Isomerism — structural (linkage, ionization, coordination, hydrate) and stereo (geometrical, optical): Worked examples

The scenario matrix

Think of isomerism problems as a grid. Each cell is a kind of situation. If we solve one worked example per cell, we have covered everything.

Cell	Scenario class	What makes it tricky	Example that hits it
C1	Structural — linkage (ambidentate ligand flips donor atom)	must know which atom bonds	Ex 1
C2	Structural — ionization / hydrate (inside ↔ outside sphere)	counting free ions	Ex 2
C3	Structural — coordination (two complex ions swap ligands)	both ions must be complex	Ex 3
C4	Stereo — geometrical, square planar $M A_{2} B_{2}$	cis vs trans, only 2 exist	Ex 4
C5	Degenerate case: tetrahedral $M A_{2} B_{2}$	looks like it should have cis/trans — it does not	Ex 5
C6	Stereo — geometrical, octahedral $M A_{3} B_{3}$	fac vs mer (a 3-vs-3 twist)	Ex 6
C7	Stereo — optical (chirality present)	no symmetry plane → enantiomers	Ex 7
C8	Boundary case: chirality absent (trans has a mirror plane)	proving something is achiral	Ex 7 (both parts)
C9	Counting limit: total stereoisomers of a mixed complex	combine geometrical + optical without double-counting	Ex 8
C10	Real-world / word problem: qualitative lab test identifies isomer	translate a test result into a structure	Ex 9
C11	Exam twist: one formula, multiple isomerism types at once	classify systematically	Ex 10

Before the tricky bits, three words we will lean on constantly, defined from zero.

Ex 1 — Cell C1 · Linkage isomerism

Forecast: Guess now — does swapping the bonding atom (N vs O) change the colour? Why would just flipping one atom do that?

Figure — Isomerism — structural (linkage, ionization, coordination, hydrate) and stereo (geometrical, optical)

Figure s01: two octahedra with a navy $C o$ centre and five violet $N H_{3}$ ligands. Left: the sixth ligand is $N O_{2}^{-}$ gripping the metal through its nitrogen atom (magenta bond to N) — the nitro isomer, tagged yellow. Right: the same $N O_{2}^{-}$ grips through an oxygen atom (magenta bond to O) — the nitrito isomer, tagged red.

Identify the ambidentate ligand. Why this step? Linkage isomerism only happens when a ligand can donate through two atoms. Here $N O_{2}^{-}$ can bond via N or via O. The five $N H_{3}$ groups are ordinary (bond only through N), so they play no part.
Write isomer A — bonds through nitrogen (nitro). $[C o (N H_{3})_{5} (N O_{2})]^{2 +}$ , the nitro isomer. Systematic name: pentaamminenitrito-κN-cobalt(III) ion (older style: pentaamminenitrocobalt(III)). Why this step? We keep the ligand inside the sphere; we only change which atom faces the metal. This is the defining move of linkage isomerism.
Write isomer B — bonds through oxygen (nitrito). $[C o (N H_{3})_{5} (O N O)]^{2 +}$ , the nitrito isomer. Systematic name: pentaamminenitrito-κO-cobalt(III) ion (older style: pentaamminenitritocobalt(III)). Why this step? This is the partner isomer — same atoms, same coordination sphere, but the ligand now faces the metal with its oxygen. Writing it explicitly lets us line the two up and see that only the bonding atom differs, which is exactly the definition of linkage isomerism.
Predict colours using field strength. Why this step? N is a stronger-field donor than O. A stronger field means a larger d-orbital splitting $Δ_{o}$ (the energy gap defined just above). Larger $Δ_{o}$ ⇒ the complex absorbs higher-energy (shorter-wavelength) light ⇒ we see a colour further toward yellow. So:
- nitro (N-bonded, bigger $Δ_{o}$ ) → yellow
- nitrito (O-bonded, smaller $Δ_{o}$ ) → red

Verify: Absorbing shorter wavelength (say $λ \approx 430 nm$ , blue-violet) leaves us seeing yellow; absorbing longer wavelength (say $λ \approx 500 nm$ , green) leaves us seeing red. Since $Δ_{o} \propto 1/ λ$ , the N-isomer having the larger $Δ_{o}$ = smaller $λ$ = yellow is internally consistent. ✓

Ex 2 — Cell C2 · Ionization / hydrate isomerism (counting free ions)

Forecast: Guess: of the three chlorines in the formula, how many are truly "free"? Of the six waters, how many are removable?

Figure s02: an octahedron with a navy $C r$ centre gripping five violet $H_{2} O$ and one magenta $C l$ inside the coordination sphere; drawn outside the brackets, floating free, are two magenta $C l^{-}$ counter-ions and one orange lattice $H_{2} O$ . An orange dividing line separates "inside — bonded" from "outside — free."

Split the formula into inside vs outside the sphere. Why this step? Only species outside the brackets are free ions/lattice molecules; inside-sphere species are bonded and do not react as simple ions.
- Inside $[]$ : $C r$ , five $H_{2} O$ , one $C l$ (bonded).
- Outside: two $C l^{-}$ (counter-ions) and one $H_{2} O$ (lattice water of crystallisation).
Count precipitating chloride. Why this step? $A g N O_{3}$ precipitates only ionic, free chloride as $A g C l$ . The one $C l$ inside the sphere is covalently held to $C r$ and stays put. $free C l^{-} = 2 \Rightarrow 2 mol A g C l per mol complex.$
Count removable water. Why this step? Only lattice water (outside) leaves over a drying agent; coordinated water (bonded to $C r$ ) does not. $removable H_{2} O = 1.$

Verify (charge balance): Inside sphere charge $= (+ 3) + 5 (0) + (- 1) = + 2$ , i.e. $[C r (H_{2} O)_{5} C l]^{2 +}$ . It needs two $C l^{-}$ outside to be neutral — matches step 1. Total Cl $= 1 + 2 = 3$ ✓ and total water $= 5 + 1 = 6$ ✓ (a hexahydrate).

Ex 3 — Cell C3 · Coordination isomerism

Forecast: Guess what the partner isomer looks like — which metal ends up wearing which ligand?

Figure s03: two side-by-side ion pairs. Left isomer: a navy $C o$ octahedron wearing six violet $N H_{3}$ next to a magenta $C r$ octahedron wearing six orange $C N$ . Right isomer: the ligand sets are swapped — $C r$ now wears the six violet $N H_{3}$ and $C o$ wears the six orange $C N$ . A curved orange arrow between the panels shows the ligand-set swap.

Confirm both ions are complexes. Why this step? Coordination isomerism = redistributing ligands between two metal centres. If only one ion were a complex, there would be no second metal to hand ligands to. Here cation $[C o (N H_{3})_{6}]^{3 +}$ and anion $[C r (C N)_{6}]^{3 -}$ are both complexes. ✓
Swap the ligand sets between the metals. Why this step? Keep the metals, keep the ligands, keep the total formula — just change which metal holds which ligand set. $[C r (N H_{3})_{6}] [C o (C N)_{6}]$
Confirm same overall formula. Both give $C o, C r, (N H_{3})_{6}, (C N)_{6}$ overall. Why this step? Two structures only qualify as isomers if they share the same molecular formula. Checking this proves we have made a genuine isomer and not accidentally added or lost a ligand — a swap must conserve every atom.

Verify (charge neutrality of each isomer): Isomer 1: $[C o (N H_{3})_{6}]^{3 +} [C r (C N)_{6}]^{3 -}$ → $+ 3 - 3 = 0$ ✓. Isomer 2: $[C r (N H_{3})_{6}]^{3 +} [C o (C N)_{6}]^{3 -}$ → $+ 3 - 3 = 0$ ✓. Both are legitimate neutral compounds.

Ex 4 — Cell C4 · Square planar geometrical isomerism

Forecast: Guess the number: 1, 2, or 3?

Figure s04: two squares side by side, each with a navy Pt at the centre and four ligands at the corners. Left square (cis): the two magenta $C l$ sit on adjacent corners, $90°$ apart, with the two violet $N H_{3}$ filling the other two corners. Right square (trans): the two magenta $C l$ sit on opposite corners across the diagonal, $180°$ apart. Orange angle labels mark $90°$ (cis) and $180°$ (trans).

Draw the square. Why this step? A square planar complex puts the four ligands at the corners of a square, all in one plane, $90°$ apart. Geometry decides "adjacent" vs "opposite."
Place the two $C l$ (the two like ligands). Why this step? Geometrical isomerism needs two different kinds of ligand ( $M A_{2} B_{2}$ : two $N H_{3}$ , two $C l$ ). We ask: can the two $C l$ be next to each other or across from each other?
- cis: the two $C l$ occupy adjacent corners ( $90°$ apart) — see the left square, red $C l$ 's side by side.
- trans: the two $C l$ occupy opposite corners ( $180°$ apart) — right square, red $C l$ 's across the diagonal.
Count. Exactly two geometrical isomers: cis and trans. Why this step? Any other way of dropping two $C l$ on the four corners is just a rotation of one of these two pictures — pick a corner for the first $C l$ , then the second $C l$ is either next to it (cis) or across from it (trans); there is no third relationship on a square. So the enumeration is complete and closes at two.
Identify the drug. Why this step? The two $C l$ are the "leaving groups" that let Pt grab DNA. Only when they are cis (close together) can they reach two neighbouring DNA sites (see Cisplatin and Bioinorganic Chemistry). So cis = cisplatin (active), trans = inactive.

Verify: Any other placement of two $C l$ on a square is just a rotation of cis or of trans, so no third isomer exists. Count $= 2$ ✓.

Ex 5 — Cell C5 · Degenerate trap: tetrahedral $M A_{2} B_{2}$

Forecast: It has two $N H_{3}$ and two $C l$ , just like Ex 4 — so surely it has cis and trans too... or does it?

Figure s05: a single tetrahedron with a navy $Z n$ at its centre and four ligands at the four vertices — two magenta $C l$ and two violet $N H_{3}$ . An orange caption notes that every pair of vertices is separated by the same $109.5°$ angle, so there is no $180°$ "opposite" seat anywhere.

Recall what "trans" means. Why this step? trans requires a unique opposite position — a ligand directly across from another at $180°$ . Cis–trans isomerism lives or dies on whether "opposite" exists.
Look at the tetrahedron's angles. Why this step? In a tetrahedron every vertex sits at $109.5°$ from every other vertex. No vertex is $180°$ from another — there is no "opposite" corner at all.
Conclude. Since every pair of positions is equivalent (all adjacent at the same angle), you cannot distinguish a "near" from a "far" arrangement of the two $C l$ . Rotating the molecule maps any placement onto any other. Why this step? Having shown no $180°$ seat exists (step 2), the only honest conclusion is that "cis" and "trans" collapse into a single arrangement — so we state the verdict rather than inventing a distinction that the geometry forbids. $Tetrahedral M A_{2} B_{2} has NO geometrical isomers.$

Verify: The number of distinct ways to choose 2 of 4 equivalent vertices, up to the full tetrahedral rotation symmetry, is exactly 1. Compare Ex 4's square (which has distinguishable adjacent/opposite pairs) giving 2. The difference is entirely the missing $180°$ position. ✓

Ex 6 — Cell C6 · Octahedral $M A_{3} B_{3}$ · fac vs mer

Forecast: With three $N H_{3}$ and three $C l$ , how can the same three like ligands be arranged in two genuinely different ways?

Figure s06: two octahedra side by side, each with a navy $C o$ centre and six ligands ( $\pm x, \pm y, \pm z$ seats). Left (fac): the three magenta $C l$ occupy three corners that share one triangular face, each $C l$ at $90°$ to the others; the three violet $N H_{3}$ fill the opposite face. Right (mer): the three magenta $C l$ lie in one meridian plane through the metal, so two of them sit at $180°$ (an orange "180° trans" label marks this pair) while the third is at $90°$ to both.

Set up the octahedron. Why this step? An octahedron has 6 positions: think of $\pm x, \pm y, \pm z$ axes. Positions come in trans pairs (each axis has two ends at $180°$ ).
Arrangement 1 — all three $C l$ on one triangular face (fac). Why this step? If the three $C l$ occupy three corners that share a single face, each $C l$ is cis ( $90°$ ) to the other two. This is the facial (fac) isomer.
Arrangement 2 — three $C l$ around a "meridian" (mer). Why this step? If the three $C l$ lie in one plane through the metal (a great-circle "meridian"), then two of them are $180°$ apart (trans) while the third is cis to both. This is the meridional (mer) isomer.
Count. Exactly two: fac and mer. Why this step? Three like ligands on an octahedron can differ only in how many of them sit in trans pairs: either zero trans pairs (all three share a face ⇒ fac) or one trans pair (they line up along a meridian ⇒ mer). Two trans pairs is impossible with only three $C l$ (that would need four ligands filling two full axes). Since the trans-pair count can be only 0 or 1, the enumeration closes at exactly two — this counting argument guarantees we have missed nothing.

Verify (adjacency test):

fac: every $C l$ – $C o$ – $C l$ angle is $90°$ (0 trans pairs of $C l$ ).
mer: exactly one $C l$ – $C o$ – $C l$ pair is $180°$ (1 trans pair). Since the count of trans pairs differs (0 vs 1), the two are genuinely distinct. ✓

Ex 7 — Cells C7 + C8 · Optical isomerism present and absent

First, from zero:

Forecast: Guess: is it the cis form, the trans form, both, or neither that splits into two mirror-image forms?

Figure s07: two octahedra with a navy $C o$ centre. Left (trans): the two magenta $C l$ are axial ( $180°$ apart) and the two orange $e n$ chelates wrap the equator; a dashed navy line marks the mirror plane that makes it achiral. Right (cis): the two magenta $C l$ are at $90°$ and the two $e n$ take the remaining sites; a magenta " $Δ$ / $Λ$ " label flags that this form has no mirror plane and so exists as two non-superimposable enantiomers.

Part A — trans (Cell C8, boundary "achiral"):

Place the two $C l$ trans (axial, $180°$ apart); the two $e n$ wrap around the equator. Why this step? We test the high-symmetry case first — it usually turns out achiral.
Find a mirror plane. Why this step? If a plane of symmetry exists, the molecule equals its own mirror image ⇒ achiral. Here the plane through the two axial $C l$ and the metal reflects the two $e n$ onto each other.
Conclude: trans- $[C o (e n)_{2} C l_{2}]^{+}$ is optically inactive (achiral).

Part B — cis (Cell C7, "chiral"):

Place the two $C l$ cis ( $90°$ apart); the two $e n$ take the remaining sites. Why this step? Lower symmetry — the likely chiral candidate.
Draw its mirror image and try to superimpose. Why this step? No plane or centre of symmetry survives; the mirror image is a left-handed propeller versus a right-handed one. They cannot be rotated onto each other.
Conclude: cis- $[C o (e n)_{2} C l_{2}]^{+}$ is chiral → exists as two enantiomers $Δ$ and $Λ$ .

Verify: The two enantiomers rotate plane-polarised light by equal and opposite amounts, e.g. $+ α$ and $- α$ , so a racemic (50:50) mix gives net rotation $+ α + (- α) = 0$ . That zero net rotation is the experimental signature of an enantiomer pair. ✓

Ex 8 — Cell C9 · Counting total stereoisomers without double-counting

Forecast: Guess a whole number before reading.

List the geometrical isomers. Why this step? Start with connectivity-distinct spatial forms: cis and trans → 2 geometrical isomers.
For each, apply the chirality result from Ex 7. Why this step? Chiral ones count twice (two enantiomers); achiral ones count once.
- trans: achiral → 1 stereoisomer.
- cis: chiral → 2 stereoisomers ( $Δ$ , $Λ$ ).
Add. Why this step? Total stereoisomers = the sum over every geometrical form of its enantiomer count. This avoids double-counting because we double only the chiral form. $total = trans 1 + cis (Δ, Λ) 2 = 3.$

Verify: $3$ distinct stereoisomers — never $4$ , because the trans form is not chiral, so it is not doubled. The common wrong answer $4$ comes from doubling both; our Ex 7 boundary case prevents that. ✓

Ex 9 — Cell C10 · Real-world lab word problem

Forecast: Guess the structure from the two clues (1 free $C l$ , 2 free waters) before working it.

Use the $A g N O_{3}$ result. Why this step? $1$ mol $A g C l$ ⇒ exactly 1 free (outside-sphere) $C l^{-}$ . The other 2 chlorides must be inside, bonded to $C r$ .
Use the drying result. Why this step? $2$ mol $H_{2} O$ removed ⇒ 2 lattice (outside) waters. The remaining $6 - 2 = 4$ waters are coordinated inside.
Assemble the coordination sphere. Inside: $C r$ , four $H_{2} O$ , two $C l$ → $[C r (H_{2} O)_{4} C l_{2}]^{+}$ . Outside: one $C l^{-}$ counter-ion and two lattice $H_{2} O$ . $[C r (H_{2} O)_{4} C l_{2}] C l \cdot 2 H_{2} O (green — matches!)$

Verify (mass and charge balance):

Chlorine: inside $2$ + outside $1 = 3$ ✓ (matches $C r C l_{3}$ ).
Water: inside $4$ + outside $2 = 6$ ✓ (matches $6 H_{2} O$ ).
Charge: $[C r (H_{2} O)_{4} C l_{2}]^{+}$ needs one $C l^{-}$ → neutral ✓.
The colour green is the tabulated colour for this exact isomer. ✓

Ex 10 — Cell C11 · Exam twist: one formula, several isomerism types

Forecast: Guess which three of {linkage, ionization, coordination, geometrical, optical} apply.

Scan for an ambidentate ligand → linkage. Why this step? $N O_{2}^{-}$ is ambidentate (N-nitro or O-nitrito). Two of them are present, so the complex shows linkage isomerism.
Scan the geometry ( $M A_{2} B_{2}$ -type, octahedral with two $e n$ + two $N O_{2}$ ) → geometrical. Why this step? The two $N O_{2}$ can be cis or trans, so it shows geometrical (cis–trans) isomerism.
Test the cis form for chirality → optical. Why this step? Like Ex 7, the cis form of $[C o (e n)_{2} X_{2}]$ -type has no symmetry plane ⇒ chiral ⇒ optical isomerism ( $Δ/Λ$ ). (The trans form is achiral — the boundary case again.)
Rule out the others. Why this step? No exchangeable outside-sphere counter-ion in the bracket beyond charge balance to swap → weak ionization case; only one metal centre → no coordination isomerism. So the three clean types are linkage, geometrical, optical.

Verify: Charge check — $C o^{3 +}$ , two neutral $e n$ , two $N O_{2}^{-}$ : $+ 3 + 0 - 2 = + 1$ , matching the written $[]^{+}$ ✓. Each of the three isomerism types was justified from a distinct structural feature (ambidentate ligand; two like ligands + $180°$ seat available; cis form lacking a mirror plane), so they are independent and genuinely co-present. ✓

Recall Quick self-test (cloze)

Tetrahedral $M A_{2} B_{2}$ shows this many geometrical isomers ::: zero (no "opposite" position) Octahedral $M A_{3} B_{3}$ geometrical isomer names ::: fac and mer trans- $[C o (e n)_{2} C l_{2}]^{+}$ is optically ::: inactive (has a plane of symmetry) Total stereoisomers of $[C o (e n)_{2} C l_{2}]^{+}$ ::: 3 (trans + cis $Δ$ + cis $Λ$ ) Free $C l^{-}$ in $[C r (H_{2} O)_{4} C l_{2}] C l \cdot 2 H_{2} O$ that precipitates with $A g N O_{3}$ ::: 1