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### Theory:

Oxidation state:
The oxidation number of an element is defined as the total number of electrons that an atom gains or loses to form a chemical bond with another atom. Thus, the oxidation number is also called the Oxidation state.
1. If the oxidation number is positive, it means that the atom loses an electron, and if it is negative, it means that the atom gains electrons.
2. If it is zero, then the atom neither gains nor loses electrons.
3. The amount of oxidation numbers of all the atoms in the formula for a neutral compound is $$0$$.
4. The amount of oxidation numbers of an ion is the same as the charge on that ion.
5. A negative oxidation number in a compound of two, unlike atoms, is assigned to the more electronegative atom.
Example:
• The oxidation numbers of $$K$$ and $$Br$$ in the $$KBr$$ molecule are $$+1$$ and $$-1$$, respectively.
• The oxidation number of $$N$$ in an $$NH_3$$ molecule is $$-3$$.
• The oxidation number of $$H$$ is $$+1$$ (except for hydrides).
• In most cases, the oxygen oxidation number is $$-2$$.
Note: The neutral molecule's ON (Oxidation Number) is always zero.

Illustration $$1$$ Oxidation Number of $$H$$ and $$O$$ in $$H_2O$$:

Let us take oxidation number of $$H$$ = $$+1$$ and oxidation number of $$O$$ = $$-2$$
$$2$$ × ($$+1$$) + $$1$$ × ($$-2$$) = $$0$$
($$+2$$) + ($$-2$$) = $$0$$
Thus, the oxidation number of $$H$$ is $$+1$$, and the oxidation number of $$O$$ is $$-2$$.

Illustration $$2$$ Oxidation Number of $$S$$ in $$H_2SO_4$$:

Let oxidation number  of $$S$$ be $$x$$ and we know that oxidation number of $$H$$ = $$+1$$ and $$O$$ = $$-2$$
$$2$$ × ($$+1$$) + $$x$$ + $$4$$ × ($$-2$$) = $$0$$
($$+2$$) + $$x$$ + ($$-8$$) = $$0$$
$$x$$ = $$+6$$
Therefore, oxidation number of $$S$$ is $$+6$$.

Illustration $$3$$ Oxidation Number of $$Cr$$ in $$K_2Cr_2O_7$$:

Let oxidation number of $$Cr$$ be $$x$$ and we know that oxidation number of $$K$$ = $$+1$$ and $$O$$ = $$-2$$
$$2$$ × ($$+1$$) + $$2$$ × $$x$$ + $$7$$ × ($$-2$$) = $$0$$
($$+2$$) + $$2x$$ + ($$-14$$) = $$0$$
$$2x$$ = $$+12$$
$$x$$ = $$+6$$
Therefore, the oxidation number of $$Cr$$ in $$K_2Cr_2O_7$$ is $$+6$$.

Illustration $$4$$ Oxidation Number of $$Fe$$ in $$FeSO_4$$:

Let oxidation number of $$Fe$$ be $$x$$ and we know that oxidation number of $$S$$ = $$+6$$ and $$O$$ = $$-2$$
$$x$$ + $$1$$ × ($$+6$$) + $$4$$ × ($$-2$$) = $$0$$
$$x$$ + ($$+6$$) + ($$-8$$) = $$0$$
$$x$$ = $$+2$$
Therefore, the oxidation number of $$Fe$$ in $$FeSO_4$$ is $$+2$$.