Question

In nature, metal A is found in free state while metal B is found in the form of its compound. B of these two will be nearer to the top of the activity series of metals
A.True
B.False

Answer 1

Hint: We have to know that reactivity series (or) activity series of metals gives the order of reactivity from highest to lowest. A metal that is highly reactive will displace the other.

Complete step by step answer: We know that in the activity series of metals, metals that are highly reactive are placed in the top of the series, whereas least reactive metals lie at the bottom of the series.
Metals which are highly reactive cannot exist in their free state. Those metals which take the middle position of the series are slightly reactive and are present in the form of oxides, carbonates (or) sulfides.
From the question, we can understand that, in free state we can find metal A whereas metal B is found as one the compounds of metal A. This means that metal B is highly reactive because it has the ability to form compound and metal A is least reactive.
Because of the high reactive nature of metal B, metal B will nearer at the top of the activity series of metals. So, the given statement B will be occupying a position to the top of the activity series of metal is true.
Therefore, the option (A) is correct.

Note: We can define displacement reactions as chemical reactions in which a reactive element displaces an element that is less/least reactive. We have to know that metals as well as nonmetals take place in a displacement reaction.
For example, consider the reaction between zinc and copper sulfate.
According to the activity series of metals, zinc is more reactive when compared to copper. Therefore, zinc would displace copper from copper sulfate solution. We can write this chemical reaction as,
\[CuS{O_4}\left( {aq} \right) + Zn\left( {aq} \right)\xrightarrow{{}}ZnS{O_4}\left( {aq} \right) + Cu\left( s \right)\]
Copper sulfate on reaction zinc forms zinc sulfate and metallic copper.