Oxidation number rules — assigning, change: Worked examples

Definition The rules we lean on (restated here so you never flip back)

Rule 3 — Fluorine: in any compound, F is $- 1$ (it is the most electronegative element, so it always wins the electron tug-of-war).
Rule 4 — Oxygen: usually $- 2$ ; but $- 1$ in peroxides (an O–O bond, e.g. $H_{2} O_{2}$ ), $- \frac{1}{2}$ in superoxides (e.g. $K O_{2}$ ), and $+ 2$ when bonded to F (e.g. $O F_{2}$ ).
Rule 5 — Hydrogen: $+ 1$ with nonmetals; but $- 1$ in metal hydrides (e.g. $N a H$ ), where the metal is less greedy than H.
Rule 6 — Groups: Group 1 metals are always $+ 1$ ; Group 2 metals always $+ 2$ . These never bend.
Mistake D (from the parent): "oxidation numbers must be whole." They need not be — averages over non-equivalent atoms can be fractional (e.g. $+ \frac{8}{3}$ in $F e_{3} O_{4}$ ). It is bookkeeping, not a real per-atom charge.

The scenario matrix

Every oxidation-number question falls into one of these case classes. Read this table first — each worked example below is tagged with the cell (C1, C2, …) it lives in.

Cell	Case class	What makes it tricky	Example
C1	Neutral molecule, one unknown	Right-hand side (RHS) $= 0$	Ex 1 (Mn in $K M n O_{4}$ )
C2	Polyatomic ion, one unknown	RHS $=$ ion charge, not $0$	Ex 2 (P in $P O_{4}^{3 -}$ )
C3	Exception overrides a rule	O is not $- 2$ ; H is not $+ 1$	Ex 3 ( $O F_{2}$ ), Ex 4 ( $N a H$ )
C4	Fractional / average answer	Atoms not equivalent	Ex 5 (Fe in $F e_{3} O_{4}$ )
C5	Negative oxidation number for the target	Target is the greedy atom	Ex 6 (C in $C H_{4}$ )
C6	Same element, two different states in one formula	Must split, not average blindly	Ex 7 (N in $N H_{4} N O_{3}$ )
C7	Redox change — track before/after, balance electrons	Sign of change matters	Ex 8 (word problem: rusting)
C8	Exam twist — combine several traps at once	Peroxide/superoxide + ion + unknown	Ex 9 ( $H S O_{5}^{-}$ ), Ex 10 ( $K O_{2}$ )

Two universal moves you use in every cell:

The worked examples

Ex 1 — Cell C1: neutral molecule, one unknown

Worked example Manganese in potassium permanganate

$K M n O_{4}$ Forecast first: Mn sits between very electron-hungry oxygens and an electron-giving potassium. Guess: is Mn positive or negative, and roughly how big? (Write your guess before reading on.)

Step 1 — Answer Q1 (what is $Q$ ?). The formula has no charge shown, so it's a neutral molecule: $Q = 0$ . Why this step? Everything downstream depends on the right-hand side of the sum equation.

Step 2 — Answer Q2 (exceptions?). No O–O bond, no metal hydride, no O–F. So oxygen is the default $- 2$ (Rule 4: oxygen is usually $- 2$ ). Potassium is Group 1, so $+ 1$ (Rule 6: Group 1 always $+ 1$ ). Why this step? Fixing the known atoms leaves exactly one unknown, $x$ = Mn.

Step 3 — Write the sum equation and solve. $n_{K} 1 \cdot (+ 1) + n_{M n} 1 \cdot x + n_{O} 4 \cdot (- 2) = 0$ Why this step? Each term is $n_{i} \times (ox no)_{i}$ , exactly the master rule; the small braces mark the atom counts $n_{i}$ . $1 + x - 8 = 0 \Rightarrow x = + 7$

✅ Mn is $+ 7$ .

Verify: Plug back: $(+ 1) + (+ 7) + 4 (- 2) = 1 + 7 - 8 = 0 = Q$ . ✔ Also a sanity check: $+ 7$ is Mn's maximum possible state (it has 7 valence electrons), which matches "very oxidised" — our forecast of a big positive number was right.

Ex 2 — Cell C2: polyatomic ion, one unknown

Worked example Phosphorus in phosphate

$P O_{4}^{3 -}$ Forecast: This is an ion with charge $- 3$ . Does the RHS change your equation? Guess P's number.

Step 1 — Q1: $Q = - 3$ , because the little superscript $3 -$ is the net charge. Why this step? This is the single most common trap (mistake C in the parent note): people write $= 0$ for ions. The $- 3$ must appear on the RHS.

Step 2 — Q2: no exceptions. O is the default $- 2$ (Rule 4). Why this step? Before solving we must confirm there is no O–O bond, metal hydride, or O–F lurking; phosphate is a plain oxo-anion, so the default $- 2$ is safe and only P stays unknown.

Step 3 — Solve. $n_{P} 1 \cdot x + n_{O} 4 \cdot (- 2) = - 3$ Why this step? Same master rule, only the RHS is $- 3$ instead of $0$ . $x - 8 = - 3 \Rightarrow x = + 5$

✅ P is $+ 5$ .

Verify: $(+ 5) + 4 (- 2) = 5 - 8 = - 3 = Q$ . ✔ Contrast: if you'd wrongly used $Q = 0$ you'd get $x = + 8$ — impossible for phosphorus (only 5 valence electrons). The absurd answer is the alarm bell that you used the wrong $Q$ .

Ex 3 — Cell C3: exception overrides the oxygen rule

Worked example Oxygen in oxygen difluoride

$O F_{2}$ Forecast: Oxygen is "always $- 2$ ", right? Look at the partner atom before you commit.

Step 1 — Q2 first this time (the exception is the whole point; see the "Order note" above). Fluorine is the most electronegative element there is — see Electronegativity & periodic trends. Rule 3 says F is always $- 1$ in compounds. So here F wins the tug-of-war against O, and it is O that must go positive. Why this step? Oxidation number always hands electrons to the greedier atom. Normally O is the greedy one; against F it isn't. The rule priority (F above O) encodes exactly this.

Step 2 — Q1: $Q = 0$ (neutral molecule).

Step 3 — Solve for O ( $= x$ ). $n_{O} 1 \cdot x + n_{F} 2 \cdot (- 1) = 0 \Rightarrow x = + 2$

✅ O is $+ 2$ — the rare positive oxygen.

Verify: $(+ 2) + 2 (- 1) = 0 = Q$ . ✔ This is the named exception in Rule 4 ("with F, O $= + 2$ "). The figure below shows why: watch the shared electrons get dragged off oxygen toward the two fluorines.

Figure — Oxidation number rules — assigning, change

Ex 4 — Cell C3: exception overrides the hydrogen rule

Worked example Hydrogen in sodium hydride

$N a H$ Forecast: H is " $+ 1$ "... but sodium is a metal that loves to give electrons away. Who's greedier?

Step 1 — Q2: metal hydride exception. Na (Group 1) is far less electronegative than H. So H is the greedier atom here and takes the shared pair ⇒ H goes negative. Rule 5's exception: H is $- 1$ in metal hydrides. Why this step? Same logic as Ex 3 — the electron always goes to the greedier partner. With a metal partner, hydrogen is the greedy one.

Step 2 — Q1: $Q = 0$ ; Na is $+ 1$ (Group 1, Rule 6: Group 1 always $+ 1$ — it never goes negative). Why this step? Fixing $Q$ and the one locked atom (Na) leaves hydrogen as the sole unknown $x$ ; without pinning Na we'd have two unknowns and no solvable equation.

Step 3 — Solve for H. $n_{N a} 1 \cdot (+ 1) + n_{H} 1 \cdot x = 0 \Rightarrow x = - 1$ Why this step? The master sum rule with everything known except H turns into a one-line equation for $x$ .

✅ H is $- 1$ .

Verify: $(+ 1) + (- 1) = 0 = Q$ . ✔ Cross-check: Na must be $+ 1$ (Group 1 rules never bend), so H is forced to $- 1$ . Consistent.

Ex 5 — Cell C4: fractional / average answer

Worked example Iron in magnetite

$F e_{3} O_{4}$ Forecast: Three irons, four oxygens, neutral. Bet you can't get a whole number for Fe. Guess the fraction.

Step 1 — Q1: $Q = 0$ ; Q2: O is default $- 2$ (no O–O bond — magnetite is a normal oxide). Why this step? Fixing the RHS ( $0$ , neutral) and confirming no peroxide/superoxide lets us assign every oxygen $- 2$ , leaving iron as the only unknown to solve for.

Step 2 — Let each Fe be $x$ (the average). $n_{F e} 3 \cdot x + n_{O} 4 \cdot (- 2) = 0$ Why "average"? $F e_{3} O_{4}$ is a mixed-valence oxide: its three irons are not identical (some are $F e^{2 +}$ , some $F e^{3 +}$ ). The sum rule only sees the total charge, so it can only report the average oxidation number, which is allowed to be fractional (this is exactly Mistake D, restated at the top: oxidation numbers need not be whole).

Step 3 — Solve. $3 x - 8 = 0 \Rightarrow x = + \frac{8}{3} \approx + 2.67$ Why this step? The sum rule is now one equation in one unknown ( $x$ ); dividing gives the mean iron state directly, and it lands on a fraction rather than an integer.

✅ Average Fe is $+ \frac{8}{3}$ .

Verify: What real mixture gives this average? Magnetite is one $F e^{2 +}$ and two $F e^{3 +}$ . Check neutrality with four $O^{2 -}$ : $1 (+ 2) + 2 (+ 3) + 4 (- 2) = 2 + 6 - 8 = 0$ ✔. Now the average of those three iron states: $\frac{( + 2 ) + ( + 3 ) + ( + 3 )}{3} = \frac{8}{3}$ ✔ — it lands exactly on the sum-rule answer, confirming both the composition and the fraction.

Ex 6 — Cell C5: target atom goes negative

Worked example Carbon in methane

$C H_{4}$ Forecast: Carbon and hydrogen — which is greedier? That decides carbon's sign.

Step 1 — Q2 / electronegativity. Carbon is slightly more electronegative than hydrogen, so in each C–H bond carbon takes the pair. Four bonds ⇒ carbon gains four "extra" electron shares ⇒ carbon goes strongly negative; each H goes $+ 1$ (Rule 5: H is $+ 1$ with nonmetals). Why this step? This is the first example where our target is the greedy atom, so its number is negative. Same machinery, sign flips.

Step 2 — Q1: $Q = 0$ (neutral molecule). Why this step? With $Q$ fixed and every H pinned at $+ 1$ , carbon is the lone unknown $x$ ; without setting $Q$ the equation has no known right-hand side to solve against.

Step 3 — Solve for C ( $= x$ ). $n_{C} 1 \cdot x + n_{H} 4 \cdot (+ 1) = 0 \Rightarrow x = - 4$ Why this step? The master sum rule collapses to one equation in $x$ , and solving delivers carbon's negative state directly.

✅ C is $- 4$ (its most reduced state).

Verify: $(- 4) + 4 (+ 1) = 0 = Q$ . ✔ Sanity: $- 4$ is carbon's minimum (it can hold 4 extra electrons), the mirror of Mn's $+ 7$ maximum in Ex 1. Both extremes now covered.

Ex 7 — Cell C6: same element, two states in one formula

Worked example Nitrogen in ammonium nitrate

$N H_{4} N O_{3}$ Forecast: If you blindly average all N, you get one number. But is that the chemically true number? The nitrogen in $N H_{4}^{+}$ and the nitrogen in $N O_{3}^{-}$ live in very different neighbourhoods.

Step 1 — Recognise the salt splits into two ions. $N H_{4} N O_{3} = N H_{4}^{+} + N O_{3}^{-}$ . Treat each ion separately. Why this step? The two nitrogens are in different ions with different partners — averaging them would hide real chemistry (mistake similar to Mistake D, but here we can pin exact values).

Step 2 — N in ammonium $N H_{4}^{+}$ . $Q = + 1$ , H is $+ 1$ (Rule 5). $n_{N} 1 \cdot x + n_{H} 4 \cdot (+ 1) = + 1 \Rightarrow x = - 3$ Why this step? The ammonium ion carries $+ 1$ , so its own sum rule has RHS $= + 1$ ; with all four H fixed at $+ 1$ , nitrogen is the sole unknown and solves to $- 3$ .

Step 3 — N in nitrate $N O_{3}^{-}$ . $Q = - 1$ , O is $- 2$ (Rule 4). $n_{N} 1 \cdot x + n_{O} 3 \cdot (- 2) = - 1 \Rightarrow x - 6 = - 1 \Rightarrow x = + 5$ Why this step? The nitrate ion carries $- 1$ , a completely different RHS from ammonium; with all three O fixed at $- 2$ , nitrogen here solves to $+ 5$ — the opposite sign from the same element two lines up.

✅ N is $- 3$ in the ammonium half and $+ 5$ in the nitrate half.

Verify: Ammonium: $(- 3) + 4 (+ 1) = + 1 = Q$ ✔. Nitrate: $(+ 5) + 3 (- 2) = - 1 = Q$ ✔. Blind average would be $\frac{- 3 + 5}{2} = + 1$ — a number no single N actually has. Splitting was the right call.

Ex 8 — Cell C7: redox change (word problem)

Worked example Iron rusting:

$4 F e + 3 O_{2} \to 2 F e_{2} O_{3}$ A steel gate left in the rain turns to rust. Who is oxidised, who reduced, and do the electrons balance? Forecast: Metals give electrons. Guess Fe's direction of change and $O_{2}$ 's direction.

Step 1 — Assign before and after.

$F e$ (free element) $\to 0$ (Rule 1: free element $= 0$ ).
$O_{2}$ (free element) $\to 0$ (Rule 1).
In $F e_{2} O_{3}$ : O is $- 2$ (Rule 4), so $2 x + 3 (- 2) = 0 \Rightarrow x = + 3$ : Fe is $+ 3$ ; O is $- 2$ . Why this step? Redox tracking needs a before and an after number for each element — see Oxidising and reducing agents.

Step 2 — Compute the change (tabulate before → after).

Atom	Before	After	Change	Verdict
Fe	$0$	$+ 3$	$+ 3$	oxidised (loses $3 e^{-}$ )
O	$0$	$- 2$	$- 2$	reduced (gains $2 e^{-}$ )

Why this step? Laying before and after side by side makes the sign of each change visible at a glance — a rise means electrons lost (oxidation), a fall means electrons gained (reduction). Without the table you can't see who is the reducing agent and who the oxidising agent.

Step 3 — Balance electrons (the conservation law). Total electrons lost by iron $= 4 Fe \times 3 = 12 e^{-}$ . Total electrons gained by oxygen $= 6 O \times 2 = 12 e^{-}$ . Why this step? Electrons are moved, never created — the two totals must match. This is the same conservation that powers Redox reactions — balancing (half-reaction & ion-electron).

✅ Fe is oxidised (reducing agent); $O_{2}$ is reduced (oxidising agent); 12 e⁻ transferred, balanced.

Verify: electrons lost $= 12$ , electrons gained $= 12$ , difference $= 0$ . ✔ Also mass/atom count: 4 Fe and 6 O on both sides. ✔

The figure below draws this as two half-reactions on a single number line — watch the green arrow climb (oxidation) and the red arrow fall (reduction), with the electron count matching in the middle.

Ex 9 — Cell C8: the exam twist (peroxide + ion + unknown)

Worked example Sulfur in Caro's acid, peroxymonosulfate

$H S O_{5}^{-}$ Forecast: If you treat all five oxygens as $- 2$ , you'll get an outrageous number for S. There's a hidden peroxide bond. Guess whether the trap makes S bigger or smaller.

Step 1 — Spot the exception. $H S O_{5}^{-}$ has one O–O (peroxide) linkage: of its five oxygens, one pair is a peroxide (both O at $- 1$ , Rule 4 exception) and the other three are normal ( $- 2$ ). Why this step? Rule 4's priority: peroxide overrides the default $- 2$ . Missing this is exactly mistake A from the parent note ("O is always $- 2$ ").

Step 2 — Set up the sum with the split oxygens. $Q = - 1$ (the ion), H is $+ 1$ (Rule 5). $n_{H} 1 \cdot (+ 1) + n_{S} 1 \cdot x + peroxide O 2 \cdot (- 1) + normal O 3 \cdot (- 2) = - 1$ Why this step? Two oxygen "kinds" get two different values — that's the whole subtlety. The RHS is $- 1$ because it's an ion, and pinning H and both oxygen groups leaves S as the only unknown.

Step 3 — Solve. $1 + x - 2 - 6 = - 1 \Rightarrow x - 7 = - 1 \Rightarrow x = + 6$ Why this step? Collecting the known terms turns the sum rule into one equation in $x$ , giving sulfur's honest state.

✅ S is $+ 6$ (the same sensible sulfur maximum as in $H_{2} S O_{4}$ ).

Verify: $(+ 1) + (+ 6) + 2 (- 1) + 3 (- 2) = 1 + 6 - 2 - 6 = - 1 = Q$ ✔. Now the wrong method: all five O as $- 2$ gives $1 + x - 10 = - 1 \Rightarrow x = + 8$ — impossible (S has only 6 valence electrons). The absurd $+ 8$ is your proof the peroxide correction was needed, and $+ 6$ is the honest answer.

Ex 10 — Cell C8: the superoxide exception

Worked example Oxygen and potassium in potassium superoxide

$K O_{2}$ Forecast: Two oxygens, one potassium, neutral. If O were $- 2$ you'd need K at $+ 4$ — impossible for a Group 1 metal. So O cannot be $- 2$ here. What is it?

Step 1 — Q2: superoxide exception. $K O_{2}$ contains the superoxide ion $O_{2}^{-}$ : an O–O bond carrying one extra electron shared over two oxygens. Rule 4's superoxide exception gives each O an oxidation number of $- \frac{1}{2}$ . Why this step? K is Group 1, locked at $+ 1$ (Rule 6, never bends). With $Q = 0$ and one K at $+ 1$ , the two oxygens together must total $- 1$ , i.e. $- \frac{1}{2}$ each — a fractional value, and the second legal fraction on this page (alongside Ex 5).

Step 2 — Q1: $Q = 0$ ; let each O be $x$ . Why this step? Setting the RHS to $0$ (neutral compound) and pinning K at $+ 1$ leaves the two identical oxygens as the only unknown, so one equation solves the pair. $n_{K} 1 \cdot (+ 1) + n_{O} 2 \cdot x = 0 \Rightarrow 1 + 2 x = 0 \Rightarrow x = - \frac{1}{2}$ Why the division by 2? The sum rule gives the total oxygen contribution ( $- 1$ ); dividing by the two atoms yields each oxygen's fractional state.

✅ Each O is $- \frac{1}{2}$ ; K is $+ 1$ .

Verify: $(+ 1) + 2 (- \frac{1}{2}) = 1 - 1 = 0 = Q$ ✔. Sanity: this is exactly the superoxide value listed in Rule 4. Contrast with a peroxide like $K_{2} O_{2}$ (O at $- 1$ ) and a normal oxide $K_{2} O$ (O at $- 2$ ) — three different oxygen values for the same partner, decided purely by the O–O bonding.

Connections

1.3.06 Oxidation number rules — assigning, change (Hinglish)
Redox reactions — balancing (half-reaction & ion-electron)
Electronegativity & periodic trends
Oxidising and reducing agents
Stoichiometry — mole conservation in reactions
Electrochemistry — cell potentials