Question

Explain: A green solution of potassium manganate turns purple when \[C{O_2}\] passes through the solution.

Answer 1

Hint:Transition metal shows various colors in different oxidation states. The color is responsible for the \[d - d\] transition or charge transfer between the ligands and metal ions.

Complete step by step answer:Potassium manganate is an inorganic compound composed of potassium, manganese and oxygen. The chemical formula of potassium manganate is \[{K_2}Mn{O_4}\].
When \[C{O_2}\] is passed through water carbonic acid is generated. The carbonic acid undergoes dissociation and makes the solution acidic by releasing \[{H^ + }\] ion in the solution.
Manganese is an element in the periodic table with atomic number \[25\]. It is a transition metal and the electron configuration of manganese is \[\left[ {Ar} \right]3{d^5}4{s^2}\]. Thus manganese can exhibit a highest oxidation state of \[ + 7\].
Manganese is known to undergo a disproportionation reaction to produce other manganese compounds in acidic medium. The oxidation state of manganese in potassium manganate (\[{K_2}Mn{O_4}\] ) is \[ + 6\] . Under the acidic conditions the manganate undergoes disproportionation to produce permanganate \[ + 7\] oxidation state and manganese dioxide with \[ + 4\] oxidation state.
The colour of potassium permanganate is purple and the colour of manganese dioxide is brown solid. Hence when \[C{O_2}\] gas is passed through a green solution of potassium manganate the colour of the solution turns into purple due to the formation of potassium permanganate. The corresponding reaction is:
$3Mn{O_4}^{2 - } + 4{H^ + } \to 2Mn{O_4}^ - + Mn{O_2} + 2{H_2}O$
The colour of \[Mn{O_4}^{2 - }\] ion is purple.

Note:The ability to exhibit such higher oxidation states are common with transition metals. All the transition metals possess inner d orbitals which they use to extend their valency and form multiple bonds which have different atoms or ions.