Isomerism in Coordination Compounds

Isomers share a molecular formula but are genuinely different compounds. Coordination chemistry sorts them into two big families.

Structural isomers differ in connectivity — which atom is actually bonded to the metal, or which species sits inside versus outside the coordination sphere.

• Ionisation isomerism: a ligand and a counter-ion (outside the coordination sphere) trade places, so the two isomers release different ions in water. Example: $[\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4$ versus $[\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}$.
• Linkage isomerism: an ambidentate ligand (one with two possible donor atoms) attaches through a different atom. Example: $[\text{Co}(\text{NH}_3)_5(\text{ONO})]^{2+}$, bonded through oxygen, versus $[\text{Co}(\text{NH}_3)_5(\text{NO}_2)]^{2+}$, bonded through nitrogen.
• Coordination isomerism: in a salt built from a complex cation and a complex anion, ligands are distributed differently between the two metal centres. Example: $[\text{Co}(\text{NH}_3)_6][\text{Cr}(\text{CN})_6]$ versus $[\text{Cr}(\text{NH}_3)_6][\text{Co}(\text{CN})_6]$.
• Solvate (hydrate) isomerism: a solvent molecule, usually water, sits either as a ligand bonded to the metal or as a free molecule of crystallisation outside the sphere. Example: $[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3$ versus $[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2\cdot\text{H}_2\text{O}$.

Stereoisomers have identical connectivity — the same atoms bonded to the metal — but a different arrangement in three-dimensional space.

• Geometrical (cis–trans) isomerism arises in square planar $\text{MA}_2\text{B}_2$ complexes and in octahedral $\text{MA}_4\text{B}_2$ or $\text{MA}_3\text{B}_3$ complexes. Cis places like ligands next to each other; trans places them opposite each other. Octahedral $\text{MA}_3\text{B}_3$ gives a further split into facial (fac), where three like ligands occupy one triangular face, and meridional (mer), where they lie along a plane through the metal.
• Optical isomerism arises when a complex is chiral — its mirror image cannot be superimposed on the original. This is common in octahedral complexes built from chelating (bidentate or polydentate) ligands, most classically $[\text{Co}(\text{en})_3]^{3+}$, where en is ethylenediamine. A chiral pair of mirror-image forms are called enantiomers, labelled $d$ and $l$ (or $\Delta$ and $\Lambda$).

The figure below fixes the geometry: the same $\text{MA}_2\text{B}_2$ formula gives two genuinely different square planar molecules, distinguished only by which corners the like ligands occupy.

The goal of this section is to explain, from symmetry alone, why some coordination isomers are physically distinct compounds and why only certain stereoisomers are optically active, using nothing beyond the idea of a mirror plane and the rigidity of a fixed coordination geometry.

Step 1. Start from rigidity. A square planar or octahedral complex has a fixed geometric skeleton — the metal sits at a defined centre with ligand positions locked at defined angles (90 degrees and 180 degrees for square planar; 90 and 180 degree combinations for octahedral). Swapping which type of ligand occupies which position, without breaking any bond, produces a different arrangement only if no rotation of the whole molecule can map one onto the other. This is exactly the test that separates cis from trans: rotating the whole cis molecule in space never reproduces the trans arrangement, because the relative positions of A and B around the metal are physically different, not just differently drawn.

Step 2. Distinguish "different molecule" from "different drawing of the same molecule." Two structures represent the same compound if a rotation (or, for flexible cases, a bond-preserving distortion) superimposes them exactly. Cis and trans $\text{MA}_2\text{B}_2$ fail this test — no rotation converts adjacent A ligands into opposite ones — so they are counted as separate, isolable compounds with their own melting points, dipole moments, and reactivities.

Step 3. Bring in the mirror-image test for chirality. A molecule is optically active only if its mirror image is not superimposable on the original — equivalently, only if the molecule has no internal plane of symmetry. If a plane of symmetry exists, the molecule and its mirror image are identical, and reflecting it does nothing new. Square planar complexes are flat, and the molecular plane itself is almost always a plane of symmetry, so square planar complexes are typically optically inactive even when cis/trans isomerism exists.

Step 4. Apply the same logic to octahedral chelate complexes. In $\text{trans-}[\text{Co}(\text{en})_2\text{Cl}_2]^+$, a plane of symmetry runs through the metal and both $\text{Cl}$ atoms, so the trans form is achiral. In the cis form, the two chelating "en" rings twist the whole structure into a propeller-like shape with no such plane, so the cis form and its mirror image are genuinely different, non-superimposable objects — a chiral pair of enantiomers. Chirality, in other words, is not a property of the formula; it is a property of the specific three-dimensional arrangement, decided entirely by whether a mirror plane can be found.

Problem. $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$ is square planar. Identify its geometrical isomers, state which one is the anticancer drug cisplatin, and explain why the two isomers are not interconvertible by simple rotation.

Step 1. Set up the frame. Four coordination positions sit at the corners of a square around $\text{Pt}$, holding two $\text{NH}_3$ and two $\text{Cl}^-$ ligands — an $\text{MA}_2\text{B}_2$ type complex.

Step 2. Place the ligands two distinct ways. In the cis isomer the two $\text{NH}_3$ groups occupy adjacent corners, $90°$ apart, with the two $\text{Cl}$ groups on the other adjacent corners. In the trans isomer each $\text{NH}_3$ sits diagonally opposite the other $\text{NH}_3$, $180°$ away, so $\text{NH}_3$ and $\text{Cl}$ alternate around the square.

Step 3. Confirm these are not the same molecule redrawn. Because the square planar frame is rigid, no rotation of the whole molecule converts the cis arrangement into the trans arrangement — the relative angle between the two $\text{NH}_3$ groups ($90°$ versus $180°$) is a fixed structural fact, not an artefact of how the molecule is drawn on paper. The two isomers therefore have different dipole moments, different melting points, and different biological behaviour.

Step 4. Identify cisplatin. The cis isomer, cis-$[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$, is the form used clinically as the anticancer drug cisplatin; the trans isomer is inactive for this purpose. Answer: cis-[Pt(NH3)2Cl2] is cisplatin; the two Pt(NH3)2Cl2 isomers cannot interconvert by rotation because their ligand geometry — 90 degrees versus 180 degrees apart — is a genuine structural difference, not a drawing choice.

Competitive stretch

Problem. Does the octahedral ion $[\text{Co}(\text{en})_2\text{Cl}_2]^+$ show optical isomerism? Use its cis and trans geometrical forms to decide.

Step 1. This is an $\text{M(AA)}_2\text{B}_2$-type octahedral complex, where "en" (ethylenediamine) is a bidentate chelating ligand occupying two adjacent coordination positions each. It has cis (the two $\text{Cl}$ adjacent) and trans (the two $\text{Cl}$ opposite) geometrical forms.

Step 2. Test the trans form. A plane of symmetry passes through the metal and both $\text{Cl}$ atoms, so trans-$[\text{Co}(\text{en})_2\text{Cl}_2]^+$ is superimposable on its own mirror image — it is not optically active.

Step 3. Test the cis form. No such plane exists, because the two chelate rings twist the molecule into an asymmetric arrangement. Its mirror image cannot be superimposed on it. Answer: only the cis isomer of [Co(en)2Cl2]+ is optically active, existing as a d and l pair of enantiomers; the trans isomer is optically inactive — proof that geometrical and optical isomerism coexist in one formula but are decided by different tests.

• Treating "cis/trans" and "optical" isomerism as mutually exclusive, when a complex can show both at different levels — as with $[\text{Co}(\text{en})_2\text{Cl}_2]^+$, where the cis geometrical isomer is chiral but the trans geometrical isomer is not.
• Assuming every square planar $\text{MA}_2\text{B}_2$ complex is optically active — square planar complexes are flat with a mirror plane in the molecular plane itself, so they are usually optically inactive even when cis and trans forms exist.
• Confusing linkage isomerism (same ligand, different donor atom) with ionisation isomerism (different ligand and counter-ion swap positions) — both are described loosely as "give different ions/behaviour," but the structural change is not the same.
• Forgetting that optical isomerism in octahedral complexes generally needs chelating ligands — simple $\text{MA}_6$ or $\text{MA}_5\text{B}$ complexes with only monodentate ligands are not chiral.

Ask the tutor: if visualising which isomer has a mirror plane and which does not feels hard from a flat formula, ask the AI tutor to walk through building the 3-D cis and trans models step by step.