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.