Question

Name the gas that can readily decolorize acidified  $ KMn{O_4} $ solution:

A. $ {P_2}{O_5} $ 

B. $ C{O_2} $ 

C. $ S{O_2} $ 

D. $ N{O_2} $ 

Answer 1

Hint: The d-block elements show a transitional behavior between the highly electropositive s-block elements and weakly electropositive p-block elements and are known as transition elements. 

Complete step by step solution:

Step 1:

Potassium permanganate is the salt of an unstable acid, permanganic acid, and is one of the most important compounds of manganese.

Step 2:

Potassium permanganate is a crystalline solid having m.p. 523K. It forms needle-like crystals having a dark purple color with a greenish luster. It is moderately soluble in water and forms a purple-colored solution.

Step 3:

Potassium permanganate is a powerful oxidizing agent in neutral, alkaline, and acidic solutions. The modes of reactions are different in different types of solutions.

In neutral solution, potassium permanganate behaves as a moderate oxidizing agent due to the following reaction:

$ 2KMn{O_4} + {H_2}O\xrightarrow{{}}2KOH + 2Mn{O_2} + 3O $ 

The reaction involves the reduction of  $ Mn{O_4}^ -  $ ion into  $ Mn{O_2} $ . Due to this tendency potassium permanganate acts as an oxidizing agent in a neutral medium.

In alkaline solution, purple-colored potassium permanganate reduces to green colored potassium manganate as per the following reaction:

$ Mn{O_4}^ - + 2{H_2}O + 3{e^ - }\xrightarrow{{}}2Mn{O_2} + 4O{H^ - } $ 

Step 4:

$ {P_2}{O_5} $ ,  $ C{O_2} $ ,  $ N{O_2} $ are all oxidizing agents but themselves cannot further get oxidized.

$ S{O_2} $ acts both as an oxidizing agent as well as a reducing agent. So, sulfur dioxide gets oxidized to sulphuric acid in the presence of potassium permanganate, a strong oxidizing agent. As potassium permanganate oxidizes sulfur dioxide to sulphuric acid it gets reduced to  $ Mn{O_2} $ .

$ 2KMn{O_4} + 5S{O_2} + 2{H_2}O\xrightarrow{{}}{K_2}S{O_4} + 2MnS{O_4} + 2{H_2}S{O_4} $ 

As potassium permanganate itself gets reduced it gets decolorized.

So,  $ S{O_2} $  can readily decolorize acidified  $ KMn{O_4} $ solutions. 

So, option (C) is the correct answer.

Note: $ KMn{O_4} $ is widely used as an oxidizing agent in the laboratory as well as in industry. Most of the compounds of the transition metals are colored both in the solid-state as well as in the aqueous solution.