Question

The microwave spectrum of gaseous \(\mathrm{HF}\) consists of a series of lines separated by \(41.11\ \mathrm{cm^{-1}}\). The bond length (in \(\mathrm{\mathring{A}}\)) of \(\mathrm{HF}\) is (rounded off to two decimal places).

(Given: Atomic mass (in amu): \(\mathrm{H} = 1.008\), \(\mathrm{F} = 18.998\); \(1\ \mathrm{amu} = 1.661\times10^{-27}\ \mathrm{kg}\); \(h = 6.626\times10^{-34}\ \mathrm{J\,s}\); \(c = 2.998\times10^{8}\ \mathrm{m\,s^{-1}}\))

Hint:

Rotational lines in a rigid-rotor microwave spectrum are spaced by \(2B\). Get \(B\), then \(I\), then \(r\) from \(I=\mu r^2\).

Answer 1

Correct Answer: 0.93

Step 1: Find the rotational constant B.
In a rigid rotor, the pure rotation (microwave) spectrum has lines at \(\tilde{\nu} = 2B(J+1)\), so adjacent lines are spaced by \(2B\).
The paper gives the spacing as \(41.11\ \mathrm{cm^{-1}}\), so
\[ B = \frac{41.11}{2} = 20.555\ \mathrm{cm^{-1}} \]

Step 2: Find the reduced mass of HF.
For a diatomic \(\mathrm{HF}\), the reduced mass is
\[ \mu = \frac{m_H m_F}{m_H+m_F} \]
Using \(m_H = 1.008\) amu and \(m_F = 18.998\) amu,
\[ \mu = \frac{1.008\times18.998}{1.008+18.998} = \frac{19.150}{20.006} = 0.9572\ \mathrm{amu} \]
Convert to kg using \(1\ \mathrm{amu} = 1.661\times10^{-27}\ \mathrm{kg}\):
\[ \mu = 0.9572\times1.661\times10^{-27} = 1.5899\times10^{-27}\ \mathrm{kg} \]

Step 3: Get the moment of inertia from B.
The rotational constant in \(\mathrm{cm^{-1}}\) is related to the moment of inertia \(I\) by
\[ B = \frac{h}{8\pi^2 c I} \implies I = \frac{h}{8\pi^2 c B} \]
Here \(c\) must be in \(\mathrm{cm\,s^{-1}}\), so \(c = 2.998\times10^{10}\ \mathrm{cm\,s^{-1}}\).
\[ I = \frac{6.626\times10^{-34}}{8\pi^2\times(2.998\times10^{10})\times20.555} = \frac{6.626\times10^{-34}}{4.8656\times10^{13}} = 1.3618\times10^{-47}\ \mathrm{kg\,m^2} \]

Step 4: Find the bond length.
The moment of inertia of a diatomic is \(I = \mu r^2\), so
\[ r^2 = \frac{I}{\mu} = \frac{1.3618\times10^{-47}}{1.5899\times10^{-27}} = 8.565\times10^{-21}\ \mathrm{m^2} \]
\[ r = \sqrt{8.565\times10^{-21}} = 9.253\times10^{-11}\ \mathrm{m} = 0.9253\ \mathrm{\mathring{A}} \]

Final Answer:
Rounded to two decimal places, the bond length of HF is about \(0.93\ \mathrm{\mathring{A}}\), close to the accepted textbook value for HF.
\[ \boxed{r \approx 0.93\ \mathrm{\mathring{A}}} \]