JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
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This MCQ module is based on: Isomerism
TOPIC 20 OF 23
Isomerism
🎓 Class 12
Chemistry
CBSE
Theory
Ch 5 – Coordination Compounds
⏱ ~14 min
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Isomerism
5.5 Isomerism in Coordination Compounds
Isomers are compounds with identical chemical formulas but different arrangements of atoms. Coordination chemistry shows two broad families:
- Stereoisomerism — same connectivity, different spatial arrangement.
(i) Geometrical (cis–trans, fac–mer); (ii) Optical (enantiomers). - Structural isomerism — different connectivity (different bonds).
(i) Linkage; (ii) Coordination; (iii) Ionisation; (iv) Solvate/hydrate.
5.5.1 Geometrical Isomerism
This arises in heteroleptic complexes when ligands can occupy different relative positions. It is most common for coordination numbers 4 (square planar only — not tetrahedral) and 6 (octahedral).
Square planar [MX2L2] — cis & trans
In cis the two X ligands are adjacent (90° apart); in trans they are opposite (180°). The classic medical example is cisplatin = cis-[Pt(NH3)2Cl2] (anti-cancer drug); the trans isomer is biologically inactive.
Octahedral [MX2L4] — cis & trans
In octahedral [Co(NH3)4Cl2]+ the two Cl− can be adjacent (cis, violet) or opposite (trans, green). The same kind of cis/trans isomerism appears in [MX2(L–L)2] entities where L–L is a didentate ligand such as en, e.g. [CoCl2(en)2]+.
Octahedral [Ma3b3] — facial (fac) & meridional (mer)
If three identical ligands occupy the corners of one triangular face of the octahedron, we have the fac isomer. If they wrap around an equatorial meridian, we have the mer isomer. Example: [Co(NH3)3(NO2)3].
Tetrahedral complexes do not show geometrical isomerism. In a tetrahedron every vertex is equivalent — adjacent to all the others at the same 109.5° — so no "cis vs. trans" distinction can exist for [MX2L2] tetrahedral complexes.
Worked Example 5.4 — Why no geometric isomerism in tetrahedrons?
Why is geometrical isomerism not possible in tetrahedral complexes containing two different unidentate ligands?
In a tetrahedron, every vertex is at the same angle (≈109.5°) from every other vertex. So whichever way you arrange two A and two B ligands, the relative spatial relationship is identical — there is no "cis" or "trans" position. Hence tetrahedral [MA2B2] gives only one isomer.
5.5.2 Optical Isomerism
Optical isomers are non-superimposable mirror images (enantiomers). The species are chiral; they rotate the plane of polarised light in opposite directions — dextro (d) to the right, laevo (l) to the left. Optical isomerism is most common in octahedral complexes containing didentate ligands, e.g. [Co(en)3]3+ and cis-[CoCl2(en)2]+.
Important: In [PtCl2(en)2]2+ only the cis isomer is optically active. The trans isomer has a plane of symmetry and is therefore achiral.
Worked Example 5.5 — Identifying chirality
Which of (a) cis-[CrCl2(ox)2]3− and (b) trans-[CrCl2(ox)2]3− is optically active?
The cis form has no plane or centre of symmetry — it exists as two non-superimposable mirror images. cis-[CrCl2(ox)2]3− is chiral (optically active). The trans form has a plane of symmetry and is therefore achiral.
5.5.3 Structural Isomerism
(a) Linkage isomerism
Arises when an ambidentate ligand can attach through either of two donor atoms. Jørgensen's classic example: [Co(NH3)5(NO2)]Cl2 exists as a yellow form (Co–NO2, nitro) and a red form (Co–ONO, nitrito). NCS− can also bond either via N (M–NCS) or S (M–SCN).
(b) Coordination isomerism
Occurs when both the cation and anion of a complex salt are themselves complexes — the ligands can be swapped between the two metal centres. Example: [Co(NH3)6][Cr(CN)6] and its coordination isomer [Cr(NH3)6][Co(CN)6].
(c) Ionisation isomerism
The counter ion in a complex salt is itself a potential ligand and can swap places with one inside the sphere. Example: [Co(NH3)5(SO4)]Br vs [Co(NH3)5Br]SO4. They give different ions on dissolution and hence different reactions:
- [Co(NH3)5Br]SO4 + Ba2+ → BaSO4(s) (free SO42−)
- [Co(NH3)5(SO4)]Br + Ag+ → AgBr(s) (free Br−)
- The reverse mixings give no precipitate.
(d) Solvate/hydrate isomerism
Differs by whether a solvent molecule (usually water) sits inside the coordination sphere or outside as lattice solvent. Example: [Cr(H2O)6]Cl3 (violet), [Cr(H2O)5Cl]Cl2·H2O (grey-green) and [Cr(H2O)4Cl2]Cl·2H2O (dark green).
| Type | Diagnostic | Classic example |
|---|---|---|
| Linkage | Ambidentate ligand bonded through different atoms | [Co(NH3)5NO2]2+ vs [Co(NH3)5ONO]2+ |
| Coordination | Ligands swapped between cationic and anionic complex centres | [Co(NH3)6][Cr(CN)6] vs [Cr(NH3)6][Co(CN)6] |
| Ionisation | Counter ion exchanges with an internal ligand | [Co(NH3)5SO4]Br vs [Co(NH3)5Br]SO4 |
| Hydrate / solvate | Water molecule inside vs outside the sphere | [Cr(H2O)6]Cl3 vs [Cr(H2O)5Cl]Cl2·H2O |
🧪 Activity 5.2 — Distinguishing Ionisation Isomers (Predict → Observe → Explain)
Setup: Two unlabeled bottles each contain a red-violet solid. One is [Co(NH3)5Br]SO4, the other [Co(NH3)5SO4]Br. You add aqueous BaCl2 to fresh solutions of both.
Predict: Which solution will give a white precipitate with BaCl2? What other test would confirm the second isomer?
[Co(NH3)5Br]SO4 dissolves to give the [Co(NH3)5Br]2+ cation and free SO42−. With BaCl2 it gives a white BaSO4 precipitate.
[Co(NH3)5SO4]Br has SO42− coordinated, but Br− is free. It gives a pale-yellow AgBr precipitate with AgNO3 and no reaction with BaCl2. This pair of complementary tests confirms ionisation isomerism.
🔧 Interactive: Isomer Identifier
Choose a complex; the simulation lists every isomer (geometrical + optical, where present), counts them, and explains why.
Pick a complex above.
Worked Example 5.6 — Counting isomers of [Pt(NH3)(Br)(Cl)(py)]
How many geometrical isomers are possible for the square-planar complex [Pt(NH3)(Br)(Cl)(py)] (py = pyridine)? Which of them shows optical activity?
For a square-planar complex with four different ligands MABCD, the trans pairs can be (A,B), (A,C) or (A,D); fixing one such pair fixes the whole geometry. So we get three geometrical isomers. None of them is optically active because every square-planar arrangement has at least a plane of symmetry through the metal (the plane of the molecule itself).
🎯 Competency-Based Questions
Q1. The number of geometrical isomers of [Cr(C2O4)3]3− is: L1 Remember
Answer: (a) 0 — only one geometrical arrangement is possible (the three didentate oxalate ligands fully wrap the octahedron). However, it does have two optical isomers (Δ and Λ).
Q2. Which complex shows linkage isomerism? L1 Remember
Answer: (b). NO2− is ambidentate — bonds via N (yellow) or O (red).
Q3. Show that [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4 are ionisation isomers. L3 Apply
[Co(NH3)5SO4]Br + Ag+ → AgBr(s) (free Br−); + Ba2+ → no reaction.
[Co(NH3)5Br]SO4 + Ba2+ → BaSO4(s) (free SO42−); + Ag+ → no reaction.
The complementary precipitation patterns confirm that the SO42− and Br− have swapped between the inside and outside of the sphere — i.e. ionisation isomerism.
[Co(NH3)5Br]SO4 + Ba2+ → BaSO4(s) (free SO42−); + Ag+ → no reaction.
The complementary precipitation patterns confirm that the SO42− and Br− have swapped between the inside and outside of the sphere — i.e. ionisation isomerism.
Q4. Identify all isomerism types possible for [Co(NH3)5(NO2)](NO3)2. L4 Analyse
(i) Linkage — NO2− can bond via N (nitrito-N) or O (nitrito-O).
(ii) Ionisation — the outside NO3− can swap with the inside NO2−, giving for instance [Co(NH3)5(NO3)](NO2)(NO3).
No geometrical isomerism (only one type of ligand other than the single substituent).
(ii) Ionisation — the outside NO3− can swap with the inside NO2−, giving for instance [Co(NH3)5(NO3)](NO2)(NO3).
No geometrical isomerism (only one type of ligand other than the single substituent).
Q5. Design a ligand-set/metal combination that would give simultaneous geometric, optical and ionisation isomerism. Explain. L6 Create
A Co(III) complex such as [Co(en)2(NH3)Cl]Br fits all three criteria.
• Geometric: The two en chelates plus NH3 and Cl can sit in cis or trans relative arrangements (en–en cis is common, but pendant Cl/NH3 can be cis or trans).
• Optical: The cis form has no plane of symmetry → Δ and Λ enantiomers.
• Ionisation: The outside Br− can swap with the inside Cl− → [Co(en)2(NH3)Br]Cl as the ionisation isomer.
• Geometric: The two en chelates plus NH3 and Cl can sit in cis or trans relative arrangements (en–en cis is common, but pendant Cl/NH3 can be cis or trans).
• Optical: The cis form has no plane of symmetry → Δ and Λ enantiomers.
• Ionisation: The outside Br− can swap with the inside Cl− → [Co(en)2(NH3)Br]Cl as the ionisation isomer.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: Tetrahedral complexes do not show geometrical isomerism.
R: All four vertices of a tetrahedron are equivalent positions to one another.
Answer: (A). Both true; R correctly explains A. Equivalent vertices mean no cis/trans distinction.
A: trans-[CoCl2(en)2]+ is optically inactive but cis-[CoCl2(en)2]+ is optically active.
R: The trans isomer has a plane of symmetry passing through the two Cl ligands.
Answer: (A). Both true; R correctly explains A. A molecule with a plane of symmetry is achiral.
A: [Co(NH3)5(SO4)]Br and [Co(NH3)5Br]SO4 show identical reactions in solution.
R: They differ only in the order of writing.
Answer: (D). A is FALSE — they give different ions in solution and react differently with Ag+ and Ba2+. R is also FALSE — the formulas are genuinely different (different ligands inside the sphere). They are ionisation isomers.
Frequently Asked Questions - Isomerism
What is the main concept covered in Isomerism?
In NCERT Class 12 Chemistry Chapter 5 (Coordination Compounds), "Isomerism" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Isomerism useful in real-life or applied chemistry?
Real-life applications of "Isomerism" from NCERT Class 12 Chemistry Chapter 5 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Isomerism?
Key reactions in "Isomerism" (NCERT Class 12 Chemistry Chapter 5 Coordination Compounds) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 5?
NCERT Class 12 Chemistry Chapter 5 (Coordination Compounds) is structured so each part builds chemical understanding sequentially. "Isomerism" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Isomerism?
CBSE board questions from "Isomerism" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Isomerism" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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