Quantum Chemistry/Example 22

Make a table showing all the absorption transitions for the first 5 rotational states of HCl. Include a column that shows which are allowed and which are forbidden

The rotational energy for any molecule is all kinetic and comes from the angular momentum of the molecule. To get the rotational energy of any one molecule the moment of inertia first needs to be derived (inertia units:(kg·m²). For a diatomic molecule, the inertia is dependent on the equilibrium bond length and the mass of the two atoms in the form of reduced mass. Inertia for a diatomic:

Moment of Inertia (Diatomic) $I=\mu {\text{r}}_{e}^{2}$ , $\mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}$

Where $I$ is the moment of inertia, $\mu$ is the reduced mass, $m_{1}$ and $m_{2}$ are the masses of the two atoms, and $r_{e}$ is the equilibrium distance between the two atoms.

The rotational energy levels of a molecule is related to the reduced Planck's constant, the inertia of the molecule, and the J quantum number.

Energy Levels for Rigid Rotor $E_{J}={\frac {{\text{ħ}}^{2}}{2I}}J(J+1)$

Where $E_{J}$ is the energy of the energy state of the molecule, $I$ is the inertia, ${\text{ħ}}$ is the reduced planck's constant, and J is the second quantum number of the rigid rotor.  

The Energy of a photon is related to the wavelength, the speed of light and Planck's constant. The energy between two rotational energy states of a molecule is equal to the energy of the photon absorbed to get to the higher energy state. Therefore the wavelength of a photon can be determined from the difference in rotational energy states of a molecule.

Energy of a Photon ${\text{E}}={\frac {hc}{\lambda }}$

Where $E$ is the energy of the photon, $h$ is Planck's constant, $c$ is the speed of light and $\lambda$ is the wavelength of the photon.

Table 1. First five rotational energy states of HCl and the associated energy. and wavelengths of light associated with them. .

Rotational Absorption Transition States J → J'	Allowed/Forbidden	Energy of Transitions (${\text{×}}10^{-21}$J)	Wavelength of Light Absorbed (μm)
0 → 1	Allowed	4.75	418.5
0 → 2	Forbidden	1.42	139.5
0 → 3	Forbidden	2.85	69.8
0 → 4	Forbidden	4.75	41.9
0 → 4+n	Forbidden
1 → 2	Allowed	0.95	209.3
1 → 3	Forbidden	2.37	83.7
1 → 4	Forbidden	4.27	46.5
1 → 4+n	Forbidden
2 → 3	Allowed	1.42	139.5
2 → 4	Forbidden	3.32	59.8
2 → 4+n	Forbidden
3 → 4	Allowed	1.90	104.6
3 → 4+n	Forbidden
4 → 5	Allowed	2.37	83.7
4 → 5+n	Forbidden

Example 22

What is the wavelength of light in μm needed to excite an HCl molecule in its second rotational excited state to its third rotational excited state if it has an equilibrium bond length of 1.200 Å?

Solution:

$\mu ={\frac {m_{1}m_{2}}{m_{1}+m_{2}}}$

$\mu ={\frac {(1.0078g/mol)(35.453g/mol)}{(1.0078g/mol)+(35.453g/mol)}}$

$\mu ={\frac {0.989944{\text{g/mol}}}{6.022140857{\text{×}}10^{23}{{\text{mol}}^{-1}}}}{\text{×}}{\frac {1{\text{kg}}}{1000{\text{g}}}}$

Divide by Avogadro constant and convert to SI units:

$\mu =1.62723{\text{×}}10^{-27}{\text{kg}}$

${\text{ΔE}}={\frac {{\text{ħ}}^{2}}{2\mu {\text{r}}_{e}^{2}}}(J'(J'+1)-J(J+1))$

${\text{ΔE}}={\frac {(1.054571817{\text{×}}10^{-34}{\text{J·s}})^{2}}{2(1.62723{\text{×}}10^{-27}{\text{kg}}){(1.2{\text{×}}10^{-10}{\text{m}}})^{2}}}((2)((2)+1)-(1)((1)+1))$

${\text{ΔE}}=9.49226{\text{×}}10^{-22}{\text{J}}$

${\text{E}}={\frac {hc}{\lambda }}$

$\lambda ={\frac {hc}{\text{E}}}$

$\lambda ={\frac {(6.62607015{\text{×}}10^{-34}{\text{J·s}})(2.99792{\text{×}}10^{8}{\text{m/s}})}{(9.49226{\text{×}}10^{-22}{\text{J}})}}$

$\lambda =0.00020927m{\text{×}}10^{6}\mu m/m$

Convert Units:

$\lambda =2.09.2\mu m$