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:
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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.
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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)
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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.
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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).
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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.
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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.
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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 25 on this concept.
- AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQQ.Aqueous solutions of which of the following compounds do not form AgCl precipitate with excess AgNO3 solution? I. PtCl4.2HCl II. PtCl2.2NH3 III. CoCl3.4NH3 IV. PdCl2.4NH3 The correct answer is (only = only) (A) I , II only (B) III , IV only (C) I , II , III only (D) I , II , IV only
›Reveal solutionSolution
Only the chloride ions OUTSIDE the coordination sphere (ionisable) precipitate as AgCl; ligand (coordinated) chlorides do not. Complexes I and II have zero free Cl−.
Concept and Intuition
This is the classical Werner coordination-number test: dissolve the complex, add excess AgNO3, and see how much AgCl precipitates. Chlorine atoms bonded directly to the metal as ligands are held too tightly to react; only chloride counter-ions (outside the coordination sphere, present as free Cl− in solution) react with Ag+.
Step-by-Step Solution
- I. PtCl4.2HCl: rewrite as the ionic formula H2[PtCl6] — Pt(IV), coordination number 6, all six chlorides are ligands inside [PtCl6]2−; the two H+ are the counter-ions. Free Cl−=0 → no precipitate.
- II. PtCl2.2NH3: [Pt(NH3)2Cl2], Pt(II), coordination number 4 (square planar), both Cl are ligands, complex is neutral overall (no counter-ions at all). Free Cl−=0 → no precipitate.
- III. CoCl3.4NH3: Co(III), coordination number 6 is satisfied by 4 NH3 + 2 Cl as ligands, giving [Co(NH3)4Cl2]+; the third Cl is a free counter-ion: [Co(NH3)4Cl2]Cl. Free Cl−=1 per formula unit → gives precipitate (though only a third of the total chlorine).
- IV. PdCl2.4NH3: Pd(II) prefers coordination number 4, which the four NH3 alone already fill: [Pd(NH3)4]2+; both chlorides are then free counter-ions: [Pd(NH3)4]Cl2. Free Cl−=2 → gives precipitate (fully).
- Compounds giving no precipitate: I, II only.
Common Mistakes
- Assuming "more total Cl in the formula" means "more precipitate" — what matters is only how many Cl are ligands vs counter-ions, decided by the metal's coordination number.
- Missing that Pd(II) normally adopts coordination number 4 (not 6), which is exactly why all its chlorine ends up outside the sphere in IV.
✓Final answerThe correct option is (A) — I, II only.
ANSWER: A
- AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQQ.The metal ion, ligands present in Wilkinson catalyst are respectively (A) Rh3+ , Cl− , PPh3 (B) Rh+ , Cl− , PPh3 (C) Rh2+ , Br− , PH3 (D) Re+ , I− , P(CH3)3
›Reveal solutionSolution
Wilkinson's catalyst, [RhCl(PPh₃)₃], is built from Rh in the +1 state, a chloride ligand, and neutral triphenylphosphine ligands.
Concept and Intuition
Wilkinson's catalyst is one of the classic homogeneous hydrogenation catalysts, a square-planar 16-electron Rh(I) complex. To find the metal's oxidation state in a coordination complex, sum the charges of the ligands and set the whole species' charge (here, neutral, since it's written without any counter-ion) equal to metal charge + ligand charges.
Step-by-Step Solution
- The formula is [RhCl(PPh3)3] — an overall neutral complex.
- PPh3 (triphenylphosphine) is a neutral, two-electron-donor ligand (donates via the lone pair on P) — contributes 0 charge, and there are three of them.
- Cl here is bound as the anionic chloride ligand, Cl−.
- Charge balance: (charge on Rh) + (−1 from Cl⁻) + 3×(0 from PPh₃) = 0 ⟹ charge on Rh = +1, i.e. Rh+.
- So metal ion = Rh+; ligands = Cl− and PPh3 — matching option (B).
Common Mistakes
- Assuming Rh is in the more common +3 state (as in RhCl3) — in Wilkinson's catalyst it is specifically +1, which is what makes it active for oxidative-addition/reductive-elimination catalytic cycles.
- Confusing the ligand as PH3 or P(CH3)3 instead of the correct bulky PPh3 (triphenylphosphine), which is essential to the catalyst's steric and electronic properties.
✓Final answerThe correct option is (B) — Rh+ , Cl− , PPh3.
ANSWER: B
- AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQQ.Identify the correct set containing only ambidentate ligands (A) NO2−,CN−,SCN− (B) NH3,CN−,C2O42− (C) SO42−,SCN−,CO (D) C2O42−,(CH3)3P,CO
›Reveal solutionSolution
Ambidentate ligands offer two different donor atoms; the classic textbook trio is NO2−, CN−, SCN−.
Concept and Intuition
An ambidentate ligand has two different atoms, either of which can coordinate to the metal (though not simultaneously, unlike a chelating bidentate ligand). The identity of the coordinating atom can even change the name of the complex (e.g. nitro vs nitrito).
Step-by-Step Solution
- NO2− can bind through N (nitro) or through O (nitrito) — ambidentate.
- CN− can bind through C (cyano) or N (isocyano) — ambidentate.
- SCN− (thiocyanate) can bind through S (thiocyanato) or N (isothiocyanato) — ambidentate.
- Compare to the other sets: NH3 (monodentate, only N), C2O42− (bidentate through two different O atoms simultaneously — a chelate, not ambidentate), SO42− and CO and (CH3)3P (not ambidentate in the classic list).
- Only option (A) contains exclusively ambidentate ligands.
Common Mistakes
- Confusing a bidentate chelating ligand like oxalate (C2O42−) with an ambidentate one — chelation uses both donor atoms at once, ambidentate uses one OR the other.
✓Final answerThe correct option is (A) — NO2−,CN−,SCN−.
ANSWER: A
- AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQQ.The co-ordination number of chromium in K[Cr(H2O)2(C2O4)2] is (A) 5 (B) 4 (C) 6 (D) 3
›Reveal solutionSolution
This tests counting coordination number from a mixed-ligand complex; oxalate's bidentate nature is the key, giving coordination number 6.
Concept and Intuition
Coordination number counts the total number of donor-atom bonds to the central metal, not the number of ligand molecules/ions. A ligand like oxalate (C2O42−) is bidentate — it uses two oxygen atoms to bond to the metal simultaneously, forming a five-membered chelate ring — so each oxalate contributes 2 to the coordination number, not 1.
Step-by-Step Solution
- Identify ligands in [Cr(H2O)2(C2O4)2]−: 2 water molecules (monodentate) and 2 oxalate ions (bidentate).
- Donor atoms from water: 2×1=2.
- Donor atoms from oxalate: 2×2=4 (each oxalate binds through 2 oxygen atoms).
- Total coordination number =2+4=6.
Common Mistakes
- Counting oxalate as contributing only 1 (treating it as monodentate) instead of 2, which would wrongly give coordination number 4.
- Forgetting to add the two ligand types together before concluding.
✓Final answerThe correct option is (C) — 6.
ANSWER: C
- AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQQ.Which of the following exhibit ionization isomerism? (only = మాత్రమే) I) [Cr(NH3)4Cl2]Cl II) [Ti(H2O)5Cl](NO3)2 III) [Pt(en)(NH3)Cl]NO3 IV) [Co(NH3)4(NO3)2]NO3 (A) II & III only (B) I & II only (C) II & IV only (D) III & IV only
›Reveal solutionSolution
This tests when ionization isomerism is actually possible. The answer is (A): only II and III have two chemically different anions available to swap between the coordination sphere and the counter-ion position.
Concept and Intuition
Ionization isomers are compounds with the same overall formula that give different ions in solution because an anionic ligand and the counter-ion have exchanged positions (one moves from inside the coordination sphere to outside, and vice versa). For this to produce a genuinely different, distinguishable compound, the ligand-anion and the counter-anion must be chemically different species. If a complex only contains ONE kind of extra anion (split between 'inside' and 'outside' just to satisfy the coordination number and overall charge), then any rearrangement of identical anions gives back an indistinguishable compound — so no real isomerism is possible.
Step-by-Step Solution
- I: [Cr(NH3)4Cl2]Cl. Cr(III) is octahedral (CN 6); 4 NH3 occupy 4 sites, so exactly 2 more anionic ligands are needed to complete the sphere — and Cl− is the only anion present (3 total: 2 in, 1 out). The split (2 in / 1 out) is forced by the coordination number, and since all three anions are identical Cl−, no alternative arrangement gives a distinguishable compound. No ionization isomerism.
- II: [Ti(H2O)5Cl](NO3)2. Ti(III) is octahedral; 5 H2O + 1 Cl− fill the sphere, with 2 NO3− outside. Here TWO different anions exist (Cl− and NO3−). Swapping the ligand Cl− for one of the counter NO3− ions gives [Ti(H2O)5(NO3)]Cl(NO3) — a genuinely different, distinguishable compound. Shows ionization isomerism.
- III: [Pt(en)(NH3)Cl]NO3. Pt(II) is square planar (CN 4); en (bidentate, 2 sites) + NH3 (1) + Cl− (1) fill the sphere, with NO3− as the sole counter-ion. Cl− (ligand) and NO3− (counter-ion) are different anions, so swapping gives [Pt(en)(NH3)(NO3)]Cl — a distinguishable isomer. Shows ionization isomerism.
- IV: [Co(NH3)4(NO3)2]NO3. Co(III) is octahedral; 4 NH3 fill 4 sites, so exactly 2 more anionic ligands are needed — and NO3− is the only anion present (3 total: 2 in, 1 out), just like case I. The split is forced by the coordination number and all three anions are identical NO3−, so no distinguishable rearrangement exists. No ionization isomerism.
- Therefore only II and III exhibit ionization isomerism.
Common Mistakes
- Assuming any complex with a 'free' counter-ion automatically shows ionization isomerism — it only does if that counter-ion is chemically different from an in-sphere ligand, AND swapping them still satisfies the fixed coordination number.
- Missing that I and IV each contain only ONE type of anion overall, so any 'swap' is indistinguishable from the original compound.
✓Final answerThe correct option is (A) — II & III only.
ANSWER: A
- AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQQ.Identify the set which does not have ambidentate ligand (s) (A) NO2−,CN−,C2O42− (B) C2O42−,H2O,SO42− (C) SCN−,NH3,CH3COO− (D) CN−,SCN−,CH3NH2
›Reveal solutionSolution
Ambidentate ligands can coordinate through two different atoms (e.g. NO2−, CN−, SCN−). Only set (B) contains none of these classic ambidentate ligands.
Concept and Intuition
An ambidentate ligand has two different potential donor atoms, only one of which binds the metal at a time — e.g. NO2− can bind via N (nitro) or O (nitrito); CN− via C or N; SCN− via S (thiocyanato) or N (isothiocyanato). Ligands like C2O42− (oxalate, binds via two O atoms simultaneously — bidentate but not ambidentate), H2O, SO42−, NH3, CH3COO−, CH3NH2 are not ambidentate.
Step-by-Step Solution
- List known ambidentate ligands: NO2−, CN−, SCN−.
- Check option (A): NO2−,CN−,C2O42− — contains NO2− and CN− (both ambidentate) → has ambidentate ligands, so NOT the answer.
- Check option (B): C2O42−,H2O,SO42− — none of these three is ambidentate → this IS the set without ambidentate ligands.
- Check option (C): SCN−,NH3,CH3COO− — contains SCN− (ambidentate) → NOT the answer.
- Check option (D): CN−,SCN−,CH3NH2 — contains CN− and SCN− (both ambidentate) → NOT the answer.
- So the set with no ambidentate ligand is (B).
Common Mistakes
- Mistaking oxalate (C2O42−) for ambidentate; it is bidentate (binds via 2 O atoms at once) but not ambidentate (it doesn't switch which atom binds).
- Forgetting SCN− is a very commonly tested ambidentate ligand.
✓Final answerThe correct option is (B) — C2O42−,H2O,SO42−.
ANSWER: B
- AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQQ.The sum of coordination number and oxidation number of the metal M in the complex [M(en)2(C2O4)]Cl is (en = ethylenediamine) (A) 8 (B) 6 (C) 7 (D) 9
›Reveal solutionSolution
This tests computing both the coordination number and the oxidation state of the metal in a mixed-ligand complex, then adding them.
Concept and Intuition
The counter-ion outside the coordination sphere balances the charge on the complex ion, which lets us back out the metal's oxidation state. The coordination number is the total count of donor atoms bonded to the metal, counting each ligand's denticity (en and oxalate are both bidentate).
Step-by-Step Solution
- Since one Cl− balances the complex, the complex ion [M(en)2(C2O4)]+ carries a +1 charge.
- Ethylenediamine (en) is a neutral ligand (charge 0); oxalate (C2O42−) carries a −2 charge.
- Charge balance: (oxidation number of M) +2(0)+(−2)=+1, giving oxidation number of M=+3.
- Coordination number: en is bidentate, so 2 molecules of en contribute 2×2=4 donor atoms (N atoms); oxalate is also bidentate, contributing 2 donor atoms (O atoms). Total coordination number =4+2=6.
- Sum of coordination number and oxidation number =6+3=9.
Common Mistakes
- Treating oxalate or en as monodentate, which would undercount the coordination number.
- Forgetting to include the charge of the outer-sphere Cl− when working out the complex ion's own charge.
✓Final answerThe correct option is (D) — 9.
ANSWER: D
- AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQQ.The number of ions present in tris (ethane-1, 2-diamine) cobalt (III) sulphate is (A) 2 (B) 4 (C) 5 (D) 3
›Reveal solutionSolution
The compound is [Co(en)3]2(SO4)3; dissolving it releases 2
complex cations and 3 sulfate anions, i.e. 5 ions total.
Concept and Intuition
Ethane-1,2-diamine ("en", H2NCH2CH2NH2) is a neutral bidentate
ligand, so it does not change the oxidation state contribution when coordinated. The
"tris(en)cobalt(III)" cation is [Co(en)3]3+ (Co is +3, en contributes 0
charge each). To form a neutral salt with sulfate (SO42−), you need the
overall positive and negative charges to balance.
Step-by-Step Solution
- Complex cation: [Co(en)3]3+, charge +3.
- To balance with SO42− (charge -2), find the LCM of 3 and 2, which is 6: need 2 cations (total +6) and 3 anions (total -6).
- Formula: [Co(en)3]2(SO4)3.
- On dissolving in water, this dissociates into its constituent ions: 2 complex cations [Co(en)3]3+ + 3 sulfate anions SO42− = 5 ions total (the en ligands stay coordinated within the complex ion and do not dissociate further, since they are strong-field/chelating and part of the coordination sphere).
Common Mistakes
- Forgetting to balance charge with the LCM method, and assuming a simple 1:1 or 1:3 formula.
- Counting each nitrogen-donor atom of "en" as a separate dissociating ion (en is a neutral molecule and stays intact as one ligand within the coordination sphere).
✓Final answerThe correct option is (C) — 5.
ANSWER: C
- AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQQ.Cobalt (III) chloride forms a green coloured complex 'X' with NH3. Number of moles of AgCl formed when excess of AgNO3 solution is added to 100 mL of 1M solution of 'X' is (A) 0.3 (B) 0.2 (C) 0.1 (D) 1
›Reveal solutionSolution
The green Co(III)-ammine complex is [Co(NH3)4Cl2]Cl, with only one ionizable chloride per formula unit, so 0.1 mol of the complex gives 0.1 mol of AgCl.
Concept and Intuition
Werner complexes of cobalt(III) chloride with ammonia give differently coloured isomers depending on how many Cl− and NH3 ligands are bound inside the coordination sphere versus left outside as free (ionizable) counter-ions: [Co(NH3)6]Cl3 (yellow/orange, luteo, 3 ionizable Cl−), [Co(NH3)5Cl]Cl2 (purple, purpureo, 2 ionizable Cl−), and [Co(NH3)4Cl2]Cl (green in the trans form — praseo — or violet in cis — violeo — 1 ionizable Cl−). Only ligands OUTSIDE the coordination sphere (free counter-ions) react instantly with AgNO3 to precipitate AgCl; ligands bound directly to the metal do not.
Step-by-Step Solution
- The green colour identifies the complex as trans-[Co(NH3)4Cl2]Cl (praseocobaltic chloride), where 2 Cl− are coordinated to Co and only 1 Cl− is a free, ionizable counter-ion.
- Moles of complex in 100 mL of 1 M solution: n=0.100 L×1 mol/L=0.1 mol.
- Each formula unit releases 1 ionizable Cl−, so moles of free Cl− = 0.1 mol.
- With excess AgNO3: Cl−+Ag+→AgCl↓, so moles of AgCl formed = moles of free Cl−=0.1 mol.
Common Mistakes
- Assuming all 3 Cl atoms in the formula are ionizable and precipitating 0.3 mol AgCl — this ignores that 2 of them are coordinated directly to cobalt and don't precipitate with AgNO3.
✓Final answerThe correct option is (C) — 0.1.
ANSWER: C
- AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQQ.Impure silver ore+CN−+H2OO2[X]−+OH− [X]−+Zn→[Y]2−+Ag (pure) The co-ordination numbers of the metals in [X], [Y] are respectively (A) 3, 4 (B) 1, 4 (C) 4, 2 (D) 2, 4
›Reveal solutionSolution
This is the cyanide (Mac Arthur–Forrest) process for silver extraction: X is the dicyanoargentate(I) ion [Ag(CN)2]− (Ag coordination number 2), and Y is the tetracyanozincate(II) ion [Zn(CN)4]2− (Zn coordination number 4).
Concept and Intuition
Silver ores are leached with aerated cyanide solution, forming a soluble silver–cyanide complex; pure silver is then recovered by displacing it with a more reactive metal (zinc), which itself forms a cyanide complex.
Step-by-Step Solution
- Leaching step: 4Ag+8CN−+O2+2H2O4[Ag(CN)2]−+4OH−. Here X−=[Ag(CN)2]−; silver is bonded to two CN⁻ ligands, so its coordination number is 2.
- Displacement (cementation) step: 2[Ag(CN)2]−+Zn→[Zn(CN)4]2−+2Ag (pure). Here Y2−=[Zn(CN)4]2−; zinc is bonded to four CN⁻ ligands, so its coordination number is 4.
- Hence the coordination numbers of the metals in X and Y respectively are 2 and 4.
Common Mistakes
- Assuming silver keeps a coordination number of 4 (common for many other metal-cyanide complexes) instead of recognising [Ag(CN)2]− is linear/2-coordinate, a well-known exception.
- Reversing which complex is X and which is Y based on the equation order.
✓Final answerThe correct option is (D) — 2, 4.
ANSWER: D
- AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQQ.Which one of the following has the highest molar conductivity? (A) Diammine dichloroplatinum (II) (B) Tetraamminedichlorocobalt (III) chloride (C) Potassium hexacyano ferrate (II) (D) Hexa aquo chromium (III) chloride
›Reveal solutionSolution
Molar conductivity of an electrolyte scales with the total number of ions it dissociates into; K₄[Fe(CN)₆] gives the most ions (5) among the four choices.
Concept and Intuition
For coordination compounds, the counter ions outside the coordination sphere (square brackets) dissociate freely, while ligands inside the sphere do not contribute extra ions. More ions in solution (both in number and charge) generally means higher molar conductivity, all else being similar.
Step-by-Step Solution
- (A) [Pt(NH3)2Cl2]: both Cl are ligands (inside brackets) — complex is neutral, dissociates into 0 ions.
- (B) [Co(NH3)4Cl2]Cl: one Cl is outside as counter-ion, complex cation is +1 — dissociates into 2 ions total.
- (C) K4[Fe(CN)6]: 4 K⁺ ions plus 1 [Fe(CN)6]4− ion — 5 ions total, the most of the four.
- (D) [Cr(H2O)6]Cl3: complex cation +3 plus 3 Cl⁻ — 4 ions total.
- Highest ion count (5) → highest molar conductivity → option C.
Common Mistakes
- Counting ligands bound inside the coordination sphere as if they dissociate (they don't).
- Assuming higher molecular weight/complexity implies higher conductivity, rather than counting actual dissociated ions.
✓Final answerThe correct option is (C) — Potassium hexacyano ferrate (II).
ANSWER: C
- AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQQ.The compounds having coordinated water are CrCl3.6H2O (I) BaCl2.2H2O (II) CuSO4.5H2O (III) (A) II, III only (B) I, III only (C) I, II only (D) I, II, III
›Reveal solutionSolution
CrCl3⋅6H2O and CuSO4⋅5H2O have genuinely coordinated water; BaCl2⋅2H2O has simple water of crystallization.
Concept and Intuition
Water in a hydrate can be present in two structurally different ways: coordinated (directly bonded to the metal ion as a ligand, inside the coordination sphere) or water of crystallization (held in the crystal lattice by weaker hydrogen bonds/ionic packing forces, not bonded to the metal). Transition-metal complexes with well-defined coordination numbers (like Cr3+, Cu2+) commonly show coordinated water, whereas simple ionic salts of larger, less strongly coordinating cations like Ba2+ typically just have lattice/crystallization water.
Step-by-Step Solution
- CrCl3⋅6H2O: exists as [Cr(H2O)6]Cl3, a well-known octahedral hexaaqua complex — all six water molecules are coordinated.
- BaCl2⋅2H2O: Ba2+ is a large, weakly polarizing ion that does not form a strong discrete aqua-complex here; the two waters are water of crystallization in the lattice, not coordinated.
- CuSO4⋅5H2O: four water molecules are coordinated directly to Cu2+ (forming [Cu(H2O)4]2+), and this compound is classically cited as having coordinated water.
- So compounds with coordinated water: I and III only.
Common Mistakes
- Assuming every hydrate automatically has coordinated water — many common salts (like BaCl2⋅2H2O) have only crystallization water.
- Overlooking that CuSO4⋅5H2O's fifth water is different (hydrogen-bonded) from its first four (coordinated), but the compound as a whole is still correctly described as having coordinated water.
✓Final answerThe correct option is (B) — I, III only.
ANSWER: B
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