Oxidation number rules — step by step assignment for any compound

Question

How do we assign oxidation numbers to atoms in any compound or ion? What rules do we follow, and in what order, to handle tricky cases like $Na_2O_2$ , $OF_2$ , and $KO_2$ ?

(CBSE 11, JEE Main, NEET — oxidation number assignment is the gateway to all redox chemistry problems)

Solution — Step by Step

Apply these rules in sequence — earlier rules override later ones:

Free elements: Oxidation number = 0 (e.g., $O_2$ , $Na$ , $P_4$ , $S_8$ )
Monoatomic ions: Oxidation number = charge ( $Na^+ = +1$ , $Cl^- = -1$ , $Al^{3+} = +3$ )
Fluorine: Always $-1$ (most electronegative element, no exceptions)
Hydrogen: Usually $+1$ . Exception: $-1$ in metal hydrides ( $NaH$ , $CaH_2$ )
Oxygen: Usually $-2$ . Exceptions: $-1$ in peroxides ( $H_2O_2$ , $Na_2O_2$ ), $-1/2$ in superoxides ( $KO_2$ ), $+2$ in $OF_2$ (F takes priority)
Sum rule: The sum of oxidation numbers in a neutral compound = 0; in a polyatomic ion = charge of the ion

Example 1: Find oxidation number of Cr in $K_2Cr_2O_7$

K = +1 (alkali metal), O = -2 (standard)
$2(+1) + 2(Cr) + 7(-2) = 0$
$+2 + 2Cr - 14 = 0$
$2Cr = +12$ , so Cr = +6

Example 2: Find oxidation number of S in $Na_2S_2O_3$ (thiosulphate)

Na = +1, O = -2
$2(+1) + 2(S) + 3(-2) = 0$
$+2 + 2S - 6 = 0$
$2S = +4$ , so S = +2 (average)

Note: In $Na_2S_2O_3$ , the two S atoms actually have different oxidation states (+5 and -1) — the average is +2. JEE Advanced tests this distinction.

$Na_2O_2$ (sodium peroxide):

Na = +1, and this is a peroxide, so O = $-1$ (not $-2$ )
Check: $2(+1) + 2(-1) = 0$ . Correct.

$OF_2$ (oxygen difluoride):

F = $-1$ (always), so $O + 2(-1) = 0$ , giving O = +2
This is the rare case where oxygen has a positive oxidation number.

$KO_2$ (potassium superoxide):

K = +1, and superoxide ion is $O_2^-$ , so each O = $-1/2$
Check: $+1 + 2(-1/2) = 0$ . Correct.

flowchart TD
    A["Assign oxidation number"] --> B{"Free element?"}
    B -->|"Yes"| C["ON = 0"]
    B -->|"No"| D{"Monoatomic ion?"}
    D -->|"Yes"| E["ON = charge"]
    D -->|"No"| F["Apply priority: F=-1, H=+1*, O=-2*"]
    F --> G["Use sum rule to find unknown"]
    G --> H["Neutral compound: sum = 0"]
    G --> I["Polyatomic ion: sum = charge"]
    F --> J["*Check exceptions:<br/>H = -1 in metal hydrides<br/>O = -1 in peroxides<br/>O = +2 in OF₂"]

Why This Works

Oxidation numbers track electron ownership in a compound by assigning all shared electrons to the more electronegative atom. Fluorine, being the most electronegative, always wins the electrons (hence always $-1$ ). Oxygen usually wins (hence $-2$ ) except against fluorine. The sum rule comes from charge conservation — the total electron bookkeeping must match the actual charge on the species.

Common Mistake

The peroxide trap: in $H_2O_2$ and $Na_2O_2$ , oxygen is $-1$ , not $-2$ . Students who blindly assign $-2$ to oxygen get the wrong answer for any atom bonded to peroxide oxygen. Always check: is the compound a peroxide (contains $O-O$ bond)? If yes, use $-1$ for oxygen. This appears in JEE Main at least once every 2-3 years.

For speed: most exam questions only test the standard rules ( $O = -2$ , $H = +1$ ). Use the sum rule and solve for the unknown. Only invoke exceptions when you see the keywords: peroxide, superoxide, metal hydride, or fluorine-oxygen compound. This saves time on the 90% of questions that follow the standard pattern.

Question

Solution — Step by Step

Why This Works

Common Mistake

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