What Isomerism Means Here
Recall Werner's cobalt–ammonia series: $\ce{CoCl3.4NH3}$ exists as both a green and a violet solid with identical empirical formula but distinct properties. Werner recognised these as isomers, and used precisely this kind of evidence to deduce the octahedral geometry of six-coordinate cobalt(III). Isomerism is therefore not a footnote — historically it is the experimental backbone of coordination theory.
The classification in NCERT §5.4 splits into two principal families. Stereoisomers have the same chemical formula and the same metal–ligand bonds, differing only in spatial arrangement; these are geometrical and optical isomers. Structural isomers differ in their bonds — what is bonded to the metal, or which atom of a ligand does the bonding; these are linkage, coordination, ionisation and solvate (hydrate) isomers.
| Family | Type | What differs | Key example |
|---|---|---|---|
| Stereoisomerism (same bonds) | Geometrical | cis/trans, fac/mer positions | $\ce{[Pt(NH3)2Cl2]}$ |
| Optical | non-superimposable mirror images | $\ce{[Co(en)3]^3+}$ | |
| Structural isomerism (different bonds) | Linkage | which donor atom of an ambidentate ligand binds | $\ce{[Co(NH3)5(NO2)]Cl2}$ |
| Coordination | ligand distribution between cation/anion complexes | $\ce{[Co(NH3)6][Cr(CN)6]}$ | |
| Ionisation | ligand vs counter ion swap | $\ce{[Co(NH3)5(SO4)]Br}$ | |
| Solvate / hydrate | solvent inside vs outside the sphere | $\ce{[Cr(H2O)6]Cl3}$ |
Geometrical Isomerism: Square Planar Complexes
Geometrical isomerism arises in heteroleptic complexes because the ligands can be arranged in more than one way about the metal. For a square planar complex of the type $\ce{[MA2B2]}$ (A and B unidentate), the two A ligands can sit adjacent (cis) or opposite (trans) to each other. The textbook example is $\ce{[Pt(NH3)2Cl2]}$ — the cis form is the anticancer drug cisplatin, while the trans form is therapeutically inert, a difference that hangs entirely on geometry.
The same idea extends to chelate complexes of the type $\ce{[M(AB)2]}$ with an unsymmetrical didentate ligand, and to the type $\ce{[MABXL]}$ with four different unidentate ligands — the latter gives three isomers, two cis and one trans (NCERT explicitly notes this). For square planar geometry, then, the rule is simple: any two distinguishable ligands can be placed cis or trans.
Geometrical Isomerism: Octahedral Complexes
Octahedral complexes are the richest source of geometrical isomerism. The first case is $\ce{[MX2L4]}$, where the two X ligands can be cis (90°) or trans (180°). $\ce{[Co(NH3)4Cl2]^+}$ is the standard example, again giving exactly one cis and one trans form (NCERT Fig. 5.3).
The same cis/trans behaviour appears in $\ce{[MX2(L-L)2]}$ complexes containing two symmetrical didentate ligands such as ethane-1,2-diamine (en) — for instance $\ce{[CoCl2(en)2]^+}$ (NCERT Fig. 5.4). This complex is a NEET favourite because the cis isomer is also chiral, layering optical isomerism on top of the geometrical, as we shall see below.
Facial and Meridional Isomerism
A second, distinct kind of octahedral geometrical isomerism appears in complexes of the type $\ce{[Ma3b3]}$, such as $\ce{[Co(NH3)3(NO2)3]}$. Here three identical ligands can be placed two ways. If the three same ligands occupy the corners of one triangular face of the octahedron, all three are mutually cis — this is the facial (fac) isomer. If instead they lie along a meridian (a great circle through the metal), two are trans and one is cis to both — this is the meridional (mer) isomer.
cis/trans vs fac/mer — match the formula type first
Students apply cis/trans to $\ce{[Ma3b3]}$ and fac/mer to $\ce{[MX2L4]}$ — both wrong. The vocabulary is dictated by the stoichiometry of the coordination sphere.
$\ce{[MX2L4]}$ and $\ce{[MX4L2]}$ → cis/trans · $\ce{[Ma3b3]}$ → fac/mer · square planar $\ce{[MA2B2]}$ → cis/trans only.
Why Tetrahedral Complexes Show No Geometrical Isomerism
NCERT Example 5.4 poses exactly this question. In a tetrahedral complex, the four ligand positions are all equivalent: every corner is the same distance from every other and subtends the same angle (109.5°). There is no concept of "opposite" positions — no two ligands can be trans. Consequently, for a complex such as $\ce{[MA2B2]}$ in tetrahedral geometry, the relative positions of the ligands are identical no matter how you draw them, so only one arrangement exists and geometrical isomerism is impossible.
This is in sharp contrast to the square planar $\ce{[MA2B2]}$, which has the same formula but, because its four positions lie in a plane with distinct cis and trans relationships, does show cis/trans isomerism. The presence or absence of geometrical isomerism for a four-coordinate $\ce{[MA2B2]}$ complex is therefore a clean diagnostic of square planar versus tetrahedral geometry.
Cis/trans counting only makes sense once coordination number and ligand denticity are second nature. Revise them in Important Terms: Ligands & Coordination Number.
Optical Isomerism and Chirality
Optical isomers are mirror images that cannot be superimposed on one another; they are called enantiomers, and molecules of this kind are chiral. The two forms are labelled dextro (d) and laevo (l) by the direction in which they rotate the plane of plane-polarised light in a polarimeter — d to the right, l to the left. The operational test for chirality is the absence of any improper symmetry element: no plane of symmetry and no centre of symmetry.
Optical isomerism is most common in octahedral complexes bearing didentate ligands. The flagship case is $\ce{[Co(en)3]^3+}$, in which three en chelate rings wind around the metal like a three-bladed propeller; the left- and right-handed propellers are non-superimposable mirror images.
Optical activity also rides on top of geometrical isomerism. In $\ce{[PtCl2(en)2]^2+}$ and the analogous $\ce{[CoCl2(en)2]^+}$, only the cis isomer is optically active (NCERT Fig. 5.7). The trans isomer possesses a plane of symmetry passing through the metal and the two axial Cl ligands, making it achiral; the cis isomer has no such plane, so it resolves into a d/l pair. NCERT Example 5.5 makes the parallel point for $\ce{[CrCl2(ox)2]^3-}$: the cis form is chiral, the trans form is not.
| Complex | Geometrical isomers | Optically active? |
|---|---|---|
| $\ce{[Co(en)3]^3+}$ | none (homoleptic chelate) | Yes — d and l pair |
| cis-$\ce{[CoCl2(en)2]^+}$ | cis | Yes — chiral |
| trans-$\ce{[CoCl2(en)2]^+}$ | trans | No — has plane of symmetry |
| cis-$\ce{[CrCl2(ox)2]^3-}$ | cis | Yes — chiral |
| trans-$\ce{[CrCl2(ox)2]^3-}$ | trans | No |
Structural Isomerism
Structural (constitutional) isomers differ in their actual bonds. NCERT recognises four sub-types, and NEET 2024 (Q.65) tested all four in a single match-the-following — so each must be recognisable on sight.
Linkage isomerism
Linkage isomerism occurs only with ambidentate ligands — those with two different donor atoms, of which only one binds at a time. The nitrite ion $\ce{NO2^-}$ can bind through nitrogen (nitrito-N, $\ce{-NO2}$) or through oxygen (nitrito-O, $\ce{-ONO}$); the thiocyanate ion $\ce{SCN^-}$ can bind through sulphur ($\ce{M-SCN}$) or nitrogen ($\ce{M-NCS}$). Jørgensen's classic pair is the complex $\ce{[Co(NH3)5(NO2)]Cl2}$, isolated as a yellow N-bound form and a red O-bound form.
Ionisation isomerism
Ionisation isomerism arises when the counter ion is itself a potential ligand and can swap places with a ligand inside the sphere, so the two isomers release different ions in solution. The standard pair is $\ce{[Co(NH3)5(SO4)]Br}$ and $\ce{[Co(NH3)5Br]SO4}$. They are distinguished chemically: $\ce{[Co(NH3)5Br]SO4}$ gives a white $\ce{BaSO4}$ precipitate with $\ce{Ba^2+}$ but no reaction with $\ce{Ag+}$, whereas $\ce{[Co(NH3)5(SO4)]Br}$ gives a pale $\ce{AgBr}$ precipitate with $\ce{Ag+}$ but no reaction with $\ce{Ba^2+}$ (NCERT Answer 5.4).
Coordination isomerism
Coordination isomerism arises when both the cation and the anion are complex ions, and the ligands are distributed differently between the two metal centres. The textbook pair is $\ce{[Co(NH3)6][Cr(CN)6]}$ — in which $\ce{NH3}$ binds $\ce{Co^3+}$ and $\ce{CN^-}$ binds $\ce{Cr^3+}$ — and its coordination isomer $\ce{[Cr(NH3)6][Co(CN)6]}$, in which the ligand sets have been swapped between the two metals.
Solvate (hydrate) isomerism
Solvate isomers differ in whether a solvent molecule is bonded inside the coordination sphere or merely sits in the crystal lattice as free solvent; when that solvent is water it is called hydrate isomerism. The chromium(III) chloride hexahydrate system is the standard set: violet $\ce{[Cr(H2O)6]Cl3}$ has all six water molecules coordinated and gives 3 mol $\ce{Cl^-}$ on precipitation, while grey-green $\ce{[Cr(H2O)5Cl]Cl2.H2O}$ has one chloride coordinated and one water in the lattice, giving only 2 mol precipitable $\ce{Cl^-}$.
Ionisation vs coordination vs solvate — what is swapping?
These three look alike on paper. Diagnose by asking what exchanges places:
Ligand ↔ counter ion = ionisation · ligand ↔ solvent (in/out of lattice) = solvate · ligand sets ↔ between two complex ions = coordination.
Counting Isomers — Worked Method
NEET frequently asks "how many isomers" rather than "name the type". The reliable method: fix the geometry, enumerate geometrical isomers first, then test each for chirality.
How many geometrical isomers does $\ce{[Co(NH3)3Cl3]}$ (octahedral) have? Are any optically active?
This is an $\ce{[Ma3b3]}$ type. The three same ligands sit either on a face or on a meridian, giving exactly two geometrical isomers — fac and mer. Each possesses a plane of symmetry, so neither is optically active. This matches NCERT Exercise 5.9(ii).
Enumerate all stereoisomers of $\ce{[CoCl2(en)2]^+}$.
Geometrical: the two Cl ligands are cis or trans → 2 geometrical forms. Optical: the trans form has a plane of symmetry (achiral, 1 form); the cis form is chiral and resolves into d and l (2 enantiomers). Total = 3 stereoisomers — trans, cis-d and cis-l. This is the structure tested in NEET 2018 Q.83 (which asked only for the type — geometrical).
Why does $\ce{[Cr(C2O4)3]^3-}$ show optical but not geometrical isomerism?
It is a homoleptic tris-chelate $\ce{[M(LL)3]}$: three identical symmetrical didentate oxalate ligands. With all three chelates identical there is no cis/trans choice, so no geometrical isomers. But the three chelate rings form a propeller with no plane or centre of symmetry, so it exists as a non-superimposable d/l pair — optically active (NCERT Exercise 5.10(i)).
Does the square planar complex $\ce{[Pt(NH3)2Cl2]}$ show optical isomerism in addition to its cis/trans forms?
No. Both the cis and trans square planar isomers are planar, and any planar molecule contains the molecular plane itself as a plane of symmetry. A molecule possessing a plane of symmetry is superimposable on its mirror image and is therefore achiral. So while $\ce{[Pt(NH3)2Cl2]}$ has geometrical isomers, neither is optically active — a clean reminder that square planar geometry essentially never gives optical isomerism, whereas octahedral chelates frequently do.
How Geometry Decides Which Isomerism Appears
The single most useful takeaway for NEET is that geometry, not formula, dictates the isomerism on offer. Two complexes with identical formulae can behave completely differently once their shape is fixed — and the shape itself is set by the metal's oxidation state, d-electron count and the field strength of the ligands, as developed under valence bond and crystal field theory. The table below condenses the geometry-to-isomerism map that almost every PYQ in this area silently relies on.
| Geometry | Coordination number | Geometrical isomerism? | Optical isomerism? |
|---|---|---|---|
| Tetrahedral | 4 | No — all positions equivalent | Only with four different groups / unsymmetrical chelates (rare) |
| Square planar | 4 | Yes — cis/trans for $\ce{[MA2B2]}$, $\ce{[MABXL]}$ | Essentially no — molecular plane is a symmetry plane |
| Octahedral $\ce{[MX2L4]}$ | 6 | Yes — cis/trans | cis can be chiral with chelates |
| Octahedral $\ce{[Ma3b3]}$ | 6 | Yes — fac/mer | Usually no (both have a symmetry plane) |
| Octahedral $\ce{[M(LL)3]}$ | 6 | No — all chelates identical | Yes — d/l propeller pair |
Notice the recurring logic: the criterion for optical activity is always the absence of an improper symmetry element (a plane or a centre of symmetry), while the criterion for geometrical isomerism is the existence of more than one distinguishable spatial placement of the ligands. A complex can show one, both, or neither — $\ce{[Co(en)3]^3+}$ shows only optical isomerism, $\ce{[Co(NH3)3Cl3]}$ shows only geometrical, and cis-$\ce{[CoCl2(en)2]^+}$ shows both. Reasoning through the symmetry rather than memorising lists is what protects you from the carefully constructed distractors NEET likes to use here.
Lock these before the exam
- Stereoisomers keep the same bonds (geometrical, optical); structural isomers change the bonds (linkage, ionisation, coordination, solvate).
- Square planar $\ce{[MA2B2]}$ → cis/trans; tetrahedral $\ce{[MA2B2]}$ → no geometrical isomerism (all positions equivalent).
- Octahedral $\ce{[MX2L4]}$ → cis/trans; octahedral $\ce{[Ma3b3]}$ → fac/mer.
- Optical activity needs no plane and no centre of symmetry: $\ce{[Co(en)3]^3+}$ and cis-$\ce{[CoCl2(en)2]^+}$ are chiral; the trans forms are not.
- Linkage = ambidentate ligand ($\ce{NO2^-}$, $\ce{SCN^-}$); ionisation = ligand/counter-ion swap; coordination = swap between two complex ions; solvate = solvent in vs out of the sphere.