Effective Atomic Number (EAN) rule: Worked examples

This is a Deep Dive child of EAN rule. We take the single formula $EAN = Z - (oxidation state) + 2 \times (coordination number)$ and drive it through every kind of situation it can meet: neutral ligands, charged ligands, cationic ligands, zero charge, a metal in a negative oxidation state, complexes that obey, complexes that break the rule, a word problem, an exam twist, and a bridging-ligand case. Nothing here is new machinery — it is the parent formula, exercised until no case can surprise you.

The scenario matrix

Each row is a distinct "case class". The examples that follow are labelled with the cell they cover, and together they fill the whole table.

Cell	Case class	What is special about it	Example
A	Neutral ligands, positive metal	ligand charge $= 0$ , oxidation state $=$ complex charge	Ex 1
B	Charged (anionic) ligands	must subtract ligand charges to get oxidation state	Ex 2
C	Zero-charge complex, neutral ligands	oxidation state $= 0$ (the carbonyl case)	Ex 3
D	Metal in negative oxidation state	anionic carbonyl; the $- (OS)$ term adds electrons	Ex 4
E	Obeys formula but different noble gas (Rn, not Kr)	check you target the nearest noble gas	Ex 5
F	Disobeys EAN yet is real/stable	shows EAN is a guide, not a law	Ex 6
G	Word problem (given EAN, find something)	run the formula backwards	Ex 7
H	Exam twist: bidentate ligand / CN ≠ number of ligands	CN counts donor atoms, not molecules	Ex 8
I	Cationic ligand ( $NO^{+}$ )	ligand charge is $+ 1$ , so it raises the metal's OS	Ex 9
J	Bridging ligand in a dinuclear complex	count donors per metal; EAN is per-atom	Ex 10

Case A — neutral ligands, positive metal

Worked example Ex 1 —

$[Co(NH_{3})_{6}]^{3 +}$ (cell A) Forecast: Ammonia is neutral, cobalt is a middle transition metal. Guess: will the count land on $36$ (Kr)? Write your guess before reading.

Step 1 — $Z$ of the metal. Cobalt $\Rightarrow Z = 27$ . Why this step? EAN measures the total electrons as if starting from the neutral atom, so we need its full electron count.

Step 2 — Oxidation state. $NH_{3}$ carries charge $0$ ; six of them contribute $0$ . The bracket charge is $+ 3$ , so $x + 6 (0) = + 3 \Rightarrow x = + 3.$ Why this step? The removed electrons (oxidation state) no longer sit on the metal, so we must subtract them. See Oxidation state of central metal.

Step 3 — Electrons left on the ion $= Z - OS = 27 - 3 = 24$ . Why this step? This is the blue bar — the electrons the metal actually keeps after becoming a $+ 3$ ion. Losing 3 electrons from 27 leaves 24.

Step 4 — Donated electrons $= 2 \times CN = 2 \times 6 = 12$ . Why this step? This is the orange bar. Six $NH_{3}$ donors form 6 dative bonds, each carrying a lone pair of 2 electrons, so $6 \times 2 = 12$ .

Step 5 — EAN (add the two bars) $= 24 + 12 = 36 = Kr$ . ✔ Why this step? EAN is defined as kept plus borrowed; stacking the orange on the blue gives the total height, which we compare to the noble-gas line.

Verify: $27 - 3 + 12 = 36$ ; Kr has $Z = 36$ . The bar top meets the green line — forecast confirmed, complex is famously stable. See Noble gas configuration.

Figure — Effective Atomic Number (EAN) rule

Figure s01 — cases A, B, C side by side. Three complexes with different metals (Co, Fe, Ni), different charges (

+ 3, - 4, 0

) and different ligands (NH₃, CN⁻, CO). Read each bar as blue (kept electrons

Z - OS

) plus orange (donated

2 \times CN

); notice all three bar tops land exactly on the green Kr line at 36. The picture's message: many different inputs funnel to the same noble-gas target.

Case B — charged (anionic) ligands

Worked example Ex 2 —

$[Fe(CN)_{6}]^{4 -}$ (cell B) Forecast: Now the ligands themselves carry charge. Do you think the oxidation state equals the visible $- 4$ ? (It does not — that is the trap.) Guess the EAN.

Step 1 — $Z$ . Iron $\Rightarrow Z = 26$ . Why this step? Every EAN starts from the neutral atom's full electron count, so we look up iron's atomic number first.

Step 2 — Oxidation state. Each $CN^{-}$ is $- 1$ , six give $- 6$ . Overall bracket is $- 4$ : $x + 6 (- 1) = - 4 \Rightarrow x = - 4 + 6 = + 2.$ Why this step? Oxidation state = complex charge − sum of ligand charges. Ignoring the ligand charge is the classic error (see [!mistake] below).

Step 3 — Electrons on ion $= 26 - 2 = 24$ . Why this step? This is the blue bar: iron gave away 2 electrons to become $Fe^{2 +}$ , so 24 remain on it.

Step 4 — Donated $= 2 \times 6 = 12$ . Why this step? This is the orange bar: 6 cyanide donors, each contributing a lone pair (2 electrons), give 12.

Step 5 — EAN (add the two bars) $= 24 + 12 = 36 = Kr$ . ✔ Why this step? Total electrons = kept + borrowed; the stacked bar height is the EAN we compare to Kr.

Verify: $26 - 2 + 12 = 36$ . Same Kr target as Ex 1 — a nice sanity check that two different metals reach the same noble gas (this is the middle bar in figure s01).

Case C — zero-charge complex, neutral ligands

Worked example Ex 3 —

$[Ni(CO)_{4}]$ (cell C) Forecast: No charge anywhere. What is the oxidation state of Ni? Predict the EAN.

Step 1 — $Z$ . Nickel $\Rightarrow Z = 28$ . Why this step? We need nickel's neutral electron count as the starting height.

Step 2 — Oxidation state. CO is neutral, complex is neutral: $x + 4 (0) = 0 \Rightarrow x = 0$ . Why this step? Nothing was lost; Ni is a neutral atom sitting in the middle, so the $- OS$ term removes nothing.

Step 3 — Electrons on metal $= 28 - 0 = 28$ . Why this step? This is the blue bar: with OS $= 0$ , the metal keeps all 28 of its electrons.

Step 4 — Donated $= 2 \times 4 = 8$ . Why this step? This is the orange bar: 4 CO donors, each a lone pair (2 electrons), give 8.

Step 5 — EAN (add the two bars) $= 28 + 8 = 36 = Kr$ . ✔ Why this step? Kept plus borrowed; the bar top reaches Kr.

Verify: $28 - 0 + 8 = 36$ . This is exactly why nickel carbonyl is so stable — it hits Kr and hence 18 valence electrons (18-electron rule). It is the right-hand bar in figure s01.

Case D — metal in a negative oxidation state

Worked example Ex 4 —

$[Co(CO)_{4}]^{-}$ (cell D) Forecast: Here the complex is negative but the ligands (CO) are neutral. That forces the metal below zero. When oxidation state is negative, the term $- OS$ becomes a plus. Guess whether EAN still lands on Kr.

Step 1 — $Z$ . Cobalt $\Rightarrow Z = 27$ . Why this step? Start from cobalt's neutral electron count.

Step 2 — Oxidation state. CO neutral, bracket $- 1$ : $x + 4 (0) = - 1 \Rightarrow x = - 1$ . Why this step? Complex charge minus ligand charges gives the metal's OS; here that is negative, meaning the metal has gained an electron's worth of charge.

Step 3 — Electrons on metal $= Z - OS = 27 - (- 1) = 28$ . Why this step? This is the blue bar. Subtracting a negative adds: a negative oxidation state means the metal kept extra electrons, so its blue bar is taller than the neutral count, not shorter.

Step 4 — Donated $= 2 \times 4 = 8$ . Why this step? The orange bar: 4 CO donors × 2 electrons $= 8$ .

Step 5 — EAN (add the two bars) $= 28 + 8 = 36 = Kr$ . ✔ Why this step? Kept plus borrowed reaches the Kr line even with a negative OS.

Verify: $27 - (- 1) + 8 = 27 + 1 + 8 = 36$ . The tetracarbonylcobaltate anion is a well-known stable 18-electron species — the negative oxidation state is not a problem, it is exactly what makes the count work.

Case E — obeys, but a different noble gas (Rn)

Worked example Ex 5 —

$[W(CO)_{6}]$ (cell E) Forecast: Tungsten is a heavy metal, $Z = 74$ . The nearest noble gas above it is Xe ( $Z = 54$ )? Or Rn ( $Z = 86$ )? Predict which, then the EAN.

Step 1 — $Z$ . Tungsten $\Rightarrow Z = 74$ . Why this step? We need tungsten's full electron count; it is a 5d metal, so the numbers are large.

Step 2 — Oxidation state. CO neutral, complex neutral $\Rightarrow x = 0$ . Why this step? No charges to subtract, so the metal is neutral and nothing leaves.

Step 3 — Electrons on metal $= 74 - 0 = 74$ . Why this step? The blue bar: with OS $= 0$ , all 74 electrons stay.

Step 4 — Donated $= 2 \times 6 = 12$ . Why this step? The orange bar: 6 CO donors × 2 electrons $= 12$ .

Step 5 — EAN (add the two bars) $= 74 + 12 = 86 = Rn$ . ✔ Why this step? Kept plus borrowed $= 86$ ; we compare to the nearest noble gas, which for such a large count is Rn, not Kr.

Verify: $74 - 0 + 12 = 86$ ; Rn has $Z = 86$ . Lesson for the matrix: "nearest noble gas" is not always Kr. For a 3d metal it is Kr(36); for a 5d metal like W it is Rn(86). Always target the noble gas the count actually reaches, not a memorised 36.

Case F — disobeys EAN yet is perfectly stable

Worked example Ex 6 —

$[Ni(NH_{3})_{6}]^{2 +}$ (cell F) Forecast: Same nickel as Ex 3, but now octahedral with 6 ammonias and a $+ 2$ charge. Do you expect $36$ again? Guess a number.

Step 1 — $Z$ . Nickel $\Rightarrow Z = 28$ . Why this step? Start from nickel's neutral electron count.

Step 2 — Oxidation state. $NH_{3}$ neutral, bracket $+ 2$ : $x = + 2$ . Why this step? Neutral ligands contribute no charge, so the metal's OS equals the bracket charge.

Step 3 — Electrons on ion $= 28 - 2 = 26$ . Why this step? The blue bar: losing 2 electrons to become $Ni^{2 +}$ leaves 26.

Step 4 — Donated $= 2 \times 6 = 12$ . Why this step? The orange bar: 6 ammonia donors × 2 electrons $= 12$ .

Step 5 — EAN (add the two bars) $= 26 + 12 = 38 \neq = 36$ . ✘ Why this step? Kept plus borrowed $= 38$ ; the bar top pokes above the Kr line by 2, so the rule is not obeyed.

Verify: $28 - 2 + 12 = 38$ . This overshoots Kr by 2, yet $[Ni(NH_{3})_{6}]^{2 +}$ is a real, stable, violet complex. The rule is a heuristic, strongest for carbonyls and low-oxidation-state species, not a law. Never write "cannot exist" just because EAN $\neq =$ noble gas.

Figure s02 — obey vs. overshoot on a number line. The green dot at 36 is the Kr target. The blue marker (

[Ni(CO)_{4}]

, Ex 3) sits on the target — it obeys. The red square (

[Ni(NH_{3})_{6}]^{2 +}

, Ex 6) sits at 38, two steps to the right; the orange double-arrow measures that "+2 too many". The picture shows that missing the target is a matter of degree, and a small overshoot still leaves a perfectly stable, real complex.

Case G — word problem: run the formula backwards

Worked example Ex 7 — Detective mode (cell G)

Statement (in words): A neutral chromium carbonyl complex obeys the EAN rule (EAN $= 36$ , Kr). Chromium has $Z = 24$ and oxidation state $0$ in this complex. How many CO ligands does it have?

Forecast: You know EAN and $Z$ ; you are missing CN. Guess the number of CO's first.

Step 1 — Write the formula with the unknown. $36 = 24 - 0 + 2 \times CN .$ Why this step? Every quantity except CN is known, so we set up one equation in one unknown and solve for CN.

Step 2 — Isolate the donated term. $36 - 24 = 2 \times CN \Rightarrow 12 = 2 \times CN$ . Why this step? Moving the known blue-bar height ( $24$ ) to the left leaves only the orange donated term on the right.

Step 3 — Divide by 2. $CN = 6$ . Why this step? Each ligand supplied 2 electrons; undoing that multiplication converts "donated electrons" back into "number of donor atoms".

Step 4 — State the answer. 6 CO ligands, i.e. $[Cr(CO)_{6}]$ . Why this step? CN $= 6$ neutral CO donors on a neutral Cr is exactly the formula $[Cr(CO)_{6}]$ .

Verify: Plug forward: $24 - 0 + 2 \times 6 = 36 = Kr$ . ✔ Consistent with the parent note's forecast for $[Cr(CO)_{6}]$ .

Case H — exam twist: bidentate ligand (CN ≠ number of molecules)

Worked example Ex 8 —

$[Fe(en)_{3}]^{2 +}$ , en $=$ ethylenediamine (cell H) Forecast: "en" is one molecule but it grabs the metal with two nitrogen donor atoms (it is bidentate). If you count molecules you get CN $= 3$ ; if you count donor atoms you get CN $= 6$ . Which does the formula want? Guess the EAN.

Step 1 — $Z$ . Iron $\Rightarrow Z = 26$ . Why this step? Start from iron's neutral electron count.

Step 2 — Oxidation state. en is neutral, bracket $+ 2$ : $x = + 2$ . Why this step? Neutral ligands add no charge, so the metal's OS equals the bracket charge.

Step 3 — Coordination number. Each en donates through 2 nitrogen atoms; three en molecules give $3 \times 2 = 6$ donor atoms, so $CN = 6$ . Why this step? Coordination number counts donor atoms / dative bonds, not ligand molecules — this is the whole twist.

Step 4 — Electrons on ion $= 26 - 2 = 24$ . Why this step? The blue bar: $Fe^{2 +}$ keeps 24 electrons.

Step 5 — Donated $= 2 \times 6 = 12$ . Why this step? The orange bar: 6 nitrogen donors × 2 electrons $= 12$ (not $2 \times 3 = 6$ ).

Step 6 — EAN (add the two bars) $= 24 + 12 = 36 = Kr$ . ✔ Why this step? Kept plus borrowed reaches Kr — but only because we used the correct CN of 6.

Verify: $26 - 2 + 2 \times 6 = 36$ . Had you wrongly used CN $= 3$ (counting molecules), you would get $26 - 2 + 6 = 30$ and miss Kr entirely. The twist is the whole point: count bonds, not bottles.

Case I — cationic ligand ( $NO^{+}$ )

Worked example Ex 9 —

$[Mn(CO)_{5} (NO)]$ (cell I) Forecast: Nitrosyl here is the positive ligand $NO^{+}$ (a linear nitrosyl donates a pair as $NO^{+}$ ). A positive ligand pushes the metal's oxidation state down, unlike the anionic CN⁻ of Ex 2. The complex is neutral. Predict the EAN.

Step 1 — $Z$ . Manganese $\Rightarrow Z = 25$ . Why this step? Start from manganese's neutral electron count.

Step 2 — Oxidation state. 5 CO are neutral ( $0$ ), one $NO^{+}$ is $+ 1$ ; bracket is $0$ : $x + 5 (0) + (+ 1) = 0 \Rightarrow x = - 1.$ Why this step? Complex charge minus ligand charges: a cationic ligand carries positive charge, so subtracting it drives the metal's OS negative — the opposite direction to an anionic ligand.

Step 3 — Electrons on metal $= 25 - (- 1) = 26$ . Why this step? The blue bar: subtracting a negative OS adds, so the metal keeps 26 electrons — one more than neutral Mn.

Step 4 — Coordination number and donated electrons. 5 CO donors + 1 NO donor $= 6$ donor atoms, so $CN = 6$ and donated $= 2 \times 6 = 12$ . Why this step? The orange bar: each of the 6 dative bonds (5 from CO, 1 from NO) carries a lone pair of 2 electrons.

Step 5 — EAN (add the two bars) $= 26 + 12 = 38 \neq = 36$ . ✘… wait — recheck the NO charge. Why this step? Kept plus borrowed $= 38$ , which overshoots. This flags that our nitrosyl model needs care.

Step 6 — Correct the count. For EAN we treat NO as a 3-electron donor neutrally, but the clean EAN bookkeeping uses $NO^{+}$ as a 2-electron donor with the metal reduced to Mn(−I): $25 - (- 1) = 26$ kept, $12$ donated $= 38$ . The experimentally observed 18-electron species is actually $[Mn(CO)_{5}]$ -type fragments; the honest lesson is that nitrosyl bookkeeping is model-dependent, and different conventions give 36 vs 38. Why this step? To be truthful about a genuinely tricky ligand rather than fake a clean 36.

Verify: Using the $NO^{+}$ (2-electron) convention, $25 - (- 1) + 2 \times 6 = 38$ . Cationic ligands lower the metal's OS (here to $- 1$ ); the takeaway for the matrix is the sign direction, and that ambiguous ligands like NO can miss the noble-gas count depending on convention.

Case J — bridging ligand in a dinuclear complex

Worked example Ex 10 —

$[Mn_{2} (CO)_{10}]$ (cell J) Forecast: Two manganese atoms, ten CO total, and a Mn–Mn bond holding them together. EAN is computed per metal atom, so you must split the ligands and add the metal–metal bond electron. Predict the EAN of one Mn.

Step 1 — $Z$ . Manganese $\Rightarrow Z = 25$ . Why this step? EAN is a per-atom count, so we start from a single Mn's neutral electron count.

Step 2 — Oxidation state. All CO neutral, complex neutral, and the Mn–Mn bond is shared equally $\Rightarrow$ each Mn is $0$ . Why this step? No charged ligands and a symmetric metal–metal bond means neither metal is oxidised relative to the other.

Step 3 — Electrons on one metal $= 25 - 0 = 25$ . Why this step? The blue bar for one Mn: with OS $= 0$ it keeps all 25 electrons.

Step 4 — Terminal CO donors on this Mn. The 10 CO split 5-and-5, so each Mn has 5 terminal CO $\Rightarrow$ donated $= 2 \times 5 = 10$ . Why this step? The orange bar for one atom: only the ligands bonded to this metal count toward its EAN.

Step 5 — Add the Mn–Mn bond electron. The shared Mn–Mn bond gives each metal 1 extra electron $\Rightarrow + 1$ . Why this step? A metal–metal single bond is a shared pair; each atom "owns" one of the two electrons for its own count.

Step 6 — EAN (add all contributions) $= 25 + 10 + 1 = 36 = Kr$ . ✔ Why this step? Kept ( $25$ ) + borrowed from CO ( $10$ ) + one from the Mn–Mn bond ( $1$ ) reaches Kr exactly.

Verify: $25 + 10 + 1 = 36$ ; Kr has $Z = 36$ . This is why $[Mn_{2} (CO)_{10}]$ is stable — each manganese individually hits 36. Forgetting the Mn–Mn bond electron would give 35 and wrongly suggest instability.

Pulling every cell together

Recall Which cells share the same EAN target, and why should that not surprise you?

Cells A, B, C, D, H, and J all reached 36 (Kr) even though the metals (Co, Fe, Ni, Mn), the charges ( $+ 3, - 4, 0, - 1, + 2, 0$ ) and the ligands (NH₃, CN⁻, CO, en, bridged CO) all differed. This is not coincidence: EAN is engineered to hit a noble-gas count, and these are all electron-rich 3d systems. Cell E reached 86 (Rn) because W is a 5d metal; cells F and I missed (38) — F because a $+ 2$ 3d metal with 6 neutral σ-donors overshoots, I because nitrosyl bookkeeping is convention-dependent. Both misses are reminders the rule is a guide.

Recall Backwards use of the formula (cell G) in one line

Given EAN, $Z$ , OS $:::$ solve $2 \times CN = EAN - Z + OS$ , then divide by 2 to get CN.

Recall How does a cationic ligand differ from an anionic one for the oxidation state?

Direction of the shift $:::$ anionic ligand (CN⁻) makes the metal more positive; cationic ligand (NO⁺) makes it more negative, because OS = complex charge − sum of ligand charges.

Connections

Parent: EAN rule — the formula these examples exercise.
Coordination number — cells H and J show why CN counts donor atoms, not molecules.
Oxidation state of central metal — cells B, D, and I show the sign subtleties.
Metal carbonyls — cells C, D, E, G, I, J are all carbonyls.
18-electron rule — every "obeys" example is also an 18-electron species.
Coordinate (dative) bond — the source of the "×2".
Noble gas configuration — the stability target (Kr, Rn).
Werner's theory — the coordination framework underneath.