CUET UG Chemistry Booster Test - 2 Bonding and Applications -

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CUET UG Chemistry Booster Test - 2 Bonding and Applications

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 2 Bonding and Applications

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QUESTION 1 OF 20

What theoretical approach is primarily used in the Valence Bond Theory to yield a set of equivalent orbitals of definite geometry?

Answer Review

1 What theoretical approach is primarily used in the Valence Bond Theory to yield a set of equivalent orbitals of definite geometry?

A Degeneracy splitting

B Hybridisation

C Spectrochemical pairing

D Dipole interaction

💡 Short Explanation

�� VBT uses hybridisation. �� Equivalent orbitals are produced. �� Geometry is determined by hybridisation type.

✅ Correct Answer Explanation

Valence Bond Theory explains the geometry of coordination compounds through hybridisation of atomic orbitals. → The metal ion combines suitable orbitals to form equivalent hybrid orbitals directed in space. → These hybrid orbitals accept lone pairs from ligands. → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Degeneracy splitting is a concept of Crystal Field Theory.
�� Option C → Spectrochemical pairing is not a VBT principle.
�� Option D → Dipole interaction is associated with Crystal Field Theory.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the central concept used by VBT to explain geometry.

Final Logic:

�� VBT explains geometry through hybridisation.

🧠 Memory Trick

VBT = Hybridisation

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Valence Bond Theory

2 Consider the following regarding the limitations of early bonding models. Choose the correct statements:
1. Werner's theory successfully explained the exact geometries of molecules.
2. Werner's theory failed to explain why coordination compounds have characteristic magnetic properties.
3. VBT and CFT were introduced to answer why bonds in these compounds have directional properties.
4. Werner's secondary valency completely predicted the colour of compounds.

A 1, 4

B 2, 3

C 2, 4

D 1, 3

💡 Short Explanation

�� Werner's theory explained composition. �� Magnetic properties remained unexplained. �� VBT and CFT explained bonding and geometry.

✅ Correct Answer Explanation

Statement 1 is incorrect because Werner's theory did not explain exact orbital geometries. → Statement 2 is correct because magnetic behavior could not be explained by Werner's theory. → Statement 3 is correct because VBT and CFT were developed to explain bonding directionality and structure. → Statement 4 is incorrect because Werner's theory could not explain colour. → Therefore, statements 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Statement 4 is incorrect.
�� Option C → Statement 4 is incorrect.
�� Option D → Statement 1 is incorrect.

🎯 Strategy

Used

Elimination

Application:

�� Identify known limitations of Werner's theory.

Final Logic:

�� Only statements 2 and 3 are correct.

🧠 Memory Trick

Werner: Structure Yes, Colour & Magnetism No

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

3 What is the IUPAC name of the complex [CoF₆]³⁻ which utilizes sp³d² hybridisation?

A Hexafluoridocobalt(III) ion

B Hexafluoridocobaltate(III) ion

C Hexafluorocobalt(II) ion

D Hexafluoridocobaltate(II) ion

💡 Short Explanation

�� F⁻ is fluorido ligand. �� Complex is anionic. �� Co oxidation state is +3.

✅ Correct Answer Explanation

Let oxidation state of Co = x. → x + 6(−1) = −3 → x = +3 → Since the complex ion is negatively charged, the metal name becomes cobaltate. → Therefore, the correct IUPAC name is Hexafluoridocobaltate(III) ion.

❓ Why Other Options Are Incorrect

�� Option A → Anionic complexes use "cobaltate."
�� Option C → Incorrect oxidation state.
�� Option D → Incorrect oxidation state.

🎯 Strategy

Used

Substitution

Application:

�� Calculate oxidation state and apply IUPAC rules.

Final Logic:

�� Co = +3 and anionic complex → cobaltate(III).

🧠 Memory Trick

Anion = Cobaltate

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

4 Which of the following is true for the outer orbital complex [CoF₆]³⁻?

A It pairs all its 3d electrons due to strong ligand field.

B It uses 4d orbitals (sp³d²) for hybridisation and is paramagnetic.

C It uses 3d orbitals (d²sp³) and is diamagnetic.

D It does not utilize orbital overlap for bonding.

💡 Short Explanation

�� F⁻ is a weak-field ligand. �� Electron pairing does not occur. �� Outer orbital complex is formed.

✅ Correct Answer Explanation

Co³⁺ is a d⁶ ion. → F⁻ is a weak-field ligand and cannot force electron pairing. → Hybridisation therefore uses outer 4d orbitals giving sp³d² hybridisation. → Unpaired electrons remain, making the complex paramagnetic. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → F⁻ is not a strong-field ligand.
�� Option C → d²sp³ corresponds to an inner orbital complex.
�� Option D → VBT involves orbital overlap.

🎯 Strategy

Used

Concept Recall

Application:

�� Determine ligand strength and resulting hybridisation.

Final Logic:

�� Weak-field ligand → sp³d² → paramagnetic.

🧠 Memory Trick

Weak Field = Outer Orbital

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

5 Match List-I (Octahedral Complex) with List-II (Hybridisation & Magnetic Nature):

List I	List II
1. [Co(NH₃)₆]³⁺	a. sp³d², paramagnetic (4 unpaired e⁻)
2. [CoF₆]³⁻	b. d²sp³, paramagnetic (2 unpaired e⁻)
3. [Mn(CN)₆]³⁻	c. d²sp³, diamagnetic
4. [FeF₆]³⁻	d. sp³d², paramagnetic (5 unpaired e⁻)

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

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

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

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

💡 Short Explanation

�� NH₃ gives diamagnetic Co³⁺ complex. �� F⁻ forms outer orbital complexes. �� CN⁻ is a strong-field ligand.

✅ Correct Answer Explanation

1-c: [Co(NH₃)₆]³⁺ → d²sp³, diamagnetic. → 2-a: [CoF₆]³⁻ → sp³d², paramagnetic (4 unpaired electrons). → 3-b: [Mn(CN)₆]³⁻ → d²sp³, paramagnetic (2 unpaired electrons). → 4-d: [FeF₆]³⁻ → sp³d², paramagnetic (5 unpaired electrons). → Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect assignment for [Mn(CN)₆]³⁻.
�� Option C → Incorrect assignment for [Co(NH₃)₆]³⁺.
�� Option D → Multiple mismatches.

🎯 Strategy

Used

Option Grouping

Application:

�� Match ligand strength, hybridisation and magnetic behavior.

Final Logic:

�� Only Option A contains all correct matches.

🧠 Memory Trick

NH₃ Pairing, F⁻ Outer, CN⁻ Strong

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

6 Why is [Ni(CO)₄] diamagnetic despite having tetrahedral geometry?

A Nickel is in +2 oxidation state with all paired electrons.

B Nickel is in 0 oxidation state and the 4s electrons pair up in 3d orbitals due to strong CO ligands.

C It strictly uses dsp² hybridisation.

D Carbonyl is a weak field ligand that prevents splitting.

💡 Short Explanation

�� CO is a strong-field ligand. �� Ni is in zero oxidation state. �� Electron pairing makes the complex diamagnetic.

✅ Correct Answer Explanation

In Ni(CO)₄, nickel is in the 0 oxidation state. → Strong-field CO ligands cause pairing of electrons. → All electrons become paired before sp³ hybridisation occurs. → Therefore the complex is diamagnetic despite being tetrahedral.

❓ Why Other Options Are Incorrect

�� Option A → Nickel is not in +2 oxidation state.
�� Option C → Tetrahedral complexes use sp³ hybridisation.
�� Option D → CO is a strong-field ligand.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the electronic configuration of Ni(CO)₄.

Final Logic:

�� Strong-field CO causes electron pairing.

🧠 Memory Trick

CO Pairs, Ni(CO)₄ No Unpaired

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

7 Which of the following conditions under Valence Bond Theory guarantees a complex to be diamagnetic?

A Presence of strong field ligands only

B Complete absence of unpaired electrons in the hybridised metal d orbitals

C Use of outer d orbitals for hybridisation

D A coordination number of strictly 6

💡 Short Explanation

�� Diamagnetism requires paired electrons. �� No unpaired electrons should remain. �� Hybridisation type alone is insufficient.

✅ Correct Answer Explanation

Diamagnetic substances contain no unpaired electrons. → Under VBT, a complex is diamagnetic only when all d electrons become paired. → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Strong ligands often cause pairing but not always.
�� Option C → Outer orbital complexes may still be paramagnetic.
�� Option D → Coordination number alone does not determine magnetism.

🎯 Strategy

Used

Concept Recall

Application:

�� Apply the definition of diamagnetism.

Final Logic:

�� No unpaired electrons = diamagnetic.

🧠 Memory Trick

No Unpaired = Diamagnetic

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

8 The value of the spin-only magnetic moment for [MnBr₄]²⁻ is given as 5.9. The unit for this measurement is:

A V m⁻¹

B Debye (D)

C Bohr Magneton (BM)

D S cm² mol⁻¹

💡 Short Explanation

�� Magnetic moment is expressed in BM. �� BM relates to electron spin. �� Widely used in coordination chemistry.

✅ Correct Answer Explanation

The magnetic moment of coordination compounds is commonly expressed in Bohr Magnetons (BM). → The spin-only formula is: μ = √n(n + 2) BM → Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Unit of electric field strength.
�� Option B → Unit of dipole moment.
�� Option D → Unit related to conductivity.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Match the physical quantity with its standard unit.

Final Logic:

�� Magnetic moment is measured in BM.

🧠 Memory Trick

Magnetism = BM

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

9 CFT is often considered superior to VBT in certain aspects because VBT:

A Focuses entirely on electrostatic models.

B Fails to distinguish between weak and strong ligands.

C Accurately explains the kinetic stability of coordination compounds.

D Suggests that ligands are strict point charges.

💡 Short Explanation

�� VBT cannot explain ligand strength differences. �� CFT introduces crystal field splitting. �� Strong and weak ligands are differentiated.

✅ Correct Answer Explanation

One major limitation of VBT is that it does not provide a satisfactory basis for distinguishing strong-field and weak-field ligands. → Crystal Field Theory explains this through differences in crystal field splitting energy. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Describes CFT, not VBT.
�� Option C → VBT does not accurately explain kinetic stability.
�� Option D → Point-charge treatment belongs to CFT.

🎯 Strategy

Used

Concept Recall

Application:

�� Compare the explanatory power of VBT and CFT.

Final Logic:

�� VBT cannot satisfactorily explain ligand strength.

🧠 Memory Trick

VBT Can't Rank Ligands

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

10 Which limitation of VBT directly prevents it from predicting whether a 4-coordinate complex will form a particular geometry?

A It cannot make exact predictions regarding the tetrahedral and square planar structures without magnetic data.

B It assumes all ligands are point charges.

C It does not account for the primary valency of the metal.

D It only applies to central atoms with a d⁵ configuration.

💡 Short Explanation

�� VBT often requires experimental evidence. �� Geometry prediction is not always definitive. �� Four-coordinate complexes are a classic example.

✅ Correct Answer Explanation

VBT cannot independently predict whether a four-coordinate complex will be tetrahedral or square planar. → Experimental magnetic measurements are often needed. → This is one of the recognized limitations of the theory. → Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Point-charge assumption belongs to CFT.
�� Option C → Primary valency is explained by Werner's theory.
�� Option D → VBT applies to many electronic configurations.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the standard limitations of VBT.

Final Logic:

�� VBT cannot reliably predict tetrahedral vs square planar geometry.

🧠 Memory Trick

4-Coordinate? VBT Needs Help

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Limitations of Valence Bond Theory

📄 Passage

The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. Ligands are treated as point charges in case of anions or point dipoles in case of neutral molecules. The five d orbitals in an isolated gaseous metal atom/ion have the same energy, i.e., they are degenerate. When a negative field due to ligands surrounds the metal atom, this degeneracy is lifted resulting in splitting. The pattern of splitting depends on the crystal field's nature.

Question:
How does CFT view the fundamental nature of the metal-ligand bond?

A Purely covalent

B Ionic, arising from electrostatic interactions

C A result of synergic overlapping

D Based on the hybridization of atomic orbitals

💡 Short Explanation

�� CFT is an electrostatic model. �� Metal-ligand attraction is ionic. �� Covalent overlap is ignored.

✅ Correct Answer Explanation

According to Crystal Field Theory, the metal-ligand bond is considered purely electrostatic. → The positively charged metal ion attracts negatively charged ligands or the negative end of neutral ligand dipoles. → CFT does not involve orbital overlap or covalent bonding concepts. → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Covalent bonding is not assumed in CFT.
�� Option C → Synergic bonding is explained through Molecular Orbital Theory.
�� Option D → Hybridisation is a concept of Valence Bond Theory.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Use the exact statement provided in the passage.

Final Logic:

�� CFT treats bonding as purely electrostatic.

🧠 Memory Trick

CFT = Charges Attract

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

📄 Passage

Question:
According to the passage, how are neutral molecules treated in CFT?

A As point charges

B As point dipoles

C As degenerate orbitals

D As non-interacting spheres

💡 Short Explanation

�� Neutral ligands possess dipole moments. �� CFT treats them as dipoles. �� Examples include NH₃ and H₂O.

✅ Correct Answer Explanation

The passage clearly states that neutral molecules are treated as point dipoles in Crystal Field Theory. → Their dipole interaction with the central metal ion causes d-orbital splitting. → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Anionic ligands are treated as point charges.
�� Option C → d orbitals are degenerate, not neutral ligands.
�� Option D → Neutral molecules actively interact with the metal ion.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Extract the exact information from the passage.

Final Logic:

�� Neutral ligands are treated as point dipoles.

🧠 Memory Trick

Neutral Ligand = Dipole

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

13 In an octahedral complex, the energy of the eg set of orbitals increases by what fraction of Δo relative to the barycentre?

A 2/5

B 3/5

C 4/9

D 1/2

💡 Short Explanation

�� eg orbitals point directly toward ligands. �� They experience greater repulsion. �� Their energy increases by +0.6Δo.

✅ Correct Answer Explanation

In an octahedral crystal field, the five d orbitals split into t₂g and eg sets. → The eg orbitals (dx²−y² and dz²) experience maximum repulsion. → Their energy rises by: +3/5 Δo = +0.6Δo → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents the lowering of t₂g orbitals (−2/5Δo).
�� Option C → Relation between Δt and Δo.
�� Option D → Not the accepted value.

🎯 Strategy

Used

Formula Recall

Application:

�� Recall standard octahedral splitting values.

Final Logic:

�� eg orbitals increase by +3/5 Δo.

🧠 Memory Trick

eg Goes Up by 0.6

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

14 Why are low spin configurations rarely observed in tetrahedral complexes?

A Tetrahedral fields produce very large orbital splitting energies.

B Δt = (4/9)Δo, which means splitting energies are not sufficiently large to force pairing.

C Strong electrostatic repulsions completely dominate the pairing energy.

D Synergic bonding prevents any orbital splitting.

💡 Short Explanation

�� Tetrahedral splitting is small. �� Pairing energy is usually larger. �� High-spin configurations are favored.

✅ Correct Answer Explanation

The tetrahedral crystal field splitting energy (Δt) is only 4/9 of the corresponding octahedral splitting. → Because Δt is usually smaller than the pairing energy (P), electrons prefer to remain unpaired. → Consequently, tetrahedral complexes are generally high spin. → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Tetrahedral splitting is relatively small.
�� Option C → Not the principal reason.
�� Option D → Tetrahedral complexes still show orbital splitting.

🎯 Strategy

Used

Concept Recall

Application:

�� Compare Δt with pairing energy.

Final Logic:

�� Small Δt prevents electron pairing.

🧠 Memory Trick

Tetra = Tiny Splitting = High Spin

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

15 Arrange the following ligands in decreasing order of crystal field strength:
1. CO
2. NH₃
3. H₂O
4. Cl⁻

A 1, 2, 3, 4

B 4, 3, 2, 1

C 2, 1, 3, 4

D 1, 3, 2, 4

💡 Short Explanation

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

✅ Correct Answer Explanation

According to the spectrochemical series: CO > NH₃ > H₂O > Cl⁻ → Therefore, the decreasing order of crystal field strength is: 1 > 2 > 3 > 4 → Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Gives increasing order.
�� Option C → Places NH₃ ahead of stronger CO.
�� Option D → Places H₂O ahead of NH₃.

🎯 Strategy

Used

Ordering

Application:

�� Recall the spectrochemical series from strongest to weakest.

Final Logic:

�� CO > NH₃ > H₂O > Cl⁻.

🧠 Memory Trick

CO > NH₃ > H₂O > Cl⁻

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

16 For a d⁴ coordination entity, if Δo > P, what will be the resulting electronic configuration?

A t₂g³ eg¹

B t₂g⁴ eg⁰

C t₂g² eg²

D t₂g⁰ eg⁴

💡 Short Explanation

�� Δo > P indicates strong-field ligands. �� Electron pairing becomes favorable. �� A low-spin configuration forms.

✅ Correct Answer Explanation

When Δo exceeds pairing energy, electrons prefer pairing in the lower-energy t₂g orbitals. → For a d⁴ system: t₂g⁴ eg⁰ → This represents a low-spin configuration. → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents a high-spin d⁴ configuration.
�� Option C → Violates Aufbau filling.
�� Option D → Not possible under crystal field splitting.

🎯 Strategy

Used

Substitution

Application:

�� Compare Δo and pairing energy.

Final Logic:

�� Strong field → Low spin → t₂g⁴ eg⁰.

🧠 Memory Trick

Big Δ = Pair First

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

17 What happens when water is completely removed from [Ti(H₂O)₆]Cl₃ on heating?

A It turns deep blue due to increased field strength.

B It becomes colourless because crystal field splitting does not occur in the absence of ligands.

C It turns red due to a strong ligand field.

D It undergoes a charge transfer transition.

💡 Short Explanation

�� Colour arises from d-d transitions. �� Ligands create crystal field splitting. �� Without ligands, visible-light absorption disappears.

✅ Correct Answer Explanation

Water ligands create the crystal field responsible for d-orbital splitting. → Removal of all ligands eliminates crystal field splitting. → Without splitting, d-d transitions responsible for colour cannot occur. → Therefore, the compound becomes colourless. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → No stronger field is created.
�� Option C → No ligand field remains.
�� Option D → Charge transfer is not the reason.

🎯 Strategy

Used

Concept Recall

Application:

�� Link colour with crystal field splitting.

Final Logic:

�� No ligands → No splitting → No colour.

🧠 Memory Trick

No Ligands = No Colour

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

18 The yellow-red and blue absorption bands in emerald result in the transmission of which colour?

A Red

B Yellow

C Green

D Violet

💡 Short Explanation

�� Absorbed colours are removed. �� Remaining transmitted light determines appearance. �� Emerald appears green.

✅ Correct Answer Explanation

Emerald absorbs yellow-red and blue regions of the visible spectrum. → The remaining transmitted portion is predominantly green. → Therefore, emerald appears green. → Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Red is largely absorbed.
�� Option B → Yellow is absorbed.
�� Option D → Violet is not the dominant transmitted colour.

🎯 Strategy

Used

Elimination

Application:

�� Determine which colour remains after absorption.

Final Logic:

�� Yellow-red and blue absorbed → Green transmitted.

🧠 Memory Trick

Emerald = Green

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

19 Identify the reaction type associated with the hypo solution in black and white photography:

A Decomposition of pure nickel

B Dissolution of undecomposed AgBr to form the complex ion [Ag(S₂O₃)₂]³⁻

C Extraction of gold from solution

D Coordination of D-penicillamine to remove toxic metals

💡 Short Explanation

�� Hypo solution contains thiosulfate. �� It dissolves unreacted AgBr. �� A soluble silver-thiosulfate complex is formed.

✅ Correct Answer Explanation

Sodium thiosulfate ("hypo") is used as a fixing agent in photography. → It dissolves undecomposed AgBr by forming the soluble complex: [Ag(S₂O₃)₂]³⁻ → This removes excess silver bromide from the film. → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Unrelated to photography.
�� Option C → Refers to gold extraction.
�� Option D → Refers to chelation therapy.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the role of hypo in photographic fixing.

Final Logic:

�� Hypo dissolves AgBr through complex formation.

🧠 Memory Trick

Hypo Fixes by Complexing Silver

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

20 Which chelating ligand is utilized in the medicinal treatment of lead poisoning?

A D-penicillamine

B Desferrioxime B

C EDTA

D DMG

💡 Short Explanation

�� EDTA is a hexadentate ligand. �� Forms stable complexes with Pb²⁺. �� Used in chelation therapy.

✅ Correct Answer Explanation

EDTA strongly binds toxic lead ions through chelation. → The resulting stable complex is excreted from the body. → Therefore, EDTA is widely used in the treatment of lead poisoning. → Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Mainly used for copper poisoning.
�� Option B → Used for iron overload.
�� Option D → Used as a reagent for nickel detection.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall medicinal applications of coordination compounds.

Final Logic:

�� EDTA is the standard chelating agent for lead poisoning.

🧠 Memory Trick

EDTA Eliminates Lead

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

CUET UG Chemistry Booster Test – 2 Bonding and Applications

CUET UG Chemistry Booster Test - 2 Bonding and Applications

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