Ligand EDTA⁴⁻ is an example of a : (A) Monodentate ligand (B) Didentate ligand (C) Tridentate ligand (D) Polyd

Question

EDTA^{4-}

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★

✓ Free question

Concept understanding — Complex Formation

Complex Formation: The Intuition

Imagine you have a metal ion — say, a copper ion ( $\text{Cu}^{2+}$ ) — 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: $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ .

Now, suppose you add ammonia ( $\text{NH}_3$ ) 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: $[\text{Cu}(\text{NH}_3)_4]^{2+}$ .

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.

Note

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 ( $K_f$ ) or stability constant.

For a general reaction:

\text{M} + n\text{L} \rightleftharpoons \text{ML}_n

The overall formation constant is:

K_f = \frac{[\text{ML}_n]}{[\text{M}][\text{L}]^n}

A large $K_f$ means the complex is very stable — the ligands bind tightly and are hard to remove.

K_f = \frac{[\text{complex}]}{[\text{metal}][\text{ligand}]^n}

Key Features to Remember

Stepwise nature: Complexes don't form all at once. First one ligand binds, then another, and so on. Each step has its own constant ( $K_1, K_2, \dots$ ). The product of all stepwise constants equals the overall $K_f$ .
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 $[\text{Cu}(\text{NH}_3)_4]^{2+}$ , the coordination number is 4.
Ligand denticity: A ligand can have one donor atom (monodentate, like $\text{NH}_3$ or $\text{H}_2\text{O}$ ) 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.

Watch out

A common mistake is to think complex formation happens in one step. It doesn't. The deep blue $[\text{Cu}(\text{NH}_3)_4]^{2+}$ 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:

$Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$

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:

$M + L \rightleftharpoons ML$

Equilibrium constant: $K_1 = \frac{[ML]}{[M][L]}$

Step 2:

$ML + L \rightleftharpoons ML_2$

$K_2 = \frac{[ML_2]}{[ML][L]}$

Step 3:

$ML_2 + L \rightleftharpoons ML_3$

$K_3 = \frac{[ML_3]}{[ML_2][L]}$

... and so on up to $ML_n$ .

Each $K_i$ 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:

$M + nL \rightleftharpoons ML_n$

We can multiply the stepwise equilibria (because when you add reactions, you multiply their equilibrium constants):

$K_f = K_1 \times K_2 \times K_3 \times \dots \times K_n$

So:

\boxed{K_f = \frac{[ML_n]}{[M][L]^n}}

Why this form?

Because each step contributes one $[L]$ in the denominator and one $[ML_i]$ 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: $[ML] = K_1[M][L]$

From step 2: $[ML_2] = K_2[ML][L] = K_1K_2[M][L]^2$

From step 3: $[ML_3] = K_3[ML_2][L] = K_1K_2K_3[M][L]^3$

Continuing:

$[ML_n] = (K_1K_2\dots K_n)[M][L]^n$

Therefore:

\frac{[ML_n]}{[M][L]^n} = K_1K_2\dots K_n = K_f

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 $K_f$ 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 $K_f$ is also called the stability constant.

6. Quick Summary — The "Why" in One Line

The formula $K_f = \frac{[ML_n]}{[M][L]^n}$ 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 ( $K_1, K_2, \dots$ ) are usually not equal — often $K_1 > K_2 > K_3$ due to steric hindrance and statistical factors.
The overall constant $K_f$ 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.

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 out

A 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.

Tip

If 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 answer

The correct option is (D) Polydentate ligand .

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 23 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Mohr's salt is-$
(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, $\text{FeSO}_4\cdot(\text{NH}_4)_2\text{SO}_4\cdot 6\text{H}_2\text{O}$ . Correct option: (ii).

Concept. Mohr's salt is a double salt — a stoichiometric combination of two simple salts, ferrous sulphate $\text{FeSO}_4$ and ammonium sulphate $(\text{NH}_4)_2\text{SO}_4$ — that dissolves in water to release all its constituent ions independently ( $\text{Fe}^{2+}$ , $\text{NH}_4^{+}$ , $\text{SO}_4^{2-}$ ).

Why the other options are wrong.

(i) $\text{Fe}_2(\text{SO}_4)_3\cdot(\text{NH}_4)_2\text{SO}_4\cdot 6\text{H}_2\text{O}$ contains ferric iron ( $\text{Fe}^{3+}$ ) — that is ferric alum-type, not Mohr's salt.

(iii) $\text{MgSO}_4\cdot 7\text{H}_2\text{O}$ is Epsom salt.

(iv) $\text{FeSO}_4\cdot 7\text{H}_2\text{O}$ is green vitriol (ordinary ferrous sulphate), not the double salt.

Mohr's salt is a standard primary standard for redox titrations because $\text{Fe}^{2+}$ in it is comparatively resistant to aerial oxidation.

✓Final answer
$(ii) FeSO₄\cdot(NH₄)₂SO₄\cdot6H₂O — ferrous ammonium sulphate hexahydrate.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following is a chelating ligand ?$
(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.

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 mark
$Q. Write True or False: The Ca2+ and Mg2+ ions form stable complexes with EDTA.$
›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 answer
$True.$
CBSE 2025Set A1 mark
$Q. Write the central metal atom in [Ni(CO)4].$
›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 $3d^{10}4s^0$ (a $d^{10}$ system, consistent with its tetrahedral geometry). The central metal atom — the species all the ligands coordinate to — is nickel.

✓Final answer
$Nickel (Ni).$
CBSE 2024Set 56/2/11 markMCQ
$EDTA^{4-}$
›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 out

A 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.

Tip

If 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 answer
$The correct option is (D) Polydentate ligand .$
CBSE 2024Set 56/2/11 markMCQ
$C_2O_4^{2-}$
›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 $C_2O_4^{2-}$ 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, $C_2O_4^{2-}$

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:
$\text{M}^{n+} + C_2O_4^{2-} \rightarrow \begin{bmatrix} O-C(=O) \\ \text{M} \\ O-C(=O) \end{bmatrix}^{(n-2)+}$
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, $NO_2^-$

While $NO_2^-$ has both nitrogen and oxygen atoms, it typically coordinates through only one atom at a time — either the nitrogen (nitro, $\text{M}-NO_2$ ) or an oxygen (nitrito, $\text{M}-ONO$ ). 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, $NH_3$

Ammonia has one nitrogen donor atom with a lone pair. Like chloride, it's monodentate and cannot form chelate rings.

Tip

To 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 out

Don't confuse "ambidentate" with "chelating." An ambidentate ligand like $NO_2^-$ or $SCN^-$ can bind through different atoms in different complexes, but still uses only one donor at a time — no ring forms.

✓Final answer
$C_2O_4^{2-}$
CBSE 2024Set A11 mark
$Q. Transition metals form large number of complex compounds due to high ____________.$
›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 answer
$high ionic charge$
CBSE 2024Set B1 mark
$Q. Fill in the blank: E.D.T.A. is a ______ ligand.$
›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 answer
$Hexadentate (polydentate) ligand.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Metal present in haemoglobin is -$
(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 markMCQ
$CO$
›Reveal solutionSolution

A ligand must have at least one lone pair of electrons to donate to a metal centre. $NH_4^+$ 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:

$:C \equiv O:$

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 $NH_4^+$ (ammonium ion)

Ammonium ion forms when $NH_3$ accepts a proton ( $H^+$ ). In $NH_3$ , nitrogen has one lone pair. That lone pair is exactly what binds the $H^+$ , forming a fourth N–H bond. In $NH_4^+$ , all four electron pairs around nitrogen are used in sigma bonds — there are no lone pairs left.

Watch out

A common mistake is to think that because $NH_3$ is a ligand, $NH_4^+$ must also be one. But the lone pair that made $NH_3$ a ligand is gone — it’s now part of an N–H bond. $NH_4^+$ is a cation with no available electron pair for donation.

No lone pair → cannot be a ligand.

3. Check $NH_3$ (ammonia)

Ammonia has the familiar structure: nitrogen with three N–H bonds and one lone pair. That lone pair is available for donation — $NH_3$ is a classic ligand (e.g., in $[Cu(NH_3)_4]^{2+}$ ).

Has lone pair → can be a ligand.

4. Check $H_2O$ (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., $[Fe(H_2O)_6]^{3+}$ ).

Has lone pairs → can be a ligand.

Tip

A quick mental check: if a species can act as a Lewis base (electron pair donor), it can be a ligand. $NH_4^+$ is a Lewis acid — it can accept a pair, not donate one.

✓Final answer
$NH_4^+$
CBSE 2023Set A1 mark
$Q. Fill in the blank: The chemical name of EDTA is ______.$
›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, $(HOOCCH_2)_2NCH_2CH_2N(CH_2COOH)_2$ . 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 answer
$Ethylenediaminetetraacetic acid.$
CBSE 2023Set ANNUAL1 mark
$Q. Give an example for a didentate ligand.$
›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 answer
$Ethylenediamine (en), H₂N-CH₂-CH₂-NH₂ — a didentate ligand (two N donor atoms). Oxalate ion (C₂O₄²⁻) is another valid example.$

Q.Ligand EDTA4−EDTA^{4-}EDTA4− is an example of a : (A) Monodentate ligand (B) Didentate ligand (C) Tridentate ligand (D) Polydentate ligand

Why This Question Tests Your Understanding of Denticity

Step-by-Step Reasoning

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Q.Ligand $EDTA^{4-}$ is an example of a : (A) Monodentate ligand (B) Didentate ligand (C) Tridentate ligand (D) Polydentate ligand

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