Oxidation number rules — step by step assignment for any compound

Question

Assign the oxidation number of each element in $\text{K}_2\text{Cr}_2\text{O}_7$ , $\text{H}_2\text{SO}_4$ , and $\text{Na}_2\text{O}_2$ . What are the rules, and in what order should we apply them?

(CBSE 11, JEE Main, NEET — appears almost every year)

Solution — Step by Step

Apply these rules in order of priority:

Free element → oxidation number = 0
Monoatomic ion → oxidation number = charge
F is always $-1$
H is $+1$ (except in metal hydrides: $-1$ )
O is $-2$ (except in peroxides: $-1$ , and in $\text{OF}_2$ : $+2$ )
Sum of all oxidation numbers = charge on the species

K = $+1$ (alkali metal), O = $-2$ (normal oxide). Let Cr = $x$ .

Sum = 0 for a neutral compound:

2(+1) + 2x + 7(-2) = 0

2 + 2x - 14 = 0

x = +6

So Cr is +6 in potassium dichromate.

H = $+1$ , O = $-2$ . Let S = $x$ .

2(+1) + x + 4(-2) = 0

2 + x - 8 = 0

x = +6

S is +6 in sulfuric acid.

This is a peroxide, so O = $-1$ (not $-2$ ). Na = $+1$ .

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

O is $-1$ here. If you used $-2$ by default, you’d get Na = $+2$ , which is impossible for sodium.

Why This Works

Oxidation numbers are a bookkeeping system — they track how electrons are distributed in a compound by assigning all shared electrons to the more electronegative atom. The rules are simply a shortcut for this electronegativity-based assignment.

The priority order matters because exceptions exist. Fluorine is the most electronegative element, so it always wins $-1$ . Oxygen comes next at $-2$ , but loses to fluorine in $\text{OF}_2$ and shares equally with itself in peroxides.

graph TD
    A["Assign oxidation number"] --> B{"Free element?"}
    B -->|"Yes"| C["ON = 0"]
    B -->|"No"| D{"Monoatomic ion?"}
    D -->|"Yes"| E["ON = ion charge"]
    D -->|"No"| F{"Contains F?"}
    F -->|"Yes"| G["F = -1 always"]
    F -->|"No/Done with F"| H{"Contains H?"}
    H -->|"Yes"| I{"Metal hydride?"}
    I -->|"Yes"| I1["H = -1"]
    I -->|"No"| I2["H = +1"]
    H -->|"No/Done with H"| J{"Contains O?"}
    J -->|"Yes"| K{"Peroxide?"}
    K -->|"Yes"| K1["O = -1"]
    K -->|"No"| K2["O = -2"]
    J -->|"No/Done"| L["Use sum rule<br/>to find unknown"]

Alternative Method — Structural Formula Approach

For organic compounds and complex molecules, draw the Lewis structure and assign shared electrons to the more electronegative atom. Count lone pairs as belonging to the atom. This is slower but works when the shortcut rules fail (like in $\text{Fe}_3\text{O}_4$ where Fe has mixed oxidation states).

For MCQs: if the compound has only one unknown element, the algebraic method (Step 2-3 above) takes under 30 seconds. Save the structural approach for JEE Advanced problems with unusual compounds.

Common Mistake

The classic trap: treating all oxygen as $-2$ . In peroxides ( $\text{Na}_2\text{O}_2$ , $\text{H}_2\text{O}_2$ , $\text{BaO}_2$ ), oxygen is $-1$ . In superoxides ( $\text{KO}_2$ ), oxygen is $-1/2$ . In $\text{OF}_2$ , oxygen is $+2$ . NEET 2023 had a question asking the oxidation state of O in $\text{KO}_2$ — many students wrote $-2$ and lost the mark.

Question

Solution — Step by Step

Why This Works

Alternative Method — Structural Formula Approach

Common Mistake

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