Q.Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
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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 …
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. …
Concept: Valence Bond Theory (VBT) – Hybridisation and magnetic behaviour depend on the ligand field strength (strong vs weak field ligands).
(i) [Fe(CN)6]4−
Fe is in +2 state (3d6). CN⁻ is a strong field ligand, causing pairing. Hybridisation: d2sp3 (inner orbital). All electrons paired → diamagnetic.
(ii) [FeF6]3−
Fe is in +3 state (3d5). F⁻ is a weak field ligand, no pairing. Hybridisation: sp3d2 (outer orbital). Five unpaired electrons → paramagnetic.
(iii) [Co(C2O4)3]3−
Co is in +3 state (3d6). Oxalate (C2O42−) is a strong field ligand, causing pairing. Hybridisation: d2sp3 (inner orbital). All electrons paired → diamagnetic.
(iv) [CoF6]3− …
Valence Bond Theory explains bonding in coordination compounds by considering the hybridisation of the central metal ion’s orbitals, which depends on the ligand field strength. For the given complexes: (i) [Fe(CN)6]4− has d2sp3 hybridisation (inner orbital, low spin), (ii) [FeF6]3− has sp3d2 hybridisation (outer orbital, high spin), (iii) [Co(C2O4)3]3− has d2sp3 hybridisation (inner orbital, low spin), and (iv) [CoF6]3− has sp3d2 hybridisation (outer orbital, high spin).
Werner Coordination Theory first suggested that metal ions have primary (ionisable) and secondary (non-ionisable) valencies. Valence Bond Theory (VBT), developed by Linus Pauling, refined this by describing how the metal ion’s vacant orbitals hybridise to accommodate lone pairs from ligands. The key insight: strong field ligands (like CN⁻, C₂O₄²⁻) cause pairing of electrons in the metal’s d-orbitals, leading to inner orbital (low spin) complexes with d2sp3 hybridisation. Weak field ligands (like F⁻) do not force pairing, giving outer orbital (high spin) complexes with sp3d2 hybridisation. The number of unpaired electrons determines magnetic behaviour.
Let’s apply this step by step to each complex.
(i) [Fe(CN)6]4−
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Determine the oxidation state of iron.
CN⁻ is a neutral ligand (charge −1 each). Let Fe be x:
x+6(−1)=−4⟹x=+2. So, Fe is in the +2 state.
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Write the electronic configuration of Fe²⁺.
Fe (atomic number 26): [Ar]3d64s2.
Fe²⁺: [Ar]3d6 (the two 4s electrons are lost first).
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Identify ligand strength and decide spin state.
CN⁻ is a strong field ligand. It causes pairing of electrons in the 3d orbitals.
The six d-electrons pair up completely: three pairs occupy three d-orbitals, leaving two d-orbitals empty. This gives zero unpaired electrons (diamagnetic).
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Determine hybridisation.
The metal uses two empty 3d orbitals, one 4s, and three 4p orbitals to form six equivalent d2sp3 hybrid orbitals. These accept lone pairs from six CN⁻ ligands.
This is an inner orbital (low spin) complex.
A common mistake is to forget that CN⁻ is a strong field ligand and assume high spin for Fe²⁺. Always check the ligand series: CN⁻, CO, NH₃ (strong) vs. F⁻, Cl⁻, H₂O (weak).
(ii) [FeF6]3−
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Oxidation state of iron.
F⁻ is −1 each. Let Fe be x:
x+6(−1)=−3⟹x=+3. So, Fe is in the +3 state.
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Electronic configuration of Fe³⁺.
Fe³⁺: [Ar]3d5 (five d-electrons).
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Ligand strength and spin state.
F⁻ is a weak field ligand. It does not cause pairing.
The five d-electrons occupy all five d-orbitals singly (Hund’s rule), giving five unpaired electrons (paramagnetic).
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Hybridisation.
Since no d-orbitals are vacated by pairing, the metal uses outer orbitals: one 4s, three 4p, and two 4d orbitals to form six sp3d2 hybrid orbitals.
This is an outer orbital (high spin) complex.
For Fe³⁺ with weak field ligands, the maximum number of unpaired electrons is 5. This is a quick check: if you see F⁻, Cl⁻, or H₂O with Fe³⁺, expect high spin.
(iii) [Co(C2O4)3]3−
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Oxidation state of cobalt.
Oxalate ion (C2O42−) is a bidentate ligand with charge −2 each. Let Co be x:
x+3(−2)=−3⟹x=+3. So, Co is in the +3 state.
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Electronic configuration of Co³⁺.
Co (atomic number 27): [Ar]3d74s2.
Co³⁺: [Ar]3d6 (six d-electrons).
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Ligand strength and spin state.
Oxalate (C2O42−) is a strong field ligand (it appears high in the spectrochemical series).
The six d-electrons pair up completely: three pairs in three d-orbitals, leaving two d-orbitals empty. Zero unpaired electrons (diamagnetic). …
Method: Valence Bond Theory (VBT) for Coordination Compounds
Core idea: Metal ion uses hybrid orbitals to accept lone pairs from ligands. The hybridisation depends on the metal's oxidation state, coordination number, and ligand field strength (strong vs weak field ligands).
General Steps for VBT Analysis
- Find oxidation state of the central metal ion.
- Write electronic configuration of the metal ion (in free state).
- Decide ligand field strength — strong field ligands (CN⁻, CO, en, C₂O₄²⁻) cause pairing; weak field ligands (F⁻, Cl⁻, H₂O usually) do not force pairing.
- Determine number of unpaired electrons after pairing (if any).
- Choose hybridisation based on coordination number and geometry:
- Coordination number 6 → d2sp3 (inner orbital, low spin) or sp3d2 (outer orbital, high spin)
- Describe bonding — ligand lone pairs fill hybrid orbitals; unpaired electrons give magnetic behaviour.
(i) [Fe(CN)6]4−
Step 1: Oxidation state
Let Fe be x: x+6(−1)=−4⇒x=+2
So, Fe(II).
Step 2: Electronic configuration
Fe (atomic no. 26): [Ar]3d64s2
Fe²⁺: [Ar]3d6
Step 3: Ligand field
CN⁻ is a strong field ligand → causes pairing.
Step 4: Unpaired electrons after pairing
All 6 electrons pair up in 3d orbitals → 0 unpaired electrons (diamagnetic).
Step 5: Hybridisation
One 3d orbital is vacated → uses d2sp3 hybridisation (inner orbital complex).
Step 6: Bonding
Six CN⁻ donate lone pairs into six d2sp3 hybrid orbitals.
Geometry: Octahedral.
Magnetic behaviour: Diamagnetic.
(ii) [FeF6]3−
Step 1: Oxidation state
x+6(−1)=−3⇒x=+3
Fe(III).
Step 2: Electronic configuration
Fe³⁺: [Ar]3d5
Step 3: Ligand field
F⁻ is a weak field ligand → no pairing.
Step 4: Unpaired electrons
All 5 electrons remain unpaired → 5 unpaired electrons (high spin).
Step 5: Hybridisation
Uses outer sp3d2 hybridisation (outer orbital complex).
Step 6: Bonding
Six F⁻ donate lone pairs into six sp3d2 hybrid orbitals.
Geometry: Octahedral.
Magnetic behaviour: Paramagnetic (5 unpaired electrons).
(iii) [Co(C2O4)3]3−
Step 1: Oxidation state
Oxalate (C₂O₄²⁻) is a bidentate ligand with charge -2 each.
x+3(−2)=−3⇒x=+3
Co(III).
Step 2: Electronic configuration
Co (atomic no. 27): [Ar]3d74s2
Co³⁺: [Ar]3d6
Step 3: Ligand field
C₂O₄²⁻ is a strong field ligand → causes pairing.
Step 4: Unpaired electrons
All 6 electrons pair up → 0 unpaired electrons (diamagnetic).
Step 5: Hybridisation
One 3d orbital vacated → d2sp3 hybridisation (inner orbital).
Step 6: Bonding
Three oxalate ions (each bidentate, donating 2 lone pairs) occupy six coordination sites via d2sp3 hybrid orbitals.
Geometry: Octahedral. …
Here are the most common mistakes students make when applying Valence Bond Theory (VBT) to these Werner-type coordination entities, along with how to avoid each.
Mistake 1: Forgetting to determine the correct oxidation state of the central metal ion first
Why it happens: Students jump straight to hybridization without calculating the charge on the metal. This leads to wrong d-electron counts and wrong geometry.
How to avoid: Always start with the formula:
- For [Fe(CN)6]4−: Let oxidation state of Fe be x. x+6(−1)=−4⟹x=+2. So, Fe(II).
- For [FeF6]3−: x+6(−1)=−3⟹x=+3. So, Fe(III).
- For [Co(C2O4)3]3−: Oxalate (C2O42−) is bidentate. x+3(−2)=−3⟹x=+3. So, Co(III).
- For [CoF6]3−: x+6(−1)=−3⟹x=+3. So, Co(III).
Key rule: Write the charge balance equation first — never guess the oxidation state.
Mistake 2: Confusing strong-field vs weak-field ligands and their effect on pairing
Why it happens: Students memorize "CN⁻ is strong, F⁻ is weak" but forget to apply pairing logic to the d-orbital configuration.
How to avoid: Use the spectrochemical series:
- Strong field (low spin): CN⁻, CO, en, NH₃ (for Co³⁺), oxalate (C2O42−) — causes pairing.
- Weak field (high spin): F⁻, Cl⁻, H₂O (for Fe²⁺/Fe³⁺) — no pairing.
Apply to each:
- [Fe(CN)6]4−: Fe²⁺ has d6. CN⁻ is strong → low spin → all 6 electrons paired → t2g6, eg0.
- [FeF6]3−: Fe³⁺ has d5. F⁻ is weak → high spin → 5 unpaired electrons → t2g3, eg2.
- [Co(C2O4)3]3−: Co³⁺ has d6. Oxalate is strong → low spin → all paired → t2g6, eg0.
- [CoF6]3−: Co³⁺ has d6. F⁻ is weak → high spin → 4 unpaired electrons → t2g4, eg2.
Key rule: Pairing only happens with strong-field ligands. Weak-field ligands keep electrons unpaired.
Mistake 3: Choosing the wrong hybridization (e.g., sp3d2 vs d2sp3)
Why it happens: Students forget that inner orbital (low spin) uses (n−1)d orbitals, while outer orbital (high spin) uses nd orbitals.
How to avoid:
- Low spin (paired electrons) → inner d-orbitals are available → d2sp3 (octahedral, inner orbital complex).
- High spin (unpaired electrons) → inner d-orbitals are occupied → uses outer d → sp3d2 (octahedral, outer orbital complex).
Apply:
- [Fe(CN)6]4−: low spin → d2sp3.
- [FeF6]3−: high spin → sp3d2.
- [Co(C2O4)3]3−: low spin → d2sp3.
- [CoF6]3−: high spin → sp3d2.
Key rule: Low spin = d2sp3 (inner). High spin = sp3d2 (outer).
Mistake 4: Ignoring the magnetic property (paramagnetic vs diamagnetic)
Why it happens: Students calculate hybridization but forget to state the magnetic behavior — a common exam requirement.
How to avoid: After determining unpaired electrons:
- 0 unpaired → diamagnetic.
- 1 or more unpaired → paramagnetic.
Check:
- [Fe(CN)6]4−: 0 unpaired → diamagnetic.
- [FeF6]3−: 5 unpaired → paramagnetic.
- [Co(C2O4)3]3−: 0 unpaired → diamagnetic.
- [CoF6]3−: 4 unpaired → paramagnetic. …
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). …
- 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 …
- 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. …
- 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. …
- 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. …
- 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. …
- 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. …
- 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 …
- 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. …
- 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. …
- 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. …
- 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. …
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