Effective Atomic Number (EAN) rule: Exercises — Chemistry deep dive

Before any exercise, let's build the whole idea from a single picture. A metal atom sitting in a complex is like a small nucleus surrounded by shells of electrons. Some of those electrons are its own; some are lent to it by the molecules stuck onto it. If we add up all of them, we get one number that predicts stability. That number is the Effective Atomic Number (EAN).

The figure below shows the counting, one arrow at a time — read it left to right before reading the formula.

Figure — Effective Atomic Number (EAN) rule

The blue box (left) is $Z$ : the electron count of the neutral metal — every electron it owns before anything happens.
The pink box is the oxidation state (OS): electrons the metal gave away to become a charged ion. We subtract these because they left — the pink arrow points out of the metal.
The yellow box is the coordination number (CN) times two: each attached molecule (a ligand) hands the metal a pair of electrons (2), and the yellow arrows point into the metal.
The white box (right) sums them: that total is EAN. We compare it against the noble-gas numbers on the ledger below it.

Level 1 — Recognition

Goal: read a formula off the parts. No traps yet, just correct counting.

Recall Solution L1.1

WHAT we're doing: picking out the three numbers the formula eats, nothing more.

$Z$ (blue box): the metal is cobalt. Its atomic number is $Z = 27$ .
Oxidation state (pink box): ammonia $NH_{3}$ is a neutral ligand (charge $0$ ). So the metal alone must carry the whole $+ 3$ charge shown on the bracket ⇒ OS $= + 3$ .
CN (yellow box): count the donor atoms — six $NH_{3}$ , each bonds once ⇒ CN $= 6$ .

Answer: $Z = 27, OS = + 3, CN = 6$ .

Recall Solution L1.2

A complex is especially stable when its EAN equals the atomic number of the nearest noble gas. The targets are: $He = 2, Ne = 10, Ar = 18, Kr = 36, Xe = 54, Rn = 86.$ This is a guide (Sidgwick's heuristic), not a law — see the mistakes below.

Level 2 — Application

Goal: run the full formula, including the oxidation-state subtraction with charged ligands.

Recall Solution L2.1

Step 1 — $Z$ : iron, $Z = 26$ . Step 2 — OS (the careful step): each cyanide $CN^{-}$ has charge $- 1$ ; six of them sum to $- 6$ . Let the metal charge be $x$ . The parts must add to the bracket charge: $x + 6 (- 1) = - 4 \Rightarrow x = - 4 + 6 = + 2.$ So Fe is $+ 2$ , not $- 4$ . Step 3 — electrons left on the ion: $Z - OS = 26 - 2 = 24$ . Step 4 — donated electrons: $2 \times CN = 2 \times 6 = 12$ . Step 5 — EAN: $24 + 12 = 36 = Kr$ . ✔ Obeys the rule.

Recall Solution L2.2

$Z (Ni) = 28$ .
CO is neutral; the complex is neutral ⇒ OS $= 0$ .
Electrons on metal $= 28 - 0 = 28$ .
Donated $= 2 \times 4 = 8$ .
EAN $= 28 + 8 = 36 = Kr$ . ✔ One line: hitting the Kr count (a full noble-gas shell, see Noble gas configuration) makes $Ni(CO)_{4}$ exceptionally stable — the textbook success of the EAN rule.

Level 3 — Analysis

Goal: find a missing quantity, or judge disobedience.

Recall Solution L3.1

$Z (Ni) = 28$ ; $NH_{3}$ neutral ⇒ OS $= + 2$ ; CN $= 6$ .
$EAN = (28 - 2) + 2 \times 6 = 26 + 12 = 38.$
$38 \neq = 36$ , so it misses the Kr target by $2$ . ✘ Why still stable: EAN is a heuristic, strongest for low-oxidation-state carbonyls, not a law of existence. Octahedral $Ni^{2 +}$ amine complexes are common and robust. The rule "predicts extra stability when hit," but does not forbid stability when missed.

Recall Solution L3.2

WHAT we're doing: running the formula backwards to solve for the unknown $Z$ .

OS $= 0$ , CN $= 5$ ⇒ donated $= 2 \times 5 = 10$ .
Require $EAN = 36$ : $Z - 0 + 10 = 36 \Rightarrow Z = 26.$
$Z = 26$ is iron. The molecule is $Fe(CO)_{5}$ . ✔ (Indeed a famous stable carbonyl.)

Level 4 — Synthesis

Goal: combine EAN with the 18-electron rule and with unknowns on two sides.

Before the problems, look at the shell picture below. It shows why "EAN $= 36$ " and "18 valence electrons" are the same statement seen twice. The buried inner shells (the argon core, 18 electrons) are drawn faint; the outer working shells ( $4 s, 3 d, 4 p$ ) are drawn bright. EAN counts every dot; the 18-electron rule counts only the bright outer dots.

Recall Solution L4.1

WHAT: Kr has $Z = 36$ . Its electron configuration is an argon core (18 electrons, the faint inner region of the figure) plus $4 s^{2} 3 d^{10} 4 p^{6}$ (the bright outer region).

Valence electrons (the bright $4 s, 3 d, 4 p$ set) $= 2 + 10 + 6 = 18$ .
Core (Ar) electrons $= 18$ .
So $EAN - (noble-gas core) = 36 - 18 = 18 =$ the valence count. ✔ General bridge: EAN counts all electrons and aims at the noble-gas $Z$ ; the 18-electron rule counts only valence electrons and aims at 18. Same physics, two ledgers. See 18-electron rule.

Recall Solution L4.2

EAN target = Kr = 36. $24 - 0 + 2 x = 36 \Rightarrow 2 x = 12 \Rightarrow x = 6.$ So the molecule is $[Cr(CO)_{6}]$ . 18-electron cross-check: neutral Cr has $6$ valence electrons ( $4 s^{1} 3 d^{5}$ ); each CO donates $2$ ; total valence $= 6 + 2 \times 6 = 6 + 12 = 18$ . ✔ Both ledgers agree.

Level 5 — Mastery

Goal: build a full multi-constraint argument from scratch.

Recall Solution L5.1

Set-up: $Z (Co) = 27$ , CN $= 6$ (octahedral, see Werner's theory), so donated $= 12$ . (a) OS from EAN: require EAN $= 36$ : $27 - OS + 12 = 36 \Rightarrow 39 - OS = 36 \Rightarrow OS = + 3.$ (b) Overall charge: ligands are neutral, so the complex charge equals the metal charge: $n = OS + 6 \times (0) = + 3.$ (c) Example: $[Co(NH_{3})_{6}]^{3 +}$ — matches all three constraints. ✔

Recall Solution L5.2

WHAT we're doing: turning the formula into a general statement, not one number.

Neutral carbonyl ⇒ OS $= 0$ . EAN condition: $Z - 0 + 2 k = 36 \Rightarrow Z = 36 - 2 k .$
$36$ is even and $2 k$ is even, so $Z = 36 - 2 k$ is even for every whole number $k$ . ∎ Sanity list: $k = 4 \Rightarrow Z = 28$ (Ni ✔), $k = 5 \Rightarrow Z = 26$ (Fe ✔), $k = 6 \Rightarrow Z = 24$ (Cr ✔). All even, all real. This is exactly why the classic Kr-carbonyls belong to even- $Z$ metals.

Recall Master mnemonic (carry into the exam)

"Start with $Z$ , subtract the metal's charge, add two electrons for every ligand — then check the total against a noble gas." The six targets are just a list to compare against (2, 10, 18, 36, 54, 86); they are not multiplied together — the dots below are only separators, not a product: $targets = {2, 10, 18, 36, 54, 86} .$ Backwards-solving? Just isolate whichever letter is missing; the equation $Z - OS + 2 CN = (noble gas)$ has one unknown at a time.

Connections

Parent: EAN rule — the theory these drills test.
Coordination number — the CN in every problem.
Oxidation state of central metal — the trap of L1/L2.
18-electron rule — the valence-only ledger of L4.
Metal carbonyls — the star performers (L2, L5).
Coordinate (dative) bond — source of the "×2".
Noble gas configuration — the target we aim at.
Werner's theory — octahedral geometry in L5.