Question

The oxidation number of Phosphorus in $M{g_2}{P_2}{O_7}$ is:
A.$+ 3$
B.$+ 2$
C.$+ 5$
D.$- 3$

Answer 1

In chemistry, it is common that elements lose or donate their electrons easily to get stability. During the oxidation and reduction atom lose or gain an electron and due to this, we can get the oxidation number of a particular substance.

Complete step by step answer:
Let's discuss the oxidation and reduction reaction.
Oxidation - When an electron loses during the reaction is known as an oxidation reaction.
Example - $Ni \to N{i^{2 + }} + 2{e^ - }$
Reduction - When electron gain during the reaction is known as reduction reaction.
example - $C{o^{3 + }} + {e^ - } \to C{o^{2 + }}$
Redox reaction - When oxidation and reduction both reactions occur simultaneously is known as a redox reaction.
Example - ${H_2} + {F_2} \to 2HF$
Oxidation Number - It is also called an oxidation state. It is the degree of oxidation and reduction, which means the number of electrons lost or gained during the reaction is known as oxidation number.
Here, we will find the oxidation number of Phosphorus.
given - $M{g_2}{P_2}{O_7}$
Firstly, let the oxidation number of Phosphorous is $x$.
oxidation number of $Mg$ - $+ 2$
oxidation number of $O$ - $- 2$
Now, put these values in the given element
$2( + 2) + 2x + 7( - 2) = 0$
$2x - 10 = 0$
$2x = 10$
$x = + 5$
Now, we got the oxidation state of Phosphorus.

Therefore, option C is the correct answer here.

Note: Oxidizing agent - when a chemical substance accepts the electron of another or oxidizes another is known as an oxidizing agent. Reducing agent - When a chemical substance loses the electron is known as a reducing agent.
To balance redox reactions in an acidic medium, we need to add water and ${H^ + }$ ion. Likewise, in basic medium, we need to add water and $O{H^ - }$ ion. Redox reaction involves many biological reactions like cellular respiration.