Why does the color of copper sulphate change when an iron nail is kept in it? Justify your answer.

Hint:Some metals are chemically very reactive whereas other metals are less reactive or unreactive. On reaction the more reactive metal becomes a part of the salt whereas the less reactive metal is set free.

Complete step by step answer:
When a more reactive metal is placed in the salt solution of a less reactive metal, then the more reactive displaces (pushes out) the less reactive metal from its salt solution. These types of reactions are known as displacement reactions. General form of single displacement reaction is often of the type,
\[A+BC\to B+AC\]
In this reaction, element A is replaced by element B in the compound BC.
When an iron nail is placed in copper sulphate (\[CuS{{O}_{4}}\]) solution for some time, then the blue color of copper sulphate solution fades and a red-Brown coating of copper metal is deposited on the iron nail. This reaction can be written as:
\[\underset{\left( Blue\text{ }solution \right)}{\mathop{CuS{{O}_{4}}~}}\,+~\underset{\left( grey \right)}{\mathop{Fe}}\,\to \underset{~\left( greenish \right)}{\mathop{FeS{{O}_{4}}}}\,+\underset{\left( red-brown \right)}{\mathop{Cu}}\,\]
In this case, the solution turns greenish from its original blue color due to the formation of iron sulphate. Iron lies above in the reactivity series and thus more reactive than copper. As a result, the more reactive metal (iron) displaces a less reactive metal (copper) from its salt solution (copper sulphate solution).
Hence, the change in color.

Note: We can also check the reactivity of a metal in reactivity series, where the metals have been arranged in a group (or series) according to their chemical reactivities. The metals which are most reactive are placed at the top whereas the least reactive metals are placed at the bottom.