text{ amu} \]

Compared to HI, DI's heavier reduced mass reflects in a lower vibration frequency. This reduction in frequency is observable in IR spectroscopy and demonstrates how isotopic substitution affects molecular vibrations.

By studying DI, we learn how swapping lighter hydrogen with heavier deuterium shifts vibration frequencies. This knowledge becomes vital in scenarios like hydrogen-deuterium exchange reactions, which are used to study reaction mechanisms and molecular structures.