Q.[Fe(CN)6]4− and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Complex Formation
Complex Formation: The Intuition
Imagine you have a metal ion — say, a copper ion (Cu2+) — floating in water. It's positively charged, so it attracts anything negative or electron-rich nearby. Water molecules themselves have lone pairs of electrons on oxygen, so they crowd around the copper ion, each one donating a pair of electrons to form a coordinate bond. That cluster — the metal ion surrounded by water molecules — is already a complex ion: [Cu(H2O)6]2+.
Now, suppose you add ammonia (NH3) to the solution. Ammonia also has a lone pair on nitrogen, and it's a stronger electron donor than water. One by one, the ammonia molecules push the water molecules aside, replacing them. You end up with a deep blue complex: [Cu(NH3)4]2+.
That process — the stepwise replacement of one set of molecules (or ions) around a central metal atom by another set — is complex formation. The central metal is the Lewis acid (electron-pair acceptor), and the molecules or ions that attach to it are ligands (Lewis bases, electron-pair donors). The resulting species is a coordination compound or complex.
The word "complex" doesn't mean complicated. It just means a central atom (usually a metal) bonded to surrounding molecules or ions.
The Precise Statement
Complex formation is the reversible, stepwise reaction in which a central metal atom or ion (usually a transition metal) accepts electron pairs from one or more ligands to form a coordination entity. Each step has its own equilibrium constant, and the overall stability of the complex is measured by the formation constant (Kf) or stability constant.
For a general reaction:
M+nL⇌MLn
The overall formation constant is:
Kf=[M][L]n[MLn]
A large Kf means the complex is very stable — the ligands bind tightly and are hard to remove.
Kf=[metal][ligand]n[complex]
Key Features to Remember
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Stepwise nature: Complexes don't form all at once. First one ligand binds, then another, and so on. Each step has its own constant (K1,K2,…). The product of all stepwise constants equals the overall Kf.
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Coordination number: The number of ligand donor atoms directly bonded to the metal. Common values are 4 (tetrahedral or square planar) and 6 (octahedral). For [Cu(NH3)4]2+, the coordination number is 4.
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Ligand denticity: A ligand can have one donor atom (monodentate, like NH3 or H2O) or multiple donor atoms (polydentate, like EDTA, which wraps around the metal with six donor atoms). Polydentate ligands form especially stable complexes — this is the chelate effect.
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Reversibility: Complex formation is an equilibrium. Change the concentration of ligand, pH, or temperature, and the complex can break apart or form a different one. …
Why this formula?
Complex Formation: Understanding the Why Behind the Key Formulas
Complex formation is a fundamental concept in coordination chemistry and equilibrium. Let's build the reasoning step-by-step, starting from the simplest idea.
1. What is Complex Formation?
A complex forms when a central metal ion (Lewis acid) accepts electron pairs from surrounding molecules or ions called ligands (Lewis bases).
Example:
Cu2++4NH3⇌[Cu(NH3)4]2+
The key question: Why do we get a specific formula for the equilibrium constant?
2. The Stepwise Formation (The Core Reason)
Complex formation does not happen in one giant leap. It occurs in successive, reversible steps — each step adding one ligand.
For a metal M and ligand L:
Step 1:
M+L⇌ML
Equilibrium constant: K1=[M][L][ML]
Step 2:
ML+L⇌ML2
K2=[ML][L][ML2]
Step 3:
ML2+L⇌ML3
K3=[ML2][L][ML3]
... and so on up to MLn.
Each Ki is called a stepwise formation constant.
3. The Overall Formation Constant (The Key Formula)
Now, what if we want the equilibrium constant for the overall reaction:
M+nL⇌MLn
We can multiply the stepwise equilibria (because when you add reactions, you multiply their equilibrium constants):
Kf=K1×K2×K3×⋯×Kn
So:
Kf=[M][L]n[MLn]
Why this form?
Because each step contributes one [L] in the denominator and one [MLi] in the numerator, but intermediate species cancel out when multiplied.
4. Why the Denominator Has [L]n (Not n[L])
This is a common confusion. Let's derive it explicitly:
From step 1: [ML]=K1[M][L]
From step 2: [ML2]=K2[ML][L]=K1K2[M][L]2
From step 3: [ML3]=K3[ML2][L]=K1K2K3[M][L]3
Continuing:
[MLn]=(K1K2…Kn)[M][L]n
Therefore:
[M][L]n[MLn]=K1K2…Kn=Kf
The exponent n comes from repeated multiplication, not addition. Each ligand adds one factor of [L] in the denominator.
5. The Stability Connection
A larger Kf means:
- The complex is more stable
- Equilibrium lies far to the right (products favoured) …
The key idea is that the colour of a coordination compound arises from d-d transitions — the energy gap between split d-orbitals depends on the ligand field strength.
Reasoning:
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In [Fe(CN)6]4−, the ligand CN⁻ is a strong field ligand. It causes a large crystal field splitting (Δo), so the d-d transition requires high-energy (blue/violet) light. The complex appears yellow.
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In [Fe(H2O)6]2+, the ligand H₂O is a weak field ligand. It causes a small Δo, so the d-d transition absorbs lower-energy (red/orange) light. The complex appears pale green. …
The colour difference arises because the two complexes have different ligand field strengths (CN⁻ is a strong-field ligand, H₂O is a weak-field ligand), which causes different d-orbital splitting energies (Δ). This leads to absorption of different wavelengths of visible light, producing complementary colours.
The colour of a transition metal complex in solution is a direct consequence of electronic transitions between d-orbitals. When white light falls on the complex, certain wavelengths are absorbed to promote an electron from a lower-energy d-orbital to a higher-energy one. The colour we see is the complementary colour of the absorbed light.
For octahedral complexes of Fe2+ (d⁶ configuration), the key factor is the magnitude of the crystal field splitting energy, Δoct. This Δ value depends strongly on the nature of the ligand.
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Ligand strength determines Δ.
CN⁻ is a strong-field ligand (high up in the spectrochemical series), while H₂O is a weak-field ligand.
Δoct(CN−)≫Δoct(H2O)
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Electron configuration differs.
For [Fe(CN)6]4−, the large Δ forces the six d-electrons to pair up in the three lower t2g orbitals — this is a low-spin complex (t2g6eg0).
For [Fe(H2O)6]2+, the small Δ means electrons occupy all five d-orbitals singly before pairing — this is a high-spin complex (t2g4eg2).
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Energy of the d-d transition differs.
In the low-spin cyano complex, the energy gap between the filled t2g and empty eg orbitals is large. The complex absorbs higher-energy light towards the blue-violet end of the spectrum and transmits yellow light.
In the high-spin aqua complex, the energy gap is much smaller. It absorbs lower-energy light towards the red end of the spectrum and transmits pale green light. …
Concept: Complex Formation & Colour
The colour difference arises from Crystal Field Theory (CFT) — specifically, the nature of the ligand and its effect on d-orbital splitting.
Method: Crystal Field Theory — Ligand Spectrochemical Series
Step 1: Identify the metal ion and its d-electron count
- Both complexes contain Fe²⁺ (iron in +2 oxidation state).
- Fe²⁺ has the electronic configuration: [Ar]3d6.
Step 2: Determine the geometry and ligand field
- Both complexes are octahedral (coordination number 6).
- In an octahedral field, the 3d orbitals split into:
- t2g (lower energy: dxy,dxz,dyz)
- eg (higher energy: dz2,dx2−y2)
Step 3: Apply the Spectrochemical Series
The spectrochemical series ranks ligands by their ability to split d-orbitals:
Weak field⟶Strong field
I−<Br−<Cl−<F−<OH−<H2O<NH3<en<NO2−<CN−
- H2O is a weak field ligand → small splitting (Δoct is small).
- CN− is a strong field ligand → large splitting (Δoct is large).
Step 4: Determine electron configuration in each case
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[Fe(H2O)6]2+ (weak field, small Δ):
- Electrons fill according to Hund's rule (high-spin).
- Configuration: t2g4eg2 (4 unpaired electrons).
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[Fe(CN)6]4− (strong field, large Δ):
- Electrons pair up in lower t2g orbitals (low-spin).
- Configuration: t2g6eg0 (0 unpaired electrons).
Step 5: Relate to colour …
Here are the common mistakes students make when explaining the colour difference between [Fe(CN)6]4− and [Fe(H2O)6]2+, along with how to avoid each.
Mistake 1: Saying the colour is due to the metal ion alone
The error:
Students often say “Iron(II) is green” or “Iron(II) is yellow,” implying the colour comes only from the Fe2+ ion. This ignores the role of the ligand.
Why it’s wrong:
The colour of a transition metal complex arises from d-d transitions — electrons moving between split d-orbitals. The energy gap (Δ) depends on the ligand field strength, not just the metal ion.
How to avoid:
Always mention that the ligand determines the splitting energy. For the same metal ion (Fe2+), different ligands give different colours.
- CN− is a strong field ligand → large Δ → absorbs higher energy light (violet/blue) → appears yellow.
- H2O is a weak field ligand → small Δ → absorbs lower energy light (red region) → appears pale green (the complementary colour).
Key point: Colour is a complex property, not a metal-only property.
Mistake 2: Confusing the oxidation state of iron
The error:
Some students think [Fe(CN)6]4− contains Fe3+ because cyanide is often associated with ferricyanide ([Fe(CN)6]3−).
Why it’s wrong:
The charge on the complex is 4−. Cyanide (CN−) has a charge of −1 each.
Let x be the oxidation state of Fe:
x+6(−1)=−4⟹x=+2
Both complexes contain Fe2+ (d⁶ configuration).
How to avoid:
Always calculate the oxidation state using the overall charge and ligand charges. Write it down step-by-step in exams.
Mistake 3: Ignoring the spin state (high-spin vs low-spin)
The error:
Students treat both complexes as having the same electron arrangement in d-orbitals.
Why it’s wrong:
- CN− is a strong field ligand → causes large splitting → electrons pair up in lower orbitals → low-spin (t₂g⁶, eg⁰).
- H2O is a weak field ligand → small splitting → electrons remain unpaired as per Hund’s rule → high-spin (t₂g⁴, eg²).
Different spin states mean different d-orbital occupancy, which changes the energy of d-d transitions and thus the colour.
How to avoid:
Mention the spectrochemical series and decide spin state before discussing colour. For Fe2+:
- Strong field → low-spin
- Weak field → high-spin
Mistake 4: Forgetting to mention d-d transitions
The error:
Giving vague answers like “because of different ligands” without explaining the mechanism.
Why it’s wrong:
The examiner expects you to link ligand strength → splitting energy → wavelength of light absorbed → complementary colour observed.
How to avoid:
Use this chain in your answer:
Ligand strength → Crystal Field Splitting Energy (Δ) → Energy of absorbed photon → Colour observed (complementary)
For example:
- [Fe(CN)6]4−: strong field → large Δ → absorbs violet → appears yellow. …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.Among the following select the most stable complex(a) [Fe(H2O)6]3+(b) [Fe(C2O4)3]3-(c) [Fe(NH3)6]3+(d) [FeCl6]3-
›Reveal solutionSolution
Chelating (polydentate) ligands form much more stable complexes than monodentate ligands of similar donor strength — this is the chelate effect.
Among [Fe(H2O)6]3+, [Fe(C2O4)3]3-, [Fe(NH3)6]3+, and [FeCl6]3-, all four ligand types (H2O, oxalate, NH3, Cl⁻) coordinate through similar donor atoms (O, O, N, Cl), but oxalate (C2O4²⁻) is bidentate — each oxalate ion forms a 5-membered chelate ring with the metal, using two donor oxygen atoms per ligand.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.EDTA is used in treatment of _____ poisoning.(a) Pb(b) Pt(c) Ag(d) Cu
›Reveal solutionSolution
EDTA forms a very stable hexadentate chelate with Pb2+ ions, allowing it to be safely excreted from the body - this is the basis of EDTA chelation therapy for lead poisoning.
EDTA (ethylenediaminetetraacetic acid) is a hexadentate ligand that wraps around a metal ion using its two N atoms and four -COO- oxygen atoms, forming a very stable octahedral chelate complex (the chelate effect makes this complex thermodynamically very stable).
…
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.Among the following select the most stable complex(a) [Fe(H2O)6]3+(b) [Fe(OX)3]3-(c) [Fe(NH3)6]3+(d) [FeCl6]3-
›Reveal solutionSolution
Chelating ligands like oxalate (a bidentate ligand) give much more thermodynamically stable complexes than monodentate ligands of similar donor strength - this is the chelate effect.
Comparing the four Fe(III) complexes: [Fe(H2O)6]3+, [Fe(OX)3]3- (OX = oxalate, C2O4^2-), [Fe(NH3)6]3+ and [FeCl6]3- - all use monodentate ligands (H2O, NH3, Cl-) except oxalate, which is bidentate.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.Amongest the following, the most stable complex is ________.(a) [Fe(C2O4)3]3-(b) [Fe(NH3)6]3+(c) [Fe(H2O)6]3+(d) [FeCl6]3-
›Reveal solutionSolution
Chelating (multidentate) ligands form more thermodynamically stable complexes than comparable monodentate ligands, due to the favourable entropy of the chelate effect.
Among the given Fe3+ complexes, [Fe(C2O4)3]3- uses oxalate (C2O4^2-), a bidentate chelating ligand, forming three stable 5-membered chelate rings around the Fe3+ centre. This chelate effect makes it substantially more stable than the complexes with the monodentate ligands NH3, H2O, or Cl- ([Fe(NH3)6]3+, [Fe(H2O)6]3 …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.Which is correct formula of Wilkinson catalyst?(a) [(Me3As)3RhCl](b) [(Me3P)3RhCl](c) [(Ph3P)3RhCl](d) [(Ph3As)3RhCl]
›Reveal solutionSolution
Wilkinson's catalyst = [(Ph3P)3RhCl].
Wilkinson's catalyst is chlorotris(triphenylphosphine)rhodium(I), [(Ph3P)3RhCl] - a homogeneous catalyst used for the hydrogenation of alkenes. The ligand must be triphenylphosphine (Ph3P), not th …
- GSEB Higher Secondary Certificate (HSC) Examination 2020Set ANNUAL1 markMCQQ.Which of the following is the most stable complex?(a) [Fe(C2O4)3]3-(b) [Fe(NH3)6]3+(c) [Fe(H2O)6]3+(d) [FeCl6]3-
›Reveal solutionSolution
Complexes formed with a chelating (multidentate) ligand like oxalate are markedly more stable than analogous complexes of monodentate ligands such as water, ammonia or chloride - this is the well-known 'chelate effect'.
Oxalate (C2O4^2-) is a bidentate ligand that forms two Fe-O bonds per oxalate ion, creating stable five-membered chelate rings; with three oxalates wrapped around Fe3+, [Fe(C2O4)3]3- (ferrioxalate) has an exceptionally high formation/stability constant compared to complexes with only monodentate ligands (H2O, NH3, Cl-), even though those ligands individually may bind with compar …
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.Which of the following complex has sp3 hybridization?(a) K4[Fe(CN)6](b) [Ni(NH3)2Cl2](c) K2[Ni(CN)4](d) K4[Ni(CN)4]
›Reveal solutionSolution
[Ni(NH3)2Cl2] is a tetrahedral, sp3-hybridised Ni(II) complex.
Examine the hybridisation:
- K4[Fe(CN)6]: Fe2+ with strong-field CN-, octahedral, d2sp3.
- K2[Ni(CN)4]: Ni2+ with strong-field CN-, square planar, dsp2. …
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.Which of the following complex is useful in the dehydrogenation of alkanes?(a) [(Ph3P)3 Rh2 Cl](b) [(Ph3P)3 Rh Cl](c) [(Ph3P) Rh Cl](d) (Ph3P)3 Rh Cl2]
›Reveal solutionSolution
The correct formula of Wilkinson's catalyst is [(Ph3P)3RhCl] = option (b).
Wilkinson's catalyst is chlorotris(triphenylphosphine)rhodium(I), [(Ph3P)3RhCl] (Rh in +1). It is a famous homogeneous catalyst for hydrogenation of alkenes/alkynes; among the given options only (b) has the correc …
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