CUET UG Chemistry Booster Test

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CUET UG Chemistry Booster Test -3 Fundamentals

⏱️ 30 Minutes

❓ 20 Questions

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CUET UG Chemistry Booster Test -3 Fundamentals

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QUESTION 1 OF 20

In modern coordination chemistry terminology, the non-ionisable secondary linkages originally described by Werner are structurally recognized as:

Answer Review

1 In modern coordination chemistry terminology, the non-ionisable secondary linkages originally described by Werner are structurally recognized as:

A Ionic bonds

B Covalent or coordinate bonds

C Metallic bonds

D Hydrogen bonds

💡 Short Explanation

�� Secondary valences correspond to coordination bonds. �� Ligands donate electron pairs. �� Metal-ligand bonds are coordinate covalent bonds.

✅ Correct Answer Explanation

Werner's secondary valences represent the bonds formed between the central metal ion and ligands. Modern coordination chemistry interprets these as coordinate covalent (dative covalent) bonds, formed when ligands donate lone pairs to the metal ion.

❓ Why Other Options Are Incorrect

�� Option A → Primary valence is more closely associated with ionic character.
�� Option C → Metallic bonding is not involved.
�� Option D → Hydrogen bonds are not Werner's secondary valences.

🎯 Strategy

Used

Conceptual Recall

Application:

Relate Werner's terminology to modern bonding concepts.

Final Logic:

Secondary valence corresponds to coordinate covalent bonding.

🧠 Memory Trick

Secondary Valence = Coordinate Bond

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

2 Regarding the analytical and medicinal applications of coordination chemistry, consider these statements. Choose the correct statement:
1. D-penicillamine is used to remove excess copper and iron.
2. Desferrioxime B is used to remove toxic metals via chelation.
3. EDTA is used in the treatment of lead poisoning.
4. Cis-platin effectively inhibits the growth of tumours.

A 1, 3, 4

B 2, 3, 4

C 1, 2, 3, 4

D 1, 2, 3

💡 Short Explanation

�� Chelating agents remove toxic metals. �� Cis-platin is an anticancer drug. �� EDTA treats lead poisoning.

✅ Correct Answer Explanation

Statement 1 is correct because D-penicillamine is used to remove excess copper and iron. Statement 2 is correct because Desferrioxime B chelates excess iron and toxic metal ions. Statement 3 is correct because EDTA is used in lead poisoning treatment. Statement 4 is correct because cis-platin is an important antitumour drug. Therefore all four statements are correct.

❓ Why Other Options Are Incorrect

�� Option A → Omits correct Statement 2.
�� Option B → Omits correct Statement 1.
�� Option D → Omits correct Statement 4.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate Statements 1, 2, 3 and 4 individually.

Final Logic:

All four statements are correct.

🧠 Memory Trick

EDTA–Lead, Cisplatin–Cancer, Penicillamine–Copper

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

3 Identify the specific biological process or reaction role facilitated by carboxypeptidase A and carbonic anhydrase:

A Catalysts of biological systems (enzymes)

B Reversible oxygen transport

C Capturing energy in photosynthesis

D Hydrogenation of biological alkenes

💡 Short Explanation

�� Both are metalloenzymes. �� They catalyse biochemical reactions. �� They contain coordinated metal ions.

✅ Correct Answer Explanation

Carboxypeptidase A and carbonic anhydrase are metalloenzymes containing metal ions at their active sites. They function as biological catalysts and accelerate essential biochemical reactions.

❓ Why Other Options Are Incorrect

�� Option B → Reversible oxygen transport is performed by haemoglobin.
�� Option C → Photosynthesis involves chlorophyll.
�� Option D → Not their biological function.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall biological roles of coordination compounds.

Final Logic:

Both compounds function as enzymes.

🧠 Memory Trick

Carbonic Anhydrase = Enzyme

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

4 Match the catalyst/reagent (List I) with its specific application (List II):

List I	List II
1. Wilkinson catalyst [(Ph₃P)₃RhCl]	a. Photography fixing
2. Hypo solution [Na₂S₂O₃]	b. Extraction of gold/silver
3. Cyanide process [NaCN]	c. Hydrogenation of alkenes
4. EDTA	d. Estimation of water hardness

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

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

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

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

💡 Short Explanation

�� Wilkinson catalyst hydrogenates alkenes. �� Hypo is used in photography. �� EDTA estimates water hardness.

✅ Correct Answer Explanation

1 → c : Wilkinson catalyst is used in alkene hydrogenation. 2 → a : Hypo dissolves AgBr in photography. 3 → b : Cyanide process extracts gold and silver. 4 → d : EDTA is used in hardness determination.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect application mapping.
�� Option C → Multiple incorrect pairings.
�� Option D → Incorrectly assigns all major applications.

🎯 Strategy

Used

Option Grouping

Application:

Match each reagent with its standard use.

Final Logic:

Only Option A contains all correct matches.

🧠 Memory Trick

Wilkinson–Hydrogenation, Hypo–Photo, EDTA–Hardness

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

5 According to Werner's theory, the primary valences are usually satisfied by negative ions. In a modern context, the primary valence corresponds precisely to the:

A Coordination number of the metal

B Number of unpaired d-electrons

C Oxidation state of the metal

D Denticity of the inner ligand

💡 Short Explanation

�� Primary valence is ionisable. �� Corresponds to metal charge. �� Modern term is oxidation state.

✅ Correct Answer Explanation

Werner's primary valence represents the charge satisfied by anions and corresponds directly to the oxidation state of the metal ion in modern terminology.

❓ Why Other Options Are Incorrect

�� Option A → Coordination number corresponds to secondary valence.
�� Option B → Not related to Werner's primary valence.
�� Option D → Denticity refers to ligands.

🎯 Strategy

Used

Conceptual Recall

Application:

Translate Werner's terminology into modern terminology.

Final Logic:

Primary valence = oxidation state.

🧠 Memory Trick

Primary Valence = Oxidation Number

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

6 The numerical value of the secondary valence of Palladium in the compound PdCl₂·4NH₃, given that 2 moles of AgCl precipitate per mole of compound, is:

A 2

B 4

C 6

D 0

💡 Short Explanation

�� Two chloride ions are outside the coordination sphere. �� Formula becomes [Pd(NH₃)₄]Cl₂. �� Secondary valence equals coordination number.

✅ Correct Answer Explanation

Since 2 mol of AgCl precipitate, both chloride ions are outside the coordination sphere. The formulation is: [Pd(NH₃)₄]Cl₂ Palladium is coordinated to four NH₃ molecules. Therefore: Secondary valence = Coordination number = 4

❓ Why Other Options Are Incorrect

�� Option A → Coordination number is not 2.
�� Option C → No six ligands are attached.
�� Option D → Palladium clearly has coordinated ligands.

🎯 Strategy

Used

Substitution

Application:

Use AgCl precipitation to determine coordination sphere.

Final Logic:

Four NH₃ ligands imply secondary valence 4.

🧠 Memory Trick

Secondary Valence = Number of Attached Ligands

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

📄 Passage

Alfred Werner prepared and characterised a large number of coordination compounds. In a series of compounds of cobalt(III) chloride with ammonia, it was found that adding excess silver nitrate solution in cold precipitated some chloride ions as AgCl, while some remained in solution. 1 mol CoCl₃·6NH₃ gave 3 mol AgCl; 1 mol CoCl₃·5NH₃ gave 2 mol AgCl; 1 mol CoCl₃·4NH₃ gave 1 mol AgCl. These precipitation observations confirmed the formulation of coordination entities. The complex CoCl₃·6NH₃ is formulated as [Co(NH₃)₆]Cl₃ and acts as a 1:3 electrolyte.

Question:
Based on the passage's precipitation observations, which formulation corresponds to the complex that gives exactly 2 moles of AgCl precipitate per mole of the compound?

A [Co(NH₃)₆]Cl₃

B [CoCl(NH₃)₅]Cl₂

C [CoCl₂(NH₃)₄]Cl

D [CoCl₃(NH₃)₃]

💡 Short Explanation

�� AgCl forms from free chloride ions. �� Two free chloride ions give 2 mol AgCl. �� Two chlorides must lie outside the coordination sphere.

✅ Correct Answer Explanation

In [CoCl(NH₃)₅]Cl₂, one chloride is coordinated and two chloride ions remain outside the coordination sphere. These two free chloride ions react with AgNO₃ to form: 2Cl⁻ + 2Ag⁺ → 2AgCl Thus 2 mol AgCl precipitate per mole of compound.

❓ Why Other Options Are Incorrect

�� Option A → Produces 3 mol AgCl.
�� Option C → Produces 1 mol AgCl.
�� Option D → Produces 0 mol AgCl.

🎯 Strategy

Used

Substitution

Application:

Count chloride ions outside the coordination sphere.

Final Logic:

Two free chloride ions produce 2 mol AgCl.

🧠 Memory Trick

Outside Cl⁻ = AgCl

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

📄 Passage

Question:
Based on the passage's conductivity studies, the complex formulated as [CoCl₂(NH₃)₄]Cl will exhibit a solution conductivity corresponding to which type of electrolyte?

A 1:3 electrolyte

B 1:2 electrolyte

C 1:1 electrolyte

D Non-electrolyte

💡 Short Explanation

�� One chloride ion lies outside the coordination sphere. �� Dissociation gives one cation and one anion. �� Therefore it behaves as a 1:1 electrolyte.

✅ Correct Answer Explanation

The dissociation is: [CoCl₂(NH₃)₄]Cl ⇌ [CoCl₂(NH₃)₄]⁺ + Cl⁻ One complex cation and one chloride anion are produced. Hence the compound behaves as a 1:1 electrolyte.

❓ Why Other Options Are Incorrect

�� Option A → Requires four ions in solution.
�� Option B → Requires three ions in solution.
�� Option D → Ionisation clearly occurs.

🎯 Strategy

Used

Substitution

Application:

Determine the number of ions formed in solution.

Final Logic:

One cation + one anion = 1:1 electrolyte.

🧠 Memory Trick

One Free Cl⁻ → 1:1 Electrolyte

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

9 The IUPAC name of the complex K₃[Fe(C₂O₄)₃] is:

A Potassium trioxalatoferrate(III)

B Potassium trioxalatoferrate(II)

C Tripotassium trioxalatoiron(III)

D Potassium tris(oxalato)ferrate(IV)

💡 Short Explanation

�� Oxalato ligand charge = −2. �� Complex ion is anionic. �� Iron oxidation state is +3.

✅ Correct Answer Explanation

For [Fe(C₂O₄)₃]³⁻: x + 3(−2) = −3 x − 6 = −3 x = +3 Since the complex is anionic, iron becomes ferrate. Hence the name is Potassium trioxalatoferrate(III).

❓ Why Other Options Are Incorrect

�� Option B → Incorrect oxidation state.
�� Option C → Incorrect metal naming.
�� Option D → Iron oxidation state is not +4.

🎯 Strategy

Used

Substitution

Application:

Calculate oxidation state and identify anionic complex naming.

Final Logic:

Fe = +3 and anionic complex → ferrate(III).

🧠 Memory Trick

Negative Fe Complex = Ferrate

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

10 Arrange the following entities in increasing order of the metal's oxidation state:
1. [Ni(CO)₄]
2. [Cu(CN)₄]³⁻
3. [PtCl₄]²⁻
4. [Cr(H₂O)₆]³⁺

A 1, 2, 3, 4

B 2, 3, 4, 1

C 3, 4, 1, 2

D 4, 3, 2, 1

💡 Short Explanation

�� Calculate oxidation state of each metal. �� Arrange from smallest to largest. �� Use ligand charges and overall charge.

✅ Correct Answer Explanation

For [Ni(CO)₄]: Ni = 0 For [Cu(CN)₄]³⁻: x + 4(−1) = −3 x = +1 For [PtCl₄]²⁻: x + 4(−1) = −2 x = +2 For [Cr(H₂O)₆]³⁺: x = +3 Therefore: 0 < +1 < +2 < +3 So the increasing order is: [Ni(CO)₄] < [Cu(CN)₄]³⁻ < [PtCl₄]²⁻ < [Cr(H₂O)₆]³⁺

❓ Why Other Options Are Incorrect

�� Option B → Places oxidation state +1 before 0.
�� Option C → Begins with +2 instead of 0.
�� Option D → Gives decreasing order.

🎯 Strategy

Used

Substitution

Application:

Calculate oxidation state of each complex.

Final Logic:

0 < +1 < +2 < +3.

🧠 Memory Trick

CO = 0, CN = +1, Cl = +2, H₂O = +3

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Oxidation State Calculations

11 In heteroleptic complexes, the central metal atom or ion is defined as being bound to:

A Only one kind of donor group

B More than one kind of donor groups

C Only neutral molecules

D Only ionisable counter ions

💡 Short Explanation

�� Heteroleptic complexes contain different ligands. �� More than one donor group is attached. �� Central metal coordinates with unlike ligands.

✅ Correct Answer Explanation

A heteroleptic complex contains a central metal ion bonded to more than one type of ligand (donor group). For example, [Co(NH₃)₅Cl]²⁺ contains both NH₃ and Cl⁻ ligands and is therefore heteroleptic.

❓ Why Other Options Are Incorrect

�� Option A → Describes homoleptic complexes.
�� Option C → Neutral ligands alone are not required.
�� Option D → Counter ions are not directly bonded to the metal.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall the distinction between homoleptic and heteroleptic complexes.

Final Logic:

Heteroleptic means more than one kind of ligand.

🧠 Memory Trick

Hetero = Different Ligands

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

12 Consider the following statements regarding ligand denticity and the central atom. Choose the correct statements:
1. EDTA⁴⁻ is a hexadentate ligand.
2. EDTA⁴⁻ coordinates through two nitrogen and four oxygen atoms.
3. Ethane-1,2-diamine is a unidentate ligand.
4. C₂O₄²⁻ is a didentate ligand.

A 1, 2, 3

B 1, 2, 4

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� EDTA is hexadentate. �� EDTA uses 2 N and 4 O donor atoms. �� Oxalate is bidentate.

✅ Correct Answer Explanation

Statement 1 is correct because EDTA⁴⁻ is a hexadentate ligand. Statement 2 is correct because EDTA coordinates through two nitrogen atoms and four oxygen atoms. Statement 4 is correct because oxalate (C₂O₄²⁻) is a bidentate ligand. Statement 3 is incorrect because ethane-1,2-diamine (en) is a bidentate ligand, not unidentate.

❓ Why Other Options Are Incorrect

�� Statement 3 → Ethane-1,2-diamine has two donor nitrogen atoms.
�� Option A → Contains incorrect Statement 3.
�� Option C → Contains incorrect Statement 3.
�� Option D → Contains incorrect Statement 3.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate Statements 1, 2, 3 and 4 independently.

Final Logic:

Statements 1, 2 and 4 are correct.

🧠 Memory Trick

EDTA = 2N + 4O = 6 Donors

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

13 Which of the following ligands contains several donor atoms but acts specifically by donating through two nitrogen and four oxygen atoms to a single metal ion?

A N(CH₂CH₂NH₂)₃

B EDTA⁴⁻

C Ethane-1,2-diamine

D C₂O₄²⁻

💡 Short Explanation

�� EDTA contains six donor atoms. �� Coordinates through 2 N and 4 O atoms. �� Forms highly stable chelates.

✅ Correct Answer Explanation

EDTA⁴⁻ is a hexadentate ligand that binds through two nitrogen donor atoms and four oxygen donor atoms simultaneously to the same metal ion, forming stable chelate complexes.

❓ Why Other Options Are Incorrect

�� Option A → Tetradentate ligand.
�� Option C → Bidentate ligand.
�� Option D → Bidentate ligand.

🎯 Strategy

Used

Conceptual Recall

Application:

Identify the donor atoms present in common ligands.

Final Logic:

Only EDTA donates through 2 nitrogen and 4 oxygen atoms.

🧠 Memory Trick

EDTA = 6 Donors (2N + 4O)

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: EDTA as a Hexadentate Ligand

14 Arrange the following ligands based on their denticity (from highest denticity to lowest):
1. EDTA⁴⁻
2. N(CH₂CH₂NH₂)₃
3. C₂O₄²⁻
4. Cl⁻

A 1, 2, 3, 4

B 4, 3, 2, 1

C 2, 1, 3, 4

D 1, 3, 2, 4

💡 Short Explanation

�� EDTA is hexadentate. �� N(CH₂CH₂NH₂)₃ is tetradentate. �� Oxalate is bidentate. �� Chloride is unidentate.

✅ Correct Answer Explanation

Denticities are: 1. EDTA⁴⁻ → 6 2. N(CH₂CH₂NH₂)₃ → 4 3. C₂O₄²⁻ → 2 4. Cl⁻ → 1 Therefore decreasing order of denticity is: 1 > 2 > 3 > 4

❓ Why Other Options Are Incorrect

�� Option B → Completely reversed order.
�� Option C → Places tetradentate ligand above hexadentate ligand.
�� Option D → Places bidentate ligand above tetradentate ligand.

🎯 Strategy

Used

Option Grouping

Application:

Assign denticity values and arrange them.

Final Logic:

6 > 4 > 2 > 1.

🧠 Memory Trick

EDTA > Tren > Oxalate > Chloride

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

15 In the coordination compound [Co(NH₃)₅Cl]²⁺, the ligands NH₃ and Cl⁻ function respectively as:

A Didentate and Unidentate

B Both Unidentate

C Both Chelating ligands

D Both Ambidentate

💡 Short Explanation

�� NH₃ donates through one nitrogen atom. �� Cl⁻ donates through one chlorine atom. �� Both possess one donor site.

✅ Correct Answer Explanation

NH₃ contains one donor nitrogen atom and forms a single coordinate bond. Chloride ion also coordinates through a single donor atom. Therefore both are unidentate ligands.

❓ Why Other Options Are Incorrect

�� Option A → NH₃ is not didentate.
�� Option C → Neither forms chelate rings.
�� Option D → Neither is ambidentate.

🎯 Strategy

Used

Conceptual Recall

Application:

Count donor atoms in each ligand.

Final Logic:

One donor atom means unidentate.

🧠 Memory Trick

One Donor = Unidentate

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

16 Provide the correct IUPAC name for the polydentate complex [Cr(H₂O)₂(C₂O₄)₂]⁻:

A Diaquadioxalatochromate(III)

B Diaquadioxalatochromium(III)

C Dioxalatodiaquachromate(III)

D Dioxalatodiaquachromium(II)

💡 Short Explanation

�� Oxalato ligand charge = −2. �� Complex is anionic. �� Chromium oxidation state is +3.

✅ Correct Answer Explanation

For [Cr(H₂O)₂(C₂O₄)₂]⁻: Let oxidation state of Cr = x x + 2(0) + 2(−2) = −1 x − 4 = −1 x = +3 Since the complex is anionic, chromium becomes chromate. Ligands are named alphabetically: aqua before oxalato. Hence the name is Diaquadioxalatochromate(III).

❓ Why Other Options Are Incorrect

�� Option B → Uses chromium instead of chromate.
�� Option C → Incorrect ligand order.
�� Option D → Incorrect oxidation state.

🎯 Strategy

Used

Substitution

Application:

Calculate oxidation state and identify anionic complex naming.

Final Logic:

Anionic Cr(III) complex → chromate(III).

🧠 Memory Trick

Negative Complex = -ate

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

17 Chelate formation is generally favored thermodynamically. Complexes containing chelate rings tend to be:

A Less stable than those with unidentate ligands

B More stable than similar complexes containing only unidentate ligands

C Highly paramagnetic compared to unidentate analogues

D Restricted solely to square planar geometry

💡 Short Explanation

�� Chelate effect increases stability. �� Ring formation is favored. �� Multidentate ligands stabilize complexes.

✅ Correct Answer Explanation

Chelate complexes formed by di- or polydentate ligands are generally more stable than analogous complexes containing unidentate ligands. This enhanced stability is called the chelate effect.

❓ Why Other Options Are Incorrect

�� Option A → Opposite of the chelate effect.
�� Option C → Paramagnetism is unrelated.
�� Option D → Chelation occurs in many geometries.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall the chelate effect.

Final Logic:

Chelation increases complex stability.

🧠 Memory Trick

Chelate Ring = Stable Thing

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

18 Consider the following complex ions:
1. [Ni(EDTA)]²⁻
2. [Ni(en)₃]²⁺
3. [Ni(NH₃)₆]²⁺
Which statement properly reflects the chelate effect?

A Complex 3 is the most stable because it has the most ligands.

B Complexes 1 and 2 are expected to be more stable than 3 due to chelation.

C Complex 2 is least stable due to ring strain.

D All three have identical stabilities because the central ion is Ni²⁺ in all cases.

💡 Short Explanation

�� EDTA and en are multidentate ligands. �� NH₃ is unidentate. �� Chelation enhances stability.

✅ Correct Answer Explanation

Complexes 1 and 2 contain chelating ligands (EDTA and ethane-1,2-diamine) that form rings with the metal ion. These complexes are generally more stable than complex 3 containing only unidentate NH₃ ligands.

❓ Why Other Options Are Incorrect

�� Option A → Number of ligands alone does not determine stability.
�� Option C → Chelate complexes are generally stabilized.
�� Option D → Stability depends on ligands as well as metal ions.

🎯 Strategy

Used

Option Grouping

Application:

Compare chelating and non-chelating ligands.

Final Logic:

Chelating ligands produce more stable complexes.

🧠 Memory Trick

EDTA & en Beat NH₃

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

19 The thiocyanate ion (SCN⁻) can coordinate through either the sulphur or the nitrogen atom. Such dual-site ligands are directly responsible for which type of isomerism?

A Coordination isomerism

B Ionisation isomerism

C Linkage isomerism

D Geometrical isomerism

💡 Short Explanation

�� SCN⁻ is ambidentate. �� Different donor atoms can attach. �� Different attachment points create isomers.

✅ Correct Answer Explanation

SCN⁻ can coordinate through sulphur (thiocyanato-S) or nitrogen (thiocyanato-N). These different modes of attachment produce linkage isomers.

❓ Why Other Options Are Incorrect

�� Option A → Involves exchange of ligands between cationic and anionic complexes.
�� Option B → Involves exchange of counter ions and ligands.
�� Option D → Depends on spatial arrangement.

🎯 Strategy

Used

Conceptual Recall

Application:

Recall the isomerism associated with ambidentate ligands.

Final Logic:

Different donor atoms produce linkage isomerism.

🧠 Memory Trick

Ambidentate → Linkage Isomer

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

20 Match the specific linkage isomer with its binding atom characteristics:

List I	List II
1. Red form of [Co(NH₃)₅(NO₂)]Cl₂	a. Bound through nitrogen (-NO₂)
2. Yellow form of [Co(NH₃)₅(NO₂)]Cl₂	b. Bound through oxygen (-ONO)
3. M–NCS	c. Bound through sulphur
4. M–SCN	d. Bound through nitrogen (isothiocyanato)

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

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

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

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

💡 Short Explanation

�� Red form is nitrito-O. �� Yellow form is nitro (nitrito-N). �� NCS and SCN indicate donor atoms directly.

✅ Correct Answer Explanation

1 → b : Red form corresponds to O-bonded nitrito complex. 2 → a : Yellow form corresponds to N-bonded nitro complex. 3 → d : M–NCS indicates nitrogen coordination (isothiocyanato). 4 → c : M–SCN indicates sulphur coordination (thiocyanato). Therefore the correct matching is: 1-b, 2-a, 3-d, 4-c

❓ Why Other Options Are Incorrect

�� Option B → Reverses red and yellow linkage isomers.
�� Option C → Incorrectly matches NCS and NO₂ derivatives.
�� Option D → Multiple incorrect donor assignments.

🎯 Strategy

Used

Option Grouping

Application:

Identify the donor atom directly attached to the metal.

Final Logic:

NCS → N donor, SCN → S donor, red → O-bound, yellow → N-bound.

🧠 Memory Trick

NCS = N, SCN = S; Red = O, Yellow = N

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

CUET UG Chemistry Booster Test – 3 Fundamentals

CUET UG Chemistry Booster Test -3 Fundamentals

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