Question

What is the oxidation state of $Mn$ in $M{n_2}{O_3}?$

Answer 1

Hint: We know that the oxidation state is also known as the oxidation number. The oxidation state can be found by equating the oxidation number of each molecule by the overall charge of the complex. It may be negative, zero or positive. In the above question we can determine the oxidation state by using the known oxidation state of oxygen . Here we have $2$ manganese atoms and three oxygen atoms.

Complete answer:
As we know that oxidation number is the number of electrons lost or gained from a molecule in order to form a chemical bond with another molecule. If the molecule loses an electron, then it means that it lost a negative charge ,it became positively charged.
Similarly if the molecule gains an electron then it means it gained a negative charge, so we can say that it is positively charged.
Here we have to find the oxidation state of manganese in $M{n_2}{O_3}$. Let us consider the oxidation number on manganese to be $x$. We have to keep in mind that we have $2$ manganese atoms and three oxygen atoms.
So we can write : $2 \times x + 3( - 2) = 0 \Rightarrow 2x - 6 = 0$.
 On further solving we have $2x = 6 \Rightarrow x = \dfrac{6}{2} = + 3$.
Hence the oxidation state of $Mn$ in $M{n_2}{O_3}$ is $ + 3$.

Note:
We should know that the decrease in the number of atoms undergoing a chemical change is referred to as reduction. However for pure elements, the oxidation number is zero, they are neutral. They are called the neutral compounds. It is important to remember the charge on atoms in order to find the oxidation state.