Question

Linear, triatomic ${\mathbf{CO}}_2$vibrates by symmetric stretch, bend, and asymmetric stretch with frequencies of $\mathbf{4}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$, $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$and$\mathbf{7}\mathbf{.}\mathbf{05}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$ respectively.

1. In what region of the electromagnetic spectrum are these frequencies?
2. calculate the energy (in J) of each vibration. Which takes the least energy?

Short answer

1. The region of the electromagnetic spectrum for these frequencies is the IR region.
2. $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$takes the least energy.

Step-by-step solution

region of the electromagnetic spectrum

From given frequencies, we can calculate the wavelength of each frequency by the formula $\mathbf{\nu }\mathbf{=}\frac{c}{\lambda }$. where$\mathbf{\lambda }$is called wavelength,$\mathbf{\nu }$is called frequency, and c is called speed of light.

Since

For frequency

$\mathbf{4}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$

$\begin{array}{c}\mathbf{\nu }\mathbf{=}\frac{c}{\lambda }\\ \mathbf{\lambda }\mathbf{=}\frac{\mathbf{3}\mathbf{\times }{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{4}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}}\\ \mathbf{=}\mathbf{7}\mathbf{.}\mathbf{463}\mathbf{\times }{\mathbf{10}}^{-6}\mathbf{m}\\ =7463\mathrm{nm}\end{array}$

$\begin{array}{c}\mathbf{wavenumber}\mathbf{(}\mathbf{c}{\mathbf{m}}^{-1}\mathbf{)}\mathbf{=}\frac{{\mathbf{10}}^7}{\mathrm{wavelength}\left(\mathrm{nm}\right)}\\ \mathbf{=}\frac{{\mathbf{10}}^7}{7463\mathrm{nm}}\\ \mathbf{=}\mathbf{1339}\mathbf{.}\mathbf{94}{\mathbf{cm}}^{-1}\end{array}$

For frequency $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$

$\begin{array}{c}\mathbf{\nu }\mathbf{=}\frac{c}{\lambda }\\ \mathbf{\lambda }\mathbf{=}\frac{\mathbf{3}\mathbf{\times }{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}}\\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{-5}\mathbf{m}\\ =15000\mathrm{nm}\end{array}$

$\begin{array}{c}\mathbf{wavenumber}\mathbf{(}\mathbf{c}{\mathbf{m}}^{-1}\mathbf{)}\mathbf{=}\frac{{\mathbf{10}}^7}{\mathrm{wavelength}\left(\mathrm{nm}\right)}\\ \mathbf{=}\frac{{\mathbf{10}}^7}{15000\mathrm{nm}}\\ \mathbf{=}\mathbf{666}\mathbf{.}\mathbf{67}{\mathbf{cm}}^{-1}\end{array}$

For frequency$\mathbf{7}\mathbf{.}\mathbf{05}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$

$\begin{array}{c}\mathbf{\nu }\mathbf{=}\frac{c}{\lambda }\\ \mathbf{\lambda }\mathbf{=}\frac{\mathbf{3}\mathbf{\times }{\mathbf{10}}^8\mathbf{m}\mathbf{/}\mathbf{s}}{\mathbf{7}\mathbf{.}\mathbf{05}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}}\\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{255}\mathbf{\times }{\mathbf{10}}^{-6}\mathbf{m}\\ =4225\mathrm{nm}\end{array}$

$\begin{array}{c}\mathbf{wavenumber}\mathbf{(}\mathbf{c}{\mathbf{m}}^{-1}\mathbf{)}\mathbf{=}\frac{{\mathbf{10}}^7}{\mathrm{wavelength}\left(\mathrm{nm}\right)}\\ \mathbf{=}\frac{{\mathbf{10}}^7}{4225\mathrm{nm}}\\ \mathbf{=}\mathbf{2366}\mathbf{.}\mathbf{86}{\mathbf{cm}}^{-1}\end{array}$

Since, the range of IR region of electromagnetic radiation is 600- $\mathbf{4000}{\mathbf{cm}}^{-1}$. So, vibrational motions have frequencies in IR region of electromagnetic spectrum. As a result, the indicated frequencies exist in the IR area.

energy of each frequency

Calculate the energy of each frequency by using the formula$\mathbf{E}\mathbf{=}\mathbf{h\nu }$

For frequency$\mathbf{4}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$

$\begin{array}{c}{\mathbf{E}}_1\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{\times }{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{s}\mathbf{\times }\mathbf{4}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}\\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{663}\mathbf{\times }{\mathbf{10}}^{-20}\mathbf{J}\end{array}$

For frequency $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$

$\begin{array}{c}{\mathbf{E}}_2\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{\times }{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{s}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}\\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3252}\mathbf{\times }{\mathbf{10}}^{-20}\mathbf{J}\end{array}$

For frequency$\mathbf{7}\mathbf{.}\mathbf{05}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$

$\begin{array}{c}{\mathbf{E}}_3\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{626}\mathbf{\times }{\mathbf{10}}^{-34}\mathbf{J}\mathbf{.}\mathbf{s}\mathbf{\times }\mathbf{7}\mathbf{.}\mathbf{05}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}\\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{671}\mathbf{\times }{\mathbf{10}}^{-20}\mathbf{J}\end{array}$

From the calculations it is clear that $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{13}{\mathbf{s}}^{-1}$ has least energy.