Question

The oxidation state of Barium in $Ba{{\left( {{H}_{2}}P{{O}_{2}} \right)}_{2}}$ is:
(A) +3
(B) +2
(C) +1
(D) -1

Answer 1

The oxidation state of an atom in a molecule is calculated by a simple formula which would be discussed further in the upcoming section which would probably help you reach the final answer for the given illustration.

Complete step-by-step answer: Let us know oxidation state in detail;
Oxidation state or number-
The degree of oxidation of an atom in the chemical compound is known as oxidation number or state of that atom.
It is calculated as;
For example, take a molecule $X{{Y}_{2}}$ which is neutral and the oxidation number of Y is +1. Thus, oxidation number of X is;
$\begin{aligned} & x+2\left( +1 \right)=0 \\\ & x=-2 \\\ \end{aligned}$
Using this basic knowledge, let us move towards the illustration given,
The given molecule is $Ba{{\left( {{H}_{2}}P{{O}_{2}} \right)}_{2}}$ i.e. neutral molecule.
And we know that,
Oxidation number of P = +1
Oxidation number of H = +1
Oxidation number of O = -2
Thus, the oxidation number of Ba is calculated as;
$\begin{aligned} & x+2\left( 2\left( +1 \right)+\left( +1 \right)+2\left( -2 \right) \right)=0 \\\ &\Rightarrow x+2\left( -1 \right)=0 \\\ &\therefore x=+2 \\\ \end{aligned}$
Hence, the oxidation state or number of Barium in $Ba{{\left( {{H}_{2}}P{{O}_{2}} \right)}_{2}}$ is +2.

Therefore, option (B) is correct.

Note: Do note that the neutral molecule will have zero on the other side of the equal to but for other molecules i.e. cationic or anionic molecules; the value will be either a positive or negative real number respectively.