Coordination Compounds — Chemistry Mnemonic
Target: mnemonic for coordination compounds IUPAC naming ligands
Why is this hard to memorize?
Coordination compounds consist of a central metal atom/ion bonded to surrounding molecules or ions called ligands. IUPAC naming follows strict rules: name ligands alphabetically before the metal, use prefixes (di, tri, tetra for simple; bis, tris, tetrakis for complex ligands), anionic ligands end in -o (Cl⁻ = chlorido), neutral ligands keep their name (exceptions: H₂O = aqua, NH₃ = ammine, CO = carbonyl, NO = nitrosyl), cation is named before anion, and oxidation state of metal is shown in Roman numerals. JEE tests naming and isomerism heavily.
Classic mnemonics you should know
"Ligands alphabetically + Metal(oxidation state). Anionic complex → metal ends in "-ate". [Co(NH₃)₆]Cl₃ = Hexaamminecobalt(III) chloride"
Step 1: Name cation, then anion. Step 2: Within the complex, list ligands alphabetically (ignore prefixes for alphabetizing). Step 3: Add metal name + Roman numeral oxidation state. If complex is anionic, metal gets "-ate" suffix (ferrate, cuprate, cobaltate).
"H₂O = aqua, NH₃ = ammine (double m!), CO = carbonyl, NO = nitrosyl, CN⁻ = cyanido, Cl⁻ = chlorido"
Most neutral ligands keep their molecular name, but four have special names: aqua (water), ammine (ammonia — note the double m to distinguish from "amine"), carbonyl (CO), nitrosyl (NO). Anionic ligands: change ending to -ido or -o.
"CN = number of bonds from ligands to metal. CN 2 = linear. CN 4 = tetrahedral or square planar. CN 6 = octahedral."
Count the total number of donor atoms bonded to the central metal. Monodentate ligands contribute 1 each, bidentate (en, ox) contribute 2 each. CN 6 is most common for transition metals. Geometry depends on CN and metal.
The complete list
- Ligands named alphabetically before metal
- Anionic ligands: -ido suffix (chlorido, cyanido)
- Neutral: aqua(H₂O), ammine(NH₃), carbonyl(CO)
- Prefixes: di/tri/tetra (simple), bis/tris (complex)
- Anionic complex: metal gets -ate suffix
- Oxidation state in Roman numerals
- CN 4: tetrahedral or square planar
- CN 6: octahedral (most common)
Frequently asked questions
How do I find the oxidation state of the metal in a coordination compound?
Set up the equation: charge of complex = oxidation state of metal + sum of charges of all ligands. Example: [CoCl₂(NH₃)₄]⁺ → charge = +1, Cl⁻ contributes 2(−1)=−2, NH₃ is neutral. So: Co + (−2) + 0 = +1 → Co = +3. The counter ions help: if K₂[PtCl₆], the complex is [PtCl₆]²⁻, so Pt + 6(−1) = −2 → Pt = +4.
What is the difference between a monodentate and bidentate ligand?
Monodentate ligands bond through ONE donor atom (Cl⁻, NH₃, H₂O, CN⁻). Bidentate ligands have TWO donor atoms and form a ring with the metal (ethylenediamine/en has 2 N donors, oxalate/ox has 2 O donors). Polydentate ligands like EDTA have 6 donor atoms. Bidentate and polydentate ligands form chelate complexes (extra stable due to chelate effect).
When is the geometry tetrahedral vs square planar for CN=4?
For CN=4: Tetrahedral is the default (sp³ hybridization). Square planar occurs with d⁸ metals in strong-field ligand environments: Pt²⁺, Pd²⁺, Ni²⁺ (with strong ligands), Au³⁺. These d⁸ ions have a large crystal field splitting that makes square planar (dsp² hybridization) energetically favourable over tetrahedral. Quick rule: Pt(II) and Pd(II) complexes are almost always square planar.
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