Quantum Chemistry/Example 23

Create a table showing all the possible absorption transitions between the first 5 vibrational states of HCl. Include columns for the energies and frequencies. The last column should indicate if the transition is allowed or forbidden.



Solution:

The vibrational frequency of a linear diatomic molecule depends on its reduced mass and vibrational spring constant (k). The spring constant for 1H35Cl was measured and reported as 515.825 N/m.

The reduced mass was calculated as follows: $$\mu_{HCl}=\frac{m_1\cdot m_2}{m_1+m_2} =\frac{1.00783\text{ amu}\cdot 34.9689\text{ amu}}{1.00783\text{ amu}+34.9689\text{ amu}} =0.979597 \text{ amu}$$

The reduced mass was converted to SI units as follows:

$$\bigl(0.979597 \text{ amu} \bigr) \bigl(1.66054\cdot 10^{-27}\frac{\text{ kg}}{\text{amu}}\bigr) =1.62666\cdot 10^{-27} \text{ kg}$$

The fundamental vibrational frequency was calculated as follows:

$$v_0=\frac{1}{2\pi}\biggl(\frac{k}{\mu}\biggr)^{\frac{1}{2}} =\frac{1}{2\pi}\biggl( \frac{515.825 \frac{\text{N}}{\text{m}}} {1.62666 \cdot 10^{-27} \text{ kg}}\biggr)^{\frac{1}{2}} =8.961237 \cdot 10^{13} \text{ s}^{-1} $$

The fundamental vibrational frequency was converted to wavenumbers as follows:

$$\tilde{v}_0=\frac{v_0}{c}=\frac{8.961237 \cdot 10^{13} \text{ s}^{-1}} {2.99792458\cdot 10^{10} \frac{\text{cm}}{\text{s}}} =2989.52 \text { cm}^{-1} $$

The energy for a given vibrational state can be calculated as follows:

$$E_\vee=hv_0 \bigl( \vee+\frac{1}{2} \bigr) =hc\tilde{v}_0 \bigl( \vee+\frac{1}{2} \bigr) $$

For example, the energy of the vibrational ground state:

$$E_0=hv_0 \bigl( 0+ \frac{1}{2} \bigr)=\frac{1}{2}hv_0 =\frac{1}{2}hc\tilde{v}_0 $$

The transition energy between the ground state and the first excited vibrational state was calculated as follows:

$$\Delta E_{1\leftarrow0}=E_1-E_0 = \frac{3}{2}hc\tilde{v}_0-\frac{1}{2}hc\tilde{v}_0 =hc\tilde{v}_0$$ $$\Delta E_{1\leftarrow0} =\bigl(6.62607015\cdot 10^{-34} \text{ J}\cdot \text{s} \bigr) \bigl(2.99792458 \cdot10^{10} \frac{\text{cm}}{\text{s}} \bigr) \bigl( 2989.52 \text{ cm}^{-1} \bigr) =5.93853\cdot 10^{-20} \text {J} $$ According to the specific selection rule, vibrational transitions are only allowed between adjacent energy levels (where $$\Delta \vee=\pm 1$$). Consequently, due to the quantization of energy levels, vibrational transitions allowed by the specific selection rule are equivalent in energy and frequency, such that $$\Delta E=hv_0=hc\tilde{v}_0$$. Conversely, overtone transitions are those for which $$\Delta \vee\neq\pm 1$$. As such, overtone frequencies can be calculated as integer multiples of the fundamental frequency. For example, the vibrational frequency for the transition between the ground state and the second excited vibrational state was calculated as follows:

$$v_{2\leftarrow0}=2 \bigl(v_0\bigr) =2 \bigl( 8.961237 \cdot 10^{13} \text{ s}^{-1} \bigr)= 1.79247 \cdot 10^{14}\text{ s}^{-1} $$

Similarly, the wavenumber for the transition between the ground state and the third excited vibrational state was calculated as follows:

$$\tilde{v}_{3\leftarrow0}= 3\bigl(\tilde{v}_0 \bigr) =8968.57 \text{ cm}^{-1} $$

Finally, the energy for the overtone transition between the first and fourth excited vibrational states was calculated as follows:

$$\Delta E_{4\leftarrow1}=E_4-E_1 = \frac{9}{2}hc\tilde{v}_0- \frac{3}{2}hc\tilde{v}_0= 3hc\tilde{v}_0 $$

$$\Delta E_{4\leftarrow1}=3 \bigl(6.62607015 \cdot 10^{-34} \text{ J}\cdot \text{s} \bigr) \bigl(2.99792458 \cdot10^{10} \frac{\text{cm}}{\text{s}} \bigr) \bigl( 2989.52 \text{ cm}^{-1} \bigr) =5.34468\cdot 10^{-19} \text {J} $$ However, because $$\Delta \vee\neq\pm 1$$, this transition is forbidden.