Discussion Forum



1)

The colour of $KMnO_{4}$  is due to 


A) $M \rightarrow L$ charge transfer transition

B) d - d transition

C) $L \rightarrow M$ charge transfer transition

D) $\sigma $ - $\sigma $ transition

Answer:

Option C

Explanation:

$KMnO_{4}   \rightarrow K^{+}+MnO_{4}^{-}$ 

 $\therefore$ In $MnO_{4}^{-}$  , Mn has +7  oxidation state having no electron in d-orbitals.

 It is considered that higher the oxidation state of metal, greater is the tendency to occur  $L \rightarrow M$  charge transfer because ligand is able to donate the electrons into the vacant d-orbital of metal.

 Since, charge transfer is Laporte as well as spin allowed, therefore, it shows colour.

 Time Saving Technique.   

 There is no need to check all four options. Just find out the oxidation state of metal ion. If oxidation state is highest and ligand present there is of electron-donating nature, gives LMCT , which shows more intense colour.