Assertion (A) : Au and Ag are extracted by leaching their ores with a dil. solution of NaCN. Reason (R) : Impu

Question

Q. Assertion (A) : Au and Ag are extracted by leaching their ores with a dil. solution of NaCN. Reason (R) : Impurities associated with these ores dissolve in NaCN. (A) Both Assertion (A) and Reason (R) are correct statements, and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is correct, but Reason (R) is incorrect statement. (D) Assertion (A) is incorrect, but Reason (R) is correct statement.

CBSECBSE Class XII Board 2020MCQ· 1mImportance★★★★★

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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. …

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) …

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Gold and silver are indeed leached with dilute NaCN, but not because impurities dissolve — rather, the metals themselves form soluble cyanide complexes while impurities remain behind. The correct option is (C).

The extraction of precious metals like gold and silver relies on a beautiful piece of coordination chemistry called the cyanide process or MacArthur-Forrest process. Understanding why this works requires grasping what happens at the molecular level when cyanide meets these noble metals.

Why cyanide leaching works

Gold and silver, despite being "noble" (unreactive), have a special affinity for cyanide ions. When crushed ore containing these metals is treated with dilute aqueous NaCN in the presence of air, the metals undergo oxidation and simultaneously form stable, water-soluble complex ions. The key reactions are:

4\text{Au} + 8\text{CN}^- + \text{O}_2 + 2\text{H}_2\text{O} \longrightarrow 4[\text{Au(CN)}_2]^- + 4\text{OH}^-

4\text{Ag} + 8\text{CN}^- + \text{O}_2 + 2\text{H}_2\text{O} \longrightarrow 4[\text{Ag(CN)}_2]^- + 4\text{OH}^-

The dicyanoaurate(I) and dicyanoargentate(I) complexes are highly soluble and can be separated from the insoluble gangue (rocky impurities). The dissolved metal is later recovered by displacement with zinc:

2[\text{Au(CN)}_2]^- + \text{Zn} \longrightarrow [\text{Zn(CN)}_4]^{2-} + 2\text{Au}

Now let's evaluate the assertion and reason.

Step-by-step analysis

1. Examining Assertion (A)

The assertion states that Au and Ag are extracted by leaching with dilute NaCN. This is absolutely correct and is the standard industrial method for extracting these metals from low-grade ores. The process has been in use since the 1880s and remains the dominant technique worldwide.

2. Examining Reason (R)

The reason claims that impurities dissolve in NaCN. This is fundamentally incorrect. The entire point of the cyanide process is that the valuable metals dissolve while the impurities (silicates, aluminosilicates, and other gangue minerals) do not. The process works precisely because of this selectivity.

If impurities dissolved instead of the metals, the process would be useless — you'd end up with a solution full of worthless material and your gold still trapped in the solid residue. …

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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. …
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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. …
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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; …

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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 …

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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. …
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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, $NO_2^-$ …
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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 …
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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) …
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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. …

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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) …
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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 …

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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.

…

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Why cyanide leaching works

Step-by-step analysis

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