Question

Which of the following reaction produces hydrogen
A. $Mg+{{H}_{2}}O$
B. $Ba{{O}_{2}}+HCl$
C. ${{H}_{2}}{{S}_{4}}{{O}_{8}}+{{H}_{2}}O$
D. $N{{a}_{2}}{{O}_{2}}+2HCl$

Answer 1

Hint: In the main group alkali and alkaline earth metals produce hydrogen when they are in contact with water, as they are all very reactive with water. Due to the electropositive character of alkali and alkaline earth metal ions, they can easily remove one electron during a chemical reaction with water.

Complete Step by Step Answer:
The alkali metals are situated in groups $1$of the periodic table and include lithium(Li), sodium(Na), potassium(K), rubidium(Rb), cesium(Cs), and francium(Fr). Whereas the group $2$ containing alkaline earth metal ions include beryllium(Be), magnesium(Mg), calcium(Ca), strontium (Sr), barium (Ba), and radium(Ra). Most of them react with water to produce hydrogen except beryllium.

 All the alkali metals have one electron and all alkaline earth metal ions have two electrons in their outermost valence shell. This valence electron is very easily lost during a chemical reaction. After losing these valence electrons from their last shell, they become positive ions. Meanwhile, each oxygen atom in a water molecule can take up its valence electrons and become a $-1$ charged hydroxide ion. Thereby the positive metal ion and negative hydroxide ions are attracted to each other and form a metal hydroxide. And extra hydrogen atoms find each other to make hydrogen gas.

For example, Na reacts with water to form sodium hydroxide and hydrogen gas.
$2Na(s)+2{{H}_{2}}O(l)\to 2NaOH(aq.)+{{H}_{2}}(g)$
Calcium reacts with water to form calcium hydroxide and hydrogen gas.
$Ca(s)+2{{H}_{2}}O(l)\to Ca{{(OH)}_{2}}(aq.)+{{H}_{2}}(g)$
Like Calcium, alkaline earth metal magnesium (Mg) also reacts with water to form magnesium hydroxide and hydrogen gas.
$Mg(s)+2{{H}_{2}}O(l)\to Mg{{(OH)}_{2}}(aq.)+{{H}_{2}}(g)$
Hence reaction (A) produces hydrogen.

Now, What about other reactions, let us see.
 From (B), ${{H}_{2}}{{S}_{2}}{{O}_{8}}+2{{H}_{2}}O\to 2{{H}_{2}}S{{O}_{4}}+{{H}_{2}}{{O}_{2}}$
From (C), $Ba{{O}_{2}}+2HCl\to BaC{{l}_{2}}+{{H}_{2}}{{O}_{2}}$
From (D), $N{{a}_{2}}{{O}_{2}}+2HCl\to 2NaCl+{{H}_{2}}{{O}_{2}}$
So, all these (B, C, and D) reactions are mainly used for the synthesis of hydrogen peroxide (${{H}_{2}}{{O}_{2}}$).
Thus option (A) is correct.

Note: Beryllium (Be) does not react with water. Because it possesses high ionisation energy, for this it is very difficult for Be to donate its valence electrons to the hydrogen atoms of a water molecule. And also beryllium’s surface is coated with a thin layer of BaO which inhibits it from reacting with water molecules at low temperatures.