Q.Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
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🔒 Start your 14-day free trial to unlock the full solution →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 …
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 key idea is Valence Bond Theory (VBT): whether a complex is inner (low-spin, d2sp3 hybridisation) or outer (high-spin, sp3d2 hybridisation) depends on the metal ion's d-electron count and the ligand's field strength.
Reasoning:
- [Co(NH3)6]3+: Co3+ has a 3d6 configuration. NH3 is a strong field ligand, causing large crystal field splitting. The six d-electrons pair up in the three t2g orbitals (t2g6eg0), leaving two d-orbitals empty. This allows d2sp3 hybridisation (inner orbital complex). …
The difference arises from the electronic configurations of Co³⁺ (d⁶) and Ni²⁺ (d⁸) in an octahedral field. Co³⁺ uses d²sp³ hybridisation (inner d-orbitals) giving a diamagnetic inner orbital complex, while Ni²⁺ uses sp³d² hybridisation (outer d-orbitals) giving a paramagnetic outer orbital complex.
The Core Idea: Valence Bond Theory and Hybridisation
Valence Bond Theory classifies complexes as inner orbital (or low-spin) and outer orbital (or high-spin) based on whether the metal uses its inner (n−1)d orbitals or outer nd orbitals for bonding. The deciding factor is the crystal field splitting energy (Δo) relative to the pairing energy (P). When Δo>P, electrons pair up in the inner d-orbitals, freeing an inner d-orbital for d2sp3 hybridisation — this is an inner orbital complex. When Δo<P, electrons remain unpaired in the outer d-orbitals, forcing the use of sp3d2 hybridisation — this is an outer orbital complex.
The ligand here is ammonia (NH3), a moderately strong field ligand. But the metal ion's charge and size also affect Δo. Let's see how this plays out for Co³⁺ and Ni²⁺.
Step-by-Step Analysis
1. Determine the oxidation state and d-electron count
For [Co(NH3)6]3+:
- Cobalt is in +3 oxidation state. Atomic number of Co = 27.
- Co³⁺: [Ar]3d6 (remove 4s² and one 3d electron).
- So Co³⁺ has a d⁶ configuration.
For [Ni(NH3)6]2+:
- Nickel is in +2 oxidation state. Atomic number of Ni = 28.
- Ni²⁺: [Ar]3d8 (remove 4s²).
- So Ni²⁺ has a d⁸ configuration.
2. Consider the ligand field and possible hybridisations
Ammonia pairs the d⁶ electrons of the highly charged Co³⁺ ion (large Δo). For Ni²⁺ (d⁸), however, no ligand field can produce an inner orbital complex: even complete pairing of the eight 3d electrons frees only ONE 3d orbital, while d2sp3 hybridisation needs TWO.
For Co³⁺ (d⁶):
- In an octahedral field, the 3d orbitals split into t2g (lower energy) and eg (higher energy).
- With strong field NH₃, Δo is large. All six electrons pair up in the three t2g orbitals: t2g6eg0.
- This leaves two empty 3d orbitals (the eg set). Both of these hybridise with one 4s and three 4p orbitals to form six d2sp3 hybrid orbitals.
- Since the bonding uses inner (n-1)d orbitals, it is an inner orbital complex. All electrons are paired → diamagnetic.
For Ni²⁺ (d⁸):
- In an octahedral field, d⁸ always has two unpaired electrons in the eg set regardless of field strength (because pairing would require promoting an electron to a higher energy level, which is unfavourable).
- The configuration is t2g6eg2 — two unpaired electrons.
- All five 3d orbitals are occupied (three t2g full, two eg half-filled). No empty 3d orbital is available for hybridisation.
- Therefore, the metal must use its outer 4d orbitals (specifically 4d, 4s, and 4p) to form six sp3d2 hybrid orbitals.
- Since bonding uses outer (n)d orbitals, it is an outer orbital complex. Two unpaired electrons → paramagnetic. …
Method: Valence Bond Theory (VBT) Analysis of Coordination Complexes
This method uses hybridisation and magnetic behaviour to classify complexes as inner or outer orbital.
Step 1 — Determine the oxidation state and electronic configuration of the central metal ion
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For [Co(NH3)6]3+
Co atomic number = 27
Co in +3 state: Co3+ = [Ar]3d6
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For [Ni(NH3)6]2+
Ni atomic number = 28
Ni in +2 state: Ni2+ = [Ar]3d8
Step 2 — Identify the ligand field strength
- NH3 is a strong field ligand (causes pairing of electrons in the d-orbitals).
Step 3 — Decide pairing and hybridisation
For [Co(NH3)6]3+:
- 3d6 with strong field NH3 → all 6 electrons pair up in the three t2g orbitals.
- This leaves two empty 3d orbitals available.
- Hybridisation: d2sp3 (inner orbital hybridisation).
- Result: Diamagnetic (no unpaired electrons).
✓ Inner orbital complex — uses (n−1)d orbitals for hybridisation.
For [Ni(NH3)6]2+:
- 3d8: all five 3d orbitals are occupied (three doubly, two singly) — even pairing the two unpaired electrons could not empty the TWO 3d orbitals that d2sp3 needs.
- So hybridisation uses outer 4d orbitals: sp3d2 (outer orbital hybridisation). …
Here are the most common mistakes students make when explaining why [Co(NH3)6]3+ is an inner orbital complex and [Ni(NH3)6]2+ is an outer orbital complex, along with how to avoid each.
Mistake 1: Confusing the oxidation state and electron count
The error: Students often forget to first determine the oxidation state of the central metal ion. They might directly use the ground state configuration of the neutral atom (Co or Ni) instead of the ion.
How to avoid:
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Always start with the oxidation state.
For [Co(NH3)6]3+:
- NH3 is neutral, so the charge comes from Co.
- Co is in +3 state.
- Co (atomic number 27): [Ar]3d74s2
- Co3+: remove 3 electrons → [Ar]3d6
For [Ni(NH3)6]2+:
- Ni (atomic number 28): [Ar]3d84s2
- Ni2+: remove 2 electrons → [Ar]3d8
Key takeaway: Write the d-electron count for the ion, not the atom.
Mistake 2: Forgetting that NH3 is a strong field ligand
The error: Some students treat NH3 as a weak field ligand (like H2O) and incorrectly predict high-spin configurations.
How to avoid:
- Memorise the spectrochemical series — NH3 sits on the stronger side of the series (well above H2O, though en, NO2−, CN− and CO are stronger still).
- Strong field ligands cause large crystal field splitting (Δo), favouring low-spin configurations for d4 to d7 ions.
For Co3+ (d6):
- Strong field → low-spin → t2g6eg0
- All 6 electrons paired → no unpaired electrons → inner orbital complex (uses (n−1)d orbitals).
For Ni2+ (d8):
- d8 always has 2 unpaired electrons regardless of field strength (Hund’s rule).
- Even with NH3, it remains t2g6eg2 → outer orbital complex (uses ns, np and nd orbitals — sp3d2).
Mistake 3: Mixing up “inner” vs “outer” orbital terminology
The error: Students think “inner orbital” means the complex is low-spin, and “outer orbital” means high-spin — but that’s only true for d4 to d7 ions. For d8, d9, d10, the spin state is fixed.
How to avoid:
- Inner orbital complex: Uses (n−1)d orbitals for hybridisation (e.g., d2sp3).
- Outer orbital complex: Uses ns, np, and nd orbitals (e.g., sp3d2).
- Rule of thumb:
- If the complex has no unpaired electrons (or fewer than maximum), it’s likely inner orbital.
- If it has the maximum possible unpaired electrons, it’s outer orbital.
For Co3+:
- d6, low-spin → 0 unpaired → d2sp3 hybridisation → inner orbital.
For Ni2+:
- d8 → 2 unpaired electrons (always) → sp3d2 hybridisation → outer orbital.
Mistake 4: Not explaining the hybridisation clearly
The error: Students state the hybridisation without showing how the orbitals are reorganised.
How to avoid:
- Show the orbital diagram step-by-step.
For [Co(NH3)6]3+:
- Co3+: 3d6
- Under strong field, electrons pair in t2g → two 3d orbitals become empty. …
Showing the 12 most recent of 66 on this concept.
- CBSE 2026Set 56/3/11 markMCQQ.According to Werner's theory, the primary valencies of the central metal atom : (A) are satisfied by neutral molecules or negative ions. (B) are equal to its coordination number. (C) are satisfied by negative ions. (D) are non-ionisable.
›Reveal solutionSolution
Werner’s theory distinguishes primary valencies (ionisable, satisfied only by negative ions, equal to oxidation state) from secondary valencies (non-ionisable, satisfied by neutral molecules or negative ions, equal to coordination number). The correct answer is (C).
Werner’s Coordination Theory was a breakthrough because it explained why compounds like CoClX3⋅6NHX3 (which we now write as [Co(NHX3)X6]ClX3) behave so differently from simple salts. The key insight: a metal ion has two kinds of bonding capacity — not just one.
Primary valency (now called oxidation state) is the metal’s ionic bonding capacity. It is satisfied only by negative ions (anions), because it arises from the metal’s need to neutralise its positive charge. These bonds are ionisable — they break apart in solution, giving the familiar conductivity and precipitation tests.
Secondary valency (now called coordination number) is the metal’s ability to bind ligands directly. It is satisfied by neutral molecules (like NHX3) or negative ions (like ClX−). These bonds are non-ionisable — they stay attached to the metal even in solution.
Now let’s apply this to the options.
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Option (A): “Primary valencies are satisfied by neutral molecules or negative ions.”
This is false. Primary valencies are satisfied only by negative ions. Neutral molecules satisfy secondary valencies. For example, in [Co(NHX3)X6]ClX3, the three ClX− ions satisfy the primary valency (Co³⁺ needs three negative charges), while six NHX3 molecules satisfy the secondary valency.
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Option (B): “Primary valencies are equal to its coordination number.”
This is false. Primary valency equals the oxidation state of the metal. Coordination number equals the secondary valency. They are often different. In [Co(NHX3)X6]ClX3, primary valency = 3, coordination number = 6.
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Option (C): “Primary valencies are satisfied by negative ions.” …
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- CBSE 2026Set 56/1/11 markMCQQ.The secondary valency of Co in the complex [Co(NH3)5(NO2)]2+ is (A) 5 (B) 1 (C) 4 (D) 6
›Reveal solutionSolution
In Werner’s coordination theory, secondary valency equals the coordination number — the number of ligand donor atoms directly bonded to the metal. For [Co(NH3)5(NO2)]2+, there are 5 NH₃ molecules (each donating one N) and one NO₂⁻ ligand (also donating one N), giving a total of 6 donor atoms. So the secondary valency is 6.
Werner’s coordination theory is the foundation here. He proposed that metal ions have two kinds of valency: primary (ionisable, corresponding to oxidation state) and secondary (non-ionisable, corresponding to coordination number). The secondary valency is simply the number of ligand donor atoms directly attached to the central metal ion — in other words, the coordination number.
For the complex [Co(NH3)5(NO2)]2+, we need to count how many atoms are actually bonded to cobalt.
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Identify the ligands and their denticity.
- NH₃ (ammine) is a monodentate ligand — it binds through the lone pair on nitrogen. Each NH₃ contributes one donor atom.
- NO₂⁻ (nitrito or nitro, depending on binding mode) is also monodentate when it binds through nitrogen (the more common nitro form) or through oxygen. In either case, it uses one donor atom per ligand. The problem doesn’t specify the linkage isomer, but that doesn’t change the count — it’s still one donor atom.
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Count the total number of donor atoms.
- Five NH₃ ligands: 5×1=5 donor atoms.
- One NO₂⁻ ligand: 1×1=1 donor atom.
- Total = 5+1=6 donor atoms.
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Relate this to secondary valency.
Werner defined secondary valency as the number of groups directly coordinated to the metal — exactly the coordination number. So here, secondary valency = 6. …
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- CBSE 2026Set DZ1 markMCQQ.What is the coordination number of Co in the complex K3[Co(C2O4)3] ?(a) 3(b) 4(c) 5(d) 6
›Reveal solutionSolution
Oxalate is a bidentate ligand; three of them occupy six coordination sites, so the coordination number of Co is 6 — option (d).
The coordination number is the number of ligand donor atoms directly bonded to the central metal ion (not the number of ligand molecules).
…
- CBSE 2026Set ANNUAL1 markMCQQ.What is the coordination number of Cobalt in [Co(en)3]Cl3 ?(a) 6(b) 5(c) 4(d) 3
›Reveal solutionSolution
Each 'en' ligand is bidentate, so [Co(en)3]Cl3 has coordination number 6.
In [Co(en)3]Cl3, ethylenediamine (en, H2NCH2CH2NH2) acts as a bidentate ligand — it has two nitrogen donor atoms, each with a lone pair, that can simultaneously coordinate to the central metal ion, forming a five-membered chelate ring with cobalt.
Since there are three 'en' ligands attached to the cobalt centre, and each contributes 2 donor atoms, the total number of coordinate bonds (and hence the coordination number) is:
3×2=6
…
- CBSE 2026Set ANNUAL1 markMCQQ.Ambidentate ligand is(a) H2O(b) NH3(c) NO2-(d) Cl-
›Reveal solutionSolution
An ambidentate ligand has two different donor atoms but binds through only one of them at a time, depending on conditions.
- H2O, NH3 and Cl- each have only one type of donor atom, so they are simple monodentate ligands, not ambidentate. …
- CBSE 2026Set ANNUAL1 markMCQQ.Coordination number of Pt in [Pt(NH3)2 Cl (NO2)] complex is(a) 3(b) 4(c) 5(d) 6
›Reveal solutionSolution
Coordination number = the total number of ligand donor atoms directly bonded to the central metal atom/ion.
…
- CBSE 2026Set ANNUAL1 markMCQQ.Identify the homoleptic complex among the following compounds:(a) [Co(NH3)4Cl2]+(b) [Ni(CN)4]2-(c) [Cr(NH3)2Cl2(en)]+(d) Both (A) and (C)
›Reveal solutionSolution
A homoleptic complex is one in which the metal is bonded to only a single kind of donor ligand; a heteroleptic complex has more than one kind.
Checking each option:
- [Co(NH3)4Cl2]+: contains two different ligands, NH3 and Cl- (heteroleptic).
- [Ni(CN)4]2-: contains only CN- ligands, all of the same type (homoleptic). …
- CBSE 2026Set ANNUAL1 markMCQQ.Which of the following is a bidentate ligand?(a) CN-(b) NH3(c) C2O4^2-(d) CO3^2-
›Reveal solutionSolution
A bidentate ligand has two donor atoms that simultaneously coordinate to the same central metal atom, forming a ring (chelate).
- CN- has one donor atom (C), so it is monodentate.
- NH3 has one donor atom (N), so it is monodentate.
- C2O4^2- (oxalate ion) has two oxygen atoms, one from each carboxylate group, that can each donate a lone pair to the metal simultaneously - this makes it bidentate, forming a five-membered chelate ring. …
- CBSE 2026Set ANNUAL1 markMCQQ.The co-ordination number of Cr in K3[Cr(C2O4)3] is: (as printed; the paper's option values are +3/+4/+2/+6)(a) +3(b) +4(c) +2(d) +6
›Reveal solutionSolution
As printed, the options (+3,+4,+2,+6) are oxidation-state values, so this question is really testing the oxidation number of Cr, which is +3. (For completeness: the true coordination number — the count of donor atoms bonded to Cr — is 6, not among the given options, since oxalate is bidentate and there are three oxalate ligands.)
Oxidation number of Cr: In K3[Cr(C2O4)3], potassium contributes +1 each (three K+ ions), and each oxalate ion C2O42− carries a charge of −2. Let the oxidation number of Cr be x:
3(+1)+x+3(−2)=0
3+x−6=0⟹x=+3
…
- CBSE 2026Set ANNUAL1 markQ.What is the oxidation number of Fe in [Fe(CN)6]4−?
›Reveal solutionSolution
Balancing the ligand charges against the overall complex-ion charge gives the oxidation state of iron.
Working it out
Let the oxidation number of Fe be x. Cyanide is an anionic ligand, CN−, contributing −1 each; there are 6 of them:
x+6(−1)=−4 (overall charge of the complex ion) …
- CBSE 2026Set ANNUAL1 markQ.What is the coordination number of Fe in [Fe(EDTA)]⁻?
›Reveal solutionSolution
EDTA is a hexadentate chelating ligand (2 N + 4 O donor atoms), so a single EDTA fills all six coordination sites — coordination number of Fe = 6.
Concept. The coordination number of a metal ion is the number of ligand donor atoms directly bonded to it (the number of coordinate bonds), not the number of ligand molecules.
…
- CBSE 2026Set ANNUAL1 markMCQQ.Assertion (A): Glycinate ion H₂NCH₂COO⁻ is a chelating ligand. Reason (R): It can link through either nitrogen or anionic oxygen.(a) Both A and R are true and R is the correct explanation of A.(b) Both A and R are true but R is not the correct explanation of A.(c) A is true but R is false.(d) A is false but R is correct.
›Reveal solutionSolution
A (glycinate is chelating) is true, but the reason given describes ambidentate behaviour (binding through one OR the other), not chelation (binding through both at once), so R is false — option (C).
Assertion (A): The glycinate ion H2N−CH2−COO− is a chelating ligand. This is true: it is a bidentate ligand that binds a metal through the nitrogen of the −NH2 group and the anionic oxygen of the −COO− group at the same time, forming a stable five-membered chelate ring.
…
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