Question

Based on the following information arrange four metal $A,B,C$ and $D$ in order of decreasing ability to act as reducing agent:-
(I) Only A , B and C react with 1MHCl to give $H_2\left(g\right)$
(II) When C is added to solution of the other metal ions metallic $B$ and $D$ are formed
(III) Metal C does not reduce $A^{n+}$

• A. $C>A>B>D$
• B. $C>A>D>B$
• C. $A>C>D>B$
• D. $A>C>B>D$

Answer 1

The correct option is D $A>C>B>D$

1)$A>C>B>D$

2)Based on "when$C$ is added to the solution of other metal ions then mettalic$B$ and $D$ are formed" this statement,$C$ can be oxidized by $B$ and $D$ as there is formation of $B$ and $D$.And as we know,any element can be oxidized by the ions listed below it in the series.

3)Therefore$C$ must be more reactive than $B$and $D$ as it can displace both of them but less reactive than $A$ because $A$ is not formed which makes $A$ most reactive.

4)And based upon "only a,b,c react with with $1$mole $HCl$to give $H_2$" this condition,metals $A,B,C$ are all above hydrogen and $D$ is below it because $D$ can not form hydrogen .Therefore$D$ is the least reactive metal.That is greater the reactivity of the element,greater would be its power to act as a reducing agent.