Visual walkthrough — Effective Atomic Number (EAN) rule

This is the visual companion to the parent note: EAN rule. Read line by line — nothing is assumed.

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Step 1 — Start with a shelf of electrons

WHAT we do: draw the neutral metal as a shelf filled with $Z$ marbles.

WHY we start here: EAN is a head-count of electrons, so we must first know the starting count. Everything after this either removes marbles or adds marbles.

WHAT IT LOOKS LIKE: look at Figure s01 — cobalt is drawn with a shelf of $Z=27$ orange marbles. The number on the shelf is $Z$.

$Z=\underset{\text{electrons on neutral Co}}{\underbrace{27}}$

Here $Z$ is the only symbol so far, and it means "the marble count of the neutral atom" — nothing more.

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Step 2 — The metal pays an entry fee (oxidation state)

WHAT we do: physically slide $\text{OS}$ marbles off the shelf.

WHY we do it: an ion of charge $+\text{OS}$ has lost $\text{OS}$ electrons. Those electrons are gone — they can no longer be counted around the metal. If we counted them we would be lying about the metal's real electron neighbourhood.

WHAT IT LOOKS LIKE: in Figure s02, three marbles are pulled off Co's shelf (grey, faded) because Co is $+3$. The shelf now holds $27-3=24$.

$\underset{\text{start}}{\underbrace{Z}}-\underset{\text{marbles paid as entry fee}}{\underbrace{\text{OS}}}=\underset{=24\text{marbles left}}{\underbrace{27-3}}$

• $Z$ — the full starting shelf.
• $\text{OS}$ — the number removed. Subtraction, because these electrons leave.
• The result, $Z-\text{OS}$, is the electron count on the ion itself.

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Step 3 — Friends arrive carrying pairs (the ligands)

WHAT we do: draw each ligand as a friend walking up holding two marbles (its lone pair).

WHY it is two and not one: a coordinate (dative) bond is made when one partner supplies the whole pair. Both electrons come from the ligand, so each incoming bond adds 2 marbles to the metal's usable pile, not 1.

WHAT IT LOOKS LIKE: Figure s03 shows one ${\text{NH}}_3$ ligand handing across a violet pair of marbles into the metal's shelf. Follow the arrow: two marbles cross, one bond forms.

$\text{one donor}⟶+2\text{electrons}$

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Step 4 — Count the friends (coordination number) and total the gifts

WHAT we do: line up all $\text{CN}$ friends; each brought 2 marbles, so together they bring $2\times \text{CN}$.

WHY we multiply: it is repeated addition. Six friends, two marbles each, is $2+2+2+2+2+2=2\times 6=12$. The multiplication is just a shortcut for "add 2 as many times as there are donors."

WHAT IT LOOKS LIKE: Figure s04 arranges the six ${\text{NH}}_3$ friends of $[{\text{Co(NH}}_3{)}_6{]}^{3+}$ in an octahedron, each arrow depositing 2 violet marbles — $12$ marbles land on the shelf.

$\underset{\text{marbles per friend}}{\underbrace{2}}\times \underset{\text{number of friends}}{\underbrace{\text{CN}}}=2\times 6=\underset{\text{donated marbles}}{\underbrace{12}}$

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Step 5 — Add the two piles: the EAN formula is born

WHAT we do: put the leftover metal marbles (Step 2) together with the donated marbles (Step 4) on one shelf, and count the total.

WHY this total matters: this grand count is what the metal appears to own — its own surviving electrons plus everything the ligands lent it. That apparent total is the Effective Atomic Number.

WHAT IT LOOKS LIKE: Figure s05 merges the 24 orange leftover marbles with the 12 violet donated marbles into one shelf of 36. The two coloured groups let you see each term of the formula.

For $[{\text{Co(NH}}_3{)}_6{]}^{3+}$:

$\text{EAN}=27-3+2\times 6=24+12=36$

$36$ is the atomic number of krypton (Kr). The metal now looks like a noble gas — the target of maximum contentment. See Noble gas configuration.

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Step 6 — The tricky case: negative ligands (getting OS right)

WHAT we do: work out OS from a tiny balance equation, using $[{\text{Fe(CN)}}_6{]}^{4-}$.

WHY we need it: each cyanide ${\text{CN}}^{-}$ carries charge $-1$. The bracket charge $-4$ is shared between the metal and these charged ligands, so we must split it correctly.

WHAT IT LOOKS LIKE: Figure s06 shows a balance beam — the metal's unknown charge $x$ on one side, the six $-1$ ligands and the total $-4$ arranged so the beam balances.

$\underset{\text{metal charge (OS)}}{\underbrace{x}}+\underset{{\text{six CN}}^{-}\text{ ligands}}{\underbrace{6\times (-1)}}=\underset{\text{whole-complex charge}}{\underbrace{-4}}$

Solve: $x-6=-4\Rightarrow x=+2$. So Fe is $+2$, not $-4$.

Now finish with the Step-5 formula:

$\text{EAN}=\underset{Z\text{of Fe}}{\underbrace{26}}-\underset{\text{OS}}{\underbrace{2}}+\underset{\text{6 donors}}{\underbrace{2\times 6}}=24+12=36=\text{Kr}✓$

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Step 7 — The degenerate case: when the count misses (and it's fine)

WHAT we do: count $[{\text{Ni(NH}}_3{)}_6{]}^{2+}$ the same honest way.

WHY show it: to prove the count can land between noble gases. If we hid this, you would wrongly believe "EAN $\ne$ noble gas $\Rightarrow$ cannot exist."

WHAT IT LOOKS LIKE: Figure s07 puts the shelf that reaches only 38 marbles right beside the Kr target line at 36. The two overshoot marbles stick out past the line.

$\text{EAN}=\underset{Z\text{Ni}}{\underbrace{28}}-\underset{\text{OS }+2}{\underbrace{2}}+\underset{{\text{6 NH}}_3}{\underbrace{2\times 6}}=26+12=38\ne 36$

The count is $38$ — it overshoots Kr by 2 marbles. Yet $[{\text{Ni(NH}}_3{)}_6{]}^{2+}$ exists and is stable. Lesson: EAN is strongest for Metal carbonyls and low-oxidation-state species, and is a heuristic elsewhere.

Recall Quick self-check on the edge cases

Why can EAN $=38$ still describe a real complex? ::: Because EAN is a stability guide, not a law; many complexes are stable without matching a noble-gas count. In $[{\text{Fe(CN)}}_6{]}^{4-}$, why isn't Fe's OS $-4$? ::: The six ${\text{CN}}^{-}$ carry $-6$ total; solving $x-6=-4$ gives $x=+2$.

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The one-picture summary

Figure s08 compresses all seven steps into a single flow of marbles: the full shelf $Z$, three marbles paid as the OS fee, twelve marbles arriving from six friends, and the final pile matching the Kr line. Read it left to right and you have re-derived the formula.

$\boxed{\text{EAN}=Z-\text{OS}+2\times \text{CN}}$

Recall Feynman retelling — the whole walkthrough in plain words

A metal atom starts holding a full shelf of marbles — that count is $Z$. To join the party it pays an entry fee by handing over some marbles; that fee is its oxidation state, so its own pile drops to $Z-\text{OS}$. Then friends (the ligands) arrive, and each friend hands over a pair of marbles through a dative bond — two each, never one. Count the friends (that's the coordination number) and you get $2\times \text{CN}$ extra marbles. Pour the leftover pile and the gifted pile together and count: that total is the Effective Atomic Number. If it exactly equals a "cool kid" noble-gas number like 36 (krypton), the metal is delighted and the complex is extra stable. Sometimes the pile lands on 38 instead — the metal is still fine, because this is friendly advice, not an iron law.

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Connections

• Coordination number — the number of friends, our "$\times 2$" multiplier.
• Oxidation state of central metal — the entry-fee marbles removed in Step 2 & 6.
• Coordinate (dative) bond — why each friend gives 2 marbles.
• Noble gas configuration — the target line the count aims for.
• 18-electron rule — the same count read as valence electrons only.
• Metal carbonyls — where EAN predicts best (Step 6).
• Werner's theory — the bonding picture behind ligands and coordination number.