CUET UG Chemistry Booster Test - 3 Bonding and Applications -

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CUET UG Chemistry Booster Test - 3 Bonding and Applications

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 3 Bonding and Applications

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QUESTION 1 OF 20

📄 Passage

Werner prepared and characterised a large number of coordination compounds and studied their physical and chemical behaviour by simple experimental techniques. He proposed the concept of a primary valence and a secondary valence for a metal ion. In a series of compounds of cobalt(III) chloride with ammonia, it was found that some of the chloride ions could be precipitated as AgCl on adding excess silver nitrate solution in cold but some remained in solution. He formulated complexes indicating secondary valences inside square brackets.

Question:
Based on Werner's experiments, which formulation represents the purple compound that precipitates exactly 2 moles of AgCl from 1 mole of the complex?

Answer Review

📄 Passage

Question:
Based on Werner's experiments, which formulation represents the purple compound that precipitates exactly 2 moles of AgCl from 1 mole of the complex?

A [Co(NH₃)₆]³⁺ 3Cl⁻

B [CoCl(NH₃)₅]²⁺ 2Cl⁻

C [CoCl₂(NH₃)₄]⁺ Cl⁻ (Green)

D [CoCl₂(NH₃)₄]⁺ Cl⁻ (Violet)

💡 Short Explanation

�� AgNO₃ precipitates only ionisable chloride ions. �� Two chloride ions are outside the coordination sphere. �� Purple complex gives 2 moles of AgCl.

✅ Correct Answer Explanation

In Werner's formulation, only chloride ions outside the coordination sphere are ionisable and precipitate with AgNO₃. → [CoCl(NH₃)₅]²⁺ 2Cl⁻ contains two chloride ions outside the coordination sphere. → Therefore, one mole of this compound produces exactly two moles of AgCl with excess AgNO₃. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Contains three ionisable chloride ions and gives 3 moles of AgCl.
�� Option C → Contains only one ionisable chloride ion.
�� Option D → Contains only one ionisable chloride ion.

🎯 Strategy

Used

Substitution

Application:

�� Count chloride ions present outside the coordination sphere.

Final Logic:

�� Two external Cl⁻ ions produce two moles of AgCl.

🧠 Memory Trick

Outside Cl⁻ = AgCl

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

2 Analyze the following statements about early coordination chemistry models. Choose the correct statements:
1. Werner postulated that secondary valences are non-ionisable and fixed for a metal.
2. Primary valences are satisfied by neutral molecules only.
3. The spatial arrangement of secondary valences yields coordination polyhedra.
4. Werner successfully explained the exact cause of d-d transitions.

A 1, 2

B 1, 3

C 2, 4

D 1, 3, 4

💡 Short Explanation

�� Secondary valencies are fixed and non-ionisable. �� Coordination polyhedra arise from their arrangement. �� Werner did not explain electronic transitions.

✅ Correct Answer Explanation

Statement 1 is correct because Werner proposed that secondary valencies are fixed and non-ionisable. → Statement 2 is incorrect because primary valencies can be satisfied by anions, not only neutral molecules. → Statement 3 is correct because the arrangement of secondary valencies determines coordination geometry. → Statement 4 is incorrect because d-d transitions were explained later by Crystal Field Theory. → Therefore, statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Statement 2 is incorrect.
�� Option C → Statements 2 and 4 are incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used

Elimination

Application:

�� Verify each statement using Werner's postulates.

Final Logic:

�� Only statements 1 and 3 are correct.

🧠 Memory Trick

Werner = Valencies & Geometry, Not Colour

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

3 For a d⁶ metal ion (like Co³⁺), under the influence of strong field ligands, maximum spin pairing occurs. This results in which hybridisation for the resulting octahedral complex?

A sp³d²

B d²sp³

C dsp²

D sp³

💡 Short Explanation

�� Strong-field ligands cause pairing. �� Inner d orbitals become available. �� Inner orbital complex forms.

✅ Correct Answer Explanation

Strong-field ligands force pairing of electrons in the 3d orbitals. → Two vacant inner d orbitals become available. → Hybridisation involves two d, one s and three p orbitals. → Therefore, the octahedral complex exhibits d²sp³ hybridisation. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents outer orbital octahedral complexes.
�� Option C → Gives square planar geometry.
�� Option D → Gives tetrahedral geometry.

🎯 Strategy

Used

Concept Recall

Application:

�� Relate strong-field ligands to electron pairing and hybridisation.

Final Logic:

�� Strong field → pairing → d²sp³.

🧠 Memory Trick

Strong Field = Inner d Used

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

4 Match the complex with its inner/outer orbital characteristics and hybridisation:

List I	List II
1. [Co(C₂O₄)₃]³⁻	a. Outer orbital, sp³d², paramagnetic (4 unpaired e⁻)
2. [MnCl₆]³⁻	b. Inner orbital, d²sp³, diamagnetic
3. [Fe(CN)₆]³⁻	c. Inner orbital, d²sp³, paramagnetic
4. [CoF₆]³⁻	d. Outer orbital, sp³d², paramagnetic (4 unpaired e⁻)

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

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

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

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

💡 Short Explanation

�� Oxalate and CN⁻ behave as strong-field ligands. �� F⁻ and Cl⁻ are weak-field ligands. �� Strong-field ligands favor inner orbital complexes.

✅ Correct Answer Explanation

1-b: [Co(C₂O₄)₃]³⁻ → d²sp³, diamagnetic. → 2-a: [MnCl₆]³⁻ → sp³d², paramagnetic (4 unpaired electrons). → 3-c: [Fe(CN)₆]³⁻ → d²sp³, paramagnetic. → 4-d: [CoF₆]³⁻ → sp³d², paramagnetic (4 unpaired electrons). → Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect assignments for oxalate and cyanide complexes.
�� Option C → Multiple mismatches.
�� Option D → Incorrect matching of weak-field complexes.

🎯 Strategy

Used

Option Grouping

Application:

�� Determine ligand strength and corresponding hybridisation.

Final Logic:

�� Only Option A contains all correct matches.

🧠 Memory Trick

CN⁻/Oxalate → Inner, F⁻/Cl⁻ → Outer

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: VBT Hybridisation and Magnetism

5 Why do tetrahedral complexes of the formula [M(A)₂(B)₂] not exhibit geometrical isomerism?

A They possess a strong centre of symmetry preventing isomers.

B The relative positions of unidentate ligands are exactly the same with respect to each other.

C They are always high spin complexes.

D Tetrahedral splitting completely forces all electrons into the t₂ level.

💡 Short Explanation

�� All tetrahedral positions are equivalent. �� No cis-trans distinction exists. �� Geometrical isomerism is impossible.

✅ Correct Answer Explanation

In tetrahedral complexes, every ligand occupies an equivalent position. → No two ligands can be uniquely classified as cis or trans. → Therefore geometrical isomerism cannot occur in tetrahedral complexes of the type [M(A)₂(B)₂]. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Tetrahedral complexes lack a center of symmetry.
�� Option C → Spin state is unrelated.
�� Option D → Incorrect description of tetrahedral splitting.

🎯 Strategy

Used

Concept Recall

Application:

�� Compare ligand positions in tetrahedral geometry.

Final Logic:

�� Equivalent positions prevent geometrical isomerism.

🧠 Memory Trick

Tetrahedral = All Positions Same

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

6 What is the IUPAC name for the geometric isomer of [Co(NH₃)₃(NO₂)₃] that has three donor atoms of the same ligand occupying adjacent positions at the corners of an octahedral face?

A meridional-isomer

B facial-isomer

C trans-isomer

D cis-isomer

💡 Short Explanation

�� Three identical ligands occupy one face. �� Adjacent positions are involved. �� This arrangement is called fac.

✅ Correct Answer Explanation

In an octahedral complex of the type MA₃B₃, when three identical ligands occupy adjacent positions on one triangular face, the arrangement is called facial (fac). → Therefore, [Co(NH₃)₃(NO₂)₃] in this arrangement is the facial isomer. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Meridional arrangement lies along a meridian.
�� Option C → Refers to opposite ligands.
�� Option D → Refers to adjacent positions for only two ligands.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall fac-mer terminology.

Final Logic:

�� Three adjacent ligands on one face = fac.

🧠 Memory Trick

Face = Fac

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Fac and Mer Isomerism

7 Identify the reaction/interaction type responsible for the strengthening of the metal-carbon bond in metal carbonyls, often stabilizing their zero oxidation state:

A Chelate effect

B Hydration interaction

C Synergic bonding (sigma donation and pi back-bonding)

D Electrostatic repulsion

💡 Short Explanation

�� CO donates electron density. �� Metal back-donates electrons. �� Bond strength increases.

✅ Correct Answer Explanation

In metal carbonyls, CO donates a lone pair to the metal through σ-bonding. → Simultaneously, the metal donates electron density into the empty π* orbitals of CO. → This combined effect is called synergic bonding. → Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Chelate effect involves multidentate ligands.
�� Option B → Not responsible for metal-carbonyl bonding.
�� Option D → Repulsion weakens bonding.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall bonding in metal carbonyl complexes.

Final Logic:

�� σ donation + π back-bonding = synergic bonding.

🧠 Memory Trick

CO Gives and Takes

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Metal Carbonyl Bonding

8 The primary indicator used in magnetic susceptibility experiments to determine if a complex forms an inner or outer orbital coordination entity is measured in which unit?

A Molarity (mol L⁻¹)

B Spectrochemical wavelength (nm)

C Bohr Magneton (BM)

D Crystal field splitting (cm⁻¹)

💡 Short Explanation

�� Magnetic moment is measured. �� BM is the standard unit. �� Helps determine unpaired electrons.

✅ Correct Answer Explanation

Magnetic susceptibility measurements provide magnetic moment values. → These values are expressed in Bohr Magnetons (BM). → The number of unpaired electrons helps distinguish inner and outer orbital complexes. → Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Unit of concentration.
�� Option B → Unit of wavelength.
�� Option D → Unit related to crystal field splitting.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Match the measured quantity with its unit.

Final Logic:

�� Magnetic moment is measured in BM.

🧠 Memory Trick

Magnetism = BM

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

9 Which of the following experimental facts points out a major limitation in Crystal Field Theory as stated in the text?

A VBT cannot explain geometrical isomerism.

B CFT assumes ligands are point charges, meaning anionic ligands should exert the greatest splitting effect, but they are actually found at the low end of the spectrochemical series.

C VBT completely denies the existence of magnetic moments.

D CFT heavily relies on the synergic bonding of carbon monoxide.

💡 Short Explanation

�� CFT treats ligands as point charges. �� Experimental observations disagree completely. �� Spectrochemical series reveals the limitation.

✅ Correct Answer Explanation

According to pure electrostatic reasoning, negatively charged ligands should produce the largest crystal field splitting. → Experimentally, many anionic ligands appear near the weak-field end of the spectrochemical series. → This contradiction highlights a limitation of simple Crystal Field Theory. → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Refers to VBT, not CFT.
�� Option C → Incorrect statement.
�� Option D → Synergic bonding belongs to MOT explanations.

🎯 Strategy

Used

Concept Recall

Application:

�� Compare theoretical prediction with experimental evidence.

Final Logic:

�� Spectrochemical observations expose a limitation of CFT.

🧠 Memory Trick

Point Charge Prediction ≠ Experimental Reality

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Limitations of Crystal Field Theory

10 The inability of VBT to interpret the thermodynamic and kinetic stabilities of coordination compounds led chemists to rely on which experimentally determined series for predicting ligand behavior?

A Reactivity series

B Spectrochemical series

C Electromotive force series

D Homologous series

💡 Short Explanation

�� VBT cannot rank ligand strengths. �� Experimental ligand ordering is used. �� Spectrochemical series provides this order.

✅ Correct Answer Explanation

Valence Bond Theory cannot satisfactorily explain ligand strengths and related properties. → Chemists therefore use the experimentally determined spectrochemical series. → The series arranges ligands according to their crystal field splitting ability. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Concerns elemental reactivity.
�� Option C → Concerns electrochemical potentials.
�� Option D → Applies to organic chemistry.

🎯 Strategy

Used

Concept Recall

Application:

�� Identify the experimentally derived ligand-strength series.

Final Logic:

�� Ligand behavior is predicted using the spectrochemical series.

🧠 Memory Trick

Ligand Strength → Spectrochemical Series

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Limitations of VBT and Spectrochemical Series

11 Consider the following statements about Crystal Field Splitting. Choose the correct statements:
1. The dx²-y² and dz² orbitals point towards the axes along the direction of the ligand in an octahedral field.
2. The dxy, dyz, and dxz orbitals are raised in energy in an octahedral field.
3. The splitting energy is denoted by Δo in octahedral complexes.
4. The eg set decreases in energy by (2/5)Δo.

A 1, 3

B 2, 3

C 1, 4

D 1, 2, 3

💡 Short Explanation

�� eg orbitals point directly toward ligands. �� t₂g orbitals are stabilized. �� Octahedral splitting is represented by Δo.

✅ Correct Answer Explanation

Statement 1 is correct because dx²-y² and dz² orbitals lie along the axes and directly face incoming ligands. → Statement 2 is incorrect because dxy, dyz and dxz form the t₂g set and decrease in energy. → Statement 3 is correct because octahedral crystal field splitting is denoted by Δo. → Statement 4 is incorrect because the eg set increases by +(3/5)Δo, not decreases by (2/5)Δo. → Therefore, statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Statement 2 is incorrect.
�� Option C → Statement 4 is incorrect.
�� Option D → Statement 2 is incorrect.

🎯 Strategy

Used

Elimination

Application:

�� Verify each statement using octahedral crystal field theory.

Final Logic:

�� Only statements 1 and 3 are correct.

🧠 Memory Trick

eg Goes Up, t₂g Goes Down

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Octahedral Crystal Field Splitting

12 The unequal repulsion between the metal d electrons and ligand electrons in an octahedral field directly causes:

A The degeneracy of the d orbitals to be removed.

B The metal to reduce its oxidation state.

C The ligands to become point dipoles exclusively.

D The formation of synergic π bonds.

💡 Short Explanation

�� All five d orbitals are initially degenerate. �� Ligand approach causes unequal repulsion. �� Energy splitting results.

✅ Correct Answer Explanation

In a free metal ion, all five d orbitals possess equal energy. → When ligands approach, orbitals directed toward ligands experience greater repulsion. → This removes degeneracy and splits the d orbitals into groups of different energies. → Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Oxidation state does not change due to splitting.
�� Option C → Neutral ligands are treated as dipoles irrespective of splitting.
�� Option D → Synergic bonding is unrelated to crystal field splitting.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the origin of crystal field splitting.

Final Logic:

�� Unequal repulsion removes degeneracy.

🧠 Memory Trick

Repulsion → Splitting

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Crystal Field Theory

13 If the pairing energy (P) is strictly less than the octahedral crystal field splitting (Δo) for a d⁴ ion, the complex formed will be:

A High spin, t₂g³ eg¹

B Low spin, t₂g⁴ eg⁰

C High spin, t₂g² eg²

D Low spin, t₂g³ eg¹

💡 Short Explanation

�� Δo > P favors pairing. �� Electrons occupy lower-energy orbitals first. �� Low-spin configuration forms.

✅ Correct Answer Explanation

Since pairing energy is smaller than crystal field splitting, electron pairing becomes energetically favorable. → The fourth electron pairs in the t₂g set rather than occupying eg. → Thus the configuration becomes: t₂g⁴ eg⁰ → This is a low-spin d⁴ complex. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents high-spin d⁴.
�� Option C → Violates Aufbau filling.
�� Option D → Incorrect low-spin configuration.

🎯 Strategy

Used

Substitution

Application:

�� Compare P and Δo.

Final Logic:

�� P < Δo → Pairing occurs → t₂g⁴ eg⁰.

🧠 Memory Trick

Big Δ = Pair First

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: High Spin and Low Spin Complexes

14 Why is the 'g' subscript (gerade) deliberately not used with energy levels (e.g., t₂ and e) in tetrahedral complexes?

A Tetrahedral complexes have a very strong center of symmetry.

B Tetrahedral complexes lack a center of symmetry.

C Tetrahedral splitting is universally larger than octahedral splitting.

D Tetrahedral complexes always form low spin complexes.

💡 Short Explanation

�� "g" denotes symmetry about a center. �� Tetrahedral geometry lacks inversion symmetry. �� Therefore "g" is omitted.

✅ Correct Answer Explanation

The symbols t₂g and eg are used only when a center of symmetry exists. → Octahedral complexes possess a center of symmetry. → Tetrahedral complexes lack such a center. → Therefore the subscript "g" is not used in tetrahedral splitting diagrams. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Opposite of the actual reason.
�� Option C → Tetrahedral splitting is smaller.
�� Option D → Tetrahedral complexes are generally high spin.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the meaning of the term "gerade."

Final Logic:

�� No inversion center → No "g."

🧠 Memory Trick

No Center, No g

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Tetrahedral Crystal Field Splitting

15 Arrange the following ligands in increasing order of field strength based strictly on the spectrochemical series:
1. edta⁴⁻
2. SCN⁻
3. en
4. CN⁻

A 2, 1, 3, 4

B 1, 3, 4, 2

C 2, 3, 1, 4

D 4, 3, 1, 2

💡 Short Explanation

�� SCN⁻ is weaker than EDTA. �� en is stronger than EDTA. �� CN⁻ is among the strongest field ligands.

✅ Correct Answer Explanation

According to the spectrochemical series: SCN⁻ < edta⁴⁻ < en < CN⁻ → Therefore, increasing field strength is: 2 < 1 < 3 < 4 → Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Places SCN⁻ incorrectly.
�� Option C → Places en before EDTA.
�� Option D → Gives nearly reverse order.

🎯 Strategy

Used

Ordering

Application:

�� Recall the spectrochemical series.

Final Logic:

�� SCN⁻ < EDTA < en < CN⁻.

🧠 Memory Trick

SCN < EDTA < en < CN

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

16 In the complex [Co(C₂O₄)₃]³⁻, despite oxalate being near the middle of the spectrochemical series, the complex behaves as an inner orbital (diamagnetic) entity. This indicates that for this specific Co(III) system, oxalate acts predominantly as a:

A Weak field ligand

B Strong field ligand

C Neutral ligand

D Point dipole

💡 Short Explanation

�� Co³⁺ favors pairing. �� Oxalate induces pairing here. �� Inner orbital complex is formed.

✅ Correct Answer Explanation

[Co(C₂O₄)₃]³⁻ is observed to be an inner orbital, diamagnetic complex. → This requires electron pairing before hybridisation. → Therefore, in this specific Co(III) complex, oxalate behaves effectively as a strong-field ligand. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Would produce an outer orbital complex.
�� Option C → Oxalate is an anionic ligand.
�� Option D → Oxalate is not a neutral point dipole.

🎯 Strategy

Used

Elimination

Application:

�� Infer ligand behavior from observed magnetic properties.

Final Logic:

�� Diamagnetic inner orbital complex implies strong-field behavior.

🧠 Memory Trick

Pairing Means Strong

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

17 When ethane-1,2-diamine (en) is progressively added to an aqueous solution of [Ni(H₂O)₆]²⁺ in a 3:1 molar ratio, what is the final complex formed and its observed colour?

A [Ni(en)₃]²⁺; violet

B [Ni(H₂O)₂(en)₂]²⁺; blue/purple

C [Ni(H₂O)₄(en)]²⁺; pale blue

D [Ni(H₂O)₆]²⁺; green

💡 Short Explanation

�� en is a bidentate ligand. �� Complete substitution occurs. �� Violet tris-chelate complex forms.

✅ Correct Answer Explanation

Three molecules of ethane-1,2-diamine replace all six water molecules. → The final complex formed is: [Ni(en)₃]²⁺ → This tris-chelate complex is violet in colour. → Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents an intermediate substitution stage.
�� Option C → Represents an earlier intermediate.
�� Option D → Original aqua complex before ligand substitution.

🎯 Strategy

Used

Substitution

Application:

�� Determine complete ligand replacement using stoichiometry.

Final Logic:

�� 3 en ligands replace 6 H₂O molecules.

🧠 Memory Trick

3 en = Full Chelation

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Colour and Ligand Substitution

18 The complex [Co(NH₃)₅(H₂O)]³⁺ absorbs light at 500 nm (Blue Green region). What is the complementary colour of this coordination entity as observed?

A Violet

B Yellow

C Red

D Pale Yellow

💡 Short Explanation

�� Observed colour is complementary. �� Blue-green absorption corresponds to red appearance. �� Complementary colour theory applies.

✅ Correct Answer Explanation

Coordination compounds appear in the colour complementary to the light absorbed. → Since the complex absorbs blue-green light near 500 nm, the transmitted/reflected colour is red. → Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not complementary to blue-green.
�� Option B → Incorrect complementary colour.
�� Option D → Also not complementary.

🎯 Strategy

Used

Concept Recall

Application:

�� Apply complementary colour relationships.

Final Logic:

�� Blue-green absorbed → Red observed.

🧠 Memory Trick

Green Gone → Red Seen

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

19 Identify the reaction type/process associated with the use of the coordination complex formed when impure nickel is converted to [Ni(CO)₄]:

A Hydrogenation of alkenes

B Purification of metals (Mond process decomposition)

C Estimation of water hardness

D Fixing in photography

💡 Short Explanation

�� Nickel forms volatile Ni(CO)₄. �� Pure nickel is recovered on decomposition. �� Used in the Mond process.

✅ Correct Answer Explanation

In the Mond process, impure nickel reacts with carbon monoxide to form volatile Ni(CO)₄. → On heating, Ni(CO)₄ decomposes to give pure nickel metal. → This process is widely used for nickel purification. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Associated with Wilkinson's catalyst.
�� Option C → Related to EDTA titration.
�� Option D → Uses thiosulfate complexes.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall industrial applications of coordination compounds.

Final Logic:

�� Ni(CO)₄ is used in the Mond purification process.

🧠 Memory Trick

Mond = Metal Purification

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

20 The extraction of gold and silver makes extensive use of complex formation. Which exact coordination entity is formed when gold combines with cyanide in the presence of oxygen and water?

A [Au(CN)₄]⁻

B [Au(CN)₂]⁻

C [Au(CN)₆]³⁻

D [Ag(CN)₂]⁻

💡 Short Explanation

�� Gold dissolves in cyanide solution. �� A stable dicyanoaurate(I) complex forms. �� Basis of cyanide extraction.

✅ Correct Answer Explanation

During cyanide extraction, gold reacts with cyanide ions in the presence of oxygen and water. → The soluble complex formed is: [Au(CN)₂]⁻ → This allows gold to be separated from the ore. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect coordination number.
�� Option C → Not formed in cyanide extraction.
�� Option D → Corresponds to silver, not gold.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the cyanide extraction process.

Final Logic:

�� Gold forms the dicyanoaurate(I) complex.

🧠 Memory Trick

Gold + CN⁻ = [Au(CN)₂]⁻

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Extraction of Gold by Complex Formation

CUET UG Chemistry Booster Test – 3 Bonding and Applications

CUET UG Chemistry Booster Test - 3 Bonding and Applications

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