Question

A black-brown coloured solid (A) when fused with KOH in the presence of air, produces a dark green coloured compound (B) which on electrolytic oxidation in alkaline medium gives a dark purple coloured compound (C). Identify (A), (B) and (C). Write the reactions involved.

Hint:

Remember the colour sequence: [ MnO_2 ;(black-brown) K_2MnO_4 ;(green) KMnO_4 ;(purple) ] This is frequently asked in board examinations.

Answer 1

Concept: Manganese exhibits a wide range of oxidation states and forms compounds having characteristic colours. Important compounds include:

• (MnO_2) : Black-brown

• (K_2MnO_4) : Dark green

• (KMnO_4) : Dark purple
The conversion of manganese dioxide into potassium permanganate involves oxidation of manganese from lower to higher oxidation states.

Step 1: Identification of compound (A) The given black-brown solid is manganese dioxide. [ A=MnO_2 ]

Step 2: Formation of dark green compound (B) When (MnO_2) is fused with KOH in the presence of atmospheric oxygen: [ 2MnO_2+4KOH+O_2 2K_2MnO_4+2H_2O ] Potassium manganate formed is dark green. [ B=K_2MnO_4 ]

Step 3: Formation of dark purple compound (C) Electrolytic oxidation of potassium manganate in alkaline medium produces potassium permanganate. [ 2K_2MnO_4+H_2O 2KMnO_4+2KOH+H_2 ] Potassium permanganate possesses a characteristic dark purple colour. [ C=KMnO_4 ]

Final Identification [ A=MnO_2 ] [ B=K_2MnO_4 ] [ C=KMnO_4 ]