Question

For $KMn{O_4}{\text{ and }}{K_2}C{r_2}{O_7}$ are colored. Justify the given statement.

Answer 1

Hint: To understand the reason for colored complex formation of the potassium dichromate and potassium permanganate, we need to first assess the charge transfer spectra theory , i.e. on what basis do these compounds possess color. In the above two complexes, although the central atom has empty d-orbital, it still shows color formation.

Complete answer:
In general, we have studied that the color of any molecule or species is due to the excitation of the outermost shell’s electron getting excited and when they return back to their normal ground state, they radiate energy which, if falling under visible spectra, shows color.
However, in the molecule of $KMn{O_4}{\text{ and }}{K_2}C{r_2}{O_7}$ , if we look at the central metal atom, they are $M{n^{ + 7}}$ and $C{r^{ + 6}}$ . The electronic configuration the above two species is:
$M{n^{ + 7}} = \left[ {Ar} \right]{\text{ }}3{d^0}{\text{ }}4{s^0}$
$C{r^{ + 6}} = \left[ {Ar} \right]{\text{ }}3{d^0}{\text{ }}4{s^0}$
Here, we can clearly see that there is no electron in the outermost shell. So, the actual reason for the colour of these compounds is given by charge transfer spectra theory. According to it, the lone pair of oxygen atoms attached to the central metal atom as ligand, they get transferred to the central metal atom and that is the reason for the color of these compounds. In these compounds, there is a transfer of charge from anion to cation. This type of charge transfer is also termed as ligand to metal charge transfer.

Note:
The color of potassium permanganate is violet-purple and the color of potassium dichromate is orange. Also, there are many other complexes which, due to the ligand to metal charge transfer, show color even though they might not be having electrons in the outermost shell. Apart from ligand to metal charge transfer, there are other types of transfer possible too.