Understanding Microwave Spectrum Spacing

The spacing between adjacent spectral lines in the microwave spectrum of a diatomic molecule is directly proportional to its rotational constant ($B$). The rotational constant is inversely proportional to the molecule's moment of inertia ($I$).

The moment of inertia ($I$) depends on the reduced mass ($\mu$) and the bond length ($r$) as $I = \mu r^2$. Thus, the spacing ($\Delta \nu$), which is proportional to $B$ and inversely proportional to $I$, is inversely proportional to the product of reduced mass and the square of bond length:

$ \Delta \nu \propto \frac{1}{\mu r^2} $

Calculating Reduced Mass Ratios

For $H^{35}Cl$, the reduced mass is $\mu_{HCl} = \frac{m_H m_{Cl}}{m_H + m_{Cl}}$.

For $D^{35}Cl$, the reduced mass is $\mu_{DCl} = \frac{m_D m_{Cl}}{m_D + m_{Cl}}$.

Using approximate atomic masses ($m_H \approx 1$ u, $m_D \approx 2$ u, $m_{Cl} \approx 35$ u):

$ \mu_{HCl} \approx \frac{1 \times 35}{1 + 35} = \frac{35}{36} \text{ u} $

$ \mu_{DCl} \approx \frac{2 \times 35}{2 + 35} = \frac{70}{37} \text{ u} $

The ratio of reduced masses is:

$ \frac{\mu_{HCl}}{\mu_{DCl}} = \frac{35/36}{70/37} = \frac{35}{36} \times \frac{37}{70} = \frac{37}{72} $

Accounting for Bond Length Variation

The bond length of $D^{35}Cl$ ($r_{DCl}$) is 5% greater than that of $H^{35}Cl$ ($r_{HCl}$).

$ r_{DCl} = 1.05 \, r_{HCl} $

This means the ratio of squared bond lengths is:

$ \left(\frac{r_{HCl}}{r_{DCl}}\right)^2 = \left(\frac{r_{HCl}}{1.05 \, r_{HCl}}\right)^2 = \left(\frac{1}{1.05}\right)^2 = \frac{1}{1.1025} $

Determining the Spacing Ratio

The ratio of the spectral line spacings is:

$ \frac{\Delta \nu_{DCl}}{\Delta \nu_{HCl}} = \left(\frac{\mu_{HCl}}{\mu_{DCl}}\right) \left(\frac{r_{HCl}}{r_{DCl}}\right)^2 $

Substituting the calculated ratios:

$ \frac{\Delta \nu_{DCl}}{\Delta \nu_{HCl}} = \left(\frac{37}{72}\right) \times \left(\frac{1}{1.1025}\right) = \frac{37}{79.38} \approx 0.46614 $

Calculating the Spacing for $D^{35}Cl$

Given $\Delta \nu_{HCl} = 6.35 \times 10^{11} \text{ Hz}$.

The spacing for $D^{35}Cl$ is:

$ \Delta \nu_{DCl} = \Delta \nu_{HCl} \times \left(\frac{\Delta \nu_{DCl}}{\Delta \nu_{HCl}}\right) $

$ \Delta \nu_{DCl} = (6.35 \times 10^{11} \text{ Hz}) \times 0.46614 $

$ \Delta \nu_{DCl} \approx 2.960589 \times 10^{11} \text{ Hz} $

Rounding to two decimal places, the value is 2.96.