1. Based on the following information, arrange the A, B, C, and D from lowest to highest
activity (in order of increasing ability to act as reducing agents).
I. Only A, C and D react with 1 mol/L HCl to give H2(e).
II. When A is added to solutions of the other metal ions, metallic B and Care
formed but not D.
a) C b) B c) B d) C e) B

Order of metals from least reactive to most reactive: B <C <A <D

Reducing agents are substances that experience oxidation  

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series  

Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au  

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent  

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction  

Let's analyze the statement in the problem

I. Only A, C and D react with 1 mol/L HCl to give H₂(e)

M + HCl ⇒ MCl + H₂(MCl : alkali, MCl₂ : alkaline earth)

A, C and D can react with 1 mol / L HCl, meaning metals A, C and D are located to the left of element H (more reactive), and B in the right of element H

II. When A is added to solutions of the other metal ions, metallic B and C are  formed but not D.

This means that metal A is more reactive than metals B and C, while D is more reactive than A, so metal D is the most reactive

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