Q9.79CP - Chemistry: Molecular Nature Of Matter And Change

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?

Step-by-Step Solution

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.

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.