Structural: ionisation, hydrate, linkage, coordination. Stereoisomerism: geometrical (cis/trans), optical (chirality). Geometric isomers in [MA₂B₂] type, [MA₃B₃] type. Optical: octahedral [M(en)₃] type.
-- NCERT Class 12 Chemistry, Ch. 5, p. 125Isomerism Coordination
8 MCQs9-step worked example
Source: NCERT Coordination CompoundsPYQ coverage: NEET 2026Official key: NTA-verifiedLast updated: 26 Sep 2026
Isomerism Coordination, explained for NEET
Isomerism in coordination compounds is a frequent source of lost marks — not because the concept is hard, but because students confuse the type of isomerism a given pair exhibits.
The core idea: Coordination compounds with the same molecular formula can differ in the arrangement of ligands around the metal (structural isomerism) or in the spatial positioning of ligands (stereoisomerism). NCERT Class 12 Chemistry Chapter 5, page 125 establishes two broad classes:
Structural isomerism includes:
- Ionisation isomerism — same formula, different ions in solution. [Co(NH₃)₅Br]SO₄ gives SO₄²⁻ in solution; [Co(NH₃)₅SO₄]Br gives Br⁻. The trap: students forget to check which ion is inside the coordination sphere versus outside.
- Linkage isomerism — ambidentate ligands (NO₂⁻ can bind through N as nitrito-N, older name nitro, or through O as nitrito-O; SCN⁻ through S as thiocyanato or N as isothiocyanato).
- Coordination isomerism — exchange of ligands between cationic and anionic coordination entities in salts like [Co(NH₃)₆][Cr(CN)₆] vs [Cr(NH₃)₆][Co(CN)₆].
- Solvate (hydrate) isomerism — water as ligand vs lattice water. The classic trio: CrCl₃·6H₂O exists as [Cr(H₂O)₆]Cl₃, [Cr(H₂O)₅Cl]Cl₂·H₂O, and [Cr(H₂O)₄Cl₂]Cl·2H₂O.
Stereoisomerism includes:
- Geometrical (cis-trans) isomerism — in square planar [Ma₂b₂] and octahedral [Ma₂b₄] type complexes. Cis = identical ligands adjacent; trans = opposite.
- Optical isomerism — non-superimposable mirror images. Octahedral complexes with bidentate ligands like [Co(en)₃]³⁺ show optical activity.
Watch out: When a question gives a molecular formula and asks "how many isomers?", first classify which types of isomerism apply. Don't jump to drawing geometrical isomers if ionisation isomers also exist — you'll undercount.
Can you answer these Isomerism Coordination MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which type of isomerism is shown by [Co(NH₃)₅(NO₂)]²⁺ and [Co(NH₃)₅(ONO)]²⁺?
Show answer and why every option is right or wrong
Answer: D. D is correct. NO₂⁻ is an ambidentate ligand — it can bind through nitrogen (nitrito-N) or through oxygen (nitrito-O). The two complexes differ only in the donor atom of the same ligand, which defines linkage isomerism (NCERT Class 12 Chemistry Chapter 5, page 127).
Why A is wrong: A is wrong. Ionisation isomerism involves exchange of ions between the coordination sphere and the counter-ion, changing which ion is released in solution. Here, the counter-ion is unchanged — only the binding atom of NO₂⁻ differs.
Why B is wrong: B is wrong. Geometrical isomerism involves different spatial arrangements (cis/trans) of different ligands. Here, the spatial arrangement is not the distinguishing feature — the donor atom of the ambidentate ligand is.
Why C is wrong: C is wrong. Coordination isomerism requires both a cationic and an anionic coordination entity that exchange ligands. This pair has a single coordination entity with the same set of ligands.
MCQ 2Easy RecallPractice
[Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br are examples of which type of isomerism?
Show answer and why every option is right or wrong
Answer: C. C is correct. The two compounds have the same molecular formula but yield different ions in solution — one releases SO₄²⁻ and the other releases Br⁻. This exchange of an ion between the coordination sphere and the counter-ion position defines ionisation isomerism (NCERT Class 12 Chemistry Chapter 5, page 127).
Why A is wrong: A is wrong. Linkage isomerism requires an ambidentate ligand binding through different donor atoms. Neither Br⁻ nor SO₄²⁻ is ambidentate — the difference here is which species is inside vs outside the coordination sphere.
Why B is wrong: B is wrong. Solvate (hydrate) isomerism specifically involves solvent molecules (typically water) distributing between the coordination sphere and the crystal lattice. This pair involves Br⁻ and SO₄²⁻, not solvent molecules.
Why D is wrong: D is wrong. Optical isomerism involves non-superimposable mirror images. This pair differs in their ionic composition in solution, not in spatial mirror-image arrangement.
MCQ 3Easy RecallPractice
The compound CrCl₃·6H₂O exists in three hydrate isomeric forms. Which of the following is NOT one of them?
Show answer and why every option is right or wrong
Answer: B. B is correct (it is NOT a known hydrate isomer). NCERT Class 12 Chemistry Chapter 5 lists three established hydrate isomers of CrCl₃·6H₂O: [Cr(H₂O)₆]Cl₃ (violet), [Cr(H₂O)₅Cl]Cl₂·H₂O (grey-green), and [Cr(H₂O)₄Cl₂]Cl·2H₂O (dark green). The form [Cr(H₂O)₃Cl₃]·3H₂O with three Cl⁻ as ligands and no ionisable chloride is not among the recognised NCERT set.
Why A is wrong: A is wrong. [Cr(H₂O)₆]Cl₃ is the violet hydrate isomer with all six water molecules coordinated and all three Cl⁻ as counter-ions. This IS a recognised form.
Why C is wrong: C is wrong. [Cr(H₂O)₅Cl]Cl₂·H₂O is the grey-green hydrate isomer with five coordinated waters, one coordinated Cl, two ionic Cl⁻, and one lattice water. This IS a recognised form.
Why D is wrong: D is wrong. [Cr(H₂O)₄Cl₂]Cl·2H₂O is the dark green hydrate isomer with four coordinated waters, two coordinated Cl, one ionic Cl⁻, and two lattice waters. This IS a recognised form.
MCQ 4Direct ApplicationPractice
The square planar complex [Pt(NH₃)₂Cl₂] can exhibit geometrical isomerism. How many geometrical isomers does it have, and what are they?
Show answer and why every option is right or wrong
Answer: D. D is correct. A square planar complex of the type [Ma₂b₂] exhibits exactly two geometrical isomers: cis (identical ligands adjacent, 90° apart) and trans (identical ligands opposite, 180° apart). Cisplatin and transplatin are the classic example (NCERT Class 12 Chemistry Chapter 5, page 125).
Why A is wrong: A is wrong. For [Ma₂b₂] square planar, there is exactly one way to place the two identical ligands adjacent (cis) and one way to place them opposite (trans) — giving two isomers, not four.
Why B is wrong: B is wrong. The 'mer' (meridional) designation applies to octahedral complexes of type [Ma₃b₃], not to square planar [Ma₂b₂]. Square planar geometry only permits cis and trans arrangements.
Why C is wrong: C is wrong. Both cis and trans forms of [Pt(NH₃)₂Cl₂] are isolable and well-characterised compounds. Cisplatin (cis form) is a widely used anticancer drug, demonstrating the stability of the cis isomer.
MCQ 5Direct ApplicationPractice
Which of the following octahedral complexes will show geometrical isomerism?
Show answer and why every option is right or wrong
Answer: C. C is correct. [Co(NH₃)₄Cl₂]⁺ is of the type [Ma₄b₂] in octahedral geometry, which can have cis (two Cl⁻ on adjacent positions) and trans (two Cl⁻ on opposite positions) isomers (NCERT Class 12 Chemistry Chapter 5, page 125).
Why A is wrong: A is wrong. [Co(NH₃)₆]³⁺ has all six ligands identical. With only one type of ligand, no cis-trans distinction is possible — all spatial arrangements are equivalent.
Why B is wrong: B is wrong. [Co(NH₃)₅Cl]²⁺ is of type [Ma₅b]. With only one different ligand, every position it occupies is equivalent in a regular octahedron — no geometrical isomerism arises.
Why D is wrong: D is wrong. [Co(en)₃]³⁺ has three identical bidentate ligands, so every arrangement of them around the octahedron is equivalent in the cis-trans sense. It shows optical isomerism (non-superimposable mirror images), not geometrical isomerism.
MCQ 6Direct ApplicationPractice
[Co(en)₃]³⁺ exhibits optical isomerism. What structural feature makes this possible?
Show answer and why every option is right or wrong
Answer: B. B is correct. Optical isomerism arises when a coordination compound and its mirror image cannot be superimposed on each other (i.e., they are chiral). [Co(en)₃]³⁺ with three bidentate ethylenediamine ligands in an octahedral arrangement lacks a plane of symmetry, producing non-superimposable mirror images (d and l forms) as described in NCERT Class 12 Chemistry Chapter 5, page 126.
Why A is wrong: A is wrong. Ambidentate ligands (like NO₂⁻ or SCN⁻) cause linkage isomerism, not optical isomerism. Ethylenediamine (en) is a bidentate ligand but NOT ambidentate — it always binds through both nitrogen atoms.
Why C is wrong: C is wrong. Different ligands in cis and trans positions describes geometrical isomerism. Optical isomerism is about chirality — the absence of a plane of symmetry making the molecule and its mirror image non-superimposable.
Why D is wrong: D is wrong. Releasing different ions in solution is the definition of ionisation isomerism, which involves structural differences in which species are in the coordination sphere vs counter-ion positions.
MCQ 7Concept TrapPractice
A coordination compound has the formula [Co(NH₃)₄Br₂]Cl. A student claims this compound exhibits both geometrical isomerism and ionisation isomerism. Is the student correct?
Show answer and why every option is right or wrong
Answer: A. A is correct. [Co(NH₃)₄Br₂]Cl is of type [Ma₄b₂] (octahedral), so it has cis and trans geometrical isomers. Additionally, Cl⁻ is currently a counter-ion and Br⁻ is coordinated — exchanging them gives [Co(NH₃)₄BrCl]Br, an ionisation isomer. The compound therefore exhibits both types simultaneously (NCERT Class 12 Chemistry Chapter 5).
Why B is wrong: B is wrong. While ionisation isomerism is possible (Br⁻/Cl⁻ exchange between coordination sphere and counter-ion), ignoring geometrical isomerism misses the cis-trans possibility from the [Ma₄b₂] arrangement.
Why C is wrong: C is wrong. While the [Ma₄b₂] octahedral type does show geometrical isomerism (cis/trans), that is not the only isomerism possible here. The Br⁻ and Cl⁻ can exchange between the coordination sphere and counter-ion position, giving ionisation isomers.
Why D is wrong: D is wrong. Linkage isomerism requires an ambidentate ligand. Neither Br⁻, Cl⁻, nor NH₃ is ambidentate. The correct types here are geometrical (cis-trans from [Ma₄b₂]) and ionisation (Br⁻/Cl⁻ exchange).
MCQ 8CalculationPractice
How many total isomers (structural + stereoisomers) are possible for [Co(en)₂Cl₂]⁺?
Show answer and why every option is right or wrong
Answer: A. A is correct. Step 1: [Co(en)₂Cl₂]⁺ is octahedral with two bidentate (en) and two monodentate (Cl⁻) ligands. Geometrical isomerism gives cis and trans forms. Step 2: The cis isomer lacks a plane of symmetry → it is chiral → it exists as d and l optical isomers. The trans isomer has a plane of symmetry → it is achiral → no optical isomers. Total: cis-d + cis-l + trans = 3 isomers. The two chelate rings are what make this different from [Co(NH₃)₄Cl₂]⁺: a ring cannot span two trans positions, so it destroys the mirror plane that an [Ma₄b₂] cis form still has. (NCERT Class 12 Chemistry Chapter 5, page 127).
Why B is wrong: B is wrong. Counting only cis and trans misses the optical isomerism of the cis form. The cis-[Co(en)₂Cl₂]⁺ is chiral (no plane of symmetry), so it splits into d and l forms, giving 3 total isomers, not 2.
Why C is wrong: C is wrong. The trans form of [Co(en)₂Cl₂]⁺ possesses a plane of symmetry, making it achiral — it does NOT have d and l optical isomers. Only the cis form is chiral. Total is 3, not 4.
Why D is wrong: D is wrong. An octahedral complex with two bidentate and two monodentate ligands clearly has both geometrical (cis/trans) and optical (for the chiral cis form) isomerism. Multiple isomers exist.
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How do you solve a Isomerism Coordination question? A worked example
- 1
Given
• Molecular formula: CoCl₃·4NH₃• Co is in +3 oxidation state (cobalt(III)), coordination number 6 (octahedral)• This can be written as [Co(NH₃)₄Cl₂]Cl (two Cl⁻ coordinated, one Cl⁻ as counter-ion)
- 2
Required
Total number and types of isomers.
- 3
Concept
We must check for both structural isomerism (ionisation, linkage, etc.) and stereoisomerism (geometrical, optical). Multiple types can coexist for the same formula.
- 4
Identify structural isomers
Ionisation isomerism: The Cl⁻ currently outside the coordination sphere can swap with a Cl⁻ inside, but since all three are the same ligand (Cl⁻), exchanging identical ions doesn't produce a new compound. However, if we consider the possibility of [Co(NH₃)₄Cl₂]Cl vs [Co(NH₃)₄Cl(H₂O)]Cl₂ — no, there is no H₂O in the formula. With only NH₃ and Cl⁻, no ionisation isomer exists beyond the [Co(NH₃)₄Cl₂]Cl formulation itself.
Linkage isomerism: Neither Cl⁻ nor NH₃ is ambidentate. Not applicable. - 5
Identify stereoisomers
Geometrical isomerism: [Co(NH₃)₄Cl₂]⁺ is of type [Ma₄b₂] in octahedral geometry → cis and trans forms exist.
Optical isomerism: The cis form — does it have a plane of symmetry? Put the two Cl⁻ at +x and +y. The plane through Co containing the z-axis and the x = y bisector reflects +x onto +y (Cl onto Cl) and −x onto −y (NH₃ onto NH₃), and leaves ±z (NH₃) untouched. That is a mirror plane, so cis-[Co(NH₃)₄Cl₂]⁺ is ACHIRAL. The trans form is plainly symmetric too. An [Ma₄b₂] octahedral complex has no optical isomers at all. - 6
Enumerate
1. cis-[Co(NH₃)₄Cl₂]Cl
2. trans-[Co(NH₃)₄Cl₂]Cl - 7
Final answer
2 total isomers — the cis and trans geometrical forms. Neither is optically active.
- 8
Common trap
The trap is the opposite of what it looks like: having learned that cis-[M(en)₂Cl₂]⁺ splits into d and l forms, students apply the same split to cis-[Co(NH₃)₄Cl₂]⁺ and answer 3. Monodentate ligands leave the bisecting mirror plane intact; a chelate ring is what destroys it. Always LOOK for the plane rather than assuming a cis form must be chiral.
- 9
Similar NEET-style question
"The complex [Cr(en)₂Cl₂]⁺ can exist in how many stereoisomeric forms?" Apply the same two-step analysis: geometrical (cis/trans) → optical check on each. Here the answer IS 3 (cis-d, cis-l, trans), because en is bidentate: the chelate rings remove the mirror plane that kept the worked example’s cis-[Co(NH₃)₄Cl₂]⁺ achiral. Same procedure, opposite outcome — which is exactly why the plane has to be checked and not assumed.
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What to remember before solving Isomerism Coordination questions
More in Coordination Compounds: 2 exam traps and mistakes · 2 formulas · 3 question patterns from its other lessons.
Isomerism Coordination questions from past NEET papers
2 questions from NEET 2026. Answers verified against NTA official keys.
NEET 2026
1Both Statement-I and Statement-II are correct.
2Both Statement-I and Statement-II are incorrect.
3Statement-I is correct but Statement-II is incorrect.
4Statement-I is incorrect but Statement-II is correct.
NTA Answer: Option 3(final)
NEET 2026
1[Cr(py)₃(Cl)₃]
2[Cr(H₂O)₆]³⁺
3[Co(NH₃)₄(H₂O)₂]³⁺
4[Ni(en)₂(H₂O)₂]²⁺
NTA Answer: Option 1(final)
Sources
NCERT refs: Class 12 Chemistry Chapter 5, p.125
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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