Class 12 Chemistry · Original EZT practice
Coordination Compounds (Class 12 Chapter 5): summary and practice
Coordination Compounds studies the complexes: a central metal ion wrapped by ligands, named by IUPAC rules, rearranged into isomers, and explained by valence bond and crystal field theories. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Werner's theory
A metal shows two valencies: primary (ionisable, satisfied by anions) and secondary (fixed in space, satisfied by ligands). The secondary valencies set the shape of the complex.
2. Ligands and coordination number
The donor atoms around the metal are ligands - unidentate like NH3 and Cl-, bidentate like ethylenediamine and oxalate, polydentate like EDTA. Their count is the coordination number.
3. IUPAC nomenclature
Ligands first in alphabetical order, then the metal with its oxidation state in Roman numerals. Anionic complexes take the -ate ending: [Fe(CN)6]4- is hexacyanidoferrate(II).
4. Isomerism
Geometric (cis/trans) and optical isomers differ in space; linkage, coordination and ionisation isomers differ in connectivity. [Co(NH3)5(NO2)]2+ shows linkage isomerism through N or O donors.
5. Bonding in complexes
Valence bond theory reads hybridisation and magnetism; crystal field theory splits the d orbitals and explains colour and the spectrochemical series.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Find the metal's oxidation state
Choose the method: Balance the charges: metal plus ligand charges equals the overall charge.
EZT practice question
Oxidation state of Co in [Co(NH3)5Cl]Cl2?
Show the worked solution
- Two Cl- outside balance a 2+ complex ion.
- Inside: five NH3 are neutral, one Cl- is -1.
- Co + (-1) = +2, so Co = +3.
Answer: Co is +3.
Check: +3 for Co, -1 for the inner chloride: total +2, matching the two outer chlorides.
Watch for this: Only the ligands INSIDE the square brackets count toward the complex's charge.
Now try it yourself
Oxidation state of Fe in [Fe(CN)6]4-?
Check your answer
+2.
Fe + 6 x (-1) = -4 gives Fe = +2.
Count the coordination number
Choose the method: Count donor atoms bound to the metal - bidentate ligands count twice.
EZT practice question
Coordination number of Fe in [Fe(C2O4)3]3-?
Show the worked solution
- Oxalate C2O4 2- is bidentate - two O donors each.
- Three oxalates give 3 x 2 donor atoms.
- Total donors around Fe: 6.
Answer: Coordination number 6.
Check: Six donor atoms fill an octahedron - the standard shape for coordination number 6.
Watch for this: Three ligands is not coordination number 3 - count DONOR ATOMS, not ligands.
Now try it yourself
Coordination number in [Co(en)3]3+?
Check your answer
6.
en is bidentate: 3 x 2 = 6.
Name a complex by IUPAC rules
Choose the method: Ligands alphabetical with prefixes, metal, oxidation state in Roman numerals; anions end in -ate.
EZT practice question
Name [Co(NH3)5Cl]Cl2.
Show the worked solution
- Complex cation: ligands ammine (a before c of chlorido).
- Five ammine: pentaammine; one chloride: chlorido.
- Co is +3, complex is a cation so the metal keeps its name.
Answer: Pentaamminechloridocobalt(III) chloride.
Check: Alphabetical order puts ammine before chlorido, and the oxidation state (III) matches the +3 found from the charge balance.
Watch for this: Prefixes di, tri, penta do NOT count in alphabetising - it is ammine vs chlorido, a before c.
Now try it yourself
Name K4[Fe(CN)6].
Check your answer
Potassium hexacyanidoferrate(II).
Anionic complex: ferrate, with Fe at +2.
Spot geometric isomerism
Choose the method: Square planar and octahedral MA2B2 or MA4B2 complexes split into cis and trans forms.
EZT practice question
Draw the geometric isomers of [Co(NH3)4Cl2]+.
Show the worked solution
- Octahedral MA4B2 type.
- cis: the two Cl- sit on adjacent corners.
- trans: the two Cl- sit opposite each other.
Answer: cis and trans forms - adjacent vs opposite chlorides.
Check: The cis form is the one whose medicinal cousin, cisplatin, works as a drug; trans does not - the geometry is real, not bookkeeping.
Watch for this: Tetrahedral MA2B2 shows NO geometric isomerism - every corner is adjacent to every other.
Now try it yourself
Does [Pt(NH3)2Cl2] show cis/trans isomerism?
Check your answer
Yes - it is square planar MA2B2.
Adjacent vs opposite positions are distinct in a square.
Identify ionisation isomers
Choose the method: A ligand and a counter-ion swap places between the bracket and outside.
EZT practice question
[Co(NH3)5(SO4)]Br and [Co(NH3)5Br]SO4 - what kind of isomers?
Show the worked solution
- First: sulfate inside the bracket, Br- outside.
- Second: Br- inside, sulfate outside.
- The ions released on dissolving differ.
Answer: Ionisation isomers.
Check: The first solution precipitates AgBr with AgNO3 (free Br-); the second does not - the swap is chemically testable.
Watch for this: This is not linkage isomerism - the same ligands swap POSITIONS, no donor atom changes.
Now try it yourself
[Co(NH3)5(NO2)]Cl2 and [Co(NH3)5(ONO)]Cl2 - which type?
Check your answer
Linkage isomers.
NO2 binds through N, ONO through O - same ligand, different donor atom.
Read colour from crystal field splitting
Choose the method: Strong-field ligands split d orbitals wide; the absorbed light's complement is the colour seen.
EZT practice question
Why is [Ti(H2O)6]3+ violet while [Zn(H2O)6]2+ is colourless?
Show the worked solution
- Ti3+ is d1 - one electron jumps across the crystal field gap, absorbing green-yellow light.
- The complement, violet, is what the eye sees.
- Zn2+ is d10 - the d shell is full, so no d-d jump is possible.
Answer: d1 Ti3+ absorbs visible light and appears violet; d10 Zn2+ has no d-d transition and is colourless.
Check: Ions with empty (d0) or full (d10) d shells - Sc3+, Zn2+ - are all colourless, exactly as the theory says.
Watch for this: The colour you SEE is the complement of the colour ABSORBED - absorb green, look violet.
Now try it yourself
Is [Sc(H2O)6]3+ coloured?
Check your answer
No - Sc3+ is d0.
No d electrons, no d-d transition.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Coordination number?
Show answer and reason
Donor atoms bound to the metal.
Bidentate ligands count twice.
2. Anionic complex ending?
Show answer and reason
Metal takes -ate: ferrate, cobaltate, argentate.
IUPAC rule for complex anions.
3. cis/trans possible in tetrahedral MA2B2?
Show answer and reason
No - all corners adjacent.
Needs square planar or octahedral.
4. Strong-field ligand effect?
Show answer and reason
Large splitting, low-spin pairing.
CN- and CO sit high in the spectrochemical series.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Oxidation state of Co in [Co(NH3)5Cl]Cl2.
- Name K4[Fe(CN)6].
- Coordination number of Fe in [Fe(C2O4)3]3-.
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Questions about this practice page
- What is the difference between a ligand and a coordination number?
- The ligand is the molecule or ion donating an electron pair to the metal; the coordination number is how many donor atoms sit around the metal. Three bidentate ligands like ethylenediamine give coordination number 6, not 3.
- How do I name a coordination compound?
- Ligands come first in alphabetical order with Greek prefixes, then the metal with its oxidation state as a Roman numeral. If the complex is an anion, the metal name ends in -ate. The cation is always named before the anion in a salt.
- Which types of isomerism does Class 12 cover?
- Two structural-space types: geometric (cis/trans) and optical isomerism; and three connectivity types: linkage, coordination and ionisation isomerism. Ionisation isomers are testable in solution because they release different ions.
- Why are many coordination compounds coloured?
- Ligands split the metal's d orbitals into two energy levels. Electrons jump the gap by absorbing visible light, and the eye sees the complement. d0 and d10 ions cannot make the jump and are colourless.
- Are these the NCERT exercise questions for Coordination Compounds (Class 12)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.