Question

The vibrational frequency (expressed in wavenumber) of \( ^1H^{35}Cl \) is 2990.6 cm\(^{-1}\). Assuming that the force constant is the same in both cases, vibrational frequency (in cm\(^{-1}\)) of \( ^2D^{35}Cl \) is .............

Hint:

When isotopes are substituted in a molecule, the vibrational frequency decreases because the reduced mass increases. The relationship is inversely proportional.

Answer 1

Correct Answer: 2100 - 2200

Step 1: Understanding the relationship between vibrational frequency and reduced mass.
The vibrational frequency of a diatomic molecule is related to the reduced mass \( \mu \) of the two atoms in the molecule, as follows: \[ v = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}}, \] where \( v \) is the vibrational frequency, \( k \) is the force constant, and \( \mu \) is the reduced mass of the two atoms, given by: \[ \mu = \frac{m_1 m_2}{m_1 + m_2}, \] where \( m_1 \) and \( m_2 \) are the masses of the two atoms. The vibrational frequency is inversely proportional to the square root of the reduced mass.

Step 2: Comparing the reduced masses.
The frequency of \( ^1H^{35}Cl \) is given as 2990.6 cm\(^{-1}\). We need to find the frequency for \( ^2D^{35}Cl \), which involves a hydrogen isotope substitution from \( ^1H \) to \( ^2D \). Since deuterium (\( ^2D \)) is twice as heavy as hydrogen (\( ^1H \)), the reduced mass for \( ^2D^{35}Cl \) will be larger. As a result, the vibrational frequency will decrease for \( ^2D^{35}Cl \).
For \( ^1H^{35}Cl \), the reduced mass is: \[ \mu_1 = \frac{m_H \times m_{Cl}}{m_H + m_{Cl}}, \] and for \( ^2D^{35}Cl \), the reduced mass is: \[ \mu_2 = \frac{m_D \times m_{Cl}}{m_D + m_{Cl}}, \] where \( m_H \) is the mass of hydrogen, \( m_D \) is the mass of deuterium, and \( m_{Cl} \) is the mass of chlorine. Since \( m_D = 2 m_H \), the reduced mass for \( ^2D^{35}Cl \) is approximately twice that of \( ^1H^{35}Cl \).

Step 3: Calculating the frequency ratio.
The ratio of the frequencies of two isotopologues is given by the square root of the inverse ratio of their reduced masses: \[ \frac{v_2}{v_1} = \sqrt{\frac{\mu_1}{\mu_2}}. \] Since \( \mu_2 \approx 2\mu_1 \), the ratio is: \[ \frac{v_2}{v_1} = \frac{1}{\sqrt{2}}. \] Therefore: \[ v_2 = v_1 \times \frac{1}{\sqrt{2}} = 2990.6 \times \frac{1}{\sqrt{2}} \approx 2119.7 \, \text{cm}^{-1}. \]

Step 4: Conclusion.
The vibrational frequency of \( ^2D^{35}Cl \) is approximately \( \boxed{2119.7} \, \text{cm}^{-1} \).