Nomenclature (IUPAC) — naming complex ions and compounds: Worked examples

Two small tools we will use everywhere

Before the examples, we lock down three things the examples lean on constantly.

Definition AANC — the four special neutral-ligand names

Most neutral ligands keep their molecule name, but four get special names you must memorise. The mnemonic AANC spells them out:

Molecule	Ligand name	Letter
$H_{2} O$	aqua	A
$NH_{3}$	ammine (two m's)	A
$NO$	nitrosyl	N
$CO$	carbonyl	C
So whenever this page says "from the AANC list", it means one of these four.

Definition The "stem" of a metal name (what "-ate" attaches to)

When a complex is an anion, the metal takes the suffix -ate. But you do not just glue "-ate" onto the full English name — you attach it to the metal's stem: the core of the name with its own ending stripped off. Think of the stem as the "root word" left after you remove the trailing letters that form a word-ending.

How to extract the stem: chop off the tail that spells the free-element ending, then add -ate.
- aluminium → stem alumin- → aluminate
- zinc → stem zinc- → zincate
- cobalt → stem cobalt- → cobaltate
Some metals switch to their Latin stem for the anion (this is memorised, not derived):

English Latin stem -ate name

iron ferr- ferrate

copper cupr- cuprate

silver argent- argentate

lead plumb- plumbate

tin stann- stannate

gold aur- aurate

So "extract the stem" always means: find the root, drop the free-element ending, add -ate — using the Latin stem when the table above says so.

The scenario matrix

Every naming problem is really "which cell am I in?" Once you spot the cell, the algorithm runs itself.

#	Cell (the scenario)	What is tricky about it	Example that covers it
A	Cationic complex (charge $> 0$ )	metal keeps its plain name	Ex 1
B	Anionic complex (charge $< 0$ )	metal takes -ate (often Latin stem)	Ex 2
C	Neutral complex (charge $= 0$ )	no counter ion, still write oxidation state	Ex 3
D	Complex ion mixed with simple counter ion	split bracket first, balance charge	Ex 1, 2
E	Ligand name already contains di/tri	use bis / tris / tetrakis + parentheses	Ex 4
F	Ambidentate ligand (two possible donor atoms)	donor atom must be named (nitrito- $N$ vs - $O$ )	Ex 5
G	Two different metals / complex cation + complex anion	name each sphere separately, cation first	Ex 6
H	Bridging ligand (joins two metals)	use the $μ$ (mu) prefix	Ex 7
I	Hydrate / water of crystallisation	that water is NOT a ligand	Ex 8
J	Zero / degenerate: oxidation state $= 0$ or a single-charge ligand set	limiting check of the charge equation	Ex 3, Ex 9

We build any missing tool (mu, ambidentate, "-ate", oxidation-state algebra) from scratch before using it.

Cell A — cationic complex

Worked example Example 1 —

$[Cr (NH_{3})_{4} Cl_{2}] Cl$ (Cells A, D) Forecast first: guess the name before reading. Is chromium plain or does it get "-ate"? What is its oxidation state?

Split the formula at the bracket. Inside: $[Cr (NH_{3})_{4} Cl_{2}]$ . Outside: one $Cl^{-}$ . Why this step? The coordination sphere behaves as one ion; the outside $Cl^{-}$ is a counter ion that only balances charge. One outside $Cl^{-}$ (charge $- 1$ ) means the complex ion is $+ 1$ .
Charge balance to get the oxidation state. $NH_{3}$ is neutral (0), each inside $Cl$ is $- 1$ . $x + 4 (0) + 2 (- 1) = + 1 \Rightarrow x = + 3.$ Why this step? We never guess the oxidation state — we derive it from charge conservation.
Alphabetise ligands by root. "ammine" (a) before "chlorido" (c). Why this step? We order by the ligand name, ignoring the prefix.
Prefixes. Four $NH_{3}$ → tetraammine; two $Cl$ → dichlorido (the "-ido" ending from our tool box above).
Metal. Complex ion is $+ 1$ (positive) → chromium stays chromium, oxidation state III.

Name: tetraamminedichloridochromium(III) chloride.

Verify: rebuild the charge: $+ 3 + 0 - 2 = + 1$ on the complex, plus outside $- 1$ → net $0$ . Balanced. ✓

Cell B — anionic complex

Worked example Example 2 —

$Na_{3} [AlF_{6}]$ (Cells B, D) Forecast: does aluminium become "aluminate" or something Latin? What is its oxidation state?

Split. Three $Na^{+}$ outside → complex ion carries $- 3$ (an anion). Why this step? Total outside charge is $+ 3$ , so inside must be $- 3$ to be neutral overall.
Oxidation state. Each $F^{-}$ is $- 1$ , six of them $= - 6$ . $x + 6 (- 1) = - 3 \Rightarrow x = + 3.$
Ligand ending. fluoride → fluorido (drop -ide, add -ido); six → hexafluorido.
Metal gets -ate (via its stem). Complex is an anion → apply the stem rule: aluminium has no Latin form, so drop the "-ium" ending → stem alumin- → add "-ate" → aluminate. Why this step? This is the "Anion → -ATE" rule, and "-ate" always attaches to the stem, not the whole name. Only the sign of the whole complex ion triggers it.
Cation first. Sodium named before the anion.

Name: sodium hexafluoridoaluminate(III).

Verify: $+ 3 + 6 (- 1) = - 3$ inside, plus $3 (+ 1)$ outside $= 0$ . ✓

Cell C / J — neutral complex, oxidation state zero

Worked example Example 3 —

$[Fe (CO)_{5}]$ (Cells C, J) Forecast: with no ions outside, what is the total charge? Can an oxidation state be zero?

Count outside ions. None. So the whole neutral molecule has charge $0$ . Why this step? No counter ion ⇒ the complex itself must be electrically neutral.
Oxidation state (degenerate/zero case). $CO$ is a neutral ligand (0). $x + 5 (0) = 0 \Rightarrow x = 0.$ Why this step? Zero is a perfectly legal oxidation state — the charge equation does not "break", it just gives $x = 0$ . We still write it.
Special ligand name. $CO$ → carbonyl (the C of the AANC list of special neutral ligands defined above). Five → pentacarbonyl.
Metal. Neutral complex → iron keeps plain name iron, with $(0)$ .

Name: pentacarbonyliron(0).

Verify: $0 + 5 (0) = 0$ . Neutral, no counter ion needed. ✓ (The same logic gives the neutral $[Ni (CO)_{4}]$ → tetracarbonylnickel(0): CO neutral, no counter ion, so nickel is $0$ .)

Cell E — ligand name already contains di/tri

Worked example Example 4 —

$[Cu (en)_{2}] SO_{4}$ (Cell E) (en = ethylenediamine, neutral, bidentate) Forecast: do we say "diethylenediamine"? Or something else? What is copper's oxidation state?

Split. One $SO_{4}^{2 -}$ outside → complex ion is $+ 2$ . Why this step? Sulfate carries $- 2$ , so the complex must be $+ 2$ to balance.
Oxidation state. en is neutral (0), and it is bidentate (it grabs the metal with two donor atoms), but denticity does not change charge. $x + 2 (0) = + 2 \Rightarrow x = + 2.$ Why this step? Only charge enters the balance; the number of donor atoms affects coordination number, not oxidation state.
Prefix collision. "ethylenediamine" already contains "di". Using "di-" would read as diethylenediamine — ambiguous. So use the multiplicative set bis, tris, tetrakis with the name in parentheses. Why this step? bis/tris/tetrakis exist precisely to avoid this collision. Two en → bis(ethylenediamine).
Metal. Complex is $+ 2$ (cation) → copper stays copper, II.

Name: bis(ethylenediamine)copper(II) sulfate.

Verify: $+ 2 + 2 (0) = + 2$ inside, $- 2$ outside → $0$ . ✓

Cell F — ambidentate ligand (donor atom matters)

Figure — Nomenclature (IUPAC) — naming complex ions and compounds

Figure 1 — The same $NO_{2}^{-}$ ion, two ways to attach. Left: nitrogen (teal) touches the cobalt → nitrito- $κ N$ . Right: an oxygen (plum) touches the cobalt → nitrito- $κ O$ . Same atoms, different donor, different name.

Worked example Example 5 —

$[Co (NH_{3})_{5} (NO_{2})] Cl_{2}$ bonded through N (Cell F) Forecast: two compounds share this formula. What single word tells them apart?

Split. Two $Cl^{-}$ outside → complex ion is $+ 2$ .
Oxidation state. $NH_{3}$ neutral, $NO_{2}^{-}$ is $- 1$ . $x + 5 (0) + (- 1) = + 2 \Rightarrow x = + 3.$
Name the donor atom. Bonded through nitrogen → nitrito- $κ N$ (equivalently the classic name nitro). Look at Figure 1, left panel: the N sits against the cobalt. Why this step? Without the $κ N$ tag the name would fit two isomers — the contract that "one compound = one name" would break.
Alphabetise. "ammine" (a) vs "nitrito" (n) → ammine first. Five $NH_{3}$ → pentaammine.
Metal. Cation → cobalt(III).

Name: ==pentaamminenitrito- $κ N$ -cobalt(III) chloride== (classic: pentaamminenitrocobalt(III) chloride).

Verify: $+ 3 + 5 (0) + (- 1) = + 2$ inside; $- 2$ outside → $0$ . ✓ The O-bonded twin would be pentaamminenitrito- $κ O$ -cobalt(III) chloride — same charges, different name. ✓

Cell G — complex cation AND complex anion together

Worked example Example 6 —

$[Co (NH_{3})_{6}] [Cr (CN)_{6}]$ (Cell G) Forecast: two metals, two brackets, no simple ions at all. Which name goes first? Which metal gets "-ate"?

Identify the two spheres. Cation sphere $[Co (NH_{3})_{6}]$ ; anion sphere $[Cr (CN)_{6}]$ . Why this step? Each bracket is its own ion; name them independently, then join cation first (like sodium chloride).
Fix the anion sphere's charge from its own ligands. The cyanide sphere is built only from $CN^{-}$ ligands, and cobalt(III)-type ammine cations pair 1:1 with a matching anion. The chromium sphere here is $[Cr(CN)_{6}]^{3 -}$ — a very common, stable ion. Take its charge as $- 3$ . Why this step? We anchor the problem on the known, stable hexacyanido sphere; with a 1:1 formula (one cation bracket, one anion bracket) the cation must then be $+ 3$ to cancel it.
Solve each metal's oxidation state from its own bracket. For chromium, with sphere charge $- 3$ and $CN^{-}$ each $- 1$ : $y + 6 (- 1) = - 3 \Rightarrow y = + 3.$ For cobalt, the cation sphere must be $+ 3$ (to cancel the $- 3$ anion in a 1:1 solid), and $NH_{3}$ is neutral: $x + 6 (0) = + 3 \Rightarrow x = + 3.$ Why this step? Each bracket has its own charge-balance equation; the neutrality of the whole solid ( $+ 3$ and $- 3$ cancelling) is what links the two spheres.
Name each sphere. Cobalt sphere is $+ 3$ (positive) → cobalt stays plain → hexaamminecobalt(III). Chromium sphere is $- 3$ (an anion) → chromium takes -ate (stem chromate) → hexacyanidochromate(III). Why this step? Only the sign of that sphere triggers "-ate"; the positive cobalt sphere keeps its plain name.
Join, cation first.

Name: hexaamminecobalt(III) hexacyanidochromate(III).

Verify: cation $+ 3$ , anion $- 3$ → net $0$ . Charge equations: $x + 6 (0) = + 3$ ✓ and $y + 6 (- 1) = - 3 \Rightarrow y = + 3$ ✓.

Cell H — bridging ligand (the μ prefix)

Figure 2 — One hydroxide (plum) sits between two cobalt centres (orange) and bonds to BOTH. Each cobalt also carries five ammines (teal). Because the OH is shared, it earns the $μ$ prefix; the two identical metal units get wrapped as bis(...).

Worked example Example 7 —

$[(NH_{3})_{5} Co - OH - Co (NH_{3})_{5}] Cl_{5}$ (Cell H) Forecast: one $OH$ sits between two cobalts. How do we flag it, and what is each cobalt's oxidation state?

Split. Five $Cl^{-}$ outside → the whole complex cation is $+ 5$ . Why this step? Five outside chlorides carry $- 5$ , so the bracketed cation must be $+ 5$ .
Count pieces inside. Two Co centres, ten $NH_{3}$ (five per metal), one bridging $OH^{-}$ .
Charge balance across both metals. $NH_{3}$ neutral, $OH^{-}$ is $- 1$ : $2 x + 10 (0) + (- 1) = + 5 \Rightarrow 2 x = 6 \Rightarrow x = + 3 (each Co) .$ Why this step? Both cobalts are identical by symmetry, so the total metal charge splits evenly between them.
Bridge notation. The hydroxide bridges → $μ$ -hydroxido. The look: in Figure 2 the plum $OH$ touches both orange cobalt spheres. Why this step? $μ$ is the only way to say "this ligand is shared between two metals"; without it the name would wrongly imply one $OH$ per metal.
Group the two identical metal units with "bis". The compound contains two identical pentaamminecobalt(III) units joined by the bridge. To say "two of this whole grouped unit" we cannot write "di" — the unit name pentaamminecobalt(III) is a composite name that already carries a prefix (penta), exactly the collision case that demands the multiplicative set bis / tris / tetrakis with parentheses. So we write bis(pentaamminecobalt(III)). Why this step? "di" is reserved for simple names; a composite unit that already carries prefixes takes bis in parentheses — the same rule as bis(ethylenediamine) in Example 4, now applied to a metal-containing unit.
Assemble. The bridging ligand (with its $μ$ ) is cited first, then the grouped metal units, then the counter ion.

Name: == $μ$ -hydroxido-bis(pentaamminecobalt(III)) chloride==.

Verify: $2 (+ 3) + 0 - 1 = + 5$ inside, $5 (- 1)$ outside → net $0$ . ✓

Cell I — hydrate (water of crystallisation is NOT a ligand)

Worked example Example 8 —

$[Cr (H_{2} O)_{4} Cl_{2}] Cl \cdot 2 H_{2} O$ (Cell I) Forecast: there are SIX waters written. How many are ligands? Where do the other two go?

Separate the lattice water. The " $\cdot 2 H_{2} O$ " after the dot is crystallisation water — not bonded to Cr. It becomes "dihydrate" at the very end. Why this step? Only atoms inside $[]$ enter the coordination-sphere name; the dot signals lattice water.
Split the rest. Inside: $[Cr (H_{2} O)_{4} Cl_{2}]$ ; outside: one $Cl^{-}$ → complex is $+ 1$ . Why this step? One outside chloride ( $- 1$ ) forces the bracketed ion to be $+ 1$ .
Oxidation state. Aqua neutral, two inside $Cl$ each $- 1$ : $x + 4 (0) + 2 (- 1) = + 1 \Rightarrow x = + 3.$
Assemble sphere. Four aqua → tetraaqua; two chlorido → dichlorido; alphabetise: aqua (a) before chlorido (c). Cation → chromium(III). Counter ion: chloride.
Append hydrate. Add "dihydrate" from step 1 to the end.

Name: tetraaquadichloridochromium(III) chloride dihydrate.

Verify: coordination number is $4 + 2 = 6$ (the four aqua + two chlorido), the standard octahedron ✓. Charge: $+ 3 + 0 - 2 = + 1$ inside, $- 1$ outside → $0$ . ✓

Cell J — the degenerate limiting case (single ligand type, sign flip check)

Worked example Example 9 —

$K [Ag (CN)_{2}]$ (Cells B, J) Forecast: silver's Latin stem? A complex with only ONE type of ligand — does the algorithm still work?

Split. One $K^{+}$ outside → complex is $- 1$ (anion). Why this step? One outside $+ 1$ ion forces the bracket to be $- 1$ .
Oxidation state (limiting: fewest pieces). Two $CN^{-}$ each $- 1$ : $x + 2 (- 1) = - 1 \Rightarrow x = + 1.$ Why this step? Even with the smallest sensible ligand set, the same charge equation holds; nothing degenerate breaks it.
Ligand. cyanide → cyanido (drop -ide, add -ido); two → dicyanido.
Metal + -ate with Latin stem. Anion → apply the stem rule: silver switches to its Latin stem argent- → argentate, oxidation state I. Why this step? Silver is one of the memorised Latin-stem metals; "-ate" attaches to argent-, not to "silver".
Cation first: potassium.

Name: potassium dicyanidoargentate(I).

Verify: $+ 1 + 2 (- 1) = - 1$ inside, plus outside $+ 1$ → net $0$ . ✓ Charge equation confirms $x = + 1$ , so silver is argentate(I). The whole compound is neutral, as every real compound must be. ✓

Active recall

Recall Which cell is each? (cover answers)

Water after a dot (·2H₂O) is named how? ::: As a hydrate (e.g. dihydrate) — it is NOT a ligand. A ligand bridging two metals gets which prefix? ::: μ (mu), e.g. μ-hydroxido. What does the $κ$ symbol tell you? ::: The donor atom through which an ambidentate ligand attaches, e.g. $κ N$ = bonded via nitrogen. Why does "-ide" become "-ido"? ::: To flag the ion is acting as a ligand (bonded to the metal), not a free ion. NO₂⁻ bound through N vs O — what distinguishes the names? ::: The donor atom: nitrito-κN (nitro) vs nitrito-κO (nitrito). Two complex spheres, no simple ions — which is named first? ::: The cation sphere, then the anion sphere (with -ate). What is the "stem" of a metal name? ::: The root left after dropping the free-element ending; "-ate" attaches to it (alumin- → aluminate, argent- → argentate). Silver in an anionic complex becomes? ::: argentate (Latin stem argent- + -ate).

Connections

Parent: the 6-step algorithm
Werner's Theory & Coordination Number
Ligands — types, denticity, chelation
Isomerism in Coordination Compounds
Oxidation State & d-electron count
Crystal Field Theory

English	Latin stem	-ate name
iron	ferr-	ferrate
copper	cupr-	cuprate
silver	argent-	argentate
lead	plumb-	plumbate
tin	stann-	stannate
gold	aur-	aurate