Question

In laboratory,${H_2}{O_2}$ is prepared by the action of:
A. cold ${H_2}S{O_4}$ and $Ba{O_2}$
B. aqueous alkali on $N{a_2}{O_2}$
C. $Mn{O_2}$ and cold ${H_2}S{O_4}$
D. dilute $HCl$ and $Mn{O_2}$

Answer 1

For solving this question, we first need to understand what is hydrogen peroxide and how it is formed. We can define Hydrogen peroxide as one of the simplest types of peroxides available in the market with oxygen-oxygen single bond.

Complete step by step answer:

As we know, hydrogen peroxide is a colorless liquid which is used in an aqueous solution for safety in the laboratory. In the laboratory, mostly hydrogen peroxide acts as a bleaching agent and can also be treated as a disinfectant. Thus, concentrated hydrogen peroxide contains species of reactive oxygen. Thus, the reactive oxygen species is used as a propellant in rockets.
Also, we know that the chemical formula of hydrogen peroxide is ${H_2}{O_2}.$
Now, let’s see the laboratory method for the preparation of ${H_2}{O_2}.$
Firstly, we take barium peroxide in the laboratory for acidification. Then, all the excess amount of water is removed by the process of evaporation at a reduced amount of pressure. Thus, at the end of this chemical reaction we obtain hydrogen peroxide.
The chemical equation of the preparation of hydrogen peroxide can be written as:
$Ba{O_2}.8{H_2}O(s) + {H_2}S{O_4}(aq) \to BaS{O_4}(s) + {H_2}{O_2}(aq) + 8{H_2}O(l)$
Therefore, in the laboratory hydrogen peroxide is prepared by the action of cold ${H_2}S{O_4}$ and $Ba{O_2}$.
Therefore, the option A is correct.

Note:
We must remember that the structure of the hydrogen peroxide used in the laboratory is non-planar. As,${H_2}{O_2}$ consists of an open book structure which has a single O-O bond and spin. The dihedral angle of ${H_2}{O_2}$ is $111^\circ$. Also, the single O-O bond has a length of $145.8{\text{ }}pm$ and the O-H bond has a length of $98.8{\text{ }}pm$.