Isomerism — structural (linkage, ionization, coordination, hydrate) and stereo (geometrical, optical): Question bank

For the mechanics of why a metal has a fixed shape, lean on Coordination Number and Geometry; for why colours change, on Crystal Field Theory; for inside vs outside the bracket, on Werner's Theory and Coordination Sphere.

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

Look at the two propellers above: same three ligands, same metal, yet no amount of rotating turns the left picture into the right — that non-superimposability is exactly what "chiral" means (see Chirality in Organic Chemistry).

For a quick visual of cis/trans and fac/mer placements referenced throughout the questions:

True or false — justify

Every item: decide true or false, then give the one-sentence reason. The reason is the whole point.

T1. Two compounds with the same molecular formula are always isomers of each other.

False — they must also have a different arrangement; identical connectivity and identical spatial layout means they are the same substance, not isomers.

T2. Linkage isomers keep the ambidentate ligand inside the coordination sphere and only change which donor atom binds.

True — nothing crosses the bracket boundary; only the atom offering its lone pair to the metal switches (e.g. N vs O of

N O_{2}^{-}

). See Ambidentate vs Polydentate Ligands.

T3.

[C o (N H_{3})_{5} (N O_{2})]^{2 +}

(nitro, yellow) and

[C o (N H_{3})_{5} (O N O)]^{2 +}

(nitrito, red) differ in colour because they have different formulas.

False — the formula is identical; the different donor atom (N vs O) gives a different field strength, hence a different octahedral splitting

Δ_{o}

(defined above) and a different absorbed wavelength via Crystal Field Theory.

T4. Ionization isomers give different ions in solution.

True — a ligand inside the bracket swaps with a counter-ion outside, so the freely-dissociating ion changes (e.g. free

S O_{4}^{2 -}

vs free

B r^{-}

), which is exactly what qualitative tests detect.

T5. Hydrate isomerism is a completely separate family from ionization isomerism.

False — it is ionization isomerism where the swapping species happens to be

H_{2} O

(coordinated water ⇄ lattice water), so the same "inside vs outside" logic applies.

T6. Coordination isomerism can occur even when only the cation is a complex ion.

False — it requires both cation and anion to be complex ions so ligands can be redistributed between two metal centres.

T7. Tetrahedral

M A_{2} B_{2}

shows cis–trans isomerism.

False — all four vertices of a tetrahedron are mutually adjacent at

109. 5^{\circ}

, so there is no "opposite" position and no cis/trans distinction.

T8. Square-planar

M A_{4}

(all four ligands identical) can show geometrical isomerism.

False — cis/trans needs at least two different kinds of ligand; with all four the same, every arrangement is identical.

T9. trans-

[C o (e n)_{2} C l_{2}]^{+}

is optically active.

False — with the two

C l

axial the molecule has a plane of symmetry, so it is superimposable on its mirror image and therefore achiral.

T10. cis-

[C o (e n)_{2} C l_{2}]^{+}

exists as two enantiomers.

True — the cis form has no plane and no centre of symmetry, so its mirror image cannot be rotated onto it, giving the

Δ

and

Λ

(right- and left-handed propeller) pair defined above; see Chirality in Organic Chemistry.

T11. Enantiomers have identical melting points, colour, and reactivity with achiral reagents.

True — they differ only in the direction they rotate plane-polarised light (and in behaviour toward other chiral things), because they are perfect mirror images.

T12. A molecule with a centre of symmetry can still be chiral.

False — a centre of symmetry, like a plane of symmetry, makes a molecule superimposable on its mirror image, so it is achiral.

T13. In

M A_{3} B_{3}

, the fac isomer has the three A ligands on one triangular face and the mer isomer has them across a meridian.

True — "fac" (facial) caps one face with all three like ligands at mutual

9 0^{\circ}

; "mer" (meridional) strings them along a curve through both poles (a

9 0^{\circ}

–

18 0^{\circ}

–

9 0^{\circ}

set), as the figure shows.

Spot the error

Each states a claim a student made. Say what is wrong and give the correction.

E1. "

[C o (N H_{3})_{5} B r] S O_{4}

and

[C o (N H_{3})_{5} (N O_{2})] S O_{4}

are ionization isomers."

Error: they don't even share a formula (Br vs

N O_{2}

), so they aren't isomers at all — ionization isomers must swap the same ion in and out of the bracket, e.g.

[C o (N H_{3})_{5} B r] S O_{4}

[C o (N H_{3})_{5} S O_{4}] B r

E2. "In

[C r (H_{2} O)_{5} C l] C l_{2} \cdot H_{2} O

all three chlorides precipitate instantly with

A g N O_{3}

Error: the

C l

inside the bracket is coordinated and won't precipitate; only the two outside-sphere

C l^{-}

give an immediate

A g C l

ppt.

E3. "Nitro and nitrito isomers are stereoisomers because the atoms point in different directions."

Error: they are structural (linkage) isomers — the actual bond changes (metal–N vs metal–O), so connectivity differs, not just spatial orientation.

E4. "

[C o (e n)_{3}]^{3 +}

has geometrical isomers."

Error: all three chelating

e n

ligands are identical, so there is no cis/trans choice; it shows optical isomerism (

Δ

Λ

), not geometrical.

E5. "Cisplatin and transplatin have different formulas, which is why cis is a drug and trans isn't."

Error: both are

[P t (N H_{3})_{2} C l_{2}]

— identical formula; the cis geometry lets the two leaving

C l

reach adjacent DNA bases, the trans geometry cannot (see Cisplatin and Bioinorganic Chemistry).

E6. "

S C N^{-}

binding through S vs

N C S^{-}

binding through N is coordination isomerism."

Error: the ligand stays in the same sphere and merely flips its donor atom (S vs N), so this is linkage isomerism, not coordination isomerism.

E7. "A complex with a mirror plane can still form enantiomers, they just interconvert fast."

Error: a mirror plane makes the molecule identical to its mirror image, so there is only one form — there are no two enantiomers to interconvert.

E8. "

[P t (N H_{3})_{3} C l_{3}]

... wait,

[C o (N H_{3})_{3} C l_{3}]

shows fac/mer isomerism, and the mer form has all three Cl on one face."

Error: the fac/mer labels are correct for this

M A_{3} B_{3}

case, but it is the fac form that puts all three

C l

on one triangular face — mer spreads them along a meridian (two trans, one perpendicular).

Why questions

W1. Why does linkage isomerism need an ambidentate ligand specifically, not just any two-atom ligand?

Because two different atoms must each carry a lone pair capable of donating to the metal; only then can the same ligand bind either way (e.g. N or O of

N O_{2}^{-}

W2. Why do only free (outside-sphere) ions show up in a quick precipitation test?

Ligands bonded inside the bracket are held tightly by the metal and don't dissociate as simple ions, so only the counter-ions outside are available to react instantly.

W3. Why does gently heating (or standing over a desiccant such as concentrated

H_{2} S O_{4}

, which absorbs moisture) remove only lattice water and not coordinated water?

Lattice (crystallisation) water sits weakly in the crystal and escapes easily, whereas coordinated water is bonded directly to the metal and needs far more energy to break off — so the mass of water lost counts exactly the outside-sphere waters.

W4. Why does cis–trans isomerism require at least two kinds of ligand?

With only one kind of ligand every position is equivalent, so "adjacent" and "opposite" describe the same molecule — you need two different ligands to make those two placements distinguishable.

W5. Why is the fac/mer distinction possible for

M A_{3} B_{3}

but not

M A_{4} B_{2}

You need three identical ligands to either cap one triangular face (fac) or spread across a meridian (mer); with only two like ligands (

M A_{4} B_{2}

) the choice collapses to simple cis/trans.

W6. Why must both ions be complex for coordination isomerism, and not just one?

The isomerism comes from swapping ligands between two metal centres; if only one ion were a complex there would be a single coordination sphere and no second metal to trade ligands with, so the redistribution — and hence the isomer — cannot exist.

W7. Why do enantiomers behave identically toward ordinary (achiral) reagents but differently in a body?

Achiral surroundings can't tell mirror images apart, but biological receptors are themselves chiral, so they fit one enantiomer better — a chiral environment breaks the tie.

W8. Why is the mirror-image test the deciding step for optical isomerism rather than just "count the different ligands"?

Because chirality depends on the whole spatial layout — you must actually draw the mirror image and try to rotate it onto the original; superimposable ⇒ achiral, not-superimposable ⇒ chiral.

Edge cases

C1. Does

[C r (H_{2} O)_{6}] C l_{3}

(all six waters coordinated) show hydrate isomerism on its own?

Not by itself — hydrate isomerism is a relationship between compounds; this violet form is one member of the trio that also includes the blue-green and green forms.

C2. Can a square-planar

M A B C D

(four different ligands) show optical isomerism?

No — a square-planar molecule is planar, and that plane itself is a mirror plane, so it is always superimposable on its mirror image (achiral), though it can show three geometrical isomers.

C3. Is tetrahedral

M ab c d

(four different ligands) chiral?

Yes — like a carbon with four different groups it has no symmetry plane, so it forms enantiomers; they are just hard to isolate because tetrahedral complexes racemise easily.

C4. Does

[P t (N H_{3})_{4}]^{2 +}

(square planar, four identical ligands) show any isomerism?

No — one kind of ligand everywhere means no cis/trans, and the planar mirror plane rules out optical isomerism.

C5. Can one compound belong to several isomerism categories at once?

Yes — e.g.

[C r (H_{2} O)_{4} C l_{2}] C l \cdot 2 H_{2} O

is a hydrate/ionization isomer and its

M A_{4} B_{2}

inner sphere shows cis–trans geometrical isomerism, so always check each category.

C6. If the trans isomer of

[M (e n)_{2} X_{2}]

is achiral, does that mean

e n

(a bidentate ligand) never causes chirality?

No — the cis form of the same complex and

[M (e n)_{3}]^{n +}

are both chiral; whether

e n

produces chirality depends on the overall arrangement, not on

e n

alone.

C7. For counting total stereoisomers, how are chiral geometrical isomers counted?

Each chiral geometrical isomer counts twice (once for each enantiomer), while achiral ones count once — so total = geometrical count with chiral members doubled.

C8. Does the fac isomer of

[C o (N H_{3})_{3} C l_{3}]

show optical isomerism?

No — the fac isomer has a mirror plane (each pair of like ligands is related by symmetry), so it is superimposable on its mirror image; neither fac nor mer of a simple

M A_{3} B_{3}

is chiral.

Recall One-line self-test

Cover every answer above and re-run the four sections; if you can state the reason (not just true/false) for each, you've internalised the traps. Inside-vs-outside the bracket ::: ionization / hydrate isomerism Flip which atom binds ::: linkage isomerism (ambidentate) Redistribute ligands between two complex ions ::: coordination isomerism Adjacent vs opposite ::: geometrical (cis/trans), needs ≥2 ligand types Three like ligands: one face vs one meridian ::: fac vs mer $Δ$ vs $Λ$ ::: right- vs left-handed propeller enantiomers Non-superimposable mirror image ::: optical (enantiomers)