Question

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

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

Answer 1

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

Reducing nature $\Rightarrow$Ability to lose electrons easily. Element with more negative reduction potential more is the tendency to lose electrons. Thus reducing nature decreases from top to bottom in the electrochemical series.
According to the question:
(I) Only A, B and C react with HCl to give $H_2$

D not give $H_2$ gas with HCl it means it has very less reducing nature among all four metals.
$\therefore A,B\text{and}C$ reduce hydrogen i.e., they come below hydrogen in spectrochemical series.
(II) & (III)
It means B and D reduced by metal C (C is more reducing in nature) however A is not reduced by C therefore A is most reducing in nature among four metals.
Decreasing order of reducing agent $\Rightarrow A>C>B>D$.