Chapter 5 · Coordination Compounds

5.2 Isomerism in Coordination Compounds

Same formula, different compound: cis and trans, fac and mer, mirror-image pairs, and the structural isomers that swap a ligand for a counter ion, a donor atom or a water molecule.

In 1965 the biophysicist Barnett Rosenberg passed an electric current between platinum electrodes in a culture of bacteria and found that the cells stopped dividing. The culprit was a trace of a platinum compound formed from the electrodes, $\mathrm{[PtCl_{2}(NH_{3})_{2}]}$. Strangely, only one form of it worked. The compound exists as two isomers: in cisplatin the two chloride ligands are next to each other, in transplatin they are opposite. Cisplatin became one of the most widely used anticancer drugs in the world; transplatin is almost inactive. Same atoms, same bonds, different arrangement in space, completely different biology.

Werner had met the same puzzle decades earlier with the green and violet forms of $\mathrm{CoCl_{3}\cdot4NH_{3}}$. Counting the isomers of a complex was in fact how he proved that six-coordinate cobalt(III) is octahedral. In this section you will learn the kinds of isomerism shown by coordination compounds, how to draw and count isomers, and how to tell structural isomers apart by simple tests.

📘 Definition 5.5 — Isomers of coordination compounds

Isomers are compounds with the same chemical formula but a different arrangement of atoms. Coordination compounds show two main kinds:

  • Stereoisomerism: the same ligands bonded to the metal through the same atoms, but arranged differently in space. Its two types are geometrical and optical isomerism.
  • Structural isomerism: the isomers differ in which bonds are present. Its four types are linkage, coordination, ionisation and solvate isomerism.

Geometrical Isomerism

When two ligands of the same kind can sit either next to each other (at $90^{\circ}$) or opposite each other (at $180^{\circ}$), two different compounds result. This needs a geometry in which some positions are adjacent and others opposite: square planar or octahedral.

📘 Definition 5.6 — Geometrical isomerism: cis/trans and fac/mer

Geometrical isomers differ in the relative positions of the ligands around the central atom. Two identical ligands that are adjacent are cis; two that are opposite are trans.

In an octahedral $\mathrm{[MA_{3}B_{3}]}$ complex, if the three B ligands occupy the corners of one triangular face of the octahedron (all mutually cis), the isomer is facial (fac); if they lie on a meridian, a line round the octahedron through the metal (two of them trans), it is meridional (mer).

Square planar complexes. $\mathrm{[MA_{2}B_{2}]}$ (cisplatin and transplatin) and $\mathrm{[MA_{2}BC]}$ each have two geometrical isomers. A square planar $\mathrm{[MABCD]}$ complex has three: each of B, C or D can be placed trans to A, for example in $\mathrm{[PtBrCl(NH_{3})(py)]}$ (py = pyridine). A square planar complex with two unsymmetrical didentate ligands, $\mathrm{[M(AB)_{2}]}$, such as bis(glycinato)platinum(II), has cis and trans forms as well.

Tetrahedral complexes show no geometrical isomerism: every position is adjacent to every other (all angles $109.5^{\circ}$), so there is no "opposite" position.

Octahedral complexes. $\mathrm{[MA_{4}B_{2}]}$ has cis and trans isomers: these are Werner's violet (cis) and green (trans) forms of $\mathrm{[CoCl_{2}(NH_{3})_{4}]^{+}}$. $\mathrm{[MA_{3}B_{3}]}$, e.g. $\mathrm{[Co(NH_{3})_{3}(NO_{2})_{3}]}$, has fac and mer isomers. Didentate ligands can only span cis positions (they are too short to reach across the metal), so $\mathrm{[M(AA)_{2}B_{2}]}$, such as $\mathrm{[CoCl_{2}(en)_{2}]^{+}}$, has the two B ligands either cis or trans.

📐 Rule 5.4 — Counting geometrical isomers

Two arrangements are the same compound if one can be turned into the other by rotating the whole complex. With that test:

TypeExampleGeometrical isomers
square planar $\mathrm{[MA_{2}B_{2}]}$, $\mathrm{[MA_{2}BC]}$$\mathrm{[PtCl_{2}(NH_{3})_{2}]}$2 (cis, trans)
square planar $\mathrm{[MABCD]}$$\mathrm{[PtBrCl(NH_{3})(py)]}$3
tetrahedral, any$\mathrm{[NiCl_{2}(PPh_{3})_{2}]}$none
octahedral $\mathrm{[MA_{5}B]}$$\mathrm{[CoCl(NH_{3})_{5}]^{2+}}$none
octahedral $\mathrm{[MA_{4}B_{2}]}$$\mathrm{[CoCl_{2}(NH_{3})_{4}]^{+}}$2 (cis, trans)
octahedral $\mathrm{[MA_{3}B_{3}]}$$\mathrm{[Co(NH_{3})_{3}(NO_{2})_{3}]}$2 (fac, mer)
octahedral $\mathrm{[M(AA)_{2}B_{2}]}$$\mathrm{[CoCl_{2}(en)_{2}]^{+}}$2 (cis, trans)
✏️ Example 1 — Square planar isomers
How many geometrical isomers are possible for the square planar complexes (a) $\mathrm{[PtCl_{2}(NH_{3})_{2}]}$, (b) $\mathrm{[PtCl_{2}(NH_{3})(py)]}$, (c) $\mathrm{[PtBrCl(NH_{3})(py)]}$? Would the answers change if the complexes were tetrahedral?
(a) $\mathrm{MA_{2}B_{2}}$: the two Cl are either adjacent (cis) or opposite (trans): 2.
(b) $\mathrm{MA_{2}BC}$ (A = Cl): the two Cl are cis or trans; once that is fixed, swapping $\mathrm{NH_{3}}$ and py gives the same molecule turned over: 2.
(c) $\mathrm{MABCD}$: fix $\mathrm{NH_{3}}$ (A); the ligand opposite it can be Br, Cl or py, and each choice fixes the rest: 3.
Tetrahedral All four positions are mutually adjacent, so no cis/trans distinction: one geometrical form each. Counting isomers is in fact a test of shape: three isomers of an MABCD complex prove it is square planar.
✏️ Example 2 — cis, trans, fac or mer?
In an octahedral complex the six positions are numbered so that 1 and 6 are opposite (top and bottom), and 2, 3, 4, 5 go round the middle with 2 opposite 4 and 3 opposite 5. Name the isomer when (a) $\mathrm{[CoCl_{2}(NH_{3})_{4}]^{+}}$ has Cl at 1 and 2; (b) Cl at 2 and 4; (c) $\mathrm{[Co(NH_{3})_{3}(NO_{2})_{3}]}$ has $\mathrm{NO_{2}}$ at 1, 2, 3; (d) $\mathrm{NO_{2}}$ at 1, 2, 6; (e) $\mathrm{NO_{2}}$ at 2, 3, 4.
(a) 1 and 2 are adjacent: cis.
(b) 2 and 4 are opposite: trans.
(c) 1, 2, 3 are mutually adjacent (one face): fac.
(d) 1 and 6 are trans, 2 is cis to both: mer.
(e) 2 and 4 are trans: mer again. Any set of three positions containing a trans pair is mer; the only other possibility is a face (fac).
Interactive 1 — Isomer Explorer: Count Them All

Choose a complex type. The explorer places the ligands in every possible way, then throws away any arrangement that is just a rotation of one already found (it tries all 24 rotations of an octahedron, 8 of a square or 12 of a tetrahedron). What is left is drawn below: every distinct stereoisomer. It also reflects each isomer in a mirror; if the mirror image is not among the rotations of the original, the isomer is chiral and its partner is shown. Didentate ligands (arcs) can only span cis positions. Press Play to spin each isomer in turn.

example—
arrangements tried—
geometrical isomers—
chiral isomers—
total stereoisomers—

Optical Isomerism

Your left and right hands are mirror images, but you cannot superimpose one on the other: a left glove does not fit a right hand. Some complexes are "handed" in the same way.

📘 Definition 5.7 — Optical isomerism

A molecule or ion that cannot be superimposed on its mirror image is chiral. A chiral complex and its mirror image are a pair of optical isomers (enantiomers). They have identical physical and chemical properties, except that they rotate the plane of plane-polarised light by equal amounts in opposite directions: the dextro isomer ($d$ or $+$) to the right, the laevo isomer ($l$ or $-$) to the left. An equimolar mixture of the two is optically inactive (a racemic mixture).

A complex that has a plane of symmetry (a mirror plane through it) is always superimposable on its mirror image, so it is not chiral. Optical isomerism is common in octahedral complexes containing didentate ligands; square planar complexes are normally achiral, because the plane of the complex is itself a mirror plane.

For $\mathrm{[PtCl_{2}(en)_{2}]^{2+}}$ (and likewise $\mathrm{[CoCl_{2}(en)_{2}]^{+}}$), the trans isomer has a mirror plane containing the two Cl atoms and the metal, and is optically inactive. The cis isomer has no mirror plane: it exists as a $d$ and an $l$ form. So $\mathrm{[CoCl_{2}(en)_{2}]^{+}}$ has three stereoisomers in all: trans, $d$-cis and $l$-cis.

✏️ Example 3 — Counting all the stereoisomers
Find the total number of stereoisomers of (a) $\mathrm{[CoCl_{2}(en)_{2}]^{+}}$, (b) $\mathrm{[Co(en)_{3}]^{3+}}$, (c) $\mathrm{[CoCl(NH_{3})(en)_{2}]^{2+}}$, (d) $\mathrm{[CoCl_{2}(NH_{3})_{4}]^{+}}$. Which of them are optically active?
(a) Geometrical: cis and trans. Trans has a mirror plane; cis is chiral (a $d$, $l$ pair). Total $1+2=\mathbf{3}$.
(b) Only one way to place three en ligands, but it is chiral: $\mathbf{2}$ (a $d$, $l$ pair).
(c) Type $\mathrm{[M(AA)_{2}BC]}$: Cl and $\mathrm{NH_{3}}$ trans (achiral, a mirror plane contains Cl, Co, N) or cis (chiral pair). Total $\mathbf{3}$.
(d) cis and trans; both have mirror planes, so neither is optically active. Total $\mathbf{2}$.
Check The isomer explorer above gives $3$, $2$ and $2$ for M(AA)₂B₂, M(AA)₃ and MA₄B₂ by enumerating every arrangement.
✏️ Example 4 — Chiral or not?
Which of these are chiral? (a) $\mathrm{[Co(en)_{3}]^{3+}}$; (b) trans-$\mathrm{[CoCl_{2}(en)_{2}]^{+}}$; (c) cis-$\mathrm{[CrCl_{2}(C_{2}O_{4})_{2}]^{3-}}$; (d) square planar $\mathrm{[PtCl_{2}(en)]}$; (e) fac-$\mathrm{[CoCl_{3}(NH_{3})_{3}]}$.
(a) Chiral: three chelate rings arranged like a propeller, no mirror plane.
(b) Not chiral: the plane containing the four N atoms and the Co is a mirror plane (it reflects one Cl onto the other).
(c) Chiral: oxalate is didentate like en, so this is a cis-$\mathrm{[M(AA)_{2}B_{2}]}$ complex.
(d) Not chiral: a planar complex lies in its own mirror plane.
(e) Not chiral: the plane containing one Cl, one $\mathrm{NH_{3}}$ and the Co, and bisecting the other two pairs, is a mirror plane. Neither fac nor mer $\mathrm{[MA_{3}B_{3}]}$ is optically active.
Interactive 2 — The Mirror Test: Can You Superimpose It?

Left: a complex. Right: its mirror image (the complex reflected left-to-right, drawn from the same viewpoint). Press Play and the mirror image is turned through each of the 24 rotations of an octahedron in turn. After each rotation every position is compared with the original: green where the ligand matches, red where it does not (for a chelate, both ends must be joined to the same partner). If some rotation gives a perfect match, the complex is achiral; if none does, it is chiral and the two forms are optical isomers.

rotations tried—
positions matching now—
best match so far—
verdict—

Structural Isomerism

Structural isomers contain different bonds. In coordination compounds this happens in four ways: a ligand can change its donor atom, ligands can swap between two complex ions, a ligand can swap with a counter ion, or water can move between the coordination sphere and the crystal.

📘 Definition 5.8 — Types of structural isomerism

Linkage isomerism arises when an ambidentate ligand binds through different donor atoms: yellow $\mathrm{[Co(NH_{3})_{5}(NO_{2})]Cl_{2}}$, pentaamminenitrito-N-cobalt(III) chloride, and red $\mathrm{[Co(NH_{3})_{5}(ONO)]Cl_{2}}$, pentaamminenitrito-O-cobalt(III) chloride.

Coordination isomerism arises when ligands are exchanged between a complex cation and a complex anion containing different metals: $\mathrm{[Co(NH_{3})_{6}][Cr(CN)_{6}]}$ and $\mathrm{[Cr(NH_{3})_{6}][Co(CN)_{6}]}$.

Ionisation isomerism arises when a counter ion is also able to act as a ligand and changes places with a ligand, so the isomers give different ions in solution: $\mathrm{[Co(NH_{3})_{5}(SO_{4})]Br}$ and $\mathrm{[CoBr(NH_{3})_{5}]SO_{4}}$.

Solvate isomerism (called hydrate isomerism when the solvent is water) arises when solvent molecules are bound to the metal in one isomer but present only as free molecules in the crystal lattice in the other. The three hydrate isomers of $\mathrm{CrCl_{3}\cdot6H_{2}O}$ are violet $\mathrm{[Cr(H_{2}O)_{6}]Cl_{3}}$, grey-green $\mathrm{[CrCl(H_{2}O)_{5}]Cl_{2}\cdot H_{2}O}$ and dark green $\mathrm{[CrCl_{2}(H_{2}O)_{4}]Cl\cdot2H_{2}O}$.

✏️ Example 5 — Telling ionisation isomers apart
Two bottles are labelled only "$\mathrm{CoBrSO_{4}\cdot5NH_{3}}$". One contains the violet isomer $\mathrm{[CoBr(NH_{3})_{5}]SO_{4}}$, the other the red isomer $\mathrm{[Co(NH_{3})_{5}(SO_{4})]Br}$. (a) Name both. (b) Describe two chemical tests that tell them apart. (c) Would conductivity measurements help?
(a) $\mathrm{[CoBr(NH_{3})_{5}]SO_{4}}$: $x-1=+2$, Co(III): pentaamminebromidocobalt(III) sulfate. $\mathrm{[Co(NH_{3})_{5}(SO_{4})]Br}$: $x-2=+1$, Co(III): pentaamminesulfatocobalt(III) bromide.
(b) Silver nitrate Only $\mathrm{[Co(NH_{3})_{5}(SO_{4})]Br}$ has free $\mathrm{Br^{-}}$: it gives a pale yellow precipitate of AgBr, $\mathrm{Ag^{+}(aq)+Br^{-}(aq)\rightarrow AgBr(s)}$.
(b) Barium chloride Only $\mathrm{[CoBr(NH_{3})_{5}]SO_{4}}$ has free sulphate: it gives a white precipitate, $\mathrm{Ba^{2+}(aq)+SO_{4}^{2-}(aq)\rightarrow BaSO_{4}(s)}$.
(c) Both give two ions per formula unit, so the number of ions is the same. ($\mathrm{[CoBr(NH_{3})_{5}]^{2+}SO_{4}^{2-}}$ carries more charge per ion and conducts somewhat more, but the precipitation tests are decisive.)
✏️ Example 6 — Which hydrate isomer?
A sample of $2.665\ \mathrm{g}$ of a green form of $\mathrm{CrCl_{3}\cdot6H_{2}O}$ ($M=266.4\ \mathrm{g\,mol^{-1}}$) is dissolved in cold water and treated at once with excess $\mathrm{AgNO_{3}(aq)}$; $2.866\ \mathrm{g}$ of AgCl precipitates. (a) Identify the isomer and name its complex ion. (b) What mass would the dry sample lose when kept over concentrated sulphuric acid, which removes only water of crystallisation?
Moles $n=\dfrac{2.665}{266.4}=1.000\times10^{-2}\ \mathrm{mol}$; $n(\mathrm{AgCl})=\dfrac{2.866}{143.32}=2.000\times10^{-2}\ \mathrm{mol}$.
(a) Two free chlorides per formula unit, so one Cl is coordinated: $\mathrm{[CrCl(H_{2}O)_{5}]Cl_{2}\cdot H_{2}O}$ (grey-green). The ion is pentaaquachloridochromium(III).
(b) One water of crystallisation per formula unit: $1.000\times10^{-2}\times18.02=0.180\ \mathrm{g}$.
Check The violet isomer would give $3\times10^{-2}\ \mathrm{mol}$ AgCl ($4.30\ \mathrm{g}$) and lose no water; the dark green one $1\times10^{-2}\ \mathrm{mol}$ ($1.433\ \mathrm{g}$) and lose $0.360\ \mathrm{g}$.
🧭 Problem-solving strategy — Finding all the isomers of a complex
  1. Shape. CN 6: octahedral. CN 4: square planar (usually $d^{8}$ ions such as $\mathrm{Pt^{2+}}$, $\mathrm{Pd^{2+}}$, often $\mathrm{Ni^{2+}}$ with strong ligands) or tetrahedral.
  2. Structural isomers. Is there an ambidentate ligand (linkage)? A counter ion that could be a ligand (ionisation)? Water of crystallisation (hydrate)? Two complex ions with different metals (coordination)?
  3. Geometrical isomers. Place the ligands that occur twice or three times: cis or trans, fac or mer. Didentate ligands span cis positions only. Discard drawings that are rotations of one another.
  4. Optical isomers. For each geometrical isomer look for a mirror plane. If there is none, it has a non-superimposable mirror image: add one more isomer.
  5. Total stereoisomers $=$ (number of geometrical isomers) $+$ (number of them that are chiral).
⚠️ Common mistakes with isomers

1. Counting rotations as isomers. For cis-$\mathrm{[MA_{4}B_{2}]}$ there are twelve ways to put two B's on adjacent corners, but they are all the same molecule turned round.

2. A didentate ligand across the metal. en, ox and gly cannot span trans positions.

3. Geometrical isomers of tetrahedral complexes. There are none. Optical isomers of square planar complexes: normally none.

4. Forgetting the mirror image. "Isomers" in JEE-style questions usually means all stereoisomers: cis-$\mathrm{[CoCl_{2}(en)_{2}]^{+}}$ counts twice.

5. Mixing up ionisation and hydrate isomers. In ionisation isomers an anion moves between the sphere and outside; in hydrate isomers a water molecule does.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — six isomers of $\mathrm{[MA_{2}B_{2}C_{2}]}$

In JEE Main, NEET and JEE Advanced 2026 (isomerism of coordination compounds; JEE Advanced lists cis–trans isomerism only); not required by the CBSE syllabus.

An octahedron has three pairs of opposite positions. In $\mathrm{[MA_{2}B_{2}C_{2}]}$, for example $\mathrm{[CoCl_{2}(H_{2}O)_{2}(NH_{3})_{2}]^{+}}$, list which ligands are trans to each other: (A–A, B–B, C–C); (A–A, B–C, B–C); (B–B, A–C, A–C); (C–C, A–B, A–B); and (A–B, B–C, C–A), in which every like pair is cis. That is five geometrical isomers. Only the last has no mirror plane, so it exists as an enantiomeric pair: six stereoisomers in all. Choose "MA₂B₂C₂ (challenge)" in the isomer explorer to see them.

Exercises

I
Geometrical isomerism.
  • 1 Draw the cis and trans isomers of square planar $\mathrm{[PtCl_{2}(NH_{3})_{2}]}$. Which one is the anticancer drug?
  • 2 Explain why a tetrahedral complex $\mathrm{[MA_{2}B_{2}]}$ shows no geometrical isomerism, while a square planar one does.
  • 3 How many geometrical isomers does the square planar complex $\mathrm{[PtCl(NH_{3})(NO_{2})(py)]}$ have?
  • 4 Draw the geometrical isomers of $\mathrm{[CoCl_{2}(NH_{3})_{4}]^{+}}$ and state the colour of each.
  • 5 Draw the fac and mer isomers of $\mathrm{[CoCl_{3}(NH_{3})_{3}]}$.
  • 6 Does $\mathrm{[Fe(CN)_{5}(NH_{3})]^{3-}}$ show geometrical isomerism? Explain.
  • 7 How many geometrical isomers has the octahedral ion $\mathrm{[CrCl_{2}(C_{2}O_{4})_{2}]^{3-}}$?
  • 8 Which isomer of $\mathrm{[CoCl_{2}(NH_{3})_{4}]^{+}}$ has no dipole moment? Explain.
1 cis: the two Cl at $90^{\circ}$; trans: at $180^{\circ}$ (see the figure). The cis isomer is cisplatin.
2 In a tetrahedron every position is adjacent to every other (all angles $109.5^{\circ}$), so any two B's are related in the same way. In a square, two B's can be adjacent ($90^{\circ}$) or opposite ($180^{\circ}$).
3 Type $\mathrm{[MABCD]}$: $\mathbf{3}$ (each of the other three ligands can be trans to Cl).
4 cis (Cl at $90^{\circ}$), violet; trans (Cl at $180^{\circ}$), green.
5 fac: the three Cl on one face (Cl–Co–Cl all $90^{\circ}$); mer: the three Cl in one plane through Co, two of them trans ($180^{\circ}$).
6 No. It is $\mathrm{[MA_{5}B]}$: wherever the single $\mathrm{NH_{3}}$ is placed, the complex looks the same after a rotation.
7 Type $\mathrm{[M(AA)_{2}B_{2}]}$: $\mathbf{2}$ (cis and trans). (The cis form is also chiral.)
8 The trans isomer: the two Co–Cl bond dipoles point in opposite directions and cancel, as do the Co–N dipoles. In the cis isomer they do not cancel.
II
Optical isomerism.
  • 1 What is meant by a chiral complex? Explain why $\mathrm{[Co(en)_{3}]^{3+}}$ is optically active.
  • 2 Explain why cis-$\mathrm{[PtCl_{2}(en)_{2}]^{2+}}$ is optically active but trans-$\mathrm{[PtCl_{2}(en)_{2}]^{2+}}$ is not.
  • 3 How many stereoisomers does $\mathrm{[Cr(C_{2}O_{4})_{3}]^{3-}}$ have?
  • 4 Find the total number of stereoisomers of $\mathrm{[CoCl_{2}(NH_{3})_{2}(en)]^{+}}$ and say which are chiral.
  • 5 Is the square planar complex $\mathrm{[PtCl_{2}(en)]}$ optically active? Explain.
  • 6 Which isomer of $\mathrm{[CoCl(NH_{3})(en)_{2}]^{2+}}$ is optically active?
  • 7 A solution of the $d$ form of a complex rotates plane-polarised light by $+12^{\circ}$. What rotation is expected for the same concentration of the $l$ form, and for a racemic mixture?
1 A chiral complex cannot be superimposed on its mirror image. In $\mathrm{[Co(en)_{3}]^{3+}}$ the three chelate rings make a propeller that can twist either way; there is no mirror plane, so the two mirror-image forms are different (a $d$, $l$ pair).
2 The trans isomer has a mirror plane (the plane of the four N atoms), so it is superimposable on its mirror image. The cis isomer has no mirror plane.
3 Type $\mathrm{[M(AA)_{3}]}$: one geometrical form, which is chiral: $\mathbf{2}$.
4 Geometrical isomers: Cl trans to Cl; $\mathrm{NH_{3}}$ trans to $\mathrm{NH_{3}}$; and both pairs cis. The first two have mirror planes; the all-cis one is chiral. Total $3+1=\mathbf{4}$.
5 No: the complex is planar, and the plane of the molecule is a mirror plane.
6 The cis isomer (Cl and $\mathrm{NH_{3}}$ adjacent). The trans isomer has a mirror plane through Cl, Co and N.
7 $l$ form: $-12^{\circ}$. Racemic mixture: $0^{\circ}$, the two rotations cancel.
III
Structural isomerism. Use $M(\mathrm{AgCl})=143.32\ \mathrm{g\,mol^{-1}}$.
  • 1 Write the formulas and names of the two linkage isomers of the pentaamminenitritocobalt(III) ion.
  • 2 Write the coordination isomer of $\mathrm{[Co(NH_{3})_{6}][Cr(CN)_{6}]}$ and name both compounds.
  • 3 Write the ionisation isomer of $\mathrm{[PtCl_{2}(NH_{3})_{4}]Br_{2}}$ and give a test to distinguish the two.
  • 4 Write the formulas of the three hydrate isomers of $\mathrm{CrCl_{3}\cdot6H_{2}O}$ and the number of moles of AgCl each gives per mole with excess $\mathrm{AgNO_{3}}$.
  • 5 What type of isomerism relates $\mathrm{[Cr(H_{2}O)_{5}(SCN)]^{2+}}$ and $\mathrm{[Cr(H_{2}O)_{5}(NCS)]^{2+}}$? Name both.
  • 6 $\mathrm{[Cu(NH_{3})_{4}][PtCl_{4}]}$ and $\mathrm{[Pt(NH_{3})_{4}][CuCl_{4}]}$ are isomers. Of what type?
  • 7 A solution containing $0.0050\ \mathrm{mol}$ of one isomer of $\mathrm{CrCl_{3}\cdot6H_{2}O}$ gives $2.150\ \mathrm{g}$ of AgCl. Identify the isomer and its colour.
  • 8 Laboratory. An unlabelled cobalt(III) compound has the composition $\mathrm{CoCl_{2}(NO_{2})\cdot5NH_{3}}$, with all five $\mathrm{NH_{3}}$ coordinated. Write the three possible isomers, state how they are related, and suggest how to identify your sample.
1 $\mathrm{[Co(NH_{3})_{5}(NO_{2})]^{2+}}$, pentaamminenitrito-N-cobalt(III) (yellow); $\mathrm{[Co(NH_{3})_{5}(ONO)]^{2+}}$, pentaamminenitrito-O-cobalt(III) (red).
2 $\mathrm{[Cr(NH_{3})_{6}][Co(CN)_{6}]}$, hexaamminechromium(III) hexacyanidocobaltate(III); the original is hexaamminecobalt(III) hexacyanidochromate(III).
3 $\mathrm{[PtBr_{2}(NH_{3})_{4}]Cl_{2}}$. With $\mathrm{AgNO_{3}}$ it gives white AgCl (soluble in dilute ammonia); the original gives pale yellow AgBr.
4 $\mathrm{[Cr(H_{2}O)_{6}]Cl_{3}}$: 3; $\mathrm{[CrCl(H_{2}O)_{5}]Cl_{2}\cdot H_{2}O}$: 2; $\mathrm{[CrCl_{2}(H_{2}O)_{4}]Cl\cdot2H_{2}O}$: 1.
5 Linkage isomerism. Pentaaquathiocyanato-S-chromium(III) and pentaaquathiocyanato-N-chromium(III).
6 Coordination isomerism (ligands exchanged between the Cu and Pt complex ions).
7 $n(\mathrm{AgCl})=\dfrac{2.150}{143.32}=0.0150\ \mathrm{mol}$, three per formula unit: $\mathrm{[Cr(H_{2}O)_{6}]Cl_{3}}$, violet.
8 Five $\mathrm{NH_{3}}$ leave one position for an anion: $\mathrm{[CoCl(NH_{3})_{5}]Cl(NO_{2})}$, $\mathrm{[Co(NH_{3})_{5}(NO_{2})]Cl_{2}}$ (yellow) and $\mathrm{[Co(NH_{3})_{5}(ONO)]Cl_{2}}$ (red). The first is an ionisation isomer of the other two; the last two are linkage isomers. With excess $\mathrm{AgNO_{3}}$, 1 mol of the first gives 1 mol AgCl, the others 2 mol; colour then separates the linkage isomers.
IV
Name the type of isomerism shown by each pair.
  • 1 cis- and trans-$\mathrm{[CoCl_{2}(en)_{2}]^{+}}$
  • 2 $d$- and $l$-$\mathrm{[Co(en)_{3}]^{3+}}$
  • 3 $\mathrm{[Co(NH_{3})_{5}(SCN)]^{2+}}$ and $\mathrm{[Co(NH_{3})_{5}(NCS)]^{2+}}$
  • 4 $\mathrm{[Cr(H_{2}O)_{6}]Cl_{3}}$ and $\mathrm{[CrCl(H_{2}O)_{5}]Cl_{2}\cdot H_{2}O}$
  • 5 $\mathrm{[Co(NH_{3})_{6}][Cr(C_{2}O_{4})_{3}]}$ and $\mathrm{[Cr(NH_{3})_{6}][Co(C_{2}O_{4})_{3}]}$
  • 6 $\mathrm{[PtCl_{2}(NH_{3})_{4}]Br_{2}}$ and $\mathrm{[PtBr_{2}(NH_{3})_{4}]Cl_{2}}$
  • 7 fac- and mer-$\mathrm{[RhCl_{3}(H_{2}O)_{3}]}$
1 Geometrical.
2 Optical.
3 Linkage (thiocyanate bound through S or N).
4 Hydrate (solvate).
5 Coordination.
6 Ionisation.
7 Geometrical (fac/mer).
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Draw the geometrical isomers of $\mathrm{[Cr(CN)_{4}(NH_{3})_{2}]^{-}}$ and name the type of each.

Idea An octahedral $\mathrm{[MA_{4}B_{2}]}$ ion (A = $\mathrm{CN^{-}}$, B = $\mathrm{NH_{3}}$): two isomers. (1 mark)
Answer cis: the two $\mathrm{NH_{3}}$ on adjacent corners ($90^{\circ}$); trans: on opposite corners ($180^{\circ}$), with the four $\mathrm{CN^{-}}$ in the remaining positions. (1 mark for both drawings)

E2. NEET Which of these is optically active? (A) trans-$\mathrm{[CoCl_{2}(en)_{2}]^{+}}$  (B) cis-$\mathrm{[CoCl_{2}(en)_{2}]^{+}}$  (C) $\mathrm{[Co(NH_{3})_{6}]^{3+}}$  (D) cis-$\mathrm{[PtCl_{2}(NH_{3})_{2}]}$

Idea Look for a mirror plane. (A) has one through the four N atoms; (C) has many; (D) is planar, so its own plane is a mirror plane.
Answer (B): cis-$\mathrm{[CoCl_{2}(en)_{2}]^{+}}$ has no mirror plane and exists as a $d$, $l$ pair.

E3. JEE Main Numerical value. The sum of the numbers of geometrical isomers of square planar $\mathrm{[PtBrCl(NH_{3})(py)]}$ and of octahedral $\mathrm{[CoCl_{3}(NH_{3})_{3}]}$ is ______.

Square planar MABCD $3$ (Br, Cl or py trans to $\mathrm{NH_{3}}$).
Octahedral MA₃B₃ $2$ (fac, mer).
Answer $3+2=\mathbf{5}$.

E4. JEE Advanced One or more correct. Which of these have exactly two stereoisomers? (A) $\mathrm{[Co(en)_{3}]^{3+}}$  (B) $\mathrm{[CoCl_{2}(en)_{2}]^{+}}$  (C) $\mathrm{[Co(NH_{3})_{3}(NO_{2})_{3}]}$  (D) $\mathrm{[CoCl_{2}(NH_{3})_{4}]^{+}}$

(A) One geometrical form, chiral: $2$ ✓.
(B) trans + $d$-cis + $l$-cis $=3$ ✗.
(C) fac and mer, both achiral: $2$ ✓.
(D) cis and trans, both achiral: $2$ ✓. Answer: (A), (C), (D).

E5. Board · 3 marks Give one example each, with formulas and names, of (a) linkage isomerism, (b) ionisation isomerism, (c) coordination isomerism.

(a) $\mathrm{[Co(NH_{3})_{5}(NO_{2})]Cl_{2}}$, pentaamminenitrito-N-cobalt(III) chloride, and $\mathrm{[Co(NH_{3})_{5}(ONO)]Cl_{2}}$, pentaamminenitrito-O-cobalt(III) chloride. (1 mark)
(b) $\mathrm{[CoBr(NH_{3})_{5}]SO_{4}}$, pentaamminebromidocobalt(III) sulfate, and $\mathrm{[Co(NH_{3})_{5}(SO_{4})]Br}$, pentaamminesulfatocobalt(III) bromide. (1 mark)
(c) $\mathrm{[Co(NH_{3})_{6}][Cr(CN)_{6}]}$, hexaamminecobalt(III) hexacyanidochromate(III), and $\mathrm{[Cr(NH_{3})_{6}][Co(CN)_{6}]}$, hexaamminechromium(III) hexacyanidocobaltate(III). (1 mark)

E6. Board · Case study Read the passage and answer. The square planar complex $\mathrm{[PtCl_{2}(NH_{3})_{2}]}$ exists as two isomers. Only one of them is used in chemotherapy: inside a cell its two chloride ligands are replaced by water and then by two nitrogen atoms of neighbouring bases on the same DNA strand, which kinks the DNA and stops the cell from dividing. (a) Draw both isomers, name them, and say which is the drug. (2 marks) (b) Using the passage, explain why only that isomer can bind two neighbouring bases. (1 mark) (c) Can either isomer show optical isomerism? Explain. (1 mark)

(a) cis- and trans-diamminedichloridoplatinum(II); the cis isomer (cisplatin) is the drug.
(b) The two leaving chlorides of the cis isomer are $90^{\circ}$ apart, close enough to bind two adjacent bases; in the trans isomer they are on opposite sides of the platinum.
(c) No: both are planar, and the plane of the complex is a mirror plane, so each is superimposable on its mirror image.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Why does the geometry of a platinum complex decide whether it is a cancer drug, and how much platinum does one dose deliver?

Why it matters. Cisplatin and its successors carboplatin and oxaliplatin are given to millions of patients worldwide for testicular, ovarian, lung and colorectal cancers. Chemists still design new platinum, ruthenium and gold complexes, and isomerism is central to that work: a drug and its isomer can differ in activity, side-effects and the way the body clears them.

What to do.

  1. Draw cis- and trans-$\mathrm{[PtCl_{2}(NH_{3})_{2}]}$ and measure (on your drawing) the angle between the two Pt–Cl bonds in each. Explain, in two sentences, why only the cis isomer can bind two adjacent bases on one DNA strand.
  2. (Illustrative dosing, constructed for this exercise.) A patient with a body surface area of $1.80\ \mathrm{m^{2}}$ is prescribed $75\ \mathrm{mg\,m^{-2}}$ of cisplatin ($M=300.0\ \mathrm{g\,mol^{-1}}$). Calculate the dose in mg, the amount in mol, and the number of platinum atoms.
  3. The drug is supplied as a solution containing $1.0\ \mathrm{mg\,mL^{-1}}$. What volume is needed?
  4. Find the structures of carboplatin and oxaliplatin. For each, name the ligands, give the coordination number and oxidation state of platinum, and say which ligands leave the platinum inside the cell.

Tools: Paper and protractor or a molecule-drawing app; a calculator; the isomer explorer in this section.

Hand in: Your two drawings with angles, the dose calculations, and a short table comparing the three drugs.

Going further. Find out why the trans isomer, although inactive on its own, has been studied again with other ligands, and what "trans-platinum" drug candidates look like.

Instructor note (the dosing example is illustrative, constructed for this exercise, not clinical guidance): angles $90^{\circ}$ (cis) and $180^{\circ}$ (trans). Dose $=75\times1.80=135\ \mathrm{mg}$; $n=\dfrac{0.135}{300.0}=4.50\times10^{-4}\ \mathrm{mol}$; $4.50\times10^{-4}\times6.022\times10^{23}=2.71\times10^{20}$ Pt atoms; volume $135\ \mathrm{mL}$. Carboplatin: two $\mathrm{NH_{3}}$ and a didentate cyclobutane-1,1-dicarboxylate; oxaliplatin: didentate cyclohexane-1,2-diamine and oxalate; both Pt(II), CN 4, square planar; the dicarboxylate or oxalate is the leaving group.

🧪 Self-check quiz

0/5 correct

Q1. The number of geometrical isomers of $\mathrm{[CoCl_{2}(NH_{3})_{4}]^{+}}$ is

Q2. Which complex shows optical isomerism?

Q3. $\mathrm{[Co(NH_{3})_{5}(NO_{2})]Cl_{2}}$ and $\mathrm{[Co(NH_{3})_{5}(ONO)]Cl_{2}}$ are

Q4. Tetrahedral complexes do not show geometrical isomerism because

Q5. $\mathrm{[Co(NH_{3})_{6}][Cr(CN)_{6}]}$ and $\mathrm{[Cr(NH_{3})_{6}][Co(CN)_{6}]}$ are an example of

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