QB_PH_SEC6_028: Molecular Spectroscopy

The fundamental vibrational absorption frequency of $^{1}\text{H}^{35}\text{Cl}$ is $\bar{\nu}_1 = 2886\text{ cm}^{-1}$. Assuming the bond force constant remains unchanged, estimate the fundamental vibrational frequency $\bar{\nu}_2$ (in $\text{cm}^{-1}$) of the deuterium-substituted molecule $^{2}\text{H}^{35}\text{Cl}$. Take atomic masses: $^{1}\text{H} = 1.0\text{ u}$, $^{2}\text{H} = 2.0\text{ u}$, $^{35}\text{Cl} = 35.0\text{ u}$. (Round off to the nearest integer)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Harmonic Vibrational Frequency Formula**: $\bar{\nu} = \frac{1}{2\pi c} \sqrt{\frac{k}{\mu}}$ 2. **Isotopic Invariance of Force Constant**: Since electronic potential curves are identical: $\frac{\bar{\nu}_2}{\bar{\nu}_1} = \sqrt{\frac{\mu_1}{\mu_2}}$ 3. **Reduced Mass Calculations**: $\mu_1 = \frac{1.0 \times 35.0}{1.0 + 35.0} = \frac{35}{36} = 0.97222\text{ u}$ $\mu_2 = \frac{2.0 \times 35.0}{2.0 + 35.0} = \frac{70}{37} = 1.89189\text{ u}$ $\sqrt{\frac{\mu_1}{\mu_2}} = \sqrt{\frac{0.97222}{1.89189}} = \sqrt{0.513889} = 0.71686$ 4. **Predicted Frequency**: $\bar{\nu}_2 = 2886\text{ cm}^{-1} \times 0.71686 = 2068.86\text{ cm}^{-1} \approx 2069\text{ cm}^{-1}$ 5. **Official Range**: $[2060, 2075]$.