Q.FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+ ion. Explain why?
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.
-
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.
A common mistake is to think complex formation happens in one step. It doesn't. The deep blue [Cu(NH3)4]2+ forms through four separate equilibria, each with its own constant. The first ammonia binds easily, the fourth one less so — because the metal is already crowded.
Why It Matters
Complex formation is everywhere in chemistry and biology:
- In your blood: Hemoglobin is an iron complex that carries oxygen.
- In water treatment: EDTA is added to bind metal ions and prevent scaling.
- In qualitative analysis: The deep blue of copper-ammonia complex is a classic test for copper ions.
- In medicine: Cisplatin, a platinum complex, is a chemotherapy drug.
The key takeaway: complex formation is about electron-pair donation from ligands to a metal, creating a stable, often colourful, coordination entity. The stability depends on the metal, the ligand, and the conditions — and it always happens step by step.
Complex formation through ligand electron-pair donation is a central theme of the NCERT/CBSE Class 12 Chemistry chapter on Coordination Compounds, and ‘complex formation important questions’ or ‘coordination complex formation examples’ are commonly searched for board exams, JEE Main and NEET. Real-world applications like EDTA and cisplatin also make this a favourite for application-based competitive-exam questions.
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 metal ion is more effectively "tied up" by the ligand
This is why Kf is also called the stability constant.
6. Quick Summary — The "Why" in One Line
The formula Kf=[M][L]n[MLn] arises because complex formation is a series of n sequential equilibria, each adding one ligand, and multiplying the stepwise constants gives the overall constant with [L]n in the denominator.
7. Exam Tip
- Stepwise constants (K1,K2,…) are usually not equal — often K1>K2>K3 due to steric hindrance and statistical factors.
- The overall constant Kf is what you use for direct calculations involving the fully formed complex.
- Always check: Is the ligand monodentate or polydentate? For polydentate ligands (like EDTA), the formula still holds — n is the number of donor atoms that bind.
Final thought: The beauty of this derivation is that it shows how a complex multi-step process collapses into a single, elegant equilibrium expression — as long as you remember the stepwise logic behind it.
The key idea is Complex Formation — the difference arises because one mixture forms a stable complex that masks the metal ion, while the other does not.
Reasoning:
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FeSO4+(NH4)2SO4 (1:1): This simply forms a double salt, Mohr's salt, FeSO4⋅(NH4)2SO4⋅6H2O. In solution, it dissociates completely into Fe2+, NH4+, and SO42− ions. The Fe2+ ion remains free and uncomplexed, so it gives its characteristic tests (e.g., the deep blue Turnbull's blue precipitate with K3[Fe(CN)6]).
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CuSO4+NH3(aq) (1:4): Ammonia acts as a ligand. Cu2+ initially forms a pale blue precipitate of Cu(OH)2, which then dissolves in excess ammonia to form the deep blue tetraamminecopper(II) complex:
Cu2++4NH3→[Cu(NH3)4]2+
This complex is very stable, so the concentration of free Cu2+ ions becomes negligible. Hence, tests that rely on free Cu2+ (e.g., the chocolate-brown precipitate of Cu2[Fe(CN)6] with K4[Fe(CN)6]) fail.
The Fe2+ ion remains free in the double salt solution, but Cu2+ is sequestered into the stable [Cu(NH3)4]2+ complex, masking its characteristic tests.
The key is complex formation: Fe2+ forms a weak, labile complex with (NH4)2SO4 that still releases free Fe2+ for its characteristic test, whereas Cu2+ forms a very stable deep-blue tetraammine complex [Cu(NH3)4]2+ that locks up virtually all Cu2+ ions, preventing them from giving the usual Cu2+ test.
The question contrasts two seemingly similar mixtures — but the chemistry of the two metal ions is fundamentally different. Let’s see why.
1. What does “test of the ion” mean?
When we say a solution “gives the test” of Fe2+ or Cu2+, we mean that the free, hydrated metal ion is present in sufficient concentration to react with a specific reagent (like K3[Fe(CN)6] for Fe2+, or NH3 itself for Cu2+) to produce a characteristic colour or precipitate. If the metal ion is tightly bound in a complex, it may not be available for that test.
2. The iron(II) case: FeSO4+(NH4)2SO4 (1:1 molar ratio)
(NH4)2SO4 provides NH4+ and SO42− ions. It does not provide free ammonia (NH3) in significant amount — ammonium ion is a weak acid (pKa≈9.25), so in neutral solution it does not release enough NH3 to form ammine complexes with Fe2+.
Fe2+ does form weak complexes with sulfate ([FeSO4]0 ion pair) and possibly with water, but these are labile — they dissociate instantly. The Fe2+ remains essentially as the hexaaqua ion [Fe(H2O)6]2+ in solution.
Even if a tiny amount of NH3 were present, Fe2+ forms only weak ammine complexes (unlike Fe3+ or Cu2+). The equilibrium heavily favours free Fe2+.
So when you add a test reagent like potassium ferricyanide, you get the deep blue Turnbull’s blue precipitate:
3Fe2++2[Fe(CN)6]3−→Fe3[Fe(CN)6]2↓
The test works because free Fe2+ is abundant.
3. The copper(II) case: CuSO4+aqueous ammonia (1:4 molar ratio)
Here, aqueous ammonia (NH3 in water) is a strong ligand and is present in excess (4 moles NH3 per mole Cu2+). Cu2+ has a strong tendency to form ammine complexes. The reaction proceeds stepwise:
[Cu(H2O)6]2++NH3⇌[Cu(NH3)(H2O)5]2++H2O
⋮
[Cu(NH3)3(H2O)3]2++NH3⇌[Cu(NH3)4(H2O)2]2++H2O
The overall formation constant for the tetraammine complex is very large:
Kf=[Cu2+][NH3]4[[Cu(NH3)4]2+]≈1012 to 1013
Cu2++4NH3⇌[Cu(NH3)4]2+Kf≈2×1012
With 1:4 stoichiometry and such a high Kf, essentially all Cu2+ ions are converted to the deep blue [Cu(NH3)4]2+ complex. The concentration of free Cu2+ drops to astronomically low levels (on the order of 10−12 M or less).
A common mistake is to think that because the solution is blue, it still contains Cu2+ ions. The blue colour is from the complex [Cu(NH3)4]2+, not from free Cu2+ (which is pale blue). The test for Cu2+ (e.g., with K4[Fe(CN)6] to give a chocolate brown precipitate of Cu2[Fe(CN)6]) requires free Cu2+ — which is virtually absent.
4. The critical difference: stability and lability
- Fe2+ with (NH4)2SO4: No strong complex forms; free Fe2+ remains.
- Cu2+ with NH3: A thermodynamically very stable complex forms (Kf≈1012), so the equilibrium leaves virtually no free Cu2+ — the masking comes from this huge formation constant. (Cu(II) complexes are actually kinetically labile — ligands exchange fast — but the equilibrium position keeps free Cu2+ negligible.)
The 1:4 molar ratio for Cu2+:NH3 is exactly the stoichiometry needed to form the tetraammine complex. If you used less NH3, some free Cu2+ would remain and the test would partially work. But at 1:4, the complexation is essentially complete.
5. Why doesn’t the same happen with Fe2+ and (NH4)2SO4?
Even if we replaced (NH4)2SO4 with actual NH3 solution, Fe2+ forms much weaker ammine complexes (Kf for [Fe(NH3)6]2+ is only about 102.2 — negligible compared to copper). Plus, (NH4)2SO4 doesn’t even provide free NH3 in the first place.
›Proof
Why (NH4)2SO4 doesn’t release NH3 significantly:
NH4+⇌NH3+H+ has Ka=5.6×10−10. In a neutral solution (pH≈7), the ratio [NH3]/[NH4+]=Ka/[H+]≈5.6×10−3. So less than 1% of ammonium is present as NH3 — far too little to complex Fe2+ even if it wanted to.
The FeSO4–(NH4)2SO4 mixture leaves Fe2+ free because no strong complex forms, while the CuSO4–NH3 (1:4) mixture completely converts Cu2+ into the stable [Cu(NH3)4]2+ complex, which does not give the characteristic Cu2+ test.
Concept: Complex Formation
Method: Ligand Substitution & Coordination Number Analysis
This method explains why certain metal ions become "hidden" from their characteristic tests due to the formation of stable coordination complexes.
Step-by-Step Explanation
Step 1: Identify the reactants and their molar ratios
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Case 1: FeSO4 + (NH4)2SO4 in 1:1 molar ratio
- Fe2+ is present, and (NH4)2SO4 provides NH4+ and SO42− ions — no strong ligand like NH3 is available in significant amount.
-
Case 2: CuSO4 + aqueous ammonia in 1:4 molar ratio
- Cu2+ is present, and NH3 (a strong ligand) is available in excess (4 moles per mole of Cu2+).
Step 2: Check coordination tendency of each metal ion
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Fe2+ has a coordination number of 6, but here the ligand is only SO42− (a weak ligand) and water. No strong ligand like NH3 is present in sufficient quantity to form a stable complex that masks Fe2+.
- Result: Fe2+ remains as free ions → gives its characteristic test.
-
Cu2+ has a coordination number of 4 (common for tetrahedral or square planar complexes with NH3).
- With 4 moles of NH3 per mole of Cu2+, the following complex forms:
Cu2++4NH3→[Cu(NH3)4]2+
This is a deep blue, stable complex.
Step 3: Consequence for ion detection
- In the FeSO4 case: No stable complex forms → Fe2+ ions are free → test positive.
- In the CuSO4 case: All Cu2+ ions are bound in [Cu(NH3)4]2+ → no free Cu2+ → test for Cu2+ (e.g., with K4[Fe(CN)6] or NaOH) fails because the ligand is already occupying all coordination sites.
Final Answer
FeSO4 with (NH4)2SO4 does not form a stable complex (no strong ligand in excess), so Fe2+ remains free and gives its test.
CuSO4 with NH3 in 1:4 ratio forms [Cu(NH3)4]2+, a stable complex that masks Cu2+ ions, so no test for Cu2+ is obtained.
Key concept: Excess of a strong ligand can completely complex a metal ion, removing it from solution as a free ion.
Common Mistakes Students Make on This Question
This is a classic coordination chemistry question from JEE/NEET. Here are the most frequent errors and how to avoid each:
Mistake 1: Thinking Both Reactions Are Similar
The error: Students assume both mixtures form simple salts or double salts, and wonder why only one gives the cation test.
Why it's wrong:
- FeSO4+(NH4)2SO4 forms a double salt — Mohr's salt FeSO4⋅(NH4)2SO4⋅6H2O
- CuSO4+NH3 forms a coordination complex — [Cu(NH3)4]SO4
How to avoid:
- Double salts dissociate completely in water into simple ions.
- Coordination complexes bind the metal ion tightly — the metal is no longer free to give its characteristic tests.
Mistake 2: Forgetting the Key Difference — Dissociation vs. Complexation
The error: Students write both as "salt formation" and miss the core concept.
Correct understanding:
| Mixture | Product | In water, it gives |
|---|---|---|
| FeSO4+(NH4)2SO4 | Mohr's salt (double salt) | Fe2+, NH4+, SO42− — all free ions |
| CuSO4+4NH3 | [Cu(NH3)4]SO4 (complex) | [Cu(NH3)4]2+, SO42− — no free Cu2+ |
How to avoid:
- Remember: Double salts exist only in solid state; in solution they break into constituent ions.
- Complexes have coordinate bonds that survive in solution — the metal is "masked."
Mistake 3: Ignoring the Molar Ratio Given
The error: Students ignore the "1:1" and "1:4" ratios and treat the problem generically.
Why it matters:
- 1:1 ratio of FeSO4:(NH4)2SO4 is exactly the stoichiometry for Mohr's salt — no excess ammonia to form a complex.
- 1:4 ratio of CuSO4:NH3 is exactly the stoichiometry for the tetraamminecopper(II) complex — all Cu2+ gets bound.
How to avoid:
- Always check if the given ratio matches the formula of a known complex or double salt.
- If excess ligand is present, complex formation is likely.
Mistake 4: Confusing the Tests for Fe2+ and Cu2+
The error: Students think both tests fail or both succeed.
Correct:
- Fe2+ test (e.g., with K3[Fe(CN)6] giving a dark blue precipitate) works because free Fe2+ exists.
- Cu2+ test (e.g., with K4[Fe(CN)6] giving a chocolate brown precipitate) fails because Cu2+ is locked in [Cu(NH3)4]2+.
How to avoid:
- Know the specific tests for each ion.
- Understand that a complexed metal ion does not respond to simple precipitation tests for the free ion.
Mistake 5: Writing Incorrect Formulae
The error: Students write FeSO4⋅(NH4)2SO4 as a complex or [Cu(NH3)4]SO4 as a double salt.
How to avoid:
- Double salt formula: FeSO4⋅(NH4)2SO4⋅6H2O — written with a dot, not brackets.
- Complex formula: [Cu(NH3)4]SO4 — square brackets indicate the coordination sphere.
Quick Summary to Remember
| Concept | Double Salt | Coordination Complex |
|---|---|---|
| Bonding | Ionic (electrostatic) | Coordinate covalent |
| In solution | Dissociates completely | Complex ion stays intact |
| Metal ion test | Positive | Negative |
| Example | Mohr's salt | [Cu(NH3)4]SO4 |
Final takeaway:
The Fe2+ test works because Mohr's salt is a double salt that breaks apart. The Cu2+ test fails because Cu2+ is tightly bound in a complex ion — no free Cu2+ remains to react.
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.
- (iv) FeSO4⋅7H2O is green vitriol (ordinary ferrous sulphate), not the double salt.
Mohr's salt is a standard primary standard for redox titrations because Fe2+ in it is comparatively resistant to aerial oxidation.
✓Final answer(ii) FeSO₄·(NH₄)₂SO₄·6H₂O — ferrous ammonium sulphate hexahydrate.
- 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-.
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NH3, H2O and Cl- are all monodentate — each donates through a single atom, so they cannot chelate.
-
Oxalate ion C2O4^2- has two carboxylate oxygen donor atoms and coordinates to the metal at two positions, forming a stable chelate ring.
✓Final answer(d) C2O4^2- is the chelating ligand.
-
- 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; this is exploited practically in complexometric titrations used to estimate the 'hardness' of water (total Ca²⁺ + Mg²⁺ content) by titrating with a standard EDTA solution.
✓Final answerTrue.
- 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 configuration 3d104s0 (a d10 system, consistent with its tetrahedral geometry). The central metal atom — the species all the ligands coordinate to — is nickel.
✓Final answerNickel (Ni).
- 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.
- Didentate: two donor atoms → no.
- Tridentate: three donor atoms → no.
- Polydentate: many donor atoms (usually ≥4) → yes.
Since EDTA⁴⁻ has six donor atoms, it falls under the umbrella of polydentate ligands. In fact, it's a specific subtype called a hexadentate ligand, but "polydentate" is the correct general answer here.
TipIf the options had included "hexadentate", that would be even more precise. But since it's not listed, "polydentate" is the only correct choice. Always pick the broadest accurate category when the specific term is absent.
4. Eliminate the wrong options
- (A) Monodentate: clearly wrong — EDTA⁴⁻ has many donor atoms.
- (B) Didentate: wrong — it has far more than two.
- (C) Tridentate: wrong — it has six, not three.
- (D) Polydentate: correct — it has multiple donor atoms.
✓Final answerThe correct option is (D) Polydentate ligand.
- 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−
While NO2− has both nitrogen and oxygen atoms, it typically coordinates through only one atom at a time — either the nitrogen (nitro, M−NO2) or an oxygen (nitrito, M−ONO). It's an ambidentate ligand (can bind through different atoms in different complexes), but not a chelating one because it doesn't use multiple donors simultaneously with the same metal.
4. Ammonia, NH3
Ammonia has one nitrogen donor atom with a lone pair. Like chloride, it's monodentate and cannot form chelate rings.
TipTo quickly identify chelating ligands, look for molecules with multiple lone pairs on different atoms separated by 2–3 atoms (optimal for five- or six-membered rings). Common examples: ethylenediamine (en), EDTA, acetylacetonate (acac⁻), and oxalate.
Watch outDon't confuse "ambidentate" with "chelating." An ambidentate ligand like NO2− or SCN− can bind through different atoms in different complexes, but still uses only one donor at a time — no ring forms.
✓Final answerThe correct option is (A) C2O42−, which forms a five-membered chelate ring through its two carboxylate oxygen donors.
- 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 in this fill-in-the-blank question is completed with the word 'ionic charge' from the printed bracket (the other listed words - Grignard reagent, C6H5N2+Cl-, collision frequency, molality, molarity - do not fit).
✓Final answerhigh ionic charge
- 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)
All six donor atoms coordinate to a single central metal ion at once, wrapping around it to form a very stable, cage-like chelate complex. Because it donates through six sites, EDTA is classified as a hexadentate (six-toothed) ligand — one of the most common polydentate/chelating ligands, widely used in complexometric titrations and in treating heavy-metal poisoning.
✓Final answerHexadentate (polydentate) ligand.
- 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.
In each haem group, an Fe2+ ion is coordinated to four nitrogen atoms of a porphyrin ring in a square-planar arrangement, with a fifth position bonded to a histidine residue of the protein and the sixth (axial) position available to reversibly bind an O2 molecule.
This is a classic example of the biological importance of coordination compounds cited in the NCERT syllabus.
✓Final answer(b) Fe.
- 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)
Ammonia has the familiar structure: nitrogen with three N–H bonds and one lone pair. That lone pair is available for donation — NH3 is a classic ligand (e.g., in [Cu(NH3)4]2+).
Has lone pair → can be a ligand.
4. Check H2O (water)
Water has two O–H bonds and two lone pairs on oxygen. Water is a common ligand in many hydrated metal complexes (e.g., [Fe(H2O)6]3+).
Has lone pairs → can be a ligand.
TipA quick mental check: if a species can act as a Lewis base (electron pair donor), it can be a ligand. NH4+ is a Lewis acid — it can accept a pair, not donate one.
✓Final answerThe species that is not expected to be a ligand is (B) NH4+.
- 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) with metal ions such as Ca²⁺, Mg²⁺, Pb²⁺ — this is the basis of EDTA titrations and its use in treating heavy-metal poisoning.
✓Final answerEthylenediaminetetraacetic acid.
- 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.
Ethylenediamine (en), H₂N–CH₂–CH₂–NH₂, has two –NH₂ groups, each with a lone pair on nitrogen, and both nitrogen atoms bind simultaneously to the metal ion — making it a didentate ligand.
(Other valid examples: the oxalate ion, C₂O₄²⁻, which binds through two oxygen atoms.)
✓Final answerEthylenediamine (en), H₂N–CH₂–CH₂–NH₂ — a didentate ligand (two N donor atoms). Oxalate ion (C₂O₄²⁻) is another valid example.
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