Isomerism

Isomers are compounds with the same molecular formula but a different arrangement of atoms or groups, so their physical and chemical properties differ. Coordination isomerism splits into two families that answer two different questions.

A. Structural isomerism — does connectivity differ? These isomers disagree on which atoms are actually bonded to the central metal.

• Ionisation isomerism. The identity of the ion inside the coordination sphere (a ligand) swaps with the ion outside it (the free counter-ion), so the compounds ionise to give different ions in solution. Example: $[Co(NH_3)_5Br]SO_4$ (releases free $SO_4^{2-}$) versus $[Co(NH_3)_5SO_4]Br$ (releases free $Br^-$).
• Hydrate (solvate) isomerism. Water sits inside the coordination sphere as a ligand in one isomer and outside as water of crystallisation in another. Classic case: the three isomers of $CrCl_3\cdot 6H_2O$ — $[Cr(H_2O)_6]Cl_3$ (violet), $[Cr(H_2O)_5Cl]Cl_2\cdot H_2O$ (pale green), $[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O$ (dark green) — distinguished by how much chloride is free outside the bracket.
• Linkage isomerism. An ambidentate ligand (a ligand that can bond through either of two different donor atoms) attaches through a different atom in each isomer. $NO_2^-$ can bind through N (nitrito-N, $-NO_2$) or through O (nitrito-O, $-ONO$); $SCN^-$ can bind through S or through N.
• Coordination isomerism. Occurs when both the cation and the anion of a salt are complex ions; the ligands redistribute between the two metal centres. Example: $[Co(NH_3)_6][Cr(CN)_6]$ versus $[Cr(NH_3)_6][Co(CN)_6]$.

B. Stereoisomerism — does geometry differ? These isomers have identical connectivity; only the spatial arrangement of the ligands around the metal differs.

• Geometrical (cis–trans) isomerism arises when two identical ligands can sit adjacent (cis, $90^\circ$ apart) or opposite (trans, $180^\circ$ apart) around the metal. It shows up in square planar complexes of type $MA_2B_2$ and in octahedral complexes of type $MA_4B_2$ or $MA_3B_3$.
• Optical isomerism arises when a complex is chiral — its mirror image cannot be superimposed on the original — giving a pair of enantiomers that rotate plane-polarised light in opposite directions. It is common in octahedral complexes with bidentate ligands, such as $[Cr(en)_3]^{3+}$, and in the cis form of $[Co(en)_2Cl_2]^+$ (the trans form of that same complex has a plane of symmetry and so is achiral).

The single question that sorts any pair of isomers into the right family is simple: do the same atoms bond to the metal in both structures? If no, it is structural isomerism (and the four subtypes above tell you how connectivity differs). If yes, it is stereoisomerism, and the geometry or chirality argument tells you which kind.

Every one of the six subtypes above traces back to one structural fact about coordination compounds: a complex has two distinct "zones" — the coordination sphere (metal plus directly bonded ligands, written inside square brackets) and everything outside it (counter-ions, water of crystallisation). Isomerism arises whenever chemically equivalent atoms or groups can be redistributed between or within these zones without changing the overall formula, and each subtype is just a different kind of redistribution.

Structural isomerism redistributes which species occupy the coordination sphere itself. Ionisation isomerism moves an entire ion between inside and outside the bracket ($Br^-$ swaps places with $SO_4^{2-}$). Hydrate isomerism does the identical trick with water molecules specifically. Linkage isomerism keeps the same ligand inside the bracket but changes which atom of that ligand touches the metal — a redistribution within a single ligand rather than between ligands. Coordination isomerism redistributes ligands between two different coordination spheres when a compound has a complex cation and a complex anion. In every case, the test for structural isomerism is the same: check whether the set of atoms directly bonded to the metal has changed.

Stereoisomerism, by contrast, never touches the coordination sphere's membership — it only asks how the fixed set of bonded atoms is arranged in three-dimensional space. Once connectivity is fixed, a coordination number and geometry (square planar, octahedral) permit only a limited number of distinct spatial arrangements. For $MA_2B_2$ or $MA_3B_3$-type formulas, the arrangements split naturally into "identical ligands adjacent" (cis) and "identical ligands opposite" (trans) — two genuinely different molecules with different bond angles between like ligands, hence different physical and chemical properties despite identical connectivity. Optical isomerism is the finer-grained case: even after geometry is fixed (say, both molecules are the cis form), the three-dimensional shape itself may lack any internal mirror plane, making the molecule and its mirror image two non-superimposable objects — enantiomers — just as a left hand and right hand are identical in connectivity but not superimposable in space.

Question: Distinguish, by a simple chemical test, between the ionisation isomers $[Co(NH_3)_5Br]SO_4$ and $[Co(NH_3)_5SO_4]Br$.

Step 1. Identify what is inside versus outside the coordination sphere in each compound. In $[Co(NH_3)_5Br]SO_4$, $Br^-$ is a ligand bonded to cobalt (inside the bracket) and $SO_4^{2-}$ is the free counter-ion (outside). In $[Co(NH_3)_5SO_4]Br$, the roles are reversed: $SO_4^{2-}$ is coordinated to cobalt and $Br^-$ is free.

Step 2. Predict what each compound releases in aqueous solution. Only ions outside the bracket dissociate freely and are instantly available to react with a test reagent; ligands bonded inside the bracket stay attached to cobalt and do not react on the timescale of a simple precipitation test.

Step 3. Apply two test reagents and read the pattern. Add $BaCl_2$ solution: the first compound (free $SO_4^{2-}$) gives an immediate white precipitate of $BaSO_4$, while the second compound (coordinated $SO_4^{2-}$) gives no precipitate. Add $AgNO_3$ solution instead: the second compound (free $Br^-$) gives an immediate pale-yellow precipitate of $AgBr$, while the first compound (coordinated $Br^-$) gives no precipitate.

Answer: the two compounds are distinguished because $[Co(NH_3)_5Br]SO_4$ gives a precipitate with $BaCl_2$ but not $AgNO_3$, while $[Co(NH_3)_5SO_4]Br$ gives a precipitate with $AgNO_3$ but not $BaCl_2$ — only the ion free outside the coordination sphere is available to react.

Competitive stretch

Question: Identify the type of stereoisomerism possible in the square planar complex $[Pt(NH_3)_2Cl_2]$, name both forms, and state one real-world consequence.

Step 1. Platinum(II) here is four-coordinate and square planar, a typical geometry for $d^8$ metals such as $Pt^{2+}$, with formula type $MA_2B_2$ ($A = NH_3$, $B = Cl$).

Step 2. In a square planar $MA_2B_2$ complex, the two $A$ ligands can sit adjacent to each other at $90^\circ$ — the cis isomer, with the two $NH_3$ groups next to each other and the two $Cl$ groups next to each other — or opposite each other at $180^\circ$ — the trans isomer, with each $NH_3$ diagonally across from a $Cl$.

Step 3. Since connectivity (which atoms bond to platinum) is identical in both forms and only spatial arrangement differs, this is geometrical (cis–trans) isomerism, not a structural isomerism.

Answer: $[Pt(NH_3)_2Cl_2]$ shows cis and trans geometrical isomerism; the cis isomer is the anticancer drug cisplatin, while the trans isomer is biologically inactive — a striking case of geometry alone changing biological activity.

• Assuming that any octahedral $MA_4B_2$ complex can show optical isomerism — geometrical (cis/trans) isomerism is common here, but optical isomerism additionally requires the molecule to lack an internal mirror plane, which the trans form of such complexes usually has and so is achiral.
• Confusing linkage isomerism (same ambidentate ligand, different donor atom, e.g. $-NO_2$ vs $-ONO$) with ionisation isomerism (different ions swap between inside and outside the bracket) — both keep the molecular formula constant but the mechanism of change is completely different.
• Forgetting that square planar $MA_2B_2$ and $MABCD$ complexes readily show cis–trans isomerism, but simple tetrahedral $MA_2B_2$ complexes generally cannot, because all four positions around a tetrahedral centre are equivalent in relative arrangement for that formula type.
• Treating hydrate isomerism as a minor curiosity rather than testing it correctly — the standard test is precipitation of the chloride that is free (outside the bracket) versus bound (inside, as a ligand, hence not immediately precipitated by $AgNO_3$).

:::faq[Why does trans-[Co(en)2Cl2]+ not show optical isomerism while the cis form does?] Optical isomerism needs the whole molecule to lack an internal mirror plane. The trans form of $[Co(en)_2Cl_2]^+$ has the two $Cl$ ligands directly opposite each other, which gives the molecule a plane of symmetry passing through both chlorines and the metal — its mirror image is superimposable on itself, so it is achiral and optically inactive. The cis form has the two $Cl$ ligands adjacent instead, which removes that mirror plane; its mirror image is a genuinely different, non-superimposable molecule, so the cis form exists as a pair of active enantiomers.

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