Visual walkthrough — Oxidation number rules — assigning, change

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Step 1 — What "sharing electrons" looks like

WHAT. Two atoms joined by a bond hold a pair of electrons between them. An electron is the tiny negative-charge particle that atoms trade and share. A bond is just that shared pair drawn as a line.

WHY start here. Oxidation number is entirely a story about who owns the shared pair. Before we can count ownership, we must see the pair sitting in the middle.

PICTURE. Look at the left panel: two identical grey atoms with a pair of dots exactly halfway between them — a fair share. On the right, one atom is drawn hungrier (bigger, magenta); the dot-pair slides toward it. That sliding is the whole subject of this page.

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Step 2 — The pretend rule: give the pair to the greedy atom completely

WHAT. We exaggerate. Instead of "the pair leans a bit toward the greedy atom," we declare: the greedy atom takes both electrons of the pair, 100%. The other atom gets nothing from that bond.

WHY exaggerate. Real charges are messy fractions we cannot easily count. But if every bond is snapped fully to one side, each atom ends up owning a whole number of electrons — countable. And because we apply the same exaggeration everywhere, it cancels consistently when we track changes.

PICTURE. The bond line is cut. Both shared dots jump entirely onto the magenta (greedy) atom. Now compare each atom's electron count to how many it started with when neutral — that comparison is the oxidation number.

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Step 3 — Turning one atom's snap into a number (water, one O–H bond)

WHAT. Take an $O-H$ bond. Oxygen has higher EN than hydrogen, so oxygen grabs the pair. Hydrogen brought $1$ electron to the bond and now owns $0$ of that pair → it is short one → $+1$. Oxygen owns the extra pair → negative.

WHY this bond first. $H_{2}O$ has two identical $O-H$ bonds. If one bond gives H a $+1$, both do — and oxygen collects the deficit from both.

PICTURE. The figure shows one $O-H$ bond snapping. The green arrow labels the electron H lost; the count box turns hydrogen's tally into $+1$. Do it twice and oxygen sits at $-2$.

$$\underset{\text{two H atoms}}{\underbrace{2}}\times \underset{{\text{each H short 1 e}}^{-}}{\underbrace{(+1)}}+\underset{\text{one O atom}}{\underbrace{1}}\times \underset{{\text{O owns 2 extra e}}^{-}}{\underbrace{(-2)}}=\underset{\text{water is neutral}}{\underbrace{0}}$$

Every symbol: the $2$ counts atoms, the $(+1)$ is that atom's score, the product is total score contributed by hydrogens, and the right side is the molecule's real net charge.

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Step 4 — Why the scores must add up to the charge (the sum rule)

WHAT. Snapping bonds only moves electrons from one atom to another. It never makes or destroys any. So if we add up every atom's surplus ($-$) and shortage ($+$), the pluses and minuses that came from shared pairs cancel in pairs — and what's left is exactly the molecule's real net charge.

WHY this is the master key. This single truth lets you find an unknown oxidation number: assign all the easy atoms by rule, call the mystery atom $x$, and the sum forces $x$.

PICTURE. Imagine a scoreboard (the figure): every snapped electron leaves a $-1$ under one atom and a $+1$ under the other. Those matched $\pm 1$ tiles annihilate. The leftover tiles = net charge.

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Step 5 — Solving for the unknown: sulfur in $H_{2}S{O}_4$

WHAT. Assign the easy atoms first (H $=+1$, O $=-2$ by the standard rules), let sulfur be $x$, and demand the sum equal the charge.

WHY these assignments. H bonds to more-electronegative O here, so it loses → $+1$. Each O is more electronegative than S, so O collects → $-2$. Sulfur is whatever the balance leaves.

PICTURE. The figure lines up the eight contributions as stacked tiles: two $+1$ (H), four $-2$ (O), and one unknown $x$ (S). The pile must total $0$ — the tile heights tell you $x$.

$$\underset{=+2\text{from H}}{\underbrace{2(+1)}}+\underset{\text{unknown S}}{\underbrace{x}}+\underset{=-8\text{from O}}{\underbrace{4(-2)}}=\underset{\text{neutral}}{\underbrace{0}}\Rightarrow x-6=0\Rightarrow \boxed{x=+6}$$

The move that matters: $+2-8=-6$, so $x$ must be $+6$ to reach $0$. Sulfur is not really carrying six plus charges — it's the bookkeeping balance.

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Step 6 — The charge is not zero: chromium in $C{r}_2{O}_7^{2-}$

WHAT. Same machine, but the species is an ion with net charge $-2$. So the right-hand side of the sum rule is $-2$, not $0$.

WHY the right side changes. $Q_{\text{species}}$ means the real net charge. For a neutral molecule that's $0$; for an ion it's the ion's charge. Forgetting this is the classic error (parent note, mistake C).

PICTURE. Two panels side by side: neutral $H_{2}S{O}_4$ balances to a flat line at $0$; the dichromate ion balances to a line dropped to $-2$. Same tiles, different target line.

$$\underset{\text{two Cr}}{\underbrace{2x}}+\underset{\text{seven O}}{\underbrace{7(-2)}}=\underset{\text{the ion’s charge}}{\underbrace{-2}}\Rightarrow 2x-14=-2\Rightarrow 2x=12\Rightarrow \boxed{x=+6}$$

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Step 7 — Edge case: the exception (peroxide) and the fraction

WHAT. Two things break the "O is always $-2$" habit:

1. Peroxide $H_2{O}_2$: there is an $O-O$ bond. In a bond between identical atoms, EN is equal — the pair is shared fairly and splits $1$ each. So neither O grabs the other's electron; each O ends at only $-1$.
2. Fractions in $N{a}_2{S}_4{O}_6$: the four sulfurs are not all in the same environment, so the sum rule gives their average.

WHY the peroxide is different. Go back to Step 1's left panel — equal atoms means the pair does not slide. Only the O–H pairs snap. Counting carefully leaves each oxygen at $-1$, not $-2$.

PICTURE. Panel A: the $O-O$ pair sits exactly in the middle (fair share) while both $O-H$ pairs snap onto oxygen — arithmetic lands on $-1$. Panel B: the tetrathionate tiles summing to force $x=+2.5$, drawn as a dashed "average" line cutting between whole numbers.

$$\text{Peroxide: }2(+1)+2x=0\Rightarrow \boxed{x=-1}$$ $$\text{Tetrathionate: }2(+1)+4x+6(-2)=0\Rightarrow 4x=10\Rightarrow \boxed{x=+2.5}$$

The $+2.5$ is real bookkeeping, not a real per-atom charge — oxidation numbers are allowed to be fractional because they're an average over non-equivalent atoms.

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Step 8 — Watching the scoreboard change: a redox reaction

WHAT. Now use oxidation numbers for their real purpose: tracking electron transfer in $Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$.

WHY this closes the loop. Steps 1–7 assigned numbers inside one species. The payoff is comparing before vs after: whichever atom's number rises lost electrons (oxidised); whichever falls gained them (reduced). See Oxidising and reducing agents and Redox reactions — balancing (half-reaction & ion-electron).

PICTURE. Two number lines. Zinc's marker climbs from $0$ to $+2$ (up-arrow, oxidation). Copper's marker drops from $+2$ to $0$ (down-arrow, reduction). The two arrows are the same length — the $2e^{-}$ zinc gives is exactly the $2e^{-}$ copper takes.

Atom	Before	After	Change
Zn	$0$	$+2$	$+2$ (↑ oxidised, reducing agent)
Cu	$+2$	$0$	$-2$ (↓ reduced, oxidising agent)

Conservation check: total change $=(+2)+(-2)=0$. Electrons are only moved, never made — the same principle that powered the sum rule (Step 4) now powers electron balance (and eventually Electrochemistry — cell potentials and Stoichiometry — mole conservation in reactions).

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The one-picture summary

Everything on one canvas: greedy atom pulls the pair (Step 1–2) → each atom's snap becomes a $+$ or $-$ tile (Step 3) → tiles must sum to the net charge (Steps 4–6) → exceptions bend the tile values (Step 7) → comparing tiles across a reaction is redox (Step 8).

Recall Feynman retelling — the whole walk in plain words

Two atoms share a candy pair. The greedier kid always snatches the whole pair — that's our pretend rule. After every snatch, we compare each kid's candy to what they started with: extra candy is a minus score, missing candy is a plus score. Because candy is only moved, never eaten, all the scores must add up to the total candy the group actually owns — that's the sum rule, and it lets us find any one kid's mystery score. Special cases: when two equal kids share, nobody snatches, so scores come out smaller (peroxide, $-1$). Sometimes several kids of the same name have different scores, so we quote the average — even a fraction like $+2.5$. Finally, in a reaction we watch the scoreboard shift: the kid whose score goes up gave candy away (oxidised, the reducing agent); the kid whose score goes down grabbed some (reduced). The up-move and down-move are always equal, because no candy vanished.

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Connections

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