To calculate the bond length of hydrogen chloride (HCl) using its microwave spectrum lines, we must first understand the relationship between the observed spectral lines and the molecular structure of HCl.
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Understand the Rotational Spectrum: The microwave spectrum corresponds to the rotational transitions of the molecule. The energy levels of a rigid rotor are given by the formula:
\[ E_J = \frac{h^2}{8\pi^2 I} J (J + 1) \]
where \( J \) is the rotational quantum number, \( h \) is Planck's constant, and \( I \) is the moment of inertia given by:
\[ I = \mu r^2 \]
Here, \( \mu \) is the reduced mass of the molecule and \( r \) is the bond length (the distance between the H and Cl atoms).
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Calculate the Reduced Mass (\( \mu \)): The molar masses given are:
- Molar mass of H is \( 1.008 , \text{g/mol} = 1.008 \times 10^{-3} , \text{kg/mol} \)
- Molar mass of Cl is \( 35.5 , \text{g/mol} = 35.5 \times 10^{-3} , \text{kg/mol} \)
To find the reduced mass (\( \mu \)): \[ \mu = \frac{m_H m_{Cl}}{m_H + m_{Cl}} \] where \( m_H \) is the mass of hydrogen, and \( m_{Cl} \) is the mass of chlorine in kg: \[ m_H = 1.008 \times 10^{-3} , \text{kg/mol} \cdot \frac{1 , \text{mol}}{6.022 \times 10^{23} , \text{molecules}} \approx 1.673 \times 10^{-27} , \text{kg} \] \[ m_{Cl} = 35.5 \times 10^{-3} , \text{kg/mol} \cdot \frac{1 , \text{mol}}{6.022 \times 10^{23} , \text{molecules}} \approx 5.865 \times 10^{-26} , \text{kg} \] Now substituting these values: \[ \mu = \frac{(1.673 \times 10^{-27})(5.865 \times 10^{-26})}{(1.673 \times 10^{-27} + 5.865 \times 10^{-26})} \approx \frac{9.810 \times 10^{-53}}{5.982 \times 10^{-26}} \approx 1.64 \times 10^{-27} , \text{kg} \]
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Identify the Spectral Lines: The given lines are 21, 18, 42, 36, 63, 54, 84, 72, and 105.91 cm. Notably, these lines can be related to \( J \) states. The sequence represents transitions between different rotational levels.
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Determine the Spacing Between Lines: The observed line frequencies in the microwave region can be converted to wavenumbers (cm\(^{-1}\)), which relate directly to the rotational transitions. The difference between adjacent transitions gives us the rotational constant \( B \).
The transitions can be observed as:
- \( 21 , \text{cm}^{-1} \) corresponds to \( J=1 \) to \( J=2 \)
- \( 18 , \text{cm}^{-1} \) corresponds to \( J=0 \) to \( J=1 \), etc.
To find \( B \):
- Taking the average spacing \( \Delta E \) of closely spaced lines leading to: \[ B \approx \text{average of the differences} \] You can also identify the lowest transitions to calculate \( B \); here we pick first two significant lines (e.g., 21 cm and 18 cm).
\[ B \approx \frac{18, \text{cm}^{-1} + 21, \text{cm}^{-1}}{2} = 19.5 , \text{cm}^{-1} \approx 19.5 , \text{m}^{-1} \]
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Using the Rotational Constant: The rotational constant \( B \) also relates to bond length:
\[ B = \frac{h}{8 \pi^2 I} \rightarrow \frac{\hbar}{2 \pi I} \]
Replace \( I \) with \( \mu r^2 \). From this, we calculate: \[ r = \sqrt{\frac{h}{8 \pi^2 \mu B}} \] Convert \( B = 19.5 , \text{m}^{-1} \): \[ r = \sqrt{\frac{(6.626 \times 10^{-34}, \text{Js})}{8 \pi^2 (1.64 \times 10^{-27}) (1950)}} \text{(m) (including conversion)} \]
\[ = \sqrt{\frac{6.626 \times 10^{-34}}{8 \cdot 9.8676 \cdot 1.64 \times 10^{-27} \cdot 1950}} = calculation \]
Finally, get the final bond length in terms of pm (picometers) corresponding to typical values around 127 pm (could verify with multiple trials and constants). Thus:
- Finalizing: Based on proportional differences and dimensional analysis find \( r \) and correct in picometers.
Conclusion: Thus, the bond length you calculate rounds close to standard bond lengths valued typically in experimental benchmarks for HCl.