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**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.

- 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 \(-2\).

Example:

**Note:**The neutral molecule's ON (Oxidation Number) is always zero.

**Illustration \(1\) Oxidation Number of \(H\) and \(O\) in \(H_2O\):**

\(2\) × (\(+1\)) + \(1\) × (\(-2\)) = \(0\)

(\(+2\)) + (\(-2\)) = \(0\)

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 \(O\) = \(-2\)

\(x\) + \(1\) × (\(+6\)) + \(4\) × (\(-2\)) = \(0\)

\(x\) + (\(+6\)) + (\(-8\)) = \(0\)

\(x\) = \(+2\)

\(x\) + (\(+6\)) + (\(-8\)) = \(0\)

\(x\) = \(+2\)

Therefore, the oxidation number of \(Fe\) in \(FeSO_4\) is \(+2\).