Stefan Hendricks

2022-01-06

The ${\left[{C}_{r}C{l}_{6}\right]}^{3}$ ion has a maximum in its absorption spectrum at 735 nm. Calculate the crystal field splitting energy (in kJ/mol) for this ion.

Carl Swisher

So here, we have the maximum absorbance of ${\left[{C}_{r}C{l}_{6}\right]}^{3-}$ is 735 nm or $735×{10}^{-7}m$. And we are required to find its crystal-field splitting energy in $k\frac{J}{m}ol$.
If we calculate the energy of the photon that is absorbed(E) we are also calculating the crystal-field splitting energy $\left({\mathrm{△}}_{0}\right)$ since they are equal. So. we have:
$E={\mathrm{△}}_{0}h\cdot \nu$
And $\nu =\frac{c}{\nu }$;
${\mathrm{△}}_{0}=h\cdot \frac{c}{\nu }$;
Where:
${\mathrm{△}}_{0}$ -crystal-field splitting energy
h-Plane's constant $\left(6.626×{10}^{-34}J-s\right)$
c-Speed of light $\left(3.0×{10}^{8}\frac{m}{s}\right)$
$\lambda$ -wavelength $\left(7.35×{10}^{7}m\right)$
${\mathrm{△}}_{0}=6.626×{10}^{-34}J\cdot s\left(\frac{3.0×{10}^{8}\frac{m}{s}}{7.35×{10}^{-7}m}\right)$
${\mathrm{△}}_{0}=2.70×{10}^{-19}J$
$2.70×{10}^{-19}J$ is the energy of only one ion.
So, to get the crystal-field splitting energy in kJ/mol. We need to convert J to kJ(I000)=IkJ) and multiply the energy to Avogadro's number which is $6.023×{10}^{23}$. So, we have
${\mathrm{△}}_{0}=2.70×{10}^{-19}J×\frac{1kJ}{1000J}\left(×\frac{6.023×{10}^{23}}{1mol}\right)$
${\mathrm{△}}_{0}=162.621k\frac{J}{m}ol$

Thomas White

${\mathrm{△}}_{0}=\frac{hc}{\lambda }=\frac{\left(6.626×{10}^{-34}J\cdot s\right)\left(2.99×{10}^{8}\frac{m}{s}\right)}{735×{10}^{-9}m}=2.70×{10}^{-19}J$
This is the energy per photon change to per mol of photons
$\left(\begin{array}{c}2.70×{10}^{-19}\frac{J}{photon}\end{array}\right)\left(\begin{array}{c}\frac{6.02214×{10}^{23}photons}{1mol}\end{array}\right)\left(\begin{array}{c}\frac{1kJ}{1000J}\end{array}\right)=162\frac{kJ}{mol}$

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