CUET UG Chemistry Booster Test

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CUET UG Chemistry Booster Test - 2 Fundamentals

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Chemistry Booster Test - 2 Fundamentals

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QUESTION 1 OF 20

The structural difference between a double salt and a coordination complex is that:

Answer Review

1 The structural difference between a double salt and a coordination complex is that:

A Double salts do not dissociate completely in water.

B Complex ions completely dissociate into simple ions in water.

C Complex ions such as [Fe(CN)₆]⁴⁻ do not dissociate into simple ions like Fe²⁺ and CN⁻ in water.

D Both double salts and complexes dissociate completely into single ions.

💡 Short Explanation

�� Double salts dissociate completely in water. �� Complex ions remain intact in solution. �� Coordination sphere is preserved.

✅ Correct Answer Explanation

Double salts such as Mohr's salt dissociate completely into constituent ions in aqueous solution. However, coordination compounds produce complex ions that remain intact and do not dissociate into simple metal ions and ligands. For example, [Fe(CN)₆]⁴⁻ remains as a coordination entity in solution.

❓ Why Other Options Are Incorrect

�� Option A → Double salts do dissociate completely.
�� Option B → Complex ions remain intact.
�� Option D → Only double salts dissociate completely into simple ions.

🎯 Strategy

Used

Odd One Out

Application:

Compare the behavior of double salts and coordination compounds in water.

Final Logic:

Complex ions remain intact whereas double salts dissociate completely.

🧠 Memory Trick

Double Salt Breaks, Complex Stays

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Difference Between Double Salts and Coordination Compounds

2 Consider the following statements regarding the analytical and industrial importance of coordination compounds. Choose the correct statements:
1. Hardness of water is estimated by titration with Na₂EDTA.
2. Ca²⁺ and Mg²⁺ form unstable complexes with EDTA.
3. Extraction of silver and gold makes use of complex formation.
4. Hypo solution dissolves AgBr to form the complex ion [Ag(S₂O₃)₂]³⁻.

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 2, 4

💡 Short Explanation

�� EDTA is used in hardness determination. �� Gold and silver extraction use complex formation. �� Hypo forms a soluble silver-thiosulphate complex.

✅ Correct Answer Explanation

Statement 1 is correct because EDTA titration is used to estimate water hardness. Statement 3 is correct because cyanide complex formation is employed in gold and silver extraction. Statement 4 is correct because AgBr dissolves in hypo (Na₂S₂O₃) forming [Ag(S₂O₃)₂]³⁻. Statement 2 is incorrect because Ca²⁺ and Mg²⁺ form stable complexes with EDTA.

❓ Why Other Options Are Incorrect

�� Statement 2 → EDTA forms stable chelates with Ca²⁺ and Mg²⁺.
�� Option A → Contains incorrect Statement 2.
�� Option C → Contains incorrect Statement 2.
�� Option D → Contains incorrect Statement 2.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate Statements 1, 2, 3 and 4 independently.

Final Logic:

Only Statements 1, 3 and 4 are correct.

🧠 Memory Trick

EDTA Measures Hardness

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Applications of Coordination Compounds

3 Identify the chemical reaction process used for the purification of nickel:

A Formation and subsequent thermal decomposition of the coordination compound [Ni(CO)₄]

B Electrolysis of the complex [Ni(CN)₄]²⁻

C Precipitation of nickel using Na₂EDTA

D Hydrogenation of nickel using Wilkinson catalyst

💡 Short Explanation

�� Mond process purifies nickel. �� Volatile nickel carbonyl is formed. �� Thermal decomposition yields pure nickel.

✅ Correct Answer Explanation

In Mond's process, impure nickel reacts with carbon monoxide to form volatile Ni(CO)₄. This compound is then heated to decompose it into pure nickel and carbon monoxide. Ni + 4CO → Ni(CO)₄ Ni(CO)₄ → Ni + 4CO Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Not used for nickel purification.
�� Option C → EDTA is not employed for nickel refining.
�� Option D → Wilkinson catalyst is used for alkene hydrogenation.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall industrial applications of coordination compounds.

Final Logic:

Nickel purification occurs through the Mond process.

🧠 Memory Trick

Mond = Nickel Carbonyl

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Mond Process

4 Match the catalyst/reagent (List I) with its specific biological or medical role (List II):

List I	List II
1. Vitamin B₁₂	a. Treatment of lead poisoning
2. EDTA	b. Removal of excess copper and iron
3. Cis-platin	c. Anti-pernicious anaemia factor
4. D-penicillamine	d. Inhibition of tumour gr

A 1-c, 2-a, 3-d, 4-b

B 1-a, 2-c, 3-d, 4-b

C 1-d, 2-b, 3-a, 4-c

D 1-b, 2-d, 3-c, 4-a

💡 Short Explanation

�� Vitamin B₁₂ treats pernicious anaemia. �� EDTA removes toxic metals. �� Cis-platin is an anticancer drug.

✅ Correct Answer Explanation

1 → c : Vitamin B₁₂ is an anti-pernicious anaemia factor. 2 → a : EDTA is used in lead poisoning treatment. 3 → d : Cis-platin inhibits tumour growth. 4 → b : D-penicillamine removes excess copper and iron. Thus the correct matching is 1-c, 2-a, 3-d, 4-b.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect matching of Vitamin B₁₂ and EDTA.
�� Option C → Incorrect assignment of all major roles.
�� Option D → Incorrect therapeutic associations.

🎯 Strategy

Used

Option Grouping

Application:

Match each compound with its medical application.

Final Logic:

Only Option A gives all correct pairings.

🧠 Memory Trick

B₁₂–Anaemia, EDTA–Lead, Cisplatin–Cancer

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Biological and Medicinal Applications

5 The modern formulation of Werner's spatial arrangements for secondary linkages is known as:

A Oxidation states

B Coordination polyhedra

C Primary valences

D Spectrochemical series

💡 Short Explanation

�� Secondary valences determine geometry. �� Ligands occupy definite positions. �� Modern description uses polyhedra.

✅ Correct Answer Explanation

Werner proposed definite spatial arrangements for secondary valences. In modern coordination chemistry, these geometrical arrangements are described using coordination polyhedra such as octahedral, tetrahedral and square planar structures.

❓ Why Other Options Are Incorrect

�� Option A → Refers to oxidation number.
�� Option C → Related to primary valence.
�� Option D → Describes ligand field strength.

🎯 Strategy

Used

Conceptual Recall

Application:

Relate Werner's theory to modern terminology.

Final Logic:

Spatial arrangements are described as coordination polyhedra.

🧠 Memory Trick

Geometry = Polyhedron

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Werner's Theory

6 The numerical value of the primary valence of the central metal ion in the complex [Co(NH₃)₅Cl]Cl₂ is:

A 2

B 3

C 6

D 5

💡 Short Explanation

�� Primary valence equals oxidation state. �� NH₃ is neutral. �� One coordinated chloride is present.

✅ Correct Answer Explanation

Let oxidation state of Co = x. x + 5(0) + (−1) = +2 x − 1 = +2 x = +3 According to Werner's theory, primary valence corresponds to oxidation state. Therefore the primary valence is 3.

❓ Why Other Options Are Incorrect

�� Option A → Not the oxidation state.
�� Option C → Represents coordination number.
�� Option D → Number of NH₃ ligands only.

🎯 Strategy

Used

Substitution

Application:

Calculate oxidation state using charge balance.

Final Logic:

Primary valence = oxidation state = +3.

🧠 Memory Trick

Primary Valence = Oxidation State

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Werner's Primary Valence

📄 Passage

A coordination entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules. For example, [CoCl₃(NH₃)₃] is a coordination entity in which the cobalt ion is surrounded by three ammonia molecules and three chloride ions. The central atoms/ions are also referred to as Lewis acids. When a ligand is bound to a metal ion through a single donor atom, as with Cl⁻, H₂O or NH₃, the ligand is said to be unidentate.

Question:
According to the passage and Werner's experimental observations, if a complex has the completely intact formulation [CoCl₃(NH₃)₃], how many moles of AgCl will precipitate upon adding excess aqueous AgNO₃?

A 3

B 2

C 1

D 0

💡 Short Explanation

�� All chloride ions are coordinated. �� No free chloride ions exist. �� AgNO₃ precipitates only free Cl⁻.

✅ Correct Answer Explanation

In [CoCl₃(NH₃)₃], all three chloride ions are inside the coordination sphere. Since no chloride ions are present outside the coordination sphere, none are available to react with AgNO₃. Therefore no AgCl precipitate is formed.

❓ Why Other Options Are Incorrect

�� Option A → Requires three free chloride ions.
�� Option B → Requires two free chloride ions.
�� Option C → Requires one free chloride ion.

🎯 Strategy

Used

Substitution

Application:

Count chloride ions outside the coordination sphere.

Final Logic:

No free chloride ions means no AgCl precipitate.

🧠 Memory Trick

Inside Bracket = No AgCl

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Werner's Experimental Evidence

📄 Passage

Question:
Based on the passage, the complex [CoCl₃(NH₃)₃] in an aqueous solution will behaviorally act as a:

A 1:3 electrolyte

B 1:2 electrolyte

C 1:1 electrolyte

D Non-electrolyte

💡 Short Explanation

�� Entire species remains neutral. �� No counter ions are present. �� No ions are produced in solution.

✅ Correct Answer Explanation

The coordination entity [CoCl₃(NH₃)₃] is electrically neutral and contains no ionisable chloride ions outside the coordination sphere. Therefore it does not produce ions in solution and behaves as a non-electrolyte.

❓ Why Other Options Are Incorrect

�� Option A → Requires four ions in solution.
�� Option B → Requires three ions in solution.
�� Option C → Requires two ions in solution.

🎯 Strategy

Used

Elimination

Application:

Check whether ionisable ions exist outside the coordination sphere.

Final Logic:

No ionisation implies non-electrolytic behavior.

🧠 Memory Trick

No Counter Ion = Non-Electrolyte

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Conductivity and Coordination Compounds

9 The IUPAC name of the complex K₂[PdCl₄] is:

A Potassium tetrachloridopalladium(II)

B Potassium tetrachloridopalladate(II)

C Potassium tetrachloridopalladate(IV)

D Dipotassium tetrachloridopalladate(II)

💡 Short Explanation

�� Complex ion is anionic. �� Palladium oxidation state is +2. �� Anionic metal names end with "-ate".

✅ Correct Answer Explanation

For [PdCl₄]²⁻: x + 4(−1) = −2 x = +2 Since the complex ion is negatively charged, palladium becomes palladate. Therefore the correct name is Potassium tetrachloridopalladate(II).

❓ Why Other Options Are Incorrect

�� Option A → Used for neutral/cationic complexes.
�� Option C → Palladium oxidation state is +2, not +4.
�� Option D → Prefix "di" is not used for counter ions.

🎯 Strategy

Used

Substitution

Application:

Calculate oxidation state and identify anionic complex naming.

Final Logic:

Anionic Pd complex → palladate(II).

🧠 Memory Trick

Negative Complex → -ate

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Nomenclature of Coordination Compounds

10 Arrange the following entities in decreasing order of the number of counter ions present:
1. K₄[Fe(CN)₆]
2. [Co(NH₃)₆]Cl₃
3. K₂[PdCl₄]
4. [Ni(CO)₄]

A 1, 2, 3, 4

B 2, 1, 3, 4

C 1, 3, 2, 4

D 4, 3, 2, 1

💡 Short Explanation

�� Count ions outside the coordination sphere. �� K⁺ and Cl⁻ act as counter ions. �� Neutral complexes have none.

✅ Correct Answer Explanation

1. K₄[Fe(CN)₆] → 4 counter ions (4 K⁺) 2. [Co(NH₃)₆]Cl₃ → 3 counter ions (3 Cl⁻) 3. K₂[PdCl₄] → 2 counter ions (2 K⁺) 4. [Ni(CO)₄] → 0 counter ions Decreasing order: 1 > 2 > 3 > 4

❓ Why Other Options Are Incorrect

�� Option B → Places 3 counter ions above 4 counter ions.
�� Option C → Places 2 counter ions above 3 counter ions.
�� Option D → Completely reversed order.

🎯 Strategy

Used

Option Grouping

Application:

Count counter ions present outside the coordination sphere.

Final Logic:

4 > 3 > 2 > 0.

🧠 Memory Trick

Count Outside-Bracket Ions

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Coordination Sphere and Counter Ions

11 In the coordination entity [Fe(CN)₆]³⁻, the metal center acts as the:

A Fe²⁺ ion

B Fe³⁺ ion

C CN⁻ ion

D Neutral Fe atom

💡 Short Explanation

�� CN⁻ has a charge of −1. �� Six CN⁻ ligands contribute −6 charge. �� Overall charge is −3.

✅ Correct Answer Explanation

Let oxidation state of Fe = x. x + 6(−1) = −3 x − 6 = −3 x = +3 Therefore the central metal ion present in the coordination entity is Fe³⁺.

❓ Why Other Options Are Incorrect

�� Option A → Fe²⁺ would give an overall charge of −4.
�� Option C → CN⁻ is the ligand, not the metal center.
�� Option D → Iron is not present in the neutral state.

🎯 Strategy

Used

Substitution

Application:

Calculate oxidation number using overall charge.

Final Logic:

Fe oxidation state is +3.

🧠 Memory Trick

−6 + Fe = −3 ⇒ Fe = +3

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Oxidation Number of Central Metal Ion

12 Consider the following statements regarding central metal ions. Choose the correct statements:
1. They accept electron pairs from ligands.
2. They are referred to as Lewis acids.
3. They are referred to as Lewis bases.
4. Complexes where the metal is bound to more than one kind of donor group are heteroleptic.

A 1, 2, 4

B 1, 3

C 2, 3, 4

D 1, 2

💡 Short Explanation

�� Metal ions accept electron pairs. �� Electron-pair acceptors are Lewis acids. �� Different ligands form heteroleptic complexes.

✅ Correct Answer Explanation

Statement 1 is correct because metal ions accept lone pairs from ligands. Statement 2 is correct because electron-pair acceptors are Lewis acids. Statement 4 is correct because complexes containing more than one kind of ligand are called heteroleptic complexes. Statement 3 is incorrect because the metal ion acts as a Lewis acid, not a Lewis base.

❓ Why Other Options Are Incorrect

�� Statement 3 → Lewis bases donate electron pairs; metal ions accept them.
�� Option B → Contains incorrect Statement 3.
�� Option C → Contains incorrect Statement 3.
�� Option D → Omits correct Statement 4.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate Statements 1, 2, 3 and 4 separately.

Final Logic:

Statements 1, 2 and 4 are correct.

🧠 Memory Trick

Metal Accepts → Acid; Mixed Ligands → Heteroleptic

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Central Metal Ion and Heteroleptic Complexes

13 What structural type of ligand is the ethylenediaminetetraacetate ion (EDTA⁴⁻)?

A Unidentate

B Didentate

C Tetradentate

D Hexadentate

💡 Short Explanation

�� EDTA contains six donor atoms. �� Forms multiple coordinate bonds. �� Produces highly stable chelates.

✅ Correct Answer Explanation

EDTA⁴⁻ coordinates through six donor atoms (two nitrogen atoms and four oxygen atoms). Therefore it is classified as a hexadentate ligand.

❓ Why Other Options Are Incorrect

�� Option A → Contains only one donor atom.
�� Option B → Contains two donor atoms.
�� Option C → Contains four donor atoms.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall denticity of common ligands.

Final Logic:

EDTA possesses six donor sites.

🧠 Memory Trick

EDTA = 6 Hands

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Hexadentate Ligands

14 Arrange the following ligands in decreasing order of their field strength based on the spectrochemical series:
1. CO
2. NH₃
3. H₂O
4. I⁻

A 1, 2, 3, 4

B 4, 3, 2, 1

C 1, 3, 2, 4

D 2, 1, 3, 4

💡 Short Explanation

�� CO is a strong-field ligand. �� NH₃ is stronger than H₂O. �� I⁻ is a weak-field ligand.

✅ Correct Answer Explanation

From the spectrochemical series: CO > NH₃ > H₂O > I⁻ CO produces the largest crystal field splitting, while iodide is among the weakest field ligands. Therefore the decreasing order is: CO > NH₃ > H₂O > I⁻

❓ Why Other Options Are Incorrect

�� Option B → Completely reversed order.
�� Option C → Places H₂O above NH₃ incorrectly.
�� Option D → Places NH₃ above CO incorrectly.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall the spectrochemical series.

Final Logic:

CO is strongest and I⁻ is weakest.

🧠 Memory Trick

CO > NH₃ > H₂O > I⁻

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Spectrochemical Series

15 When a ligand is bound to a metal ion through a single donor atom, it is called unidentate. Which of the following is NOT a unidentate ligand?

A Cl⁻

B H₂O

C C₂O₄²⁻

D NH₃

💡 Short Explanation

�� Oxalate contains two donor oxygen atoms. �� It binds through both simultaneously. �� Therefore it is bidentate.

✅ Correct Answer Explanation

Oxalate ion (C₂O₄²⁻) coordinates through two oxygen donor atoms and forms chelate rings. Hence it is a bidentate ligand and not a unidentate ligand.

❓ Why Other Options Are Incorrect

�� Option A → Cl⁻ has one donor atom.
�� Option B → H₂O coordinates through one oxygen atom.
�� Option D → NH₃ coordinates through one nitrogen atom.

🎯 Strategy

Used

Odd One Out

Application:

Identify the ligand with more than one donor atom.

Final Logic:

Oxalate is the only bidentate ligand listed.

🧠 Memory Trick

Oxalate = Two Oxygens = Bidentate

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Denticity

16 The IUPAC name of the polydentate-containing complex [Co(H₂NCH₂CH₂NH₂)₃]₂(SO₄)₃ is:

A Tris(ethane-1,2-diamine)cobalt(III) sulphate

B Tris(ethane-1,2-diamine)cobalt(II) sulphate

C Bis[tris(ethane-1,2-diamine)cobalt(III)] sulphate

D Tris(ethylenediamine)cobaltate(III) sulphate

💡 Short Explanation

�� Ethane-1,2-diamine is neutral. �� Sulphate is the counter ion. �� Cobalt oxidation state is +3.

✅ Correct Answer Explanation

The complex cation is: [Co(en)₃]³⁺ where en = ethane-1,2-diamine. Since sulphate ions balance the charge: 2(+3) + 3(−2) = 0 The oxidation state of cobalt is +3. Therefore the IUPAC name is Tris(ethane-1,2-diamine)cobalt(III) sulphate.

❓ Why Other Options Are Incorrect

�� Option B → Oxidation state is not +2.
�� Option C → Incorrect naming convention for the salt.
�� Option D → "Cobaltate" is used for anionic complexes.

🎯 Strategy

Used

Substitution

Application:

Determine oxidation state and identify the complex cation.

Final Logic:

Neutral ligand en gives Co oxidation state +3.

🧠 Memory Trick

en Neutral → Co = +3

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Nomenclature of Coordination Compounds

17 The number of ligating groups used by a di- or polydentate ligand simultaneously to bind a single metal ion and form a ring is explicitly called its:

A Primary valence

B Oxidation number

C Denticity

D Coordination number

💡 Short Explanation

�� Denticity counts donor atoms. �� Determines ligand binding capacity. �� Important in chelation.

✅ Correct Answer Explanation

Denticity refers to the number of donor atoms of a ligand that simultaneously coordinate to the same metal ion. It determines whether a ligand is unidentate, bidentate, tetradentate, hexadentate, etc.

❓ Why Other Options Are Incorrect

�� Option A → Refers to Werner's primary valence.
�� Option B → Indicates oxidation state.
�� Option D → Refers to the total number of coordinate bonds around the metal.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall the definition of denticity.

Final Logic:

Denticity equals the number of coordinating donor atoms.

🧠 Memory Trick

Denticity = Number of Teeth (Donors)

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Denticity

18 Which of the following statements about the chelate effect is/are correct?
1. Chelate complexes are more stable than similar complexes with unidentate ligands.
2. The presence of a chelating ligand decreases overall complex stability.
3. Didentate and polydentate ligands form stable chelate rings.
4. Ambidentate ligands always form chelate complexes.

A 1, 3

B 2, 4

C 1, 2, 3

D 1, 4

💡 Short Explanation

�� Chelate effect increases stability. �� Multidentate ligands form rings. �� Ambidentate ligands need not form chelates.

✅ Correct Answer Explanation

Statement 1 is correct because chelate complexes are generally more stable than analogous complexes containing unidentate ligands. Statement 3 is correct because di- and polydentate ligands form stable chelate rings. Statement 2 is incorrect because chelation increases stability. Statement 4 is incorrect because ambidentate ligands coordinate through only one donor atom at a time and do not necessarily form chelates.

❓ Why Other Options Are Incorrect

�� Statement 2 → Opposite of the chelate effect.
�� Statement 4 → Ambidentate ligands do not always form chelate rings.
�� Option B → Contains incorrect Statements 2 and 4.
�� Option C → Contains incorrect Statement 2.
�� Option D → Contains incorrect Statement 4.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate Statements 1, 2, 3 and 4 separately.

Final Logic:

Only Statements 1 and 3 are correct.

🧠 Memory Trick

Chelate Ring = Greater Stability

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Chelate Effect

19 Ambidentate ligands possessing dual sites can uniquely lead to which structural isomerism type in coordination compounds?

A Geometrical isomerism

B Optical isomerism

C Linkage isomerism

D Coordination isomerism

💡 Short Explanation

�� Ambidentate ligands have two donor atoms. �� Coordination can occur through either donor atom. �� Different attachment modes create isomers.

✅ Correct Answer Explanation

Ambidentate ligands such as NO₂⁻ and SCN⁻ can coordinate through different donor atoms. This results in compounds having the same composition but different points of attachment, producing linkage isomerism.

❓ Why Other Options Are Incorrect

�� Option A → Depends on ligand arrangement in space.
�� Option B → Depends on chirality.
�� Option D → Involves exchange between complex cation and anion.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall the special isomerism associated with ambidentate ligands.

Final Logic:

Alternative donor atoms lead to linkage isomerism.

🧠 Memory Trick

Ambidentate → Linkage Isomer

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Linkage Isomerism

20 Match the bound complex (List I) with its IUPAC naming convention for the ambidentate ligand (List II):

List I	List II
1. M–NO₂	a. thiocyanato-S
2. M–ONO	b. nitrito-O
3. M–SCN	c. thiocyanato-N
4. M–NCS	d. nitrito-N

A 1-d, 2-b, 3-a, 4-c

B 1-b, 2-d, 3-c, 4-a

C 1-d, 2-a, 3-b, 4-c

D 1-c, 2-b, 3-d, 4-a

💡 Short Explanation

�� NO₂ indicates N-bonding. �� ONO indicates O-bonding. �� SCN and NCS specify donor atoms directly.

✅ Correct Answer Explanation

1 → d : M–NO₂ = nitrito-N 2 → b : M–ONO = nitrito-O 3 → a : M–SCN = thiocyanato-S 4 → c : M–NCS = thiocyanato-N Thus the correct matching is 1-d, 2-b, 3-a, 4-c.

❓ Why Other Options Are Incorrect

�� Option B → Reverses nitrito and thiocyanato donor assignments.
�� Option C → Incorrectly matches ONO and SCN.
�� Option D → Incorrect assignment of donor atoms.

🎯 Strategy

Used

Option Grouping

Application:

Identify the donor atom attached directly to the metal.

Final Logic:

The atom written adjacent to M indicates the donor atom.

🧠 Memory Trick

NO₂ → N, ONO → O, SCN → S, NCS → N

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Ambidentate Ligands and Linkage Isomerism

CUET UG Chemistry Booster Test – 2 Fundamentals

CUET UG Chemistry Booster Test - 2 Fundamentals

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