Question

The complex $\left[Fe\right(CN{)}_6{]}^{4-}$ absorbs a radiation of 356.40 nm. The pairing energy of the complex is 230 kJ/mol. Then what is the magnitude of the total crystal field splitting energy (in kJ/mol) of the complex?
(Given, $\text{Plank constant}=6.6\times {10}^{-34}Js,N_A=6\times {10}^{23},c=3.0\times {10}^{8}m/s$)

Answer 1

Given, that the complex absorbs a radiation of 356.40 nm.
So, ${\Delta }_o=\frac{hc}{\lambda }=\frac{(6.6\times {10}^{-34})(3.0\times {10}^8)}{356.4\times {10}^{-9}}\times 6\times {10}^{23}\times {10}^{-3}kJ/mol=333.33kJ/mol$

In the complex $\left[Fe\right(CN{)}_6{]}^{4-}$, $Fe$ is in +2 oxidation state. So all the electrons are paired up, and hence the electronic configuration will be $t_{2g}^6{e}_g^0$.
$\therefore CFSE=-6\times 0.4\times {\Delta }_o+3P$
$\Rightarrow CFSE=(-6)\times 0.4\times 333.33+3\times 230kJ/mol$
$\Rightarrow CFSE=-799.99+690=-109.99$