Q.Explain the bonding in coordination compounds in terms of Werner's postulates.
Concept understanding — Werner Coordination Theory
Werner Coordination Theory: The Idea That Changed Inorganic Chemistry
Imagine you're looking at a salt like cobalt(III) chloride. The formula is written as CoClX3, and when you dissolve it in water, you expect to find CoX3+ and ClX− ions. But something strange happens: when you add silver nitrate (which precipitates chloride ions), only some of the chlorine comes out as silver chloride. Not all of it. And the amount that precipitates depends on how you made the compound.
This was the puzzle that faced chemists in the late 1800s. Compounds like CoClX3⋅6NHX3 (orange-yellow) and CoClX3⋅5NHX3 (purple) had the same metal and the same ligands (ammonia), but different colours, different conductivities in solution, and different numbers of chloride ions that could be precipitated. The old ideas of fixed valency couldn't explain it.
Alfred Werner proposed a radical solution in 1893. He said: a metal ion has two kinds of valency.
The Core Intuition
Think of a metal ion like a king in a castle. The king has two types of relationships:
-
Primary valency (today: oxidation state) — this is the king's royal authority. It's fixed, non-directional, and satisfied by negative ions. For cobalt(III), this is +3. It's like the king's crown: it doesn't change.
-
Secondary valency (today: coordination number) — this is the king's personal bodyguard. The king can have a fixed number of guards (usually 4 or 6) who stand in specific positions around him. These guards can be neutral molecules (like ammonia) or negative ions (like chloride). The key: these guards are directly attached to the metal, forming a stable cluster called the coordination sphere.
The revolutionary idea: the chloride ions that act as bodyguards (inside the coordination sphere) do not behave like free ions. They don't precipitate with silver nitrate. They don't conduct electricity. They are "locked" to the metal.
The Precise Statement
Werner Coordination Theory (1893)
-
Every metal atom has two types of valency:
- Primary valency (ionisable): corresponds to the oxidation state. It is satisfied by negative ions. These ions are outside the coordination sphere and behave as free ions in solution.
- Secondary valency (non-ionisable): corresponds to the coordination number. It is satisfied by neutral molecules or negative ions directly bonded to the metal. These are inside the coordination sphere and do not dissociate.
-
The secondary valencies are directional — they point to fixed positions in space around the metal, giving the complex a definite geometry (e.g., octahedral for coordination number 6, square planar for 4).
-
The primary valency is non-directional — it is just a number, not a spatial arrangement.
How It Explains the Puzzle
Take the compound CoClX3⋅6NHX3 (orange-yellow). Werner said:
- Cobalt has primary valency +3 (needs three negative charges to satisfy it).
- Cobalt has secondary valency 6 (can hold six ligands around it).
- The six ammonia molecules satisfy all six secondary valencies. So the chloride ions cannot be inside the coordination sphere — they must be outside, as free ions.
- Structure: [Co(NHX3)X6]ClX3. All three chlorides precipitate with AgNOX3.
Now take CoClX3⋅5NHX3 (purple):
- Again, primary valency +3, secondary valency 6.
- Five ammonia molecules satisfy five secondary valencies. One chloride ion must fill the sixth spot — it becomes a ligand inside the sphere.
- The other two chlorides are outside as free ions.
- Structure: [Co(NHX3)X5Cl]ClX2. Only two chlorides precipitate.
The number of free ions in solution determines the conductivity and the number of precipitable chlorides. Werner's theory predicted exactly these numbers — and experiments confirmed them.
The Geometry Insight
Werner didn't just count ligands — he placed them in space. For coordination number 6, he proposed an octahedral arrangement (ligands at the six corners of an octahedron). This explained why [Co(NHX3)X4ClX2]+ exists as two different compounds (isomers): one where the two chlorides are next to each other (cis) and one where they are opposite (trans). No other geometry could produce exactly two isomers.
Werner's theory was the first to show that complexes have definite three-dimensional structures. This was decades before X-ray crystallography could confirm it directly.
What It Replaced
Before Werner, chemists thought bonding was simple: each atom had a fixed valency (like carbon always forms four bonds). They tried to write chain structures for coordination compounds (like organic molecules), but it failed — you couldn't explain why CoClX3⋅6NHX3 and CoClX3⋅5NHX3 were different compounds with the same metal and ligands.
Werner's key break: the metal can bond to more species than its oxidation state would suggest, and those bonds are not all the same type.
The Legacy
Werner won the Nobel Prize in 1913. His theory:
- Introduced the concept of coordination number and coordination sphere.
- Explained isomerism in complexes (geometric, optical).
- Laid the foundation for modern coordination chemistry, crystal field theory, and ligand field theory.
- Showed that inorganic compounds could have complex, predictable geometries — not just simple salts.
When you see a formula like [Co(NHX3)X6]ClX3, read the square brackets as "the castle walls". Everything inside is tightly bound to the metal; everything outside is free. That's Werner's idea in a nutshell.
Werner's coordination theory is the historical and conceptual foundation of the NCERT/CBSE Class 12 Chemistry chapter on Coordination Compounds, and ‘Werner's theory of coordination compounds’ is one of the most frequently asked important questions in board exams, JEE Main and NEET. Understanding primary and secondary valency as Werner defined them is essential groundwork for every other topic in this chapter.
Why this formula?
Werner Coordination Theory: Why the Key Formulas Hold
Werner Coordination Theory (1893) revolutionized inorganic chemistry by explaining how metal ions bind ligands. Let's build the reasoning from first principles — not just memorize formulas.
1. The Core Observation: Primary vs. Secondary Valence
Werner noticed that metal compounds had two types of bonding capacity:
- Primary valence (now oxidation state): Satisfies the metal's charge — ionic in nature.
- Secondary valence (now coordination number): Determines how many ligands attach — directional, spatial in nature.
Why this distinction?
Consider CoClX3 ⋅6NHX3 (one of Werner's classic compounds).
- The compound is electrically neutral overall.
- Adding AgNOX3 precipitates all 3 Cl⁻ as AgCl — meaning all chlorides are free ions.
- Therefore, the NHX3 molecules must be directly bonded to Co, not the chlorides.
This forces the idea: Co has a fixed capacity for direct ligand attachment (secondary valence = 6 here), separate from its charge balance (primary valence = +3).
2. The Key Formula: Coordination Number = Number of Ligands Attached
Formula:
Coordination number=number of donor atoms directly bonded to the metal
Why this holds:
- Werner's experiments showed that only a fixed number of ligands could be replaced without breaking the compound's identity.
- For CoClX3 ⋅6NHX3, adding acid doesn't remove NHX3 easily — they are coordinated.
- The maximum number of such tightly bound ligands is the coordination number — a property of the metal ion, not the counterions.
Derivation from data:
If you have [Co(NHX3)X6]ClX3, conductivity measurements show 4 ions in solution ([Co(NHX3)X6]X3+ + 3 Cl⁻).
If you had [Co(NHX3)X5Cl]ClX2, conductivity shows 3 ions.
The number of chlorides inside the coordination sphere (non-precipitable) plus those outside must sum to the total chlorides. This gives the coordination number directly.
3. The Geometry Formula: Coordination Number Determines Shape
Werner proposed that secondary valences are directed in space — leading to specific geometries.
| Coordination Number | Geometry | Why? |
|---|---|---|
| 2 | Linear | Minimizes repulsion between 2 ligands |
| 4 | Tetrahedral or Square planar | 4 points in space — two arrangements possible |
| 6 | Octahedral | 6 ligands at 90° angles — most symmetric |
Why octahedral for 6?
- 6 ligands around a central atom must be placed to maximize separation.
- The octahedron (6 vertices, all equidistant from center, 90° between adjacent bonds) is the only regular polyhedron with 6 vertices.
- This explains why [Co(NHX3)X6]X3+ is octahedral — no other arrangement gives equal bond angles and distances.
4. The Isomer Counting Formula: Why 2n or n! Appears
Werner used isomer counts to confirm geometry. For an octahedral complex [MaX2bX2cX2]:
Number of geometrical isomers = 5 (not 6, not 4)
Why this formula?
- Place the two 'a' ligands: they can be cis (90°) or trans (180°).
- For each, place 'b' and 'c' in remaining positions — but symmetry reduces duplicates.
- The count comes from combinatorial reasoning on a fixed octahedral framework, not arbitrary permutation.
General principle:
Number of isomers=symmetry factortotal arrangements
This is not a simple n! — it depends on the point group symmetry of the complex.
5. The Valence Sum Rule: Primary + Secondary = Constant?
Not a fixed sum!
Werner's theory does not say primary + secondary valence = constant.
Example:
- CoX3+ has primary valence = 3, secondary = 6 → sum = 9
- PtX4+ has primary = 4, secondary = 6 → sum = 10
Why no fixed sum?
- Primary valence depends on the metal's oxidation state (variable).
- Secondary valence depends on the metal's size and electronic configuration (also variable).
- They are independent properties — the only link is that both must be satisfied for a stable complex.
Summary: The Core Insight
Werner's formulas hold because:
- Coordination number is an experimentally determined maximum — not a theoretical guess.
- Geometry follows from minimizing ligand-ligand repulsion on a sphere.
- Isomer counts follow from symmetry constraints on a fixed polyhedron.
- Primary and secondary valences are independent — no single formula links them.
The real power of Werner's theory: it turned coordination chemistry from a list of random compounds into a predictive, spatial science — long before X-ray crystallography confirmed the geometries.
Werner’s Coordination Theory was the first successful model to explain bonding in coordination compounds. It proposed that metal ions have two types of valency: primary valency (ionisable, corresponding to oxidation state) and secondary valency (non-ionisable, corresponding to coordination number). The secondary valencies are directed in space around the metal, giving a fixed geometry.
Reasoning steps:
- Primary valency is satisfied by negative ions (e.g., Cl⁻ in [Co(NHX3)X6]ClX3), and these ions are ionisable — they precipitate with Ag⁺.
- Secondary valency is satisfied by neutral molecules or anions (e.g., NH₃ in the same complex), and these are non-ionisable — they remain bound to the metal even in solution.
- The number of secondary valencies (coordination number) is fixed for a given metal, and they are arranged in a definite stereochemistry (e.g., octahedral for Co³⁺, square planar for Pt²⁺).
Werner’s postulates explain bonding by distinguishing primary (ionisable, corresponding to oxidation state) and secondary (non-ionisable, satisfied by ligands) valencies, with the secondary valencies directed in space to give a fixed geometry and stoichiometry.
Werner’s coordination theory explains bonding in coordination compounds by proposing that metal ions have two types of valencies — primary (ionisable) and secondary (non-ionisable) — and that ligands occupy fixed positions in space around the metal, giving a definite geometry.
Werner’s theory was revolutionary because it moved beyond simple ionic or covalent bonding ideas. Before Werner, chemists struggled to explain why compounds like CoClX3⋅6NHX3 (which we now call [Co(NHX3)X6]ClX3) did not behave like a simple mixture of CoClX3 and NHX3. Werner proposed that the metal ion has two distinct kinds of bonding capacity.
Primary valency corresponds to the oxidation state of the metal — it is satisfied by negative ions and is non-directional. Secondary valency corresponds to the coordination number — it is satisfied by neutral molecules or negative ions (ligands) and is directional, pointing to fixed positions in space around the metal. The secondary valencies give the compound its geometry.
Let’s see how this applies step by step.
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Identify the central metal and its primary valency.
In [Co(NHX3)X6]ClX3, the central atom is cobalt. The primary valency of Co is 3 (since three ClX− ions are needed to neutralise the charge). This is the oxidation state of Co: +3.
-
Determine the secondary valency (coordination number).
Six NHX3 molecules are directly attached to Co — these satisfy the secondary valency. So the coordination number is 6. Werner said secondary valencies are always satisfied by ligands, and they are fixed in number for a given metal ion.
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Assign the geometry based on secondary valencies.
For coordination number 6, Werner correctly predicted an octahedral arrangement. The six ligands occupy the six corners of an octahedron around the metal. This explained why [Co(NHX3)X6]ClX3 does not show isomerism due to different ligand positions — all six positions are equivalent.
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Distinguish between ionisable and non-ionisable groups.
The three ClX− ions satisfy the primary valency and are ionisable — they precipitate as AgCl when treated with AgNOX3. The six NHX3 molecules satisfy secondary valencies and are non-ionisable — they do not precipitate. This matched experimental conductivity and precipitation data perfectly.
-
Explain the bonding in other compounds using the same logic.
For example, [Co(NHX3)X5Cl]ClX2:
- Primary valency of Co = 3 (two ClX− ions outside + one ClX− inside).
- Secondary valency = 6 (five NHX3 + one Cl).
- Only two ClX− are ionisable (precipitate with AgNOX3), confirming the third Cl is bonded directly to Co via secondary valency.
A common mistake is to think that primary valency equals the number of ligands. It does not — primary valency is the oxidation state, while secondary valency is the coordination number. They are independent.
Werner’s theory is essentially the first successful model of coordination compounds. It correctly predicted the existence of isomers (like geometrical isomers in [Co(NHX3)X4ClX2]X+) long before X-ray crystallography confirmed them.
Werner’s postulates state that metal ions possess primary (ionisable, non-directional) and secondary (non-ionisable, directional) valencies, and that secondary valencies determine the geometry — for example, in [Co(NHX3)X6]ClX3, Co has primary valency 3 and secondary valency 6, giving an octahedral structure.
Werner Coordination Theory — Bonding Explanation
Method: Werner's Postulate Approach
This method explains bonding in coordination compounds using the primary valency and secondary valency concepts proposed by Alfred Werner in 1893.
Step 1 — Identify the Central Metal Atom
- The metal atom (usually a transition metal) acts as the central atom.
- Example: In [Co(NHX3)X6]ClX3, the central atom is cobalt (Co).
Step 2 — Assign Primary Valency (Ionisable Valency)
- Primary valency corresponds to the oxidation state of the metal.
- It is satisfied by negative ions (anions) and is non-directional.
- It is written outside the coordination sphere (square brackets).
Example:
In [Co(NHX3)X6]ClX3, Co has primary valency = +3 (since three ClX− ions are outside).
Step 3 — Assign Secondary Valency (Coordination Number)
- Secondary valency corresponds to the coordination number of the metal.
- It is satisfied by neutral molecules or negative ions (ligands) inside the coordination sphere.
- It is directional and determines the geometry of the complex.
Example:
In [Co(NHX3)X6]ClX3, Co has secondary valency = 6 (six NHX3 ligands).
Step 4 — Determine Geometry from Secondary Valency
- Secondary valency fixes the spatial arrangement of ligands around the metal.
| Coordination Number | Geometry |
|---|---|
| 2 | Linear |
| 4 | Tetrahedral or Square planar |
| 6 | Octahedral |
Example:
[Co(NHX3)X6]X3+ has octahedral geometry (secondary valency = 6).
Step 5 — Distinguish Between Ionisable and Non-Ionisable Groups
- Primary valency groups are outside the bracket — they are ionisable (precipitate with suitable reagents).
- Secondary valency groups are inside the bracket — they are non-ionisable (do not precipitate).
Example:
[Co(NHX3)X6]ClX3 gives 3 moles of AgCl with AgNOX3 (all three ClX− are ionisable).
[Co(NHX3)X5Cl]ClX2 gives only 2 moles of AgCl (one ClX− is inside the sphere, non-ionisable).
Step 6 — Summarise Bonding in Terms of Postulates
| Werner's Postulate | Explanation |
|---|---|
| 1. Every metal has two types of valencies | Primary (oxidation state) and secondary (coordination number) |
| 2. Secondary valencies are directional | They determine geometry (e.g., octahedral, tetrahedral) |
| 3. Primary valencies are satisfied by anions | They are ionisable and written outside the coordination sphere |
| 4. Secondary valencies are satisfied by ligands | They are non-ionisable and written inside the coordination sphere |
Final Key Takeaway
Werner's theory explains bonding by separating the metal's oxidation state (primary valency) from its coordination number (secondary valency), with the latter dictating the complex's shape and the former determining its charge and ionisable groups.
This method is concept-first: understand why the complex has a certain formula and geometry, then apply to any given coordination compound.
Common Mistakes in Werner's Coordination Theory (and How to Avoid Them)
Werner's theory is the foundation of coordination chemistry, but students often slip on a few key points. Here are the most frequent errors and how to fix them.
Mistake 1: Confusing Primary Valency with Secondary Valency
The error: Students think primary valency is the total charge on the complex, or that secondary valency is the oxidation state.
The truth:
- Primary valency = oxidation state of the central metal ion (ionizable, satisfied by anions)
- Secondary valency = coordination number (non-ionizable, satisfied by ligands, directional)
How to avoid: Memorise the distinction with a simple example:
- In [Co(NHX3)X6]ClX3, primary valency of Co = +3 (satisfied by 3 Cl⁻ ions), secondary valency = 6 (satisfied by 6 NH₃ molecules).
Mistake 2: Forgetting That Secondary Valency Is Fixed and Directional
The error: Students treat secondary valency as variable or non-geometric.
The truth: Werner proposed that secondary valencies are fixed in number for a given metal and point to fixed positions in space — this is the origin of stereochemistry (octahedral, square planar, tetrahedral).
How to avoid: Always draw the geometry when explaining. For example, [Co(NHX3)X6]X3+ is octahedral — all six positions are equivalent.
Mistake 3: Mixing Up Ionizable vs. Non-ionizable Groups
The error: Students think all anions satisfy primary valency, or that all neutral molecules satisfy secondary valency.
The truth:
- Primary valency is satisfied by anions (Cl⁻, SO₄²⁻, etc.) — these are ionizable and precipitate with Ag⁺, Ba²⁺, etc.
- Secondary valency can be satisfied by neutral molecules (NH₃, H₂O) or anions (Cl⁻, CN⁻) — these are non-ionizable and do not precipitate.
Example: In [Co(NHX3)X5Cl]ClX2:
- One Cl⁻ satisfies secondary valency (inside coordination sphere) — does not precipitate with Ag⁺
- Two Cl⁻ satisfy primary valency (outside sphere) — precipitate with Ag⁺
How to avoid: Practise writing the complex formula with square brackets — everything inside is secondary valency, everything outside is primary.
Mistake 4: Thinking Werner Explained All Bonding (Covalent/Electrostatic)
The error: Students believe Werner's theory describes the nature of the metal-ligand bond.
The truth: Werner's theory is purely structural — it explains how many and where ligands attach, but not why (that came later with VBT, CFT, MOT).
How to avoid: State clearly: "Werner's postulates describe the number and spatial arrangement of ligands, not the electronic structure of the bond."
Mistake 5: Ignoring the Existence of Isomers
The error: Students fail to connect secondary valency directionality to isomerism.
The truth: Because secondary valencies have fixed positions, complexes can show geometrical isomerism (e.g., cis/trans in [Co(NHX3)X4ClX2]X+) and optical isomerism.
How to avoid: When explaining Werner's postulates, always mention that the fixed spatial arrangement predicts isomerism — this was a major triumph of his theory.
Mistake 6: Using the Wrong Terminology in Exams
The error: Students write "primary valency = ionic bond" or "secondary valency = covalent bond."
The truth: Werner did not use the terms ionic/covalent. He said:
- Primary valency = ionizable (satisfied by anions)
- Secondary valency = non-ionizable (satisfied by ligands, directional)
How to avoid: Use Werner's own language: "ionizable" and "non-ionizable" or "satisfied by anions" and "satisfied by ligands."
Quick Revision Checklist
| Concept | Common Mistake | Correct Understanding |
|---|---|---|
| Primary valency | = charge on complex | = oxidation state of metal |
| Secondary valency | = variable | = fixed coordination number |
| Ionizable groups | All anions are ionizable | Only those outside coordination sphere |
| Bond nature | Werner explained covalent bonds | Werner explained structure, not bond type |
| Isomerism | Not linked to theory | Direct consequence of fixed geometry |
Final tip: When answering an exam question on Werner's postulates, always:
- State the two types of valency clearly.
- Give a concrete example with a formula.
- Mention that secondary valencies have fixed spatial positions (explaining isomerism).
- Prefer Werner's own terms — "ionisable"/"non-ionisable" — rather than flatly labelling the valencies as "ionic bonds" or "covalent bonds"; equating a valency with a bond type is the classic slip.
Showing the 12 most recent of 17 on this concept.
- KCET 2026Set D31 markMCQQ.How many ions per molecule are produced from the complex [Co(NH3)6]Cl3 in solution? (A) 6 (B) 4 (C) 3 (D) 2
›Reveal solutionSolution
The number of ions per formula unit equals the complex ion plus the free counter-ions written outside the coordination sphere.
Step 1 — Identify the coordination sphere
In [Co(NH3)6]Cl3, all six NH3 ligands are bound directly to cobalt inside the square brackets, so they do not ionize; only what is written outside the brackets ionizes in solution.
Step 2 — Write the dissociation
[Co(NH3)6]Cl3→[Co(NH3)6]3++3Cl−
Step 3 — Count the total ions
This gives 1 complex cation, [Co(NH3)6]3+, plus 3 chloride ions, Cl−, totaling 4 ions per formula unit.
✓Final answerThe correct option is (B) — 4.
- KCET 2025Set D-41 markMCQQ.A ligand which has two different donor atoms and either of the two ligates with the central metal atom/ion in the complex is called (A) Chelate ligand (B) Unidentate ligand (C) Polydentate ligand (D) Ambidentate ligand
›Reveal solutionSolution
The defining phrase is "two different donor atoms, either of the two ligates" — one binding site used at a time, chosen from two candidates — which is exactly the definition of an ambidentate ligand.
Step 1 — Parse the definition given in the stem
The stem specifies three things:
- the ligand has two different donor atoms;
- either one of them can bond to the metal;
- (implicitly) only one at a time actually coordinates — hence "either", not "both".
So the ligand occupies only one coordination site, but it has a choice of which of its own atoms to use.
Step 2 — Run through the four terms and see which fits
(A) Chelate ligand — a di- or polydentate ligand that grips the metal through two or more donor atoms simultaneously, forming a ring (the "claw"). E.g. ethylenediamine (en) binds through both N atoms at once; oxalate through both O's. ✗ — this binds through both, not either.
(B) Unidentate (monodentate) ligand — donates through exactly one donor atom, and it has only that one available. E.g. NHX3 (N only), ClX−, HX2O. ✗ — it uses one site, yes, but there is no choice of two different donor atoms. That is precisely the feature the stem adds.
(C) Polydentate ligand — binds through several donor atoms at once. E.g. EDTA⁴⁻ is hexadentate (2 N + 4 O, all six coordinating). ✗ — again, all sites used together.
(D) Ambidentate ligand — from Latin ambi = "both / either of two". It has two different donor atoms but coordinates through only one of them at a time. ✓ This is exactly the stem.
Step 3 — The standard examples (worth knowing)
Ligand Donor used Name of complex prefix Example NOX2X− N nitro (−NOX2) [Co(NHX3)X5(NOX2)]2+ NOX2X− O nitrito (−ONO) [Co(NHX3)X5(ONO)]2+ SCNX− S thiocyanato (−SCN) [Fe(SCN)]2+ SCNX− N isothiocyanato (−NCS) [Co(NCS)X4]2− CNX− C or N cyano / isocyano — Why it matters: because the same ligand can attach through different atoms, ambidentate ligands give rise to linkage isomerism — two compounds with identical formulae but different metal–ligand connectivity (and often strikingly different colours, e.g. the yellow nitro vs. red nitrito pentaamminecobalt(III) ions).
Step 4 — The distinguishing test
Denticity counts how many donor atoms are attached at once. Ambidentate ligands are unidentate in practice (denticity 1) — their special feature is not how many sites they fill but which of their own atoms they choose to fill it with.
✓Final answerThe correct option is (D) — Ambidentate ligand.
ANSWER: D
- KCET 2025Set D-41 markMCQQ.In the complex ion [Fe(C2O4)3]3−, the co-ordination number of Fe is (A) 4 (B) 5 (C) 6 (D) 3
›Reveal solutionSolution
Coordination number counts donor atoms, not ligand molecules — oxalate is bidentate, so three oxalates supply 3×2=6 donor atoms.
Step 1 — The concept: coordination number ≠ number of ligands
The coordination number (CN) of the central metal is the number of donor atoms (i.e. the number of sigma bonds / ligand bonds) directly attached to it — not the number of ligand molecules or ions.
These two counts agree only when every ligand is unidentate. Whenever a chelating ligand is present, you must multiply by its denticity:
CN=∑ligands(number of that ligand)×(its denticity)
Step 2 — What is the denticity of oxalate?
The oxalate ion, CX2OX4X2− (abbreviated ox), is the conjugate base of oxalic acid:
X−X22−OX2C−COX2X−
It carries two carboxylate groups, and one oxygen from each group coordinates to the metal. Both donor atoms bind simultaneously, wrapping around the metal to close a stable five-membered chelate ring (M–O–C–C–O).
∴ oxalate is BIDENTATE(denticity=2)
This is why oxalate is a chelating ligand and why [Fe(CX2OX4)X3]3− is unusually stable (the chelate effect).
Step 3 — Count the donor atoms
The complex is [Fe(CX2OX4)X3]3− — three oxalate ligands, each bidentate:
CN=3ligands×2liganddonor atoms=6
All six donor atoms are oxygens, arranged octahedrally about the iron.
Step 4 — The trap
The distractor (D) 3 is exactly what you get by miscounting ligands instead of donor atoms — the single most common error on this topic. Remember: the subscript
3outside the bracket tells you how many oxalate ions there are, not the coordination number.Step 5 — Cross-check with the oxidation state (consistency)
Let the oxidation state of Fe be y. Each oxalate is −2, total charge −3:
y+3(−2)=−3⇒y=+3
So the complex is the tris(oxalato)ferrate(III) ion, FeX3+ (d5). Iron(III) characteristically forms six-coordinate octahedral complexes ([Fe(CN)X6]3−, [Fe(HX2O)X6]3+), which confirms CN = 6. ✓
✓Final answerThe correct option is (C) — 6.
ANSWER: C
- COMEDK 2025Set 2025-M1 markMCQQ.An aqueous solution of CrCl3.6H2O (Molar mass =266.5 g/mol ) containing 2.665 g of the solute after processing, when treated with excess of AgNO3 gave 2.87 g of AgCl (Molar mass of AgCl=143.5 g/mol ) Choose the correct formula of the compound which give these results. (A) [Cr(H2O)4Cl2]Cl⋅2H2O (B) [Cr(H2O)3Cl3]⋅3H2O (C) [Cr(H2O)5Cl]Cl2⋅H2O (D) [Cr(H2O)6]Cl3
›Reveal solutionSolution
The AgCl formed shows 2 free chloride ions per formula unit, which fixes the structure as [Cr(H2O)5Cl]Cl2⋅H2O — option (C).
Concept
In a coordination compound, chloride ions bound to the metal (written inside the square brackets) stay put, while chloride ions outside the coordination sphere are free ions that precipitate with AgNO3 as AgCl. So counting the AgCl tells us the number of ionizable chlorides.
Solution
- Moles of complex: 266.52.665=0.0100mol.
- Moles of AgCl: 143.52.87=0.0200mol.
- Free chlorides per formula unit: 0.01000.0200=2.
- Test the options for the number of chlorides outside the bracket:
- (A) [Cr(H2O)4Cl2]Cl⋅2H2O: 1 free Cl.
- (B) [Cr(H2O)3Cl3]⋅3H2O: 0 free Cl.
- (C) [Cr(H2O)5Cl]Cl2⋅H2O: 2 free Cl.
- (D) [Cr(H2O)6]Cl3: 3 free Cl.
Only option (C) releases exactly two chloride ions, matching the experiment.
Watch outWater of crystallisation outside the bracket does not add chloride ions, and chlorides written inside the bracket are coordinated and do not precipitate — count only the chlorides outside the sphere.
✓Final answerThe correct option is (C), [Cr(H2O)5Cl]Cl2⋅H2O.
ANSWER: C
- KCET 2024Set B-21 markMCQQ.On treating 100 mL of 0.1 M aqueous solution of the complex CrCl3.6H2O with excess of AgNO3, 2.86 g of AgCl was obtained. The complex is : (A) [Cr(H2O)3Cl3].3H2O (B) [Cr(H2O)4Cl2]Cl.2H2O (C) [Cr(H2O)5Cl]Cl2.H2O (D) [Cr(H2O)6]Cl3
›Reveal solutionSolution
The number of moles of AgCl precipitated tells us how many chloride ions are free (outside the coordination sphere). From the given data, 0.02 mol of AgCl forms from 0.01 mol of complex, meaning 2 Cl⁻ are free — so the complex has one Cl inside and two outside. The correct option is (C).
The key idea is that only chloride ions that are outside the coordination sphere — the counter-ions — will react with AgNO₃ to give AgCl. Chloride ions that are coordinated to the metal (inside the square brackets) do not precipitate. So the mass of AgCl tells us exactly how many free Cl⁻ ions each formula unit releases.
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Find moles of AgCl formed.
Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol.
Moles of AgCl = 143.52.86≈0.02 mol.
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Find moles of the complex taken.
Volume = 100 mL = 0.1 L, concentration = 0.1 M.
Moles of complex = 0.1×0.1=0.01 mol.
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Find the ratio of free Cl⁻ per complex molecule.
moles complexmoles AgCl=0.010.02=2.
So each formula unit of the complex gives 2 chloride ions that precipitate with Ag⁺.
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Interpret the ratio.
The complex has the formula CrCl3⋅6H2O, so there are 3 Cl atoms total. If 2 are free (outside the coordination sphere), then only 1 Cl is inside the coordination sphere (bonded to Cr). The water molecules also distribute between inside and outside.
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Match with the options.
- (A) [Cr(H2O)3Cl3]⋅3H2O — all 3 Cl are inside → 0 free Cl⁻. Wrong.
- (B) [Cr(H2O)4Cl2]Cl⋅2H2O — 1 free Cl⁻. Wrong.
- (C) [Cr(H2O)5Cl]Cl2⋅H2O — 2 free Cl⁻. Correct.
- (D) [Cr(H2O)6]Cl3 — 3 free Cl⁻. Wrong.
Watch outA common mistake is to forget that water of crystallisation (outside the brackets) does not affect the precipitation — only the free chloride ions matter. Also, don’t confuse the total number of Cl atoms with the number that are free.
TipYou can shortcut this: moles of AgCl = 0.02, moles of complex = 0.01 → ratio 2 → look for the option with exactly two Cl outside the coordination sphere. That’s option (C) instantly.
✓Final answerThe correct option is (C): [Cr(H2O)5Cl]Cl2.H2O.
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- COMEDK 2024Set 2024-A1 markMCQQ.The molar conductivity of the complex CoCl3⋅4NH3⋅2H2O is found to be the same as that of a 1:3 electrolyte. The structural formula of the compound is : (A) [Co(NH3)4(H2O)2]Cl3 (B) [Co(NH3)4Cl(H2O)]Cl2H2O (C) [Co(NH3)4(H2O)Cl2]ClH2O (D) [Co(NH3)4(H2O)2Cl]Cl2
›Reveal solutionSolution
Behaving as a 1:3 electrolyte (4 ions) requires all three Cl− outside the coordination sphere: [Co(NH3)4(H2O)2]Cl3.
A 1:3 electrolyte dissociates into 4 ions total: one complex cation and three anions. So all three chloride ions must be outside the coordination sphere (ionisable), and both water molecules coordinate to cobalt. The complex cation is
[Co(NH3)4(H2O)2]3+
(giving Co its +3 oxidation state, coordination number 6), balanced by 3Cl−:
[Co(NH3)4(H2O)2]Cl3⟶[Co(NH3)4(H2O)2]3++3Cl−.
✓Final answerThe correct option is (A) — [Co(NH3)4(H2O)2]Cl3
- COMEDK 2024Set 2024-M1 markMCQQ.A Coordination compound is represented by the formula [CoBr3(en)x]. This compound required one mole of AgNO3 to form a pale yellow precipitate of AgBr. What is the value of x in the compound? (A) 2 (B) 1 (C) 4 (D) 3
›Reveal solutionSolution
The key is that one mole of AgNO₃ precipitates one mole of free Br⁻ ions; the complex has only one ionic bromide, so the coordination sphere must contain the other two bromides and the ethylenediamine (en) ligands, giving x = 2.
Concept & Intuition
Coordination compounds can have some ligands tightly bound inside the coordination sphere (non‑ionic) and others outside as counter‑ions (ionic). When you add AgNO₃, only the free, ionic bromide ions (Br⁻) react to form pale yellow AgBr precipitate. The problem says exactly one mole of AgNO₃ is needed per mole of the compound, meaning only one Br⁻ is ionic. The rest must be inside the coordination sphere. Since the formula is written as [CoBr3(en)x], the square brackets indicate the coordination sphere. The total number of bromides is 3, but only one is outside the brackets (ionic), so the other two must be inside. That forces the charge balance and tells us how many neutral en ligands are needed.
Step‑by‑step reasoning
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Identify the ionic bromide count
One mole of AgNO₃ gives one mole of AgBr precipitate. This means the compound releases exactly one mole of free Br⁻ ions per mole of complex. So there is one ionic bromide outside the coordination sphere.
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Determine the composition of the coordination sphere
The formula is written as [CoBr3(en)x]. The square brackets enclose the coordination sphere. Since there are three bromine atoms total and only one is ionic, the other two bromine atoms must be inside the brackets as ligands. So the coordination sphere contains Co, two Br ligands, and x molecules of ethylenediamine (en).
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Balance the charge
Cobalt in this complex is typically in the +3 oxidation state (common for Co(III) ammine/amine complexes). Each bromide ligand inside the sphere carries a –1 charge, so two bromides contribute –2. The en ligand is neutral. Therefore the charge on the complex ion is:
Charge=(+3)+2(−1)=+1
So the complex ion is [CoBr2(en)x]+. To balance this, there must be one counter‑ion — and that counter‑ion is the one ionic Br⁻ we already identified. This matches perfectly: the compound is [CoBr2(en)x]Br.
- Determine x from coordination number
Cobalt(III) typically has a coordination number of 6. The ligands inside the sphere are:
- 2 bromide ions (each monodentate, occupying one coordination site)
- x molecules of ethylenediamine (each bidentate, occupying two coordination sites) Total coordination sites used: 2+2x. Set equal to 6:
2+2x=6⇒2x=4⇒x=2.
- Verify With x = 2, the complex is [CoBr2(en)2]Br. It has one ionic Br⁻, so one mole of AgNO₃ gives one mole of AgBr. The coordination number is 2 (from Br) + 4 (from two en) = 6, which is standard for Co(III). Everything fits.
Watch outA common mistake is to think all three bromides are ionic because the formula shows “Br₃”. Remember: only those outside the square brackets are free ions; those inside are coordinated and do not precipitate with Ag⁺.
TipEthylenediamine (en) is bidentate — it always occupies two coordination sites. That’s the key to linking x to the coordination number.
✓Final answerThe correct option is (A).
ANSWER: A
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- KCET 2023Set D-21 markMCQQ.The number of protons, neutrons and electrons in the ion 1632S2− respectively are (A) 16,18,16 (B) 16,16,18 (C) 18,16,16 (D) 16,16,16
›Reveal solutionSolution
The atomic number gives protons and electrons in a neutral atom; the mass number gives neutrons; the charge tells how many extra electrons are present. For 1632S2−, the counts are 16 protons, 16 neutrons, and 18 electrons — option (B).
The key is to read the notation 1632S2− piece by piece. The subscript (16) is the atomic number Z — that’s the number of protons. In any atom or ion, the number of protons never changes; it defines the element. So protons = 16.
The superscript (32) is the mass number A, which is the sum of protons and neutrons. Neutrons = A−Z=32−16=16.
Now the charge: 2− means the ion has gained two extra electrons compared to the neutral atom. A neutral sulfur atom has as many electrons as protons — 16. With a 2− charge, it has 16+2=18 electrons.
Let’s check each option:
- Protons: always 16. That eliminates (A) and (C) immediately, since they show 18 or 16 but in the wrong order.
- Neutrons: 16. Both remaining options (B) and (D) have 16 neutrons — so far so good.
- Electrons: must be 18. Option (B) gives 18; option (D) gives 16 (which would be neutral, not an ion).
So the correct set is 16 protons, 16 neutrons, 18 electrons.
Watch outA common mistake is to forget that the charge changes the electron count, not the proton count. Another is to confuse the superscript (mass number) with the atomic number — always check which is which.
✓Final answerThe correct option is (B): 16 protons, 16 neutrons, 18 electrons.
- COMEDK 2023Set 2023-E1 markMCQQ.Match the Coordination compounds given in Column I with their characteristic features listed in Column II. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} S.No. Coordination compounds S.No. Characteristic features W [Co(NH3)5Cl]Cl3 P Oxidation state =+3 Configuration =d5μ=5.92BM X K4[Mn(CN)6 Q Oxidation state =+3 Configuration =d3μ=3.87BM Y [CrCl3(py)3] R Oxidation state =+3 Configuration =d6μ=0BM Z Cs[FeCl4] S Oxidation state =+2 Configuration =d5μ=1.732BM (A) W=SX=RY=QZ=P (B) W=SX=PY=QZ=R (C) W=RX=SY=PZ=Q (D) W=RX=SY=QZ=P
›Reveal solutionSolution
[!TLDR]
Work out oxidation state, d-count and spin state for each complex; the spin-only moments identify W=R (d⁶, μ=0), X=S (d⁵ LS, μ=1.73), Y=Q (d³, μ=3.87), Z=P (d⁵ HS, μ=5.92).
Concept
Spin-only magnetic moment μ=n(n+2) BM, where n is the number of unpaired electrons. Strong-field ligands (CN⁻, NH₃) tend to pair electrons (low spin); weak-field ligands (Cl⁻) keep them unpaired (high spin) — core CBSE/NCERT coordination-chemistry ideas.
Solution
W = [Co(NH₃)₅Cl]³⁺ type, Co(III): d6. With ammine ligands it is low spin, all electrons paired, n=0, μ=0 → feature R (d6, μ=0).
X = K₄[Mn(CN)₆], Mn(II): d5. CN⁻ is strong field → low spin, one unpaired electron, n=1, μ=1⋅3=1.73 BM → feature S (OS +2, d5, μ=1.732).
Y = [CrCl₃(py)₃], Cr(III): d3, three unpaired, n=3, μ=3⋅5=3.87 BM → feature Q (d3, μ=3.87).
Z = Cs[FeCl₄], Fe(III): d5. Cl⁻ is weak field → high spin, five unpaired, n=5, μ=5⋅7=5.92 BM → feature P (d5, μ=5.92).
Therefore W=R, X=S, Y=Q, Z=P.
[!ANSWER]
(D) W=R, X=S, Y=Q, Z=P
- KCET 2022Set B-31 markMCQQ.The complex hexamine platinum (IV) chloride will give ______ number of ions on ionization. (A) 3 (B) 2 (C) 5 (D) 4
›Reveal solutionSolution
The key is to determine the correct formula of the complex and then count the ions produced when it dissociates in solution. Hexamine platinum(IV) chloride gives 5 ions on ionization.
The question asks about "complex hexamine platinum (IV) chloride". The name tells you the coordination compound directly. "Hexamine" means six ammonia (NH3) ligands are attached to the central metal. "Platinum (IV)" tells you the oxidation state of platinum is +4. "Chloride" at the end indicates that chlorine is present as a counter-ion (outside the coordination sphere), not as a ligand.
So the coordination sphere is [Pt(NH3)6]4+. To balance the +4 charge, you need four chloride ions, each with a -1 charge. The full formula is therefore [Pt(NH3)6]Cl4.
Now, when this compound is dissolved in water (ionization), the complex ion and the chloride ions separate. The coordination sphere itself does not break apart — the NH3 ligands remain firmly attached to platinum. So the ionization is:
[Pt(NH3)6]Cl4→[Pt(NH3)6]4++4Cl−
Count the ions produced: one complex cation and four chloride anions. That gives a total of 5 ions.
Watch outA common mistake is to think that the "chloride" in the name is a ligand inside the coordination sphere. But the name "hexamine" explicitly tells you all six ligands are amines (ammonia). If any chloride were a ligand, the name would be something like "chloropentaamine" or "dichlorotetraamine". Always read the prefix carefully.
TipFor coordination compounds, the number of ions on ionization equals 1+(number of counter-ions). The "1" is the complex ion itself. Here, there are 4 chloride counter-ions, so 1+4=5 ions.
✓Final answerThe correct option is (C) 5.
- KCET 2021Set B-21 markMCQQ.Homoleptic complexes among the following are (A) K3[Al(C2O4)3], (B) [CoCl2(en)2]+ (C) K2[Zn(OH)4] (A) A only (B) (A) and (B) only (C) (A) and (C) only (D) (C) only
›Reveal solutionSolution
Homoleptic = one ligand type only; the oxalato-aluminate and the tetrahydroxozincate qualify, the mixed chloro/en cobalt complex does not.
Step 1 — The definition.
In a homoleptic complex the central metal ion is bonded to only one type of donor (ligand). If two or more different ligands are attached, the complex is heteroleptic.
Step 2 — Examine each complex.
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K3[Al(C2O4)3] — the coordination entity is [Al(C2O4)3]3−. The only ligand present is the oxalate ion C2O42− (three of them, each bidentate, giving CN = 6). One ligand type ⇒ homoleptic.
-
[CoCl2(en)2]+ — here cobalt is bonded to two different ligands: two chloride ions Cl− and two ethylenediamine molecules (en). Two ligand types ⇒ heteroleptic.
-
K2[Zn(OH)4] — the coordination entity is [Zn(OH)4]2−. The only ligand is hydroxide OH− (four of them, CN = 4). One ligand type ⇒ homoleptic.
Step 3 — Collect the homoleptic ones.
The homoleptic complexes are therefore the first and the third, i.e. (A) and (C).
✓Final answerThe correct option is (C) — (A) and (C) only.
ANSWER: C
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- COMEDK 2021Set 20211 markMCQQ.Which type of ligand is EDTA? (A) Monodentate (B) Hexadentate (C) Bidentate (D) Tridentate
›Reveal solutionSolution
Total = 6 donor atoms, so EDTA is a HEXADENTATE chelating ligand; it wraps around an octahedral metal ion occupying all six coordination sites (e.g. [Ca(EDTA)]2-).
Concept: Denticity of a ligand = number of donor atoms it uses to bind the metal.
EDTA (ethylenediaminetetraacetate, EDTA4-) has:
- 2 nitrogen donor atoms (the two amine N of the ethylenediamine backbone)
- 4 oxygen donor atoms (one from each of the four carboxylate arms)
Total = 6 donor atoms, so EDTA is a HEXADENTATE chelating ligand; it wraps around an octahedral metal ion occupying all six coordination sites (e.g. [Ca(EDTA)]2-).
✓Final answerThe correct option is (B) — Hexadentate
ANSWER: B
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