Isomerism — 30 NEET-Style Questions

Suggested timing. Isomerism questions split into two speeds. The four structural types (linkage, coordination, ionisation, solvate) are pattern-recognition and should take 30 seconds. The two stereo types need you to picture the geometry, so allow 60–75 seconds. Target about 22 minutes for the full set. If a stereo question is taking two minutes, draw the octahedron rather than trying to hold it in your head.

Log your attempt before opening the solutions. Record each wrong answer against one of three causes — this is more useful than the raw score.

Section	Q	Correct	Didn't know	Misread	Too slow
Geometrical	01–08
Optical	09–15
Linkage	16–19
Coordination	20–23
Ionisation	24–27
Solvate	28–30

Geometrical Isomerism

08 Q

[ 01 ]^G NEET styleSquare planar

The number of geometrical isomers possible for the square planar complex [Pt(NH₃)₂(py)₂]²⁺ is:

(a)1
(b)2
(c)3
(d)4

Answer — (b) 2

Given

Square planar complex of the type [MA₂B₂], with A = NH₃ and B = pyridine.

Asked

Number of geometrical isomers.

Concept

In a square planar [MA₂B₂] complex the two identical ligands can be adjacent (cis, 90° apart) or opposite (trans, 180° apart). These are the only two distinct arrangements.

Formula

Square planar MA₂B₂ → 2 geometrical isomers

Baby steps

Place the first NH₃ at any corner of the square.
The second NH₃ can go either next to it or across from it.
Adjacent → cis; across → trans. Any other placement is one of these two rotated.
Number of geometrical isomers = 2.

ShortcutSquare planar MA₂B₂ is always 2. Neither isomer is optically active, because the molecular plane itself is a mirror plane.

[ 02 ]^G NEET stylefac / mer

Which of the following complexes exhibits facial–meridional isomerism?

(a)[Co(NH₃)₄Cl₂]⁺
(b)[Cr(NH₃)₃Cl₃]
(c)[Pt(NH₃)₂Cl₂]
(d)[Co(en)₃]³⁺

Answer — (b) [Cr(NH₃)₃Cl₃]

Given

Four coordination entities of different types.

Asked

The one showing fac–mer isomerism.

Concept

Facial and meridional isomerism is specific to octahedral complexes of the type [Ma₃b₃]. Three identical ligands either cap one triangular face (fac) or lie around a meridian (mer).

Formula

[Ma₃b₃] → fac + mer

Baby steps

[Co(NH₃)₄Cl₂]⁺ is Ma₄b₂ → cis/trans only.
[Cr(NH₃)₃Cl₃] is Ma₃b₃ → fac and mer.
[Pt(NH₃)₂Cl₂] is square planar MA₂B₂ → cis/trans.
[Co(en)₃]³⁺ is a homoleptic tris-chelate → optical isomerism only.

ShortcutCount the ligand pattern first. Only a 3 + 3 split gives fac/mer; a 4 + 2 split gives cis/trans.

[ 03 ]^G NEET styleOctahedral

The number of geometrical isomers of [Cr(NH₃)₄Cl₂]⁺ and the number of them that are optically active are respectively:

(a)2 and 0
(b)2 and 1
(c)3 and 1
(d)2 and 2

Answer — (a) 2 and 0

Given

Octahedral [Ma₄b₂] complex with only unidentate ligands.

Asked

Geometrical isomer count and how many are chiral.

Concept

[Ma₄b₂] gives cis and trans. Both retain a plane of symmetry because all six ligands are unidentate — chirality in octahedral complexes needs chelate rings.

Formula

Chiral ⇒ no plane and no centre of symmetry

Baby steps

The two Cl⁻ may be adjacent (cis, 90°) or opposite (trans, 180°) → 2 geometrical isomers.
In the trans form, the plane containing the four NH₃ is a mirror plane.
In the cis form, the plane containing both Cl⁻ and two NH₃ is a mirror plane.
Both are achiral → 0 optically active.

ShortcutOctahedral complexes with only unidentate ligands are essentially never chiral. Look for a chelate before predicting optical activity.

[ 04 ]^G NEET styleHomoleptic

Which of the following cannot show geometrical isomerism?

(a)[Co(NH₃)₄Cl₂]⁺
(b)[Co(NH₃)₃Cl₃]
(c)[Co(NH₃)₆]³⁺
(d)[CoCl₂(en)₂]⁺

Answer — (c) [Co(NH₃)₆]³⁺

Given

Four octahedral cobalt(III) entities.

Asked

The one with no geometrical isomerism.

Concept

Geometrical isomerism requires a heteroleptic complex — there must be at least two different kinds of ligand for any rearrangement to produce something new.

Formula

Homoleptic ⇒ no geometrical isomerism

Baby steps

[Co(NH₃)₆]³⁺ has six identical ligands → every arrangement is the same by rotation.
The other three are heteroleptic: Ma₄b₂, Ma₃b₃ and [MX₂(AA)₂].
Each of those gives two geometrical isomers.
Answer = [Co(NH₃)₆]³⁺.

ShortcutHomoleptic = one ligand type = no geometrical isomerism, whatever the geometry. Check the ligand variety before anything else.

[ 05 ]^G NEET styleSquare planar

The square planar complex [Pt(NH₃)(Br)(Cl)(py)] shows how many geometrical isomers, and how many of these are optically active?

(a)3 and 0
(b)3 and 3
(c)2 and 0
(d)4 and 2

Answer — (a) 3 and 0

Given

Square planar complex of the type MABCD with four different unidentate ligands.

Asked

Geometrical isomer count and optical activity.

Concept

For square planar MABCD, fix one ligand and cycle which of the remaining three sits trans to it — that gives three distinct isomers. All square planar complexes are achiral because the molecular plane is itself a mirror plane.

Formula

Square planar MABCD → 3 geometrical, 0 optical

Baby steps

Fix NH₃ at one corner.
Br trans to NH₃ → isomer 1; Cl trans to NH₃ → isomer 2; py trans to NH₃ → isomer 3.
No other arrangement is new — the rest are rotations of these.
Every atom lies in one plane → that plane is a mirror → 0 optically active.

ShortcutSquare planar is always achiral. If an option claims a square planar complex is optically active, eliminate it immediately.

[ 06 ]^G NEET styleBond angle

In trans-[Co(NH₃)₄Cl₂]⁺, the Cl–Co–Cl bond angle is:

(a)90°
(b)109.5°
(c)120°
(d)180°

Answer — (d) 180°

Given

Trans isomer of an octahedral [Ma₄b₂] complex.

Asked

The Cl–Co–Cl angle.

Concept

In an octahedron the six ligands sit along the ±x, ±y and ±z axes. Two ligands on the same axis subtend 180°; two on different axes subtend 90°.

Formula

cis → 90°; trans → 180°

Baby steps

‘Trans’ means the two chlorides are directly opposite each other.
Place them at +z and −z, so their position vectors are (0,0,1) and (0,0,−1).
The angle between opposite vectors is 180°.
Cl–Co–Cl = 180°.

ShortcutThis is the definition, not a calculation: cis = 90°, trans = 180°. In the fac isomer all three like-ligand pairs are 90°.

[ 07 ]^G NEET stylefac / mer

In the facial isomer of [Co(NH₃)₃(NO₂)₃], the three nitro groups:

(a)Lie around the meridian with one pair 180° apart
(b)Occupy three corners of one triangular face, every pair being 90° apart
(c)Are all 180° apart from one another
(d)Lie in a straight line through the metal

Answer — (b) Occupy three corners of one triangular face, every pair being 90° apart

Given

fac isomer of an octahedral [Ma₃b₃] complex.

Asked

Spatial arrangement of the three identical ligands.

Concept

In the fac isomer the three like ligands cap one triangular face of the octahedron, so no two of them are trans. In the mer isomer they lie around a meridian, which forces exactly one pair to be trans.

Formula

fac: all three pairs 90°. mer: one pair 180°, two pairs 90°.

Baby steps

Put the three NO₂ groups at +x, +y and +z.
Angle between any two of these axes = 90°.
No pair is on a common axis, so no pair is trans.
That triangle of three adjacent corners is one face → facial isomer.

ShortcutQuickest test: count the trans pairs among the three like ligands. Zero → fac. One → mer.

[ 08 ]^G NEET styleTetrahedral

Tetrahedral complexes of the type [MA₂B₂] do not show geometrical isomerism because:

(a)All four ligand positions are adjacent, so no cis/trans distinction exists
(b)Tetrahedral complexes are always ionic
(c)The ligands rotate freely about the metal
(d)Tetrahedral complexes have a centre of symmetry

Answer — (a) All four ligand positions are adjacent, so no cis/trans distinction exists

Given

Tetrahedral [MA₂B₂] complex.

Asked

Reason for the absence of geometrical isomerism.

Concept

In a tetrahedron every vertex is adjacent to every other vertex, all at 109.5°. There is no ‘opposite’ position, so the relative arrangement of the two B ligands is always identical.

Formula

Tetrahedral ⇒ all L–M–L angles equal 109.5°

Baby steps

Choose any two of the four corners for the B ligands.
Whichever pair you choose, they subtend 109.5°.
Every such choice can be rotated onto every other → only one distinct arrangement.
Hence no geometrical isomerism.

ShortcutNote the exception: a tetrahedral complex can be optically active, but only with four different ligands (MABCD), which is rare in NEET.

Optical Isomerism

07 Q

[ 09 ]^O NEET styleTris-chelate

The number of optical isomers of [Cr(C₂O₄)₃]³⁻ is:

(a)0
(b)2
(c)3
(d)4

Answer — (b) 2

Given

Tris-chelate octahedral complex with three identical didentate oxalate ligands.

Asked

Number of optical isomers.

Concept

[M(AA)₃] complexes form a three-bladed propeller. The propeller can twist right-handed (Δ, dextro) or left-handed (Λ, laevo), and the two are non-superimposable mirror images.

Formula

[M(AA)₃] → 0 geometrical, 2 optical

Baby steps

Each oxalate spans a cis pair of sites, closing a five-membered ring.
The three rings twist about the metal like propeller blades.
There is no plane and no centre of symmetry → the ion is chiral.
Mirror image is non-superimposable → 2 optical isomers.

ShortcutAny tris-chelate ([Co(en)₃]³⁺, [Cr(ox)₃]³⁻, [Fe(ox)₃]³⁻): always 0 geometrical, always 2 optical.

[ 10 ]^O NEET stylecis / trans

Out of cis- and trans-[Co(en)₂Cl₂]⁺, the optically active form is:

(a)Only the trans form
(b)Only the cis form
(c)Both forms
(d)Neither form

Answer — (b) Only the cis form

Given

[MX₂(AA)₂] octahedral complex in both geometrical forms.

Asked

Which is chiral.

Concept

Placing the two identical X ligands trans to each other creates a mirror plane through the molecule. Placing them cis destroys every symmetry element, so the cis form is chiral.

Formula

Chiral ⇒ no plane, no centre of symmetry

Baby steps

Trans form: the two Cl⁻ lie on one axis and the two en rings lie in the perpendicular plane → a mirror plane exists → achiral.
Cis form: the two Cl⁻ are adjacent and the two chelate rings twist → no mirror plane.
The cis mirror image cannot be rotated onto the original.
Hence only the cis form is optically active.

ShortcutFor [MX₂(AA)₂]: trans is achiral, cis is chiral. Same result for [PtCl₂(en)₂]²⁺ and [CrCl₂(ox)₂]³⁻.

[ 11 ]^O NEET styleCounting

The total number of stereoisomers of [PtCl₂(en)₂]²⁺ is:

(a)2
(b)3
(c)4
(d)6

Answer — (b) 3

Given

Octahedral [MX₂(AA)₂] complex.

Asked

Total stereoisomers (geometrical plus optical).

Concept

Count the geometrical isomers first, then check each one separately for chirality. Only the chiral ones split into a pair.

Formula

Total = achiral geometrical isomers + 2 × (chiral geometrical isomers)

Baby steps

Geometrical isomers: cis and trans → 2.
Trans is achiral → contributes 1 stereoisomer.
Cis is chiral → contributes 2 (d and l).
Total = 1 + 2 = 3 stereoisomers.

ShortcutNever just multiply geometrical isomers by two. Test each geometrical isomer for a mirror plane individually.

[ 12 ]^O NEET styleCondition

A coordination compound is optically active when it possesses:

(a)A plane of symmetry only
(b)A centre of symmetry only
(c)Neither a plane nor a centre of symmetry
(d)Both a plane and a centre of symmetry

Answer — (c) Neither a plane nor a centre of symmetry

Given

General condition for chirality.

Asked

Symmetry requirement for optical activity.

Concept

A molecule is chiral, and therefore optically active, only if it is non-superimposable on its mirror image. Any plane of symmetry or centre of symmetry immediately makes the mirror image superimposable.

Formula

Chiral ⇒ no S_n element; for NEET, no plane and no centre

Baby steps

Draw the complex and look for any plane cutting it into mirror halves.
Also check for a centre of symmetry (every ligand matched by an identical one directly opposite).
If either is present, the mirror image can be rotated onto the original → achiral.
If neither is present → the complex is chiral and optically active.

ShortcutPractical exam version: look for cis + chelate rings. Trans arrangements and homoleptic unidentate complexes almost always have a symmetry plane.

[ 13 ]^O NEET styleGeometrical vs optical

[Cr(C₂O₄)₃]³⁻ shows how many geometrical and how many optical isomers respectively?

(a)2 and 2
(b)0 and 2
(c)2 and 0
(d)3 and 3

Answer — (b) 0 and 2

Given

Homoleptic tris-chelate octahedral complex.

Asked

Geometrical and optical isomer counts.

Concept

With three identical and symmetrical didentate ligands there is only one way to arrange them, so no geometrical isomerism. That single arrangement is nevertheless a chiral propeller.

Formula

[M(AA)₃] → 0 geometrical, 2 optical

Baby steps

All three ligands are the same, so no cis/trans distinction can arise → geometrical = 0.
Each oxalate spans a cis pair, producing a propeller shape.
No plane or centre of symmetry exists.
Optical isomers = 2 (Δ and Λ).

ShortcutA common trap: candidates assume ‘more ligands means more geometrical isomers’. Identical chelates give zero geometrical isomers.

[ 14 ]^O NEET styleSquare planar

Which of the following is not optically active?

(a)cis-[Co(en)₂Cl₂]⁺
(b)[Co(en)₃]³⁺
(c)cis-[Pt(NH₃)₂Cl₂]
(d)cis-[CrCl₂(ox)₂]³⁻

Answer — (c) cis-[Pt(NH₃)₂Cl₂]

Given

Three octahedral chelate complexes and one square planar complex.

Asked

The achiral one.

Concept

Every square planar complex has the molecular plane itself as a plane of symmetry, so square planar complexes are never optically active — the ‘cis’ label does not change this.

Formula

Square planar ⇒ molecular plane is a mirror ⇒ achiral

Baby steps

cis-[Co(en)₂Cl₂]⁺: octahedral, cis, chelate → chiral.
[Co(en)₃]³⁺: tris-chelate propeller → chiral.
cis-[Pt(NH₃)₂Cl₂]: square planar, all atoms coplanar → achiral.
cis-[CrCl₂(ox)₂]³⁻: octahedral, cis, chelate → chiral.

ShortcutSeeing ‘cis’ is not enough. Check the geometry first — cis matters for optical activity only in octahedral complexes with chelates.

[ 15 ]^O NEET styleProperties

A pair of enantiomers (d and l isomers) of a coordination compound differ in:

(a)Melting point and solubility in water
(b)The direction in which they rotate the plane of polarised light
(c)Magnetic moment
(d)Colour and absorption spectrum

Answer — (b) The direction in which they rotate the plane of polarised light

Given

A pair of optical isomers.

Asked

The property in which they differ.

Concept

Enantiomers have identical bonding, identical energies and identical achiral physical properties. They differ only in their interaction with chiral influences — most simply, plane polarised light.

Formula

d rotates to the right; l rotates to the left, by an equal amount

Baby steps

Both isomers have the same metal, ligands, bond lengths and bond angles.
So melting point, solubility, colour and magnetic moment are all identical.
In a polarimeter, one rotates the plane of polarised light clockwise (d, dextro).
The other rotates it anticlockwise by the same angle (l, laevo).

ShortcutAn equimolar mixture of d and l is a racemic mixture and shows zero net rotation, because the two rotations cancel exactly.

Linkage Isomerism

04 Q

[ 16 ]^L NEET styleIdentification

Which of the following complexes will show linkage isomerism?

(a)[Co(NH₃)₅(SCN)]²⁺
(b)[Co(NH₃)₅Cl]²⁺
(c)[Co(NH₃)₆]³⁺
(d)[Co(en)₃]³⁺

Answer — (a) [Co(NH₃)₅(SCN)]²⁺

Given

Four cobalt(III) complexes with different ligands.

Asked

The one showing linkage isomerism.

Concept

Linkage isomerism requires an ambidentate ligand — one with two chemically different donor atoms, only one of which coordinates at a time.

Formula

Ambidentate ligands: NO₂⁻, SCN⁻, CN⁻

Baby steps

Cl⁻ has only one kind of donor atom → no choice of attachment.
NH₃ donates only through nitrogen → unidentate, one mode.
en has two N donors but uses both together → didentate, not ambidentate.
SCN⁻ can bind through S (thiocyanato-S) or N (thiocyanato-N) → linkage isomerism.

ShortcutScan the formula for NO₂⁻, SCN⁻ or CN⁻. If none is present, linkage isomerism is impossible.

[ 17 ]^L NEET styleDonor atom

Jørgensen obtained [Co(NH₃)₅(NO₂)]Cl₂ in a red form and a yellow form. In the red form the nitrite ligand is bonded through:

(a)Nitrogen, written as –NO₂
(b)Oxygen, written as –ONO
(c)Both nitrogen and oxygen simultaneously
(d)Neither; it is a counter ion

Answer — (b) Oxygen, written as –ONO

Given

Two coloured forms of the same pentaamminenitrite complex.

Asked

The donor atom in the red form.

Concept

NO₂⁻ is ambidentate. The O-bonded nitrito form (–ONO) is red; the N-bonded nitro form (–NO₂) is yellow. The change of donor atom changes Δ_o and hence the colour.

Formula

O-bonded → –ONO (nitrito-O, red); N-bonded → –NO₂ (nitrito-N, yellow)

Baby steps

The ligand can donate through its nitrogen lone pair or through an oxygen lone pair.
N is the stronger field donor, giving larger Δ_o and the yellow form.
O is the weaker field donor, giving smaller Δ_o and the red form.
So the red form is O-bonded, written –ONO.

ShortcutRemember the colour pair by the field strength: N-bonded is stronger → larger Δ_o → yellow. O-bonded is weaker → red.

[ 18 ]^L NEET styleAmbidentate

An ambidentate ligand is one which:

(a)Binds to the metal through two donor atoms at the same time
(b)Has two different donor atoms but coordinates through only one at a time
(c)Can bind to two different metal ions simultaneously
(d)Carries two negative charges

Answer — (b) Has two different donor atoms but coordinates through only one at a time

Given

Definition question.

Asked

Meaning of ambidentate.

Concept

An ambidentate ligand offers a choice of donor atom; a didentate ligand uses two donors together and forms a chelate ring. The two terms are frequently confused.

Formula

Ambidentate → linkage isomerism; didentate → chelation

Baby steps

Option A describes a didentate ligand such as en or oxalate.
Option C describes a bridging ligand.
Option D describes charge, which is unrelated to denticity.
Ambidentate = two possible donor atoms, one used at a time → option B.

ShortcutBoth words start with a ‘two’ prefix, so fix the difference by the outcome: ambidentate causes linkage isomerism, didentate causes chelation and chirality.

[ 19 ]^L NEET styleElimination

Which of the following ligands can not give rise to linkage isomerism?

(a)NO₂⁻
(b)SCN⁻
(c)CN⁻
(d)en

Answer — (d) en

Given

Four ligands.

Asked

The one incapable of linkage isomerism.

Concept

Linkage isomerism needs two different donor atoms with only one binding. A ligand whose two donors are identical, or which binds through both simultaneously, cannot produce it.

Formula

—

Baby steps

NO₂⁻: N or O donor → ambidentate.
SCN⁻: S or N donor → ambidentate.
CN⁻: C or N donor → ambidentate (cyanido or isocyanido).
en: two identical N donors that bind together as a chelate → not ambidentate.

Shortcuten, oxalate, glycinate and EDTA are chelating, never ambidentate. Glycinate has N and O donors but uses both at once, so it still forms a ring rather than linkage isomers.

Coordination Isomerism

04 Q

[ 20 ]^C NEET styleRequirement

Coordination isomerism is possible only when:

(a)The complex contains an ambidentate ligand
(b)Both the cation and the anion are complex ions containing different metals
(c)The counter ion can act as a ligand
(d)A solvent molecule is present in the lattice

Answer — (b) Both the cation and the anion are complex ions containing different metals

Given

Conditions for coordination isomerism.

Asked

The necessary requirement.

Concept

Coordination isomerism arises from interchanging the ligand sets between the cationic and anionic complexes. This is only possible if both ions are complex and the two metals are different.

Formula

[M₁L₆][M₂L'₆] ⇌ [M₂L₆][M₁L'₆]

Baby steps

Option A is the condition for linkage isomerism.
Option C is the condition for ionisation isomerism.
Option D is the condition for solvate isomerism.
Two complex ions with different metals → coordination isomerism.

ShortcutVisual cue: two sets of square brackets means coordination isomerism. One bracket plus a bare counter ion means ionisation isomerism.

[ 21 ]^C NEET styleIdentification

The coordination isomer of [Cu(NH₃)₄][PtCl₄] is:

(a)[Pt(NH₃)₄][CuCl₄]
(b)[Cu(NH₃)₄]Cl₂ + PtCl₂
(c)[CuCl₄][Pt(NH₃)₄]Cl₂
(d)[Cu(NH₃)₂Cl₂][Pt(NH₃)₂Cl₂]

Answer — (a) [Pt(NH₃)₄][CuCl₄]

Given

[Cu(NH₃)₄][PtCl₄], with both ions complex.

Asked

Its coordination isomer.

Concept

Swap the complete ligand sets between the two metals while keeping each metal's oxidation state and the overall formula unchanged.

Formula

Total formula must remain identical

Baby steps

Currently: NH₃ is on copper, Cl⁻ is on platinum.
Exchange the ligand sets: NH₃ moves to platinum, Cl⁻ moves to copper.
Both metals stay in +2, so the charges still balance.
Isomer = [Pt(NH₃)₄][CuCl₄].

ShortcutOnly the metals swap places in the formula; the ligand groupings stay intact. Option D changes the groupings, so it is a different compound, not an isomer of this type.

[ 22 ]^C NEET styleClassification

[Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆] are:

(a)Ionisation isomers
(b)Coordination isomers
(c)Linkage isomers
(d)Optical isomers

Answer — (b) Coordination isomers

Given

Two salts, each with a complex cation and a complex anion.

Asked

Type of isomerism.

Concept

Both compounds contain the same total atoms; only the metal to which each ligand set is attached has changed. That is the definition of coordination isomerism.

Formula

—

Baby steps

In the first, NH₃ is bound to Co³⁺ and CN⁻ to Cr³⁺.
In the second, NH₃ is bound to Cr³⁺ and CN⁻ to Co³⁺.
Both ions are complex, and the two metals differ.
Type = coordination isomerism.

ShortcutThis exact pair is the NCERT worked example. If you see cobalt and chromium each in brackets, it is coordination isomerism.

[ 23 ]^C NEET styleElimination

Which of the following pairs does not represent coordination isomers?

(a)[Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆]
(b)[Pt(NH₃)₄][PdCl₄] and [Pd(NH₃)₄][PtCl₄]
(c)[Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br
(d)[Cr(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₄(CN)₂][Cr(NH₃)₂(CN)₄]

Answer — (c) [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br

Given

Four pairs of isomeric compounds.

Asked

The pair that is not coordination isomers.

Concept

Coordination isomerism needs both ions inside square brackets. If one species is a bare counter ion swapping with a ligand, the isomerism is ionisation instead.

Formula

Two brackets → coordination; one bracket + free ion → ionisation

Baby steps

Options A and B each have two complex ions with the ligand sets exchanged → coordination isomers.
Option D also has two complex ions, here with ligands distributed differently between the same two metals → still coordination isomerism.
Option C has only one complex ion; the Br⁻ and SO₄²⁻ exchange between inside and outside.
So option C is ionisation isomerism, not coordination.

ShortcutCoordination isomerism does not require different metals in every case — option D shows the same metal twice. What it always requires is two complex ions.

Ionisation Isomerism

04 Q

[ 24 ]^I NEET styleIdentification

The ionisation isomer of [Co(NH₃)₅(NO₃)]SO₄ is:

(a)[Co(NH₃)₅SO₄]NO₃
(b)[Co(NH₃)₅(ONO₂)]SO₄
(c)[Co(NH₃)₄(NO₃)₂]SO₄
(d)[Co(NH₃)₆](NO₃)(SO₄)

Answer — (a) [Co(NH₃)₅SO₄]NO₃

Given

[Co(NH₃)₅(NO₃)]SO₄, with nitrate inside and sulphate outside.

Asked

Its ionisation isomer.

Concept

Ionisation isomerism arises when the counter ion is itself a potential ligand. The isomer is formed by exchanging the ligand inside the bracket with the counter ion outside.

Formula

[ML₅X]Y ⇌ [ML₅Y]X

Baby steps

Take the coordinated NO₃⁻ out of the sphere.
Bring the SO₄²⁻ counter ion in as a ligand.
Charge check: Co(III) + five neutral NH₃ + SO₄²⁻ = +1, so one NO₃⁻ balances it.
Isomer = [Co(NH₃)₅SO₄]NO₃.

ShortcutOnly the ligand and counter ion trade places. The number of NH₃ and the coordination number must stay unchanged — that eliminates options C and D at once.

[ 25 ]^I NEET styleReagent test

Which of the following gives a white precipitate on adding BaCl₂ solution?

(a)[Co(NH₃)₅SO₄]Br
(b)[Co(NH₃)₅Br]SO₄
(c)Both give a white precipitate
(d)Neither gives a precipitate

Answer — (b) [Co(NH₃)₅Br]SO₄

Given

A pair of ionisation isomers, tested with Ba²⁺.

Asked

Which one precipitates BaSO₄.

Concept

Only ions outside the coordination sphere are free in solution and available to react. A sulphate bonded to the metal is locked in and cannot precipitate.

Formula

Free SO₄²⁻ + Ba²⁺ → BaSO₄ (white)

Baby steps

[Co(NH₃)₅SO₄]Br: sulphate is a ligand, bromide is free → no BaSO₄.
[Co(NH₃)₅Br]SO₄: bromide is a ligand, sulphate is free.
Free SO₄²⁻ reacts with Ba²⁺ → white BaSO₄.
Answer = [Co(NH₃)₅Br]SO₄.

ShortcutComplementary test: the other isomer gives pale yellow AgBr with AgNO₃. Each isomer answers exactly one of the two tests.

[ 26 ]^I NEET styleClassification

[Pt(NH₃)₄Cl₂]Br₂ and [Pt(NH₃)₄Br₂]Cl₂ are examples of:

(a)Coordination isomers
(b)Ionisation isomers
(c)Hydrate isomers
(d)Geometrical isomers

Answer — (b) Ionisation isomers

Given

Two platinum(IV) salts with chloride and bromide exchanged between inside and outside the sphere.

Asked

Type of isomerism.

Concept

One complex ion and a free counter ion, with the two halides trading places, is the standard pattern of ionisation isomerism.

Formula

—

Baby steps

In the first, Cl⁻ is coordinated and Br⁻ is the free counter ion.
In the second, the roles are reversed.
Only one species is inside square brackets, so it is not coordination isomerism.
Type = ionisation isomerism; they give different halide tests with AgNO₃.

ShortcutBoth isomers here give a precipitate with AgNO₃, but of different colours: AgCl is white, AgBr is pale yellow.

[ 27 ]^I NEET styleIon count

How many ions are produced in solution by 1 mol of [Co(NH₃)₅Br]SO₄?

(a)1
(b)2
(c)3
(d)4

Answer — (b) 2

Given

[Co(NH₃)₅Br]SO₄.

Asked

Number of ions on complete dissociation.

Concept

The entire coordination sphere dissociates as a single ion; only the species outside the bracket separate from it.

Formula

[Co(NH₃)₅Br]SO₄ → [Co(NH₃)₅Br]²⁺ + SO₄²⁻

Baby steps

Charge on the complex ion: +3 + 5(0) + (−1) = +2.
One SO₄²⁻ balances it as the counter ion.
Dissociation gives 1 complex cation + 1 sulphate anion.
Total ions = 2 (a 1:1 electrolyte).

ShortcutThe bromide inside the bracket is not counted as a separate ion. Only what lies outside the bracket dissociates.

Solvate (Hydrate) Isomerism

03 Q

[ 28 ]^S NEET styleAgCl count

The number of moles of AgCl precipitated when excess AgNO₃ is added to 1 mol of [Cr(H₂O)₄Cl₂]Cl·2H₂O is:

(a)1
(b)2
(c)3
(d)0

Answer — (a) 1

Given

[Cr(H₂O)₄Cl₂]Cl·2H₂O, the dark green hydrate isomer.

Asked

Moles of AgCl precipitated.

Concept

Only chloride written outside the square bracket is ionisable and precipitates with Ag⁺. Chloride acting as a ligand does not react.

Formula

moles AgCl = number of Cl⁻ outside [ ]

Baby steps

Inside the bracket: 4 H₂O + 2 Cl⁻ = 6 donor atoms, matching CN 6.
Outside the bracket: 1 Cl⁻, plus 2 lattice water molecules.
Only that single free Cl⁻ precipitates.
AgCl formed = 1 mol.

ShortcutThe lattice water written after the dot changes nothing chemically here — it is a spectator. Count only the chloride outside the bracket.

[ 29 ]^S NEET styleCounting

The number of possible hydrate isomers of CrCl₃·6H₂O, given that the coordination number of Cr(III) is 6, is:

(a)2
(b)3
(c)4
(d)6

Answer — (b) 3

Given

CrCl₃·6H₂O with Cr(III) coordination number 6.

Asked

Number of hydrate isomers.

Concept

Hydrate isomers differ in how many chlorides are coordinated and how many water molecules are displaced into the lattice. The coordination number stays fixed at 6 throughout.

Formula

Water inside + chloride inside = 6

Baby steps

0 Cl inside → [Cr(H₂O)₆]Cl₃, violet, gives 3 mol AgCl.
1 Cl inside → [Cr(H₂O)₅Cl]Cl₂·H₂O, grey-green, gives 2 mol AgCl.
2 Cl inside → [Cr(H₂O)₄Cl₂]Cl·2H₂O, dark green, gives 1 mol AgCl.
3 Cl inside would make the complex neutral with no counter ion, and is not among the isolated forms. Commonly quoted answer = 3.

ShortcutEach chloride moving in pushes one water out to the lattice and drops the AgCl yield by one mole. The AgCl count identifies the isomer uniquely.

[ 30 ]^S NEET styleDefinition

Solvate isomers differ from one another in:

(a)The oxidation state of the metal
(b)Whether a solvent molecule is bonded to the metal or present only in the crystal lattice
(c)The donor atom used by an ambidentate ligand
(d)The spatial arrangement of the ligands

Answer — (b) Whether a solvent molecule is bonded to the metal or present only in the crystal lattice

Given

Definition of solvate isomerism.

Asked

The distinguishing feature.

Concept

Solvate isomerism, called hydrate isomerism when the solvent is water, is closely related to ionisation isomerism — the species swapping between inside and outside the coordination sphere is a solvent molecule.

Formula

[M(H₂O)₆]Cl₃ ⇌ [M(H₂O)₅Cl]Cl₂·H₂O

Baby steps

Option A is wrong: the oxidation state is +3 in every hydrate isomer.
Option C describes linkage isomerism.
Option D describes stereoisomerism.
The real difference is whether water is coordinated or in the lattice → option B.

ShortcutSolvate isomerism is structural, not stereo — the bonds genuinely differ, since a coordinated water is bonded to the metal and a lattice water is not.