Question

The first rotational absorption of \(^{12}C^{16}O\) molecule is observed at 3.84 cm\(^{-1}\). If an isotopic substitution is made with \(^{18}O\) in the molecule, the frequency (in cm\(^{-1}\)) of first rotational absorption is ...............

Hint:

For isotopic substitution, the rotational frequency decreases with increasing mass of the substituent. Use the reduced mass formula to calculate the change in frequency.

Answer 1

The rotational frequency of a molecule is related to its moment of inertia. The moment of inertia \( I \) of a molecule is given by: \[ I = \mu r^2 \] where \( \mu \) is the reduced mass and \( r \) is the bond length.
For isotopic substitution, the reduced mass changes, affecting the rotational frequency. The reduced mass \( \mu \) for a molecule with two atoms is: \[ \mu = \frac{m_1 m_2}{m_1 + m_2} \] For \(^{12}C^{16}O\), we have: \[ \mu_{12} = \frac{m_{12} m_{16}}{m_{12} + m_{16}} \quad \text{and for} \quad ^{12}C^{18}O: \] \[ \mu_{18} = \frac{m_{12} m_{18}}{m_{12} + m_{18}} \] The ratio of the rotational frequencies for \( ^{12}C^{16}O \) and \( ^{12}C^{18}O \) will be inversely proportional to the square root of the reduced masses: \[ \frac{\nu_{18}}{\nu_{16}} = \sqrt{\frac{\mu_{16}}{\mu_{18}}} \] Since the mass of oxygen increases from 16 to 18, the frequency will decrease. The change in the frequency can be calculated using the ratio of the reduced masses. The first absorption for \( ^{12}C^{16}O \) occurs at 3.84 cm\(^{-1}\), so: \[ \nu_{18} = 3.84 \times \sqrt{\frac{m_{16}}{m_{18}}} \] Using the known masses, the frequency for \( ^{12}C^{18}O \) is approximately \( 3.71 \, \text{cm}^{-1} \). Final Answer: \[ \boxed{3.71} \]