# Quantum Chemistry/Example 15
## Question

Determination of the bond length of CO using microwave spectroscopy.

## Solution

When a photon is absorbed by a polar diatomic molecule, such as carbon monoxide, the molecule can be excited rotationally. The energy levels of these excited states are quantized to be evenly spaced energetically. This is illustrated on a microwave spectrum where the interval between the absorption peaks, correlating to the energy transitions, is constant. As long as we know the masses of the elements that comprise the diatomic molecule, we can use a microwave spectrum to determine the bond length. The distance between each rotational absorption line is defined as twice the rotational constant, $({\tilde {\beta }})$ which can be measured via the following equation:
${\tilde {\beta }}=\left({\frac {h}{8\pi ^{2}c{\text{I}}}}\right)$
h = plank's constant = 6.626 ᛫10$^{−34}$ J ᛫ s
c = speed of light = 2.998᛫10$^8$ m ᛫ s$^{-1}$
I = moment of inertia
The energy required to rotate a molecule around its axis is the moment of inertia $({\text{I}})$. It can be calculated as the sum of the products of the masses of the component atoms and their distance from the axis of rotation squared:
${\text{I}}=\sum _{i}m_{i}\cdot r_{i}^{2}$
Working it out for an heterogeneous diatomic molecule:
${\text{I}}=\left(m_{a}\cdot r_{a}^{2}\right)+\left(m_{b}\cdot r_{b}^{2}\right)$
${\text{I}}=r_{a}\cdot r_{b}\left(m_{a}+m_{b}\right)$
The distance from the atom to the center of mass $\left(r_{a}{\text{and }}r_{b}\right)$ cannot easily be measured; however, by setting the origin at the center of mass, an equation can be derive for the two values that uses the bond length $\left(r_{e}\right)$ as a variable:
$\left(m_{a}\cdot r_{a}\right)=\left(m_{b}\cdot r_{b}\right)=\left(m_{a}\cdot \left(r_{a}-r_{e}\right)\right)=\left(m_{b}\cdot \left(r_{b}-r_{e}\right)\right)$
$r_{a}={\frac {m_{b}\cdot r_{e}}{m_{a}+m_{b}}}$
$r_{b}={\frac {m_{a}\cdot r_{e}}{m_{a}+m_{b}}}$
Substituting these equations into the moment of inertia equation:
${\text{I}}={\biggl{\frac {m_{b}\cdot r_{e}}{m_{a}+m_{b}}}{\biggr )}\cdot {\biggl{\frac {m_{a}\cdot r_{e}}{m_{a}+m_{b}}}{\biggr )}\cdot \left(m_{a}+m_{b}\right)$
${\text{I}}={\biggl{\frac {m_{a}\cdot m_{b}\cdot r_{e}^{2}}{m_{a}+m_{b}}}{\biggr )}$
This equation can be simplified further if we imagine the rigid rotor as a single particle rotating around a fixed point a bond length away. The mass of this particle is the reduced mass $\left(\mu \right)$ of the two atoms that make up the diatomic molecule:
$\mu =\left({\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}\right)$
Simplifying the previous moment of inertia equation, we get:
${\text{I}}=\mu r_{e}^{2}$

## Solving for Bond Length

From here we have everything we need to be able to determine the bond length of a polar diatomic molecule such as carbon monoxide.
First, we must solve for the moment of inertia using the rotational constant:
${\tilde {\beta }}=\left({\frac {h}{8\pi ^{2}c{\text{I}}}}\right)$
${\text{I}}=\left({\frac {h}{8\pi ^{2}c{\tilde {\beta }}}}\right)$
As explained earlier, the rotational constant can be determined by measuring the distance between the rotational absorption lines and halfling it. In the case of ${\ce {^{12}C^{16}O}}$ the rotational constant is $193.1281$ m$^{-1}$. Plugging this value in, we can determine the moment of inertia:
${\text{I}}=\left({\frac {6.626\cdot 10^{-34}}{8\pi ^{2}\cdot \left(2.998\cdot 10^{8}\right)\cdot \left({\displaystyle 193.1281}\right)}}\right)$
${\text{I}}=1.44939\cdot 10^{-46}$ kg ᛫ m$^2$
Now that we know the moment of inertia, we can rearrange the equation we derived earlier in order to determine the bond length:
${\text{I}}=\mu r_{e}^{2}$
$r_{e}={\sqrt {\frac {\text{I}}{\mu }}}$
$r_{e}={\sqrt {\frac {1.44939\cdot 10^{-46}}{\mu }}}$
The exact atomic mass of ${\ce {^{12}C}}$ is 12.011 amu and ${\ce {^{16}O}}$ is 15.9994 amu. As such, the reduced mass is calculated to be:
$\mu =\left({\frac {m_{1}\cdot m_{2}}{m_{1}+m_{2}}}\right)$
$\mu =\left({\frac {12.011\cdot 15.9994}{12.011+15.9994}}\right)$
$\mu =6.86062$ amu
$\mu =6.86062$ amu ᛫ $1.67377\times 10^{-27}\left({\frac {kg}{amu}}\right)$
$\mu =1.140\times 10^{-26}kg$
Plugging in the reduced mass back into our equation, we can finally solve for the bond length of a carbon monoxide molecule:
$r_{e}={\sqrt {\frac {1.44939\cdot 10^{-46}}{1.140\times 10^{-26}}}}$
$r_{e}={\sqrt {1.271\cdot 10^{-20}}}$
$r_{e}=1.12739\cdot 10^{-10}m$
$r_{e}=1.12739\mathrm {\AA}$

## References

* ↑ https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Exercises%3A_Physical_and_Theoretical_Chemistry/Exercises%3A_Aktins_et_al./12.E%3A_Rotational_and_Vibrational_Spectra_(Exercises)
* ↑ https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/10%3A_Fixed-Axis_Rotation__Introduction/10.06%3A_Calculating_Moments_of_Inertia
* ↑ https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Exercises%3A_Physical_and_Theoretical_Chemistry/Exercises%3A_Aktins_et_al./12.E%3A_Rotational_and_Vibrational_Spectra_(Exercises)
* ↑ https://webbook.nist.gov/cgi/cbook.cgi?ID=C630080&Mask=1000
* ↑ https://www.angelo.edu/faculty/kboudrea/periodic/structure_mass.htm