Following tests are shown by
i Decolourization of acidified soln. of KMnO₄.
ii. Liberation of I₂ from an acidified soln. of KI.
iii. On treatment with dil HCl, brown fumes of NO₂ which turns FeSO₄ soln. black.

The oxidation state of Mn in KMnO₄ is +7 and it is purple (or violet) in color. It decolorizes when Mn is reduced from +7 to +2. This implies that the decolorization of KMnO₄ is a reducing reaction and this reaction takes place in the presence of a reducing agent.

The oxidation state of I in KI is -1 and in I₂, the oxidation state of I is 0. There is the oxidation of I and this reaction takes place in the presence of an oxidizing agent.

The formula of nitrite is ${\mathrm{NO}}_2^{-}.$ Here, the oxidation state of N is +3. The formula of nitrate is ${\mathrm{NO}}_3^{-}$ and the oxidation state of N in nitrate is +5.

Nitrogen shows an oxidation state from -3 to +5. As in nitrate, N is already present in its highest oxidation state, therefore it cannot go further oxidation or acts as a reducing agent. This implies that nitrate cannot decolorize KMnO₄.

Nitrite can act as both oxidizing agent and a reducing agent. 

When nitrite treats with dil. HCl then it forms nitrogen dioxide that has brown fumes and it also changes the color of FeSO₄ into black.