Question

State that \[{\text{M}}{{\text{n}}^{2 + }},{\text{ M}}{{\text{n}}^{2 + }}{\text{, M}}{{\text{n}}^{6 + }}\] shows colours respectively?
 (i) Pink, blue and green
(ii) Green, blue and yellow
(iii) Blue, yellow and green
(iv) Yellow, blue and green

Answer 1

Hint: The charges on Mn are the few of the possible oxidation numbers of manganese. Mn belongs to d-block. It has oxidation numbers ranging from +2 to +7.When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed.

Complete answer:
Formation of coloured ions:
This frequency absorbed generally lies in the visible region. The colour observed corresponding to the complementary colour of light absorbed. The frequency of light that is absorbed is determined by the nature of ligands, size of metal ions, oxidation state of metals.
Along with d-d transition, colour may be due to the charge transfer spectra and polarisation.
There are some compounds in which $d_0$ or $d_{10}$ configurations are present and in these configurations no d-d transition is expected. But still, these may be intensely coloured. This colour is due to charge transfer from ligands to metal or from metal to metal.
\[Mn{O_4}^ - \] is pink coloured due to charge transfer from ligands to metal while \[F{e_4}{\left[ {Fe{{(CN)}_6}} \right]_3}\]
is blue coloured due to charge transfer from metal to metal, AgBr and AgI are coloured due to polarisation.
Hence, the answer is option one.

Note:
Transitions elements are the one which has incompletely filled d orbitals in its ground state or in their most common oxidation state Zn, Cd and Hg are not typical transition elements because they have full $d_{10}$ configuration in their ground state which means fully filled valence orbital as well as in their common oxidation state.