Coordination Compounds — Class 12 Chemistry practice questions

Question 1

A research chemist isolates two cobalt(III) complexes, Complex X and Complex Y, both having the molecular formula CoBrCl(NH₃)(en)₂ (where en = ethylenediamine). She observes the following:
• When Complex X is treated with excess AgNO₃ solution, a pale yellow precipitate forms immediately.
• When Complex Y is treated with excess AgNO₃ solution, a white precipitate forms immediately.
• Both complexes show a magnetic moment of 0 BM.
• Complex X and Complex Y are non-superimposable mirror images of each other in one of their geometric forms.

On the basis of this information, answer the following:
(a) Identify which halide ion (Br⁻ or Cl⁻) is outside the coordination sphere in Complex X and in Complex Y. Justify your answer using the precipitate observations. (2 marks)
(b) What type of isomerism is shown between Complex X and Complex Y? Name it. (1 mark)
(c) Account for the observed magnetic moment of 0 BM for these cobalt(III) complexes. (1 mark)

Accepted Answer

(a) Identification of halide outside the coordination sphere:

When a complex is treated with AgNO₃, only the ions present OUTSIDE the coordination sphere (i.e., as free ions in solution) precipitate immediately.

• AgNO₃ + Br⁻(aq) → AgBr↓ (pale yellow precipitate)
• AgNO₃ + Cl⁻(aq) → AgCl↓ (white precipitate)

In Complex X: A pale yellow precipitate forms immediately.
∴ Br⁻ is outside the coordination sphere in Complex X.
∴ Complex X is [CoCl(NH₃)(en)₂]Br, i.e., the coordination sphere contains Cl⁻, and Br⁻ is the counter ion.

In Complex Y: A white precipitate forms immediately.
∴ Cl⁻ is outside the coordination sphere in Complex Y.
∴ Complex Y is [CoBr(NH₃)(en)₂]Cl, i.e., the coordination sphere contains Br⁻, and Cl⁻ is the counter ion.

Since X and Y have the same molecular formula but differ in which halide is inside versus outside the coordination sphere, they are ionisation isomers of each other. (1 mark for correct identification of both halides with justification; 1 mark for correct formulation of both complexes)

(b) Type of isomerism:

Complex X and Complex Y have the same molecular formula (CoBrCl(NH₃)(en)₂) but differ in which ligand is within the coordination sphere and which is the counter ion outside it.
∴ The isomerism shown between Complex X and Complex Y is Ionisation Isomerism. (1 mark)

(Note: The observation that one geometric form of these complexes shows non-superimposable mirror images indicates optical isomerism (Δ and Λ enantiomers) within that geometric form, but the relationship BETWEEN X and Y, as defined by the precipitate experiment, is ionisation isomerism.)

(c) Accounting for magnetic moment = 0 BM:

The spin-only magnetic moment is given by:
μ = √n(n+2) BM
where n = number of unpaired electrons.

For μ = 0 BM:
0 = √n(n+2) ∴ n = 0 (no unpaired electrons)

Co is in the +3 oxidation state in these complexes.
Electronic configuration of Co³⁺: [Ar] 3d⁶

The ligands present — en (ethylenediamine) and NH₃ — are STRONG FIELD ligands. They cause a large crystal field splitting energy (Δo) that exceeds the electron pairing energy.
∴ All 6 electrons in the 3d orbitals pair up completely in the lower energy t₂g set:
t₂g⁶ eg⁰

∴ There are 0 unpaired electrons, and the complex is diamagnetic, consistent with μ = 0 BM. (1 mark)

Question 2

A coordination chemist is investigating two complexes isolated from a reaction mixture:

Complex I: [Co(NH₃)₅Cl]SO₄
Complex II: [Co(NH₃)₅(SO₄)]Cl

Both complexes have cobalt in the same oxidation state. The chemist performs the following tests on fresh aqueous solutions of each complex:

• Test A: Aqueous BaCl₂ solution is added.
• Test B: Aqueous AgNO₃ solution is added.

Based on your understanding of coordination compounds, answer the following:

(a) What is the oxidation state of cobalt in each complex? Justify your answer with a calculation for Complex I.  (1 mark)
(b) Identify the primary valence (ionisable) and secondary valence (non-ionisable) groups in Complex I.  (1 mark)
(c) Predict the observation for each complex when Test A (BaCl₂) is performed. Name the type of isomerism shown by these two complexes.  (1 mark)
(d) If the coordination number of cobalt in both complexes is 6, what does the presence of a coordinated SO₄²⁻ in Complex II tell us about the flexibility of SO₄²⁻ as a ligand? Name the mode in which SO₄²⁻ acts here and state how many donor atoms it uses.  (1 mark)

Accepted Answer

(a) Oxidation state of cobalt in Complex I:

Let the oxidation state of Co = x.
NH₃ is neutral; Cl⁻ carries −1; SO₄²⁻ (counter ion) does not affect the inner sphere.
Inside the coordination sphere: x + 5(0) + (−1) = charge on complex cation.
The complex cation is [Co(NH₃)₅Cl]⁺ and the counter ion is SO₄²⁻, giving net overall neutral compound.
∴ x + 0 + (−1) = +1 → x = +3.

∴ Oxidation state of cobalt in both Complex I and Complex II = +3.

(b) In Complex I, [Co(NH₃)₅Cl]SO₄:

• Primary valence (ionisable): SO₄²⁻ — present outside the coordination sphere; ionises in solution.
• Secondary valence (non-ionisable): 5 NH₃ molecules and 1 Cl⁻ ion — present inside the coordination sphere; do not ionise.

(c) Test A — BaCl₂ solution added:

• Complex I ([Co(NH₃)₅Cl]SO₄): SO₄²⁻ is the free counter ion in solution.
  Observation: Immediate white precipitate of BaSO₄ is formed.
  Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)↓ (white)

• Complex II ([Co(NH₃)₅(SO₄)]Cl): SO₄²⁻ is coordinated inside the sphere; no free SO₄²⁻ in solution.
  Observation: No precipitate (or negligible precipitate) with BaCl₂.

∴ These two complexes are ionisation isomers — they have the same molecular formula but differ in the groups present inside and outside the coordination sphere, giving different ions in solution.

(d) In Complex II, SO₄²⁻ is coordinated to cobalt and occupies one coordination position while cobalt has coordination number 6 (five NH₃ + one SO₄²⁻).

∴ SO₄²⁻ acts as a monodentate ligand here, using only ONE donor atom (one oxygen atom) to bond to cobalt.

This shows that SO₄²⁻ is a flexible (ambidentate-type / variable-mode) ligand — it can exist as a free counter ion (ionic/outer-sphere) or coordinate through one oxygen atom as a monodentate ligand, depending on the conditions of complex formation.

Question 3

A chemistry student is studying two complexes used in analytical and industrial chemistry:

Complex P: [Ni(NH₃)₆]²⁺  (Atomic No. of Ni = 28)
Complex Q: [Ni(CN)₄]²⁻

Answer the following sub-parts:
(a) Write the IUPAC name of Complex P.  (1 mark)
(b) Calculate the spin-only magnetic moment of Complex P and state whether it is paramagnetic or diamagnetic.  (2 marks)
(c) The student observes that Complex Q is more stable than a hypothetical complex [Ni(Cl)₄]²⁻. Give one reason for this, based on the nature of the ligand.  (1 mark)

Accepted Answer

(a) IUPAC Name of Complex P: [Ni(NH₃)₆]²⁺

The ligand NH₃ is named 'ammine'. The central metal is nickel in +2 oxidation state (since the complex ion carries a 2+ charge and all NH₃ ligands are neutral). Coordination number = 6.

∴ IUPAC name: Hexaamminenickel(II) ion

(Award 1 mark for the correct complete name.)

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(b) Spin-only magnetic moment of Complex P: [Ni(NH₃)₆]²⁺

Step 1 — Electronic configuration of Ni (Z = 28):
3d⁸ 4s²

In Ni²⁺: 3d⁸ (two electrons removed from 4s)

Step 2 — Nature of ligand and electron pairing:
NH₃ is a weak-to-moderate field ligand. For an octahedral Ni²⁺ complex with NH₃, the d⁸ configuration gives:
t₂g⁶ eg²
→ Number of unpaired electrons, n = 2

Step 3 — Spin-only magnetic moment formula:
μ = √n(n + 2) BM

Substituting n = 2:
μ = √2(2 + 2) = √(2 × 4) = √8

∴ μ = 2√2 ≈ 2.83 BM

Since n = 2 (two unpaired electrons), Complex P is paramagnetic.

∴ Spin-only magnetic moment = 2.83 BM; Complex P is paramagnetic.

(Award 1 mark for correct value of μ with working; 1 mark for correct statement of paramagnetic with reason — unpaired electrons.)

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(c) Stability of Complex Q [Ni(CN)₄]²⁻ vs [Ni(Cl)₄]²⁻:

CN⁻ is a strong-field ligand and forms a much more stable complex than Cl⁻ (a weak-field ligand).

Due to the ability of CN⁻ to act as both a σ-donor and a π-acceptor (back-bonding from filled d-orbitals of Ni into vacant π* orbitals of CN⁻), the Ni–CN bond is significantly stronger and shorter, resulting in greater thermodynamic stability.

∴ [Ni(CN)₄]²⁻ is more stable than [Ni(Cl)₄]²⁻ because CN⁻ is a stronger ligand (strong-field / π-acceptor) compared to Cl⁻.

(Award 1 mark for stating CN⁻ is a stronger / strong-field ligand and linking it to greater stability.)

Question 4

A forensic chemist is analysing two unknown coordination compounds found at a crime scene. Compound X has the formula CoCl₃·4NH₃ and Compound Y has the formula CrCl₃·6NH₃. The chemist performs a conductivity test in aqueous solution and adds excess AgNO₃ solution to each compound.

Based on the above information, answer the following questions:
(a) How many ions does Compound X give in aqueous solution? Write the correct formula of Compound X using coordination sphere notation and identify the charge on the complex ion. (2 marks)
(b) How many moles of AgCl precipitate will form when 1 mole of Compound Y is treated with excess AgNO₃ solution? Give reason. (1 mark)
(c) The forensic chemist also finds a ligand that can bind to the metal through either its nitrogen atom or its oxygen atom. What is such a ligand called? Give one example. (1 mark)

Accepted Answer

(a) Compound X is [Co(NH₃)₄Cl₂]Cl.

In aqueous solution, the compound dissociates as:
[Co(NH₃)₄Cl₂]Cl → [Co(NH₃)₄Cl₂]⁺ + Cl⁻

∴ Compound X gives 2 ions in aqueous solution (one complex cation and one Cl⁻ anion).

The charge on the complex ion [Co(NH₃)₄Cl₂]⁺ is +1.

(Value points: correct formula with coordination sphere notation — 1 mark; correct number of ions and charge on complex ion — 1 mark)

(b) Compound Y is [Cr(NH₃)₆]Cl₃.

In aqueous solution:
[Cr(NH₃)₆]Cl₃ → [Cr(NH₃)₆]³⁺ + 3Cl⁻

All three Cl⁻ ions are outside the coordination sphere (in the outer sphere), so all three are free ions in solution.

∴ 1 mole of Compound Y gives 3 moles of AgCl precipitate when treated with excess AgNO₃.

Reason: Only the ionisable chloride ions present outside the coordination sphere react with AgNO₃ to form AgCl precipitate; chloride ligands inside the coordination sphere do not precipitate.

(c) A ligand that can bind to the central metal atom through two different donor atoms (but coordinates through only one at a time) is called an ambidentate ligand.

Example: NO₂⁻ (nitrito ligand) — can coordinate through nitrogen (–NO₂, nitro) or through oxygen (–ONO, nitrito).

(Alternate accepted example: SCN⁻ / thiocyanate — bonds through S or through N.)

Question 5

A chemist working in a pharmaceutical laboratory prepares two coordination complexes from cobalt(III) salts:

Complex P: [Co(en)₂Cl₂]Br
Complex Q: [Co(NH₃)₄Cl₂]NO₃

On the basis of the above information, answer the following questions:

(a) Identify which of the two complexes, P or Q, contains a chelating ligand. Name the chelating ligand and explain why it is classified as a chelate-forming ligand.  (2 marks)

(b) Write the IUPAC name of Complex Q.  (1 mark)

(c) How many ions are produced when one formula unit of Complex P is dissolved in water? Identify the counter ion.  (1 mark)

Accepted Answer

(a) Complex P contains the chelating ligand.

The chelating ligand is ethane-1,2-diamine (en), i.e., H₂N–CH₂–CH₂–NH₂.

It is classified as a chelate-forming ligand because it is a bidentate ligand — it possesses two donor nitrogen atoms that simultaneously coordinate to the same central metal ion (Co³⁺), forming a five-membered stable ring. A ligand that forms one or more rings with a single central metal atom/ion by donating electron pairs through two or more donor atoms is called a chelating ligand, and the resulting complex is called a chelate complex.

∴ en forms two such five-membered chelate rings with Co³⁺ in Complex P.

(b) IUPAC name of Complex Q: [Co(NH₃)₄Cl₂]NO₃

The complex ion is [Co(NH₃)₄Cl₂]⁺.
Ligands (alphabetical order): tetraammine (NH₃ × 4), dichloro (Cl⁻ × 2).
Oxidation state of Co: let it be x.
x + 4(0) + 2(−1) = +1  ∴  x = +3.

∴ IUPAC name: tetraamminedichloridocobalt(III) nitrate

(c) Complex P: [Co(en)₂Cl₂]Br

The coordination sphere is [Co(en)₂Cl₂]⁺ and the counter ion outside the sphere is Br⁻.

When one formula unit dissolves in water:
[Co(en)₂Cl₂]Br  →  [Co(en)₂Cl₂]⁺  +  Br⁻

∴ Two ions are produced per formula unit.
The counter ion is the bromide ion (Br⁻).

Question 6

Which of the following complex ions has a square planar geometry?

Accepted Answer

(B) [Ni(CN)₄]²⁻

Explanation: CN⁻ is a strong-field ligand. In [Ni(CN)₄]²⁻, Ni is in the +2 oxidation state with configuration 3d⁸. The strong-field CN⁻ ligands cause pairing of the two 3d electrons, giving a dsp² hybridisation, which corresponds to a square planar geometry. In contrast, [Ni(NH₃)₄]²⁺ and [NiCl₄]²⁻ involve weak-field ligands where Ni²⁺ adopts sp³ hybridisation giving a tetrahedral geometry, and [Ni(CO)₄] involves Ni in the zero oxidation state (3d¹⁰), also sp³ tetrahedral.

Question 7

Which of the following ligands is bidentate?

Accepted Answer

(C) Ethylenediamine (en)

Explanation: A bidentate ligand donates two lone pairs to the central metal atom/ion, forming two coordinate bonds simultaneously. Ethylenediamine (H₂N–CH₂–CH₂–NH₂) has two –NH₂ donor atoms and therefore coordinates through both nitrogen atoms, making it bidentate. Cl⁻ and NH₃ are monodentate (one donor atom each), while EDTA⁴⁻ is hexadentate (six donor atoms).

Question 8

(a) What is meant by the term 'coordination number' of a central metal ion in a complex? Give the coordination number of the central metal ion in [Fe(CN)₆]⁴⁻.
(b) Name the type of isomerism shown by the pair: [Cr(NH₃)₄Cl₂]⁺ (violet) and [Cr(NH₃)₄Cl₂]⁺ (green).

Accepted Answer

(a) The coordination number of a central metal ion is the total number of ligand donor atoms directly bonded (coordinated) to it.

In [Fe(CN)₆]⁴⁻, six CN⁻ ligands are coordinated to Fe²⁺, each donating one donor atom.
∴ Coordination number of Fe²⁺ = 6.

(b) Both complexes have the same molecular formula and the same ligands bonded to the same central metal ion, but differ in the spatial arrangement of those ligands around the metal centre.
∴ The type of isomerism exhibited is Geometrical (cis–trans) isomerism — the violet form is the cis isomer (both Cl⁻ on the same side) and the green form is the trans isomer (both Cl⁻ on opposite sides).

[Marking split: (a) definition ½ + coordination number with reason ½ = 1 mark; (b) correct name of isomerism 1 mark]

Question 9

What is meant by 'coordination number' of a central metal ion in a coordination compound? State the coordination number of Co in [Co(en)₂Cl₂]⁺.

Accepted Answer

Coordination number is the total number of ligand donor atoms directly bonded to the central metal ion in a coordination entity. (½ mark)

In [Co(en)₂Cl₂]⁺:
• Each ethylenediamine (en) is a bidentate ligand, contributing 2 donor atoms → 2 × 2 = 4 donor atoms.
• Each Cl⁻ is a monodentate ligand, contributing 1 donor atom → 2 × 1 = 2 donor atoms.

∴ Coordination number of Co = 4 + 2 = 6. (½ mark)

Question 10

Define the following terms with a suitable example in each case:
(i) Coordination isomers
(ii) Chelate effect

Accepted Answer

(i) Coordination Isomers (1 mark)
Coordination isomers are compounds that have the same molecular formula but differ in the distribution of ligands between the cation and the anion of the coordination entity.
Example: [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆] are coordination isomers — in the first, Co³⁺ carries the NH₃ ligands and Cr³⁺ carries CN⁻; in the second, the ligands are interchanged.

(ii) Chelate Effect (1 mark)
The tendency of polydentate ligands (chelating ligands) to form more stable complexes compared to monodentate ligands of comparable donor atoms is called the chelate effect. The increased stability arises due to the gain in entropy when a chelate ring is formed (one polydentate ligand displaces two or more monodentate ligands, increasing the number of free particles in solution).
Example: [Ni(en)₃]²⁺ (en = ethane-1,2-diamine, a bidentate ligand) is significantly more stable than [Ni(NH₃)₆]²⁺, even though both involve six N-donor atoms coordinated to Ni²⁺.

Question 11

(a) Among the following, identify the species that CANNOT act as a ligand and give reason:

CN⁻, BF₃, en, H₂O

(b) Write the IUPAC name of the complex [Pt(NH₃)₂Cl₂].

Accepted Answer

(a) BF₃ cannot act as a ligand.

A ligand must possess at least one lone pair of electrons to donate to the central metal ion. BF₃ is a Lewis acid — it has an empty p-orbital on boron and has no lone pair available for donation to the metal. Hence it acts as an electron-pair acceptor, not a donor, and cannot function as a ligand.                                                        (1 mark)

(b) IUPAC name of [Pt(NH₃)₂Cl₂]:

Diamminedichloridoplatinum(II)                                                                                                        (1 mark)

[Naming logic: two NH₃ ligands → diammine; two Cl⁻ ligands → dichlorido; platinum in +2 oxidation state → platinum(II). Ligands are listed alphabetically: 'a' (ammine) before 'c' (chlorido).]

Question 12

Answer the following sub-parts:
(a) Write the IUPAC name of the complex [Co(NH₃)₄Cl₂]⁺.
(b) Give one reason why [Fe(CN)₆]⁴⁻ is diamagnetic whereas [Fe(H₂O)₆]²⁺ is paramagnetic.
(c) State the type of isomerism shown by the pair:
[Co(NH₃)₅Br]SO₄  and  [Co(NH₃)₅SO₄]Br

Accepted Answer

(a) IUPAC name of [Co(NH₃)₄Cl₂]⁺:
Tetraammminedichloridocobalt(III) ion
(Ligands named alphabetically: ammine before chlorido; oxidation state of Co = +3 as the complex carries a 1+ charge and 2 Cl⁻ ligands are present.)

(b) CN⁻ is a strong-field ligand (high Δ₀) whereas H₂O is a weak-field ligand (low Δ₀).
In [Fe(CN)₆]⁴⁻, Fe is in +2 oxidation state with configuration 3d⁶. Because CN⁻ causes strong splitting, electrons pair up in the t₂g level → t₂g⁶ eg⁰ → 0 unpaired electrons → diamagnetic.
In [Fe(H₂O)₆]²⁺, H₂O causes weak splitting; electrons remain unpaired (high-spin) → t₂g⁴ eg² → 4 unpaired electrons → paramagnetic.

(c) The pair [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br shows ionisation isomerism.
In the first complex, SO₄²⁻ is outside the coordination sphere (gives white precipitate with BaCl₂); in the second, Br⁻ is outside (gives pale-yellow precipitate with AgNO₃). They have the same molecular formula but differ in the ions present inside and outside the coordination sphere.

Question 13

A research chemist isolates two cobalt(III) complexes, Complex P and Complex Q, both with the formula CoBr(NH₃)₅SO₄. When Complex P is treated with AgNO₃ solution, a pale yellow precipitate forms immediately; when treated with BaCl₂ solution, no precipitate forms. When Complex Q is treated with BaCl₂ solution, a white precipitate forms immediately; when treated with AgNO₃ solution, no precipitate forms.

(a) Identify the coordination sphere (formula in square brackets) and the counter ion for both Complex P and Complex Q.  (2 marks)

(b) Name the type of isomerism exhibited by Complex P and Complex Q, and state the basis on which these two complexes differ.  (1 mark)

(c) Write the IUPAC name of Complex P.  (1 mark)

Accepted Answer

(a)

Complex P: AgNO₃ gives a pale yellow precipitate → Br⁻ is outside the coordination sphere (AgBr↓ is pale yellow). BaCl₂ gives no precipitate → SO₄²⁻ is inside the coordination sphere.

∴ Complex P: [Co(NH₃)₅SO₄]Br
   Coordination sphere: [Co(NH₃)₅SO₄]⁺       Counter ion: Br⁻

Complex Q: BaCl₂ gives a white precipitate → SO₄²⁻ is outside the coordination sphere (BaSO₄↓ is white). AgNO₃ gives no precipitate → Br⁻ is inside the coordination sphere.

∴ Complex Q: [Co(NH₃)₅Br]SO₄
   Coordination sphere: [Co(NH₃)₅Br]²⁺       Counter ion: SO₄²⁻

(b)

Complex P and Complex Q exhibit ionisation isomerism.

Basis: The two complexes have the same molecular formula (CoBr(NH₃)₅SO₄) but differ in which anion (Br⁻ or SO₄²⁻) is present inside the coordination sphere and which is the free counter ion outside it — they therefore produce different ions in solution.

(c)

Complex P is [Co(NH₃)₅SO₄]Br.

Ligands present: five ammine (NH₃) ligands and one sulphato (SO₄²⁻) ligand; central metal: Co.

Oxidation state of Co: let x + 0×5 + (−2) = +1 (overall charge of cation)  ∴ x = +3.

IUPAC name of Complex P:

Pentaammine(sulphato)cobalt(III) bromide

Question 14

Define the following terms with a suitable example in each case:
(i) Homoleptic complex
(ii) Ionisation isomerism
(iii) Chelate ligand

Accepted Answer

(i) Homoleptic complex:
A complex in which the central metal ion is coordinated to only one type of donor ligand is called a homoleptic complex.
Example: [Co(NH₃)₆]³⁺ — cobalt(III) is bonded to six identical ammonia ligands.

(ii) Ionisation isomerism:
Isomers that give different ions in solution are called ionisation isomers. They arise when a ligand and a counter ion exchange places — one moves inside the coordination sphere while the other moves outside.
Example: [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br are ionisation isomers. The first gives SO₄²⁻ in solution (confirmed by BaCl₂ giving a white precipitate of BaSO₄), while the second gives Br⁻ in solution (confirmed by AgNO₃ giving a pale yellow precipitate of AgBr).

(iii) Chelate ligand:
A ligand that contains two or more donor atoms and uses them simultaneously to coordinate to the same central metal ion, forming a ring structure, is called a chelate (or chelating) ligand. The resulting ring is called a chelate ring.
Example: Ethane-1,2-diamine (en), H₂N–CH₂–CH₂–NH₂, is a bidentate chelate ligand. In [Co(en)₃]³⁺ each en molecule uses both its –NH₂ nitrogen atoms to bind cobalt, forming three five-membered chelate rings.

Question 15

Define the following terms giving one suitable example each:
(i) Linkage isomerism
(ii) Ionisation isomerism
(iii) Coordination isomerism

Accepted Answer

(i) Linkage Isomerism
This type of isomerism arises in coordination compounds containing ambidentate ligands — ligands that can coordinate to the central metal atom through two different donor atoms.

Example: [Co(NH₃)₅(NO₂)]²⁺ and [Co(NH₃)₅(ONO)]²⁺
In the first complex, NO₂⁻ is coordinated through nitrogen (nitro), and in the second, through oxygen (nitrito). These two are linkage isomers of each other.

(1 mark)

(ii) Ionisation Isomerism
This type of isomerism arises when the counter ion in a complex salt is itself a potential ligand, and it can exchange place with a ligand present inside the coordination sphere, giving different ions upon ionisation in solution.

Example: [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br
The first complex gives SO₄²⁻ ions in solution (confirmed by BaCl₂ test), while the second gives Br⁻ ions (confirmed by AgNO₃ test).

(1 mark)

(iii) Coordination Isomerism
This type of isomerism arises in coordination compounds containing both a complex cation and a complex anion, where the distribution of ligands between the two coordination spheres differs between the isomers.

Example: [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆]
In the first, NH₃ ligands are around Co³⁺ and CN⁻ ligands are around Cr³⁺; in the second, the distribution is reversed.

(1 mark)

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