Variation of oxidation state across the table: Visual walkthrough

We start by agreeing on the only thing we need: an atom is a nucleus (positive) surrounded by electrons (negative), and the outermost electrons are the ones that do chemistry.

Step 1 — What "oxidation state" even counts (the marble picture)

WHAT. Imagine each atom holds some marbles = electrons. When two atoms bond, they share a marble pair. To assign an oxidation state we stop pretending it's shared and give both marbles to the greedier atom.

WHY this rule and not "real charge"? Real bond charges are messy fractions. Oxidation state is a bookkeeping trick: an integer we can add up. The whole subject is "did an atom lose marbles (positive) or gain marbles (negative)?" — so we force a clean yes/no on every pair.

PICTURE. Two atoms, one shared marble pair sitting on the bond. The greedier atom (red, high electronegativity) takes both marbles; it ends up more negative, the other more positive.

Figure — Variation of oxidation state across the table

The number on each atom is its oxidation state: $- 1$ for the taker, $+ 1$ for the giver in this single-bond case.
$+ 1$ means "handed away one marble I owned."
$- 1$ means "grabbed one marble that wasn't mine."

Step 2 — Where valence electrons live (why only the outer shell plays)

WHAT. We split an atom's electrons into two piles: inner (core) electrons and valence electrons. Call the valence count $n_{v}$ .

WHY split them? Because chemistry can only move the reachable marbles. The core electrons are so tightly bound (huge Ionisation Energy to remove) that no ordinary reaction touches them. This split is the entire reason a maximum oxidation state exists — we'll cash that in next step.

PICTURE. A target-diagram atom: a deep core well (grey, "locked") and a shallow valence ring (yellow, "in play"). $n_{v}$ = how many yellow marbles.

$n_{v}$ comes straight from Electronic Configuration — it is the count of $s + p$ electrons in the outermost shell (equal to the old group number for main-group atoms).

Reveal check:

How many valence electrons does chlorine have?

3 s^{2} 3 p^{5} = 7

, so

n_{v} = 7

Step 3 — Deriving the MAXIMUM positive state (give away every yellow marble)

WHAT. Line the valence marbles up and hand them, one at a time, to a super-greedy partner (usually oxygen or fluorine). Count how many you gave: that's the biggest positive number possible.

WHY it caps here. After the last yellow marble is gone, the atom is a bare core. Removing a core marble would cost a colossal Ionisation Energy — chemistry simply cannot pay it. So the ceiling is exactly $n_{v}$ .

PICTURE. A staircase: start at $0$ , each step "gives one electron", the height climbing $+ 1, + 2, \dots, + n_{v}$ , then a wall (locked core) blocking further climb.

Step 4 — Deriving the MINIMUM (most negative) state (grab until the pocket is full)

WHAT. Instead of giving marbles, a non-metal fills its pocket to the magic number 8. It can only grab the empty slots, and there are $8 - n_{v}$ of them.

WHY the floor is here. Once the shell is full (8), an extra marble would have to enter the next, higher shell — energetically unfavourable (Electron Affinity turns unhelpful). So grabbing stops exactly when the octet completes.

PICTURE. The same shell drawn as 8 slots; $n_{v}$ already filled (yellow), $8 - n_{v}$ empty (dashed). Green arrows drop one grabbed marble into each empty slot; the count of green arrows = the size of the negative number.

Step 5 — Across a period: the max climbs by +1 each step

WHAT. Walk Na → Mg → Al → Si → P → S → Cl across period 3 and plot each element's maximum.

WHY a straight line? Because $n_{v}$ literally increases by 1 per step (group number rises by 1), and max $= + n_{v}$ . A "+1 per step" input gives a "+1 per step" output — a straight ramp.

PICTURE. A bar/line chart: elements on the x-axis, max positive OS on the y-axis, climbing $+ 1 \to + 7$ . A second (green) trace shows the negative floor $- (8 - n_{v})$ appearing only on the right, where atoms are close to a full octet.

Left (metals): low IE, so they only give — positive only, equal to group number.
Right (non-metals): show both a high positive (bonding to O/F) and the negative floor.

Step 6 — Down a group: the inert-pair dip (why heavy metals go low)

WHAT. Compare group 14 top-to-bottom: carbon/silicon happily reach $+ 4$ ; lead prefers $+ 2$ .

WHY. Removing all $n_{v}$ marbles means also prying out that stubborn $n s^{2}$ pair. Down the group that costs more than it pays back, so the compound where the pair stays put (the lower state) is the stable one. This is the Inert Pair Effect.

PICTURE. Two atoms side by side: light C with its $s^{2}$ pair loose (easily given, tall $+ 4$ bar); heavy Pb with the $s^{2}$ pair clamped (red lock, short $+ 2$ bar dominant).

Step 7 — Transition metals: the two-close-ladders picture

WHAT. Contrast a main-group atom (its $s$ and $p$ levels far apart → states jump by 2) with a transition metal (its $d$ and $s$ levels nearly level → states step by 1).

WHY the step size differs. Big energy gap ⇒ you can't stop halfway, so you spend electrons in pairs (jump 2). Tiny gap ⇒ you can remove them singly (step 1). It's purely a picture of ladder-rung spacing.

PICTURE. Two energy ladders. Left: main-group, rungs far apart, big jumps. Right: manganese, $3 d$ and $4 s$ rungs almost touching, tiny steps giving $+ 2, + 3, + 4, + 6, + 7$ .

Step 8 — The degenerate cases (where the tidy rules bend)

Case A — F and O never reach their group max. To be positive, an atom must hand marbles to something greedier. Nothing out-greeds fluorine, and only fluorine out-greeds oxygen. So F is only ever $- 1$ or $0$ ; O is positive only in $OF_{2}$ ( $+ 2$ ).

Case B — averages can be fractional. Oxidation state is a per-molecule average. In $N_{2} O$ : $2 x + (- 2) = 0 \Rightarrow x = + 1,$ a value nowhere near N's extremes of $- 3$ or $+ 5$ — because it's the mean of atoms in different environments. Peroxides ( $O = - 1$ ) and superoxides ( $O = - \frac{1}{2}$ ) are the same story for oxygen.

PICTURE. A "greed ranking" bar (F tallest, then O) with an arrow showing marbles only ever flow toward the greedier atom — visually explaining why F can't be positive; plus a small $N_{2} O$ tally showing the $+ 1$ average.

The one-picture summary

One diagram folds in the whole walkthrough: a valence-marble bank in the middle, a give-away staircase climbing to $+ n_{v}$ on the right, a grab-to-octet slide falling to $- (8 - n_{v})$ on the left, the period ramp above, the inert-pair "heavy = low" dip and the transition-metal close-ladder inset.

Recall Feynman retelling of the whole walkthrough

Every atom is a kid with marbles in an outer pocket. Only the outer marbles are reachable — the inner ones are glued down (Step 2). If you hand away all your outer marbles you hit your biggest plus number, and you can't go higher because the next marble is glued: that's max $= + n_{v}$ (Step 3). If instead you grab marbles until your pocket holds the lucky number 8, you hit your biggest minus number, $- (8 - n_{v})$ (Step 4). Walk right along a row and every kid has one more marble to give, so the plus-record climbs by one each step (Step 5). Walk down a column and the heaviest kids clamp two marbles ( $n s^{2}$ ) tight and refuse to spend them, so they settle two rungs lower — the inert-pair dip (Step 6). Transition-metal kids keep marbles in two pockets at almost the same height, so they hand them out one at a time and show a whole rainbow of numbers (Step 7). And two rules bend: the greediest kids of all (F, and O behind it) can never be made positive, and when we average kids in different situations we can even get in-between numbers like $+ 1$ in $N_{2} O$ (Step 8). That's the entire map of oxidation states, from empty pockets up.

Connections

Electronegativity — decides who takes the shared marble (Steps 1, 8).
Ionisation Energy — the wall that caps the max at $n_{v}$ (Step 3).
Electron Affinity — the floor that stops grabbing at the octet (Step 4).
Electronic Configuration — where $n_{v}$ and the $d$ – $s$ spacing come from (Steps 2, 7).
Inert Pair Effect — the heavy-element dip (Step 6).
Transition Metal Chemistry — the close-ladder rainbow (Step 7).
Redox Reactions — where these numbers get used.