Q.What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S
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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 complex formation: the copper ion binds strongly with cyanide ligands, changing its chemical identity and removing free Cu²⁺ from solution.
- When excess aqueous KCN is added to CuSO₄, Cu²⁺ first reduces to Cu⁺ (cyanide acts as a reducing agent), and the Cu⁺ then coordinates with CN⁻ to form a stable, soluble complex.
- The reaction is: 2CuSO4+10KCN→2K3[Cu(CN)4]+2K2SO4+(CN)2↑
- The coordination entity formed is the tetracyanidocuprate(I) ion, [Cu(CN)4]3−. …
Excess KCN converts Cu2+ into the stable, soluble complex [Cu(CN)4]3− (tetracyanidocuprate(I)), which has such a low dissociation constant that H2S cannot produce enough Cu2+ to form CuS precipitate.
Why this happens — the concept of complex formation
When you add a ligand like cyanide (CN−) to a metal ion in solution, the metal ion can bind several ligands to form a coordination complex. This is not just a physical mixing — it’s a chemical equilibrium that dramatically changes the metal ion’s properties. The key idea: a stable complex “locks up” the metal ion so tightly that its free concentration becomes vanishingly small.
For precipitation reactions (like forming CuS), you need a minimum concentration of free Cu2+ ions. If the complex is stable enough, that free concentration falls below the threshold needed for precipitation — even though the total copper is still present.
Step-by-step reasoning
1. What happens when KCN is added to CuSO4 solution?
Aqueous CuSO4 gives blue Cu2+ ions. When you add KCN, the CN− ligand first reduces Cu2+ to Cu+ (because CN− is a reducing agent as well as a ligand), and then forms a complex with Cu+.
The overall reaction is:
2Cu2++10CN−→2[Cu(CN)4]3−+(CN)2
The cyanogen gas (CN)2 escapes, and the copper ends up as the tetracyanidocuprate(I) ion, [Cu(CN)4]3−.
A common mistake is to write [Cu(CN)4]2− with Cu2+. But Cu2+ is reduced to Cu+ by CN− — the complex has copper in the +1 oxidation state. The charge on the complex is 3−.
2. Why is this complex so stable?
The stability constant (formation constant) of [Cu(CN)4]3− is enormous — roughly Kf≈2×1030. That means the equilibrium
Cu++4CN−⇌[Cu(CN)4]3−
lies almost entirely to the right. The free Cu+ concentration in solution is extremely low.
Kf=[Cu+][CN−]4[[Cu(CN)4]3−]≈2×1030
3. What happens when H2S is passed through this solution?
H2S dissociates slightly in water:
H2S⇌2H++S2− …
Method: Complex Formation & Stability Analysis
This problem is solved by understanding ligand substitution and the stability of coordination complexes.
Step 1: Identify the reaction
When excess aqueous KCN is added to aqueous CuSO4, the cyanide ion (CN−) acts as a strong ligand and replaces water molecules around the copper ion.
- Initially, Cu2+ in water exists as [Cu(H2O)6]2+ (blue colour).
- With excess KCN, a complex formation occurs:
2CuSO4+10KCN→2K3[Cu(CN)4]+2K2SO4+(CN)2↑
The coordination entity formed is:
[Cu(CN)4]3−
Note: Copper is in +1 oxidation state here because CN− reduces Cu2+ to Cu+ during complexation (cyanide acts as both ligand and reducing agent — the equation above shows the cyanide oxidised to cyanogen, (CN)2).
Step 2: Why no precipitate with H2S?
Normally, H2S reacts with Cu2+ to give black CuS precipitate:
Cu2++H2S→CuS↓+2H+
However, in the complex [Cu(CN)4]3−:
- The cyanide ligands are very strongly bound to copper.
- The complex has a very high stability constant (Kf is extremely large). …
🧪 The Correct Chemistry First
When excess aqueous KCN is added to CuSO4 solution:
- Initially, unstable copper(II) cyanide forms and rapidly decomposes to copper(I) cyanide (CuCN), evolving cyanogen — this is where Cu2+ is reduced to Cu+.
- With excess KCN, the CuCN dissolves to form the soluble complex:
[Cu(CN)4]3−(tetracyanidocuprate(I) ion)
Notice: copper is reduced from Cu²⁺ to Cu⁺ by cyanide ion (CN⁻ acts as a reducing agent here).
When H2S gas is passed through this solution, no CuS precipitate forms because:
- The complex [Cu(CN)4]3− is extremely stable (high formation constant).
- The concentration of free Cu+ ions is too low to exceed the solubility product of Cu2S.
✗ Common Mistake #1: Writing the wrong oxidation state of copper
Mistake: Students write [Cu(CN)4]2− (assuming Cu remains in +2 state).
Why it’s wrong: CN⁻ is a strong reducing agent. In excess, it reduces Cu2+ to Cu+. The complex formed is with Cu+, not Cu2+.
✓ How to avoid: Remember the redox behaviour of CN⁻ with Cu²⁺. Always check:
- CN⁻ can reduce Cu²⁺ → Cu⁺ (and itself oxidises to cyanogen, (CN)2).
- The final complex has coordination number 4 and charge 3− for Cu⁺.
✗ Common Mistake #2: Forgetting the reduction step entirely
Mistake: Writing the complex as [Cu(CN)6]4− or similar, ignoring the change in oxidation state.
Why it’s wrong: Cu²⁺ typically forms hexacoordinate complexes (e.g., [Cu(H2O)6]2+), but with CN⁻, the reduction to Cu⁺ changes the coordination geometry to tetrahedral (CN⁻ is a strong field ligand).
✓ How to avoid:
- For Cu⁺, common coordination number is 4 (tetrahedral).
- For Cu²⁺, common coordination number is 6 (octahedral) or 4 (square planar).
- When CN⁻ is in excess, always check if reduction occurs — it does for Cu²⁺.
✗ Common Mistake #3: Saying “no precipitate because CuS is soluble”
Mistake: Claiming that CuS does not form because it is soluble in the complex solution.
Why it’s wrong: CuS is highly insoluble (Ksp≈8×10−37). The real reason is that free Cu⁺ ions are virtually absent due to the high stability of the complex.
✓ How to avoid:
- Use the concept of formation constant (Kf). For [Cu(CN)4]3−, Kf is extremely large (~1030).
- This means the equilibrium:
[Cu(CN)4]3−⇌Cu++4CN−
lies far to the left.
- So [Cu+] is negligible — insufficient to form any precipitate with S2−.
✗ Common Mistake #4: Writing the wrong precipitate formula …
Showing the 12 most recent of 23 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.Mohr's salt is-(a)(i) Fe₂(SO₄)₃.(NH₄)₂SO₄.6H₂O(b)(ii) FeSO₄.(NH₄)₂SO₄.6H₂O(c)(iii) MgSO₄.7H₂O(d)(iv) FeSO₄.7H₂O
›Reveal solutionSolution
Mohr's salt is ferrous ammonium sulphate hexahydrate, FeSO4⋅(NH4)2SO4⋅6H2O. Correct option: (ii).
Concept. Mohr's salt is a double salt — a stoichiometric combination of two simple salts, ferrous sulphate FeSO4 and ammonium sulphate (NH4)2SO4 — that dissolves in water to release all its constituent ions independently (Fe2+, NH4+, SO42−).
Why the other options are wrong.
- (i) Fe2(SO4)3⋅(NH4)2SO4⋅6H2O contains ferric iron (Fe3+) — that is ferric alum-type, not Mohr's salt.
- (iii) MgSO4⋅7H2O is Epsom salt. …
- CBSE 2026Set ANNUAL1 markMCQQ.Which of the following is a chelating ligand ?(a) NH3(b) H2O(c) Cl-(d) C2O4^2-
›Reveal solutionSolution
A chelating ligand grips the metal at more than one point; oxalate binds through two O atoms, forming a five-membered ring. Answer: (d) C2O4^2-.
- NH3, H2O and Cl- are all monodentate — each donates through a single atom, so they cannot chelate. …
- CBSE 2025Set A1 markQ.Write True or False: The Ca2+ and Mg2+ ions form stable complexes with EDTA.
›Reveal solutionSolution
EDTA is a hexadentate ligand that forms very stable chelate complexes with both Ca²⁺ and Mg²⁺.
EDTA (ethylenediaminetetraacetate) has six donor atoms (two N and four O, from its two amine groups and four carboxylate groups) that can simultaneously bind a single metal ion, wrapping around it to form a highly stable ring (chelate) structure — this is the chelate effect. Both Ca²⁺ and Mg²⁺ form such stable 1:1 octahedral EDTA complexes; …
- CBSE 2025Set A1 markQ.Write the central metal atom in [Ni(CO)4].
›Reveal solutionSolution
In [Ni(CO)4], the central metal atom to which all four ligands are directly bonded is nickel.
[Ni(CO)4] (tetracarbonylnickel(0)) is a classic coordination/organometallic compound in which a single nickel atom is surrounded by four neutral carbon monoxide (CO) ligands, each donating a lone pair from carbon to the metal. Since CO is a neutral ligand and the complex overall is neutral, nickel here is in the zero oxidation state, Ni(0), with electron …
- CBSE 2024Set 56/2/11 markMCQQ.Ligand EDTA4− is an example of a : (A) Monodentate ligand (B) Didentate ligand (C) Tridentate ligand (D) Polydentate ligand
›Reveal solutionSolution
EDTA⁴⁻ has six donor atoms (two N, four O) that can coordinate to a metal ion, making it a polydentate ligand. The correct option is (D).
Why This Question Tests Your Understanding of Denticity
The term "denticity" comes from the Latin dens (tooth) — it tells you how many "teeth" a ligand uses to bite into a metal ion. A monodentate ligand (like NH₃ or Cl⁻) grabs on with just one donor atom. A didentate ligand (like ethylenediamine, H₂N–CH₂–CH₂–NH₂) uses two. A tridentate uses three. And a polydentate ligand uses many — typically four or more.
The trick here is that EDTA⁴⁻ is not just any polydentate ligand; it's a classic example of a hexadentate ligand (six teeth). But the options don't list "hexadentate" — they list "polydentate" as the broad category. So the question is really: does EDTA⁴⁻ belong to the class of ligands that have many donor atoms? Yes.
Let's break down why.
Step-by-Step Reasoning
1. Identify the structure of EDTA⁴⁻
EDTA is ethylenediaminetetraacetic acid. In its fully deprotonated form (EDTA⁴⁻), it looks like this:
- A central ethylenediamine backbone: –CH₂–CH₂–, with a nitrogen atom at each end.
- Each nitrogen is attached to two –CH₂–COO⁻ groups.
So the molecule has:
- Two nitrogen atoms (each with a lone pair, so they can donate).
- Four carboxylate oxygen atoms (each negatively charged and carrying lone pairs).
That gives a total of six donor atoms.
2. Count how many of these can coordinate simultaneously
When EDTA⁴⁻ wraps around a metal ion (like Ca²⁺, Fe³⁺, or Co³⁺), all six donor atoms typically bind to the metal. The geometry is octahedral: the two N atoms and four O atoms occupy the six coordination sites.
Watch outA common mistake is to think EDTA is tridentate because it has three –COO⁻ groups per nitrogen, or to miscount the nitrogens. Always draw the structure: two N + four O = six.
3. Classify by denticity
- Monodentate: one donor atom → no. …
- CBSE 2024Set 56/2/11 markMCQQ.Which of the following ligand forms chelate complex ? (A) C2O42− (B) Cl− (C) NO2− (D) NH3
›Reveal solutionSolution
A chelate complex requires a ligand with two or more donor atoms that can simultaneously bind to the same metal center, forming a ring. Only oxalate ion C2O42− satisfies this criterion.
Understanding Chelation
A chelate complex forms when a single ligand attaches to a metal ion at multiple coordination sites, creating a ring structure. The word "chelate" comes from the Greek chele (claw), reflecting how these ligands "grab" the metal like a claw.
The key requirement: the ligand must be polydentate — it needs at least two donor atoms positioned so they can both reach the same metal center. When both donors bind, they close a ring that includes the metal ion. This ring formation is what distinguishes chelates from ordinary complexes.
Why does this matter? Chelate complexes are thermodynamically more stable than analogous complexes with monodentate ligands (the "chelate effect"). Once one donor atom binds, the second is already nearby and has a much higher probability of binding before the ligand diffuses away.
Examining Each Ligand
Let's evaluate each option systematically:
1. Oxalate ion, C2O42−
The structure is −O−C(=O)−C(=O)−O−. This ion has two oxygen donor atoms (one on each carboxylate group) separated by a two-carbon bridge. When oxalate binds to a metal, both oxygens coordinate simultaneously:
Mn++C2O42−→O−C(=O)MO−C(=O)(n−2)+
This forms a stable five-membered ring (metal + two oxygens + two carbons). Oxalate is a classic bidentate chelating ligand.
2. Chloride ion, Cl−
Chlorine has only one donor atom (itself). It can donate one lone pair to form a coordinate bond, but it cannot form a ring because there's no second donor site. Cl− is strictly monodentate.
3. Nitrite ion, NO2− …
- CBSE 2024Set A11 markQ.Transition metals form large number of complex compounds due to high ____________.
›Reveal solutionSolution
Transition-metal ions form many complexes because of their high ionic charge (high charge density on a small cation) together with vacant d-orbitals available to accept ligand lone pairs; the word from the printed bank is 'ionic charge'.
Transition-metal cations are small and carry a fairly high positive charge, i.e. they have a high ionic charge / charge density, which strongly attracts and polarises ligands; together with vacant d-orbitals of suitable energy to accept lone pairs, this is why they form a large number of complex compounds. The blank …
- CBSE 2024Set B1 markQ.Fill in the blank: E.D.T.A. is a ______ ligand.
›Reveal solutionSolution
EDTA has six donor atoms (two N and four O from its carboxylate groups) that can bind a single metal ion simultaneously, so it is a hexadentate ligand.
Ethylenediaminetetraacetic acid (EDTA), used as its tetra-anion form (EDTA4-), has:
- 2 nitrogen atoms (from the two amine groups of the ethylenediamine backbone)
- 4 oxygen atoms (from the four -COO- carboxylate groups) …
- CBSE 2024Set ANNUAL1 markMCQQ.Metal present in haemoglobin is -(a) Mn(b) Fe(c) Co(d) Ni
›Reveal solutionSolution
Haemoglobin is a coordination compound of iron - each haem unit has a central Fe2+ ion bound to a porphyrin ring, and this iron is what binds molecular oxygen.
Haemoglobin is the oxygen-carrying protein in red blood cells; it consists of four polypeptide (globin) chains, each associated with a haem group. …
- CBSE 2023Set 56/3/11 markMCQQ.Which of the following species is not expected to be a ligand? (A) CO (B) NH4+ (C) NH3 (D) H2O
›Reveal solutionSolution
A ligand must have at least one lone pair of electrons to donate to a metal centre. NH4+ has no lone pair — all four electron pairs are used in N–H bonds — so it cannot act as a ligand. The correct answer is (B).
Why this question is about lone pairs
In coordination chemistry, a ligand is any molecule or ion that donates a pair of electrons to a central metal atom or ion, forming a coordinate bond. The essential requirement is a lone pair of electrons — an unshared pair that can be offered to the metal. Without a lone pair, no donation is possible, and the species cannot function as a ligand.
So the task reduces to checking each option for the presence of at least one lone pair.
1. Check CO (carbon monoxide)
Carbon monoxide has the Lewis structure:
:C≡O:
The carbon atom has a lone pair, and the oxygen has two lone pairs. CO is a well-known ligand in metal carbonyls (e.g., Ni(CO)4).
Has lone pairs → can be a ligand.
2. Check NH4+ (ammonium ion)
Ammonium ion forms when NH3 accepts a proton (H+). In NH3, nitrogen has one lone pair. That lone pair is exactly what binds the H+, forming a fourth N–H bond. In NH4+, all four electron pairs around nitrogen are used in sigma bonds — there are no lone pairs left.
Watch outA common mistake is to think that because NH3 is a ligand, NH4+ must also be one. But the lone pair that made NH3 a ligand is gone — it’s now part of an N–H bond. NH4+ is a cation with no available electron pair for donation.
No lone pair → cannot be a ligand.
3. Check NH3 (ammonia) …
- CBSE 2023Set A1 markQ.Fill in the blank: The chemical name of EDTA is ______.
›Reveal solutionSolution
EDTA is the common abbreviation for ethylenediaminetetraacetic acid, a hexadentate chelating ligand widely used in coordination chemistry and titrations.
EDTA's full chemical name is ethylenediaminetetraacetic acid, (HOOCCH2)2NCH2CH2N(CH2COOH)2. It has two nitrogen donor atoms and four carboxylate oxygen donor atoms, making it a hexadentate ligand that forms very stable complexes (chelates …
- CBSE 2023Set ANNUAL1 markQ.Give an example for a didentate ligand.
›Reveal solutionSolution
A didentate (bidentate) ligand has two donor atoms that can simultaneously bind to the same central metal ion; ethylenediamine is a classic example.
A ligand is classified by the number of donor atoms it uses to bind to the central metal atom/ion. A didentate (bidentate) ligand has exactly two donor atoms, each with a lone pair, that coordinate to the metal at the same time, usually forming a stable 5- or 6-membered chelate ring.
…
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