Q. how to calculate oxidation state

Definition. The oxidation state (also called oxidation number) of an atom in a compound is a bookkeeping number that represents the hypothetical charge the atom would have if all bonds were completely ionic. The basic calculation method is to assign oxidation numbers to all atoms according to standard rules, then use the fact that the sum of oxidation numbers in a neutral compound is 0, and in a polyatomic ion equals the ion charge.

Common rules, stated clearly and one by one.

1. An atom in its elemental form has oxidation state 0. Examples: \( \mathrm{O_2} \) , \( \mathrm{N_2} \) , \( \mathrm{S} \) are all 0.

2. For a simple (monoatomic) ion, the oxidation state equals the ion charge. Example: \( \mathrm{Na^+} \) is \(+1\), \( \mathrm{Cl^-} \) is \(-1\).

3. Alkali metals (group 1) in compounds are \(+1\). Alkaline earth metals (group 2) are \(+2\).

4. Oxygen is usually \(-2\) in compounds, except in peroxides where it is \(-1\), and in superoxides where it is \(-\tfrac{1}{2} \), and when bonded to fluorine it can be positive.

5. Hydrogen is usually \(+1\) when bonded to nonmetals, and \(-1\) when bonded to metals (metal hydrides such as \( \mathrm{NaH} \) ).

6. Halogens (F, Cl, Br, I) are usually \(-1\) when they are the most electronegative element in the compound, unless they are bonded to oxygen or a more electronegative halogen.

7. The sum rule. For a neutral compound, the sum of all oxidation states equals 0. For an ion, the sum equals the ion charge. This is the key algebraic rule used to solve for unknown oxidation states.

Step-by-step algorithm to calculate an unknown oxidation state.

Step 1. Write the chemical formula clearly.

Step 2. Assign oxidation numbers using the rules above to atoms whose oxidation numbers are known by rule.

Step 3. Let the unknown oxidation state be a variable, commonly \(x\) , for the element you want to find.

Step 4. Write the equation: the sum of all oxidation states (each multiplied by the number of those atoms in the formula) equals the overall charge (0 for neutral molecules, or the ionic charge for ions). Use algebra to solve for \(x\) .

Step 5. Report the oxidation state and note any exceptions or special cases (peroxide, superoxide, coordination compounds with formal charge differences, etc.).

Worked examples with full detail.

Example 1. Water, \( \mathrm{H_2O} \) . Find the oxidation state of oxygen.

Step A. Known rules: hydrogen is usually \(+1\) when bonded to nonmetals. There are two hydrogen atoms, so their total contribution is \(2\times(+1)\).

Step B. Let the oxidation state of oxygen be \(x\) . The molecule is neutral, so the sum of oxidation states equals 0. Write the equation:

\[
2(+1) + x = 0
\]

Step C. Solve for \(x\) .

\[
2 + x = 0
\]
\[
x = -2
\]

Conclusion. Oxygen in \( \mathrm{H_2O} \) has oxidation state \(-2\) .

Example 2. Carbon dioxide, \( \mathrm{CO_2} \) . Find the oxidation state of carbon.

Step A. Oxygen is usually \(-2\). There are two oxygens, total contribution \(2\times(-2) = -4\) .

Step B. Let carbon be \(x\) . The molecule is neutral, so

\[
x + 2(-2) = 0
\]

Step C. Solve.

\[
x – 4 = 0
\]
\[
x = +4
\]

Conclusion. Carbon in \( \mathrm{CO_2} \) is \(+4\) .

Example 3. Potassium permanganate, \( \mathrm{KMnO_4} \) . Find the oxidation state of manganese.

Step A. Potassium, an alkali metal, is \(+1\) . Oxygen is \(-2\) and there are four oxygens, giving \(4\times(-2) = -8\) .

Step B. Let manganese be \(x\) . The compound is neutral, so

\[
(+1) + x + 4(-2) = 0
\]

Step C. Solve.

\[
1 + x – 8 = 0
\]
\[
x – 7 = 0
\]
\[
x = +7
\]

Conclusion. Manganese in \( \mathrm{KMnO_4} \) is \(+7\) .

Example 4. Iron(III) oxide, \( \mathrm{Fe_2O_3} \) . Find the oxidation state of iron.

Step A. Oxygen is \(-2\) , three oxygens contribute \(3\times(-2) = -6\) . There are two iron atoms, each with the same oxidation state \(x\) . The compound is neutral.

\[
2x + 3(-2) = 0
\]

Step B. Solve.

\[
2x – 6 = 0
\]
\[
2x = 6
\]
\[
x = +3
\]

Conclusion. Iron in \( \mathrm{Fe_2O_3} \) is \(+3\) .

Example 5. Ammonium ion, \( \mathrm{NH_4^+} \) . Find the oxidation state of nitrogen.

Step A. Hydrogen is \(+1\) when bonded to nonmetals. Four hydrogens give \(4\times(+1) = +4\) . Let nitrogen be \(x\) . The ion has overall charge \(+1\) , so

\[
x + 4(+1) = +1
\]

Step B. Solve.

\[
x + 4 = 1
\]
\[
x = -3
\]

Conclusion. Nitrogen in \( \mathrm{NH_4^+} \) is \(-3\) .

Example 6. Dichromate ion, \( \mathrm{Cr_2O_7^{2-}} \) . Find the oxidation state of chromium.

Step A. Oxygen is \(-2\) and there are seven oxygens, contributing \(7\times(-2) = -14\) . Let each chromium be \(x\) . The ion charge is \(-2\) , therefore

\[
2x + 7(-2) = -2
\]

Step B. Solve.

\[
2x – 14 = -2
\]
\[
2x = 12
\]
\[
x = +6
\]

Conclusion. Chromium in \( \mathrm{Cr_2O_7^{2-}} \) is \(+6\) .

Special-case reminders and tips.

– Peroxides: oxygen is \(-1\) in peroxides such as \( \mathrm{H_2O_2} \) . Write \(2(+1) + 2x = 0\) for peroxide to get \(x = -1\) for each oxygen.

– Superoxides: oxygen has average oxidation state \(-\tfrac{1}{2}\) .

– Fluorine is always \(-1\) in compounds. When oxygen is bonded to fluorine, oxygen can be positive.

– Oxidation states can be fractional when averaging over identical atoms in unusual structures, but for integer stoichiometric formulas you will usually get integer oxidation numbers for each element type.

Summary of the practical procedure, condensed into four lines.

1. Assign known oxidation numbers by rule. 2. Introduce variable \(x\) for the unknown. 3. Write the sum equation: sum of (oxidation number × atom count) = overall charge. 4. Solve algebraically for \(x\) .