CUET UG Chemistry Booster Test

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CUET UG Chemistry Booster Test - 2 Isomerism

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 2 Isomerism

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QUESTION 1 OF 20

Why do two coordination entities with the exact identical empirical formula, such as CoCl₃·4NH₃, exhibit completely distinct color properties (e.g., green and violet)?

Answer Review

1 Why do two coordination entities with the exact identical empirical formula, such as CoCl₃·4NH₃, exhibit completely distinct color properties (e.g., green and violet)?

A They inherently possess different oxidation states of the central Cobalt.

B They have different spatial arrangements of the coordinated ligands.

C They are composed of distinct elements.

D One acts as a double salt while the other is a complex ion.

💡 Short Explanation

�� Same empirical formula can produce isomers. �� Different ligand arrangements alter crystal field splitting. �� Different electronic transitions produce different colors.

✅ Correct Answer Explanation

Coordination compounds having the same empirical formula may exist as isomers. → Different arrangements of ligands around the metal ion modify the crystal field experienced by d-electrons. → This changes the wavelengths of light absorbed and therefore the observed color. → Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Both isomers may have the same oxidation state.
�� Option C → The elemental composition remains identical.
�� Option D → Color difference is due to isomerism, not because one is a double salt.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate options that change composition or oxidation state.

Final Logic:

�� Same formula but different arrangement indicates isomerism.

🧠 Memory Trick

Same Formula, Different Space → Different Color

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

2 Match the complex formulation to the number of ions it produces in aqueous solution based on Werner's primary and secondary valences.

List I	List II
1. [Co(NH₃)₆]Cl₃	a. 4 ions
2. [Co(NH₃)₅Cl]Cl₂	b. 3 ions
3. [Co(NH₃)₄Cl₂]Cl	c. 2 ions
4. [Pt(NH₃)₂Cl₂]	d. 0 ions (Non-electrolyte)

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

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

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

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

💡 Short Explanation

�� Outside-sphere ions dissociate. �� Coordinated ligands do not dissociate. �� Number of free ions determines conductivity.

✅ Correct Answer Explanation

1. [Co(NH₃)₆]Cl₃ → [Co(NH₃)₆]³⁺ + 3Cl⁻ = 4 ions → a → 2. [Co(NH₃)₅Cl]Cl₂ → [Co(NH₃)₅Cl]²⁺ + 2Cl⁻ = 3 ions → b → 3. [Co(NH₃)₄Cl₂]Cl → [Co(NH₃)₄Cl₂]⁺ + Cl⁻ = 2 ions → c → 4. [Pt(NH₃)₂Cl₂] → Neutral complex → 0 ions → d Therefore: 1-a, 2-b, 3-c, 4-d

❓ Why Other Options Are Incorrect

�� Option B → Incorrect ion count assignments.
�� Option C → Completely mismatched dissociation pattern.
�� Option D → Compound 2 gives 3 ions, not 2 ions.

🎯 Strategy

Used

Option Grouping

Application:

�� Count ions produced after dissociation.

Final Logic:

�� Correct matching is 1-a, 2-b, 3-c, 4-d.

🧠 Memory Trick

3 Cl⁻ = 4 ions, 2 Cl⁻ = 3 ions, 1 Cl⁻ = 2 ions

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

3 When evaluating [Co(NH₃)₅(SO₄)]Br and [Co(NH₃)₅Br]SO₄, the disparity in bonding involves the interchange of the coordinated sulfate and bromide. Identify the category of structural isomerism:

A Coordination isomerism

B Ionisation isomerism

C Linkage isomerism

D Solvate isomerism

💡 Short Explanation

�� Ligand and counter ion exchange places. �� Different ions are produced in solution. �� Characteristic of ionisation isomerism.

✅ Correct Answer Explanation

In one compound Br⁻ is outside the coordination sphere. → In the other compound SO₄²⁻ is outside the coordination sphere. → Exchange between coordinated ligand and counter ion produces ionisation isomerism. → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Requires both complex cation and complex anion.
�� Option C → Requires an ambidentate ligand.
�� Option D → Involves solvent molecules.

🎯 Strategy

Used

Elimination

Application:

�� Identify the species undergoing interchange.

Final Logic:

�� Ligand-counter ion exchange indicates ionisation isomerism.

🧠 Memory Trick

Counter Ion ↔ Ligand = Ionisation

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

4 Evaluate the following statements regarding structural isomers. Choose the correct statements:
1. They possess fundamentally different chemical bonds.
2. Linkage isomerism is recognized as a specific sub-type.
3. They include geometrical fac-mer isomerism.
4. Coordination isomerism is a specific sub-type.

A 1, 2, 4

B 2, 3, 4

C 1, 3

D 1, 2, 3, 4

💡 Short Explanation

�� Structural isomers differ in bonding. �� Linkage and coordination are structural isomerisms. �� Fac-mer belongs to stereoisomerism.

✅ Correct Answer Explanation

Statement 1 is correct because structural isomers differ in connectivity and bonding. → Statement 2 is correct because linkage isomerism is a structural isomerism. → Statement 3 is incorrect because fac-mer is a type of geometrical isomerism, which belongs to stereoisomerism. → Statement 4 is correct because coordination isomerism is a structural isomerism.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 3.
�� Option C → Omits correct Statements 2 and 4.
�� Option D → Statement 3 is incorrect.

🎯 Strategy

Used

Elimination

Application:

�� Classify each statement under structural or stereoisomerism.

Final Logic:

�� Only 1, 2 and 4 are correct.

🧠 Memory Trick

Structure = Bond Change, Stereo = Space Change

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

5 Which of the following specific ligands would trigger linkage isomerism if systematically substituted into a homoleptic coordination complex?

A H₂O

B SCN⁻

C NH₃

D en (ethane-1,2-diamine)

💡 Short Explanation

�� SCN⁻ is ambidentate. �� It can coordinate through S or N. �� Different donor atoms create linkage isomers.

✅ Correct Answer Explanation

Linkage isomerism occurs only with ambidentate ligands. → SCN⁻ can bind through sulfur (thiocyanato-S) or nitrogen (isothiocyanato-N). → This ability creates linkage isomers. → Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → H₂O coordinates through oxygen only.
�� Option C → NH₃ coordinates through nitrogen only.
�� Option D → en is bidentate but not ambidentate.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall examples of ambidentate ligands.

Final Logic:

�� SCN⁻ possesses two possible donor atoms.

🧠 Memory Trick

SCN = S or N

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

6 Determine the proper IUPAC name of the red form of the linkage isomer [Co(NH₃)₅(NO₂)]Cl₂ where the nitrite ligand is exclusively bound through the oxygen atom.

A Pentaamminenitrito-O-cobalt(III) chloride

B Pentaamminenitrito-N-cobalt(III) chloride

C Pentaammineoxynitritocobalt(III) chloride

D Pentaamminenitrosylcobalt(III) chloride

💡 Short Explanation

�� Nitrite is ambidentate. �� O-bonded form is nitrito-O. �� N-bonded form is nitrito-N.

✅ Correct Answer Explanation

In the red isomer, nitrite coordinates through oxygen. → According to IUPAC nomenclature, O-bonded nitrite is named nitrito-O. → Therefore the complex is named pentaamminenitrito-O-cobalt(III) chloride.

❓ Why Other Options Are Incorrect

�� Option B → Represents N-bonded nitrite.
�� Option C → Not the accepted IUPAC name.
�� Option D → Nitrosyl refers to NO ligand.

🎯 Strategy

Used

Concept Recall

Application:

�� Identify the donor atom attached to cobalt.

Final Logic:

�� O-coordination requires the suffix nitrito-O.

🧠 Memory Trick

O → nitrito-O ; N → nitrito-N

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

7 In order for coordination isomerism to physically manifest, the precursor coordination compound must consist of:

A A complex cation paired with a simple monoatomic anion

B A simple cation paired with a complex coordination anion

C A complex cationic entity paired with a complex anionic entity

D An entirely neutral coordination sphere

💡 Short Explanation

�� Two coordination entities are required. �� Ligands exchange between them. �� Both ions must be complexes.

✅ Correct Answer Explanation

Coordination isomerism occurs due to ligand exchange between a complex cation and a complex anion. → Therefore both ionic species must themselves be coordination complexes. → Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Contains only one complex ion.
�� Option B → Contains only one complex ion.
�� Option D → No ionic exchange is possible.

🎯 Strategy

Used

Elimination

Application:

�� Check whether two coordination entities are present.

Final Logic:

�� Coordination isomerism requires a complex cation and complex anion.

🧠 Memory Trick

Coordination Isomerism = Two Complexes

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

8 Arrange the following transition metals in decreasing order of their atomic numbers.
1. Cobalt (Co)
2. Nickel (Ni)
3. Iron (Fe)
4. Chromium (Cr)

A 2, 1, 3, 4

B 1, 2, 3, 4

C 4, 3, 1, 2

D 3, 4, 2, 1

💡 Short Explanation

�� Ni = 28 �� Co = 27 �� Fe = 26 �� Cr = 24

✅ Correct Answer Explanation

Atomic numbers: → 2 (Ni) = 28 → 1 (Co) = 27 → 3 (Fe) = 26 → 4 (Cr) = 24 → Decreasing order: 2 > 1 > 3 > 4 Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Places Co before Ni.
�� Option C → Gives increasing trend.
�� Option D → Entire sequence is incorrect.

🎯 Strategy

Used

Ordering

Application:

�� Compare atomic numbers directly.

Final Logic:

�� 28 > 27 > 26 > 24 gives 2, 1, 3, 4.

🧠 Memory Trick

Ni(28) > Co(27) > Fe(26) > Cr(24)

📚 Class XI – Chemistry Part I – Chapter: Classification of Elements and Periodicity – Topic: Atomic Numbers of Transition Elements

9 How would a chemist definitively distinguish between the ionisation isomers [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅(SO₄)]Br in an aqueous laboratory setting?

A By adding BaCl₂ solution to precipitate BaSO₄ from the first isomer.

B By adding AgNO₃ solution to precipitate AgBr from the first isomer.

C By strictly measuring their optical rotation values.

D By evaluating their distinct geometrical fac/mer shapes.

💡 Short Explanation

�� Free SO₄²⁻ reacts with Ba²⁺. �� BaSO₄ gives a white precipitate. �� Identifies sulfate as counter ion.

✅ Correct Answer Explanation

In [Co(NH₃)₅Br]SO₄, sulfate exists as a free counter ion. → Addition of BaCl₂ produces insoluble BaSO₄. → In [Co(NH₃)₅(SO₄)]Br, sulfate is coordinated and does not readily precipitate. → Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → AgBr precipitates from the second isomer, not the first.
�� Option C → Optical rotation is unrelated.
�� Option D → Ionisation isomers are not distinguished by fac-mer geometry.

🎯 Strategy

Used

Elimination

Application:

�� Identify the free ion present in solution.

Final Logic:

�� Free sulfate gives BaSO₄ precipitate.

🧠 Memory Trick

Ba²⁺ + SO₄²⁻ → White BaSO₄

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

10 Identify the specific qualitative reaction that validates the presence of the bromide counter ion in the ionisation isomer [Co(NH₃)₅(SO₄)]Br.

A Addition of BaCl₂ yields a thick white precipitate

B Addition of AgNO₃ yields a distinct pale yellow precipitate

C Addition of excess aqueous ammonia yields a blue solution

D Addition of KCN yields a dark green precipitate

💡 Short Explanation

�� Free Br⁻ reacts with Ag⁺. �� AgBr precipitate forms. �� Indicates bromide counter ion.

✅ Correct Answer Explanation

In [Co(NH₃)₅(SO₄)]Br, Br⁻ is present outside the coordination sphere. → Addition of AgNO₃ provides Ag⁺ ions. → Ag⁺ reacts with Br⁻ to form pale yellow AgBr precipitate. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Detects sulfate ions.
�� Option C → Not a bromide test.
�� Option D → Unrelated to bromide identification.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall characteristic qualitative tests for halide ions.

Final Logic:

�� AgNO₃ confirms free bromide ions through AgBr formation.

🧠 Memory Trick

Ag⁺ + Br⁻ → AgBr (Pale Yellow)

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Ionisation Isomerism and Qualitative Analysis

11 Analyze the following statements regarding the solvate isomers [Cr(H₂O)₆]Cl₃ and [Cr(H₂O)₅Cl]Cl₂·H₂O:
1. The first isomer is characteristically violet in color.
2. The second isomer presents as grey-green in color.
3. They effectively represent a form of ionisation isomerism that incorporates a solvent molecule.
4. They differ strictly in the number of coordinated vs lattice water molecules.

A 1, 2, 3, 4

B 1, 2, 4

C 2, 3, 4

D 1, 3

💡 Short Explanation

�� Solvate isomerism involves solvent distribution. �� Water may be coordinated or present in the lattice. �� Solvate isomerism is distinct from ionisation isomerism.

✅ Correct Answer Explanation

Statement 1 is correct. [Cr(H₂O)₆]Cl₃ is violet. → Statement 2 is correct. [Cr(H₂O)₅Cl]Cl₂·H₂O is grey-green. → Statement 3 is incorrect. Solvate isomerism is a separate type of structural isomerism, not ionisation isomerism. → Statement 4 is correct because the isomers differ in the distribution of coordinated and lattice water molecules. → Therefore, the correct combination is 1, 2 and 4. Note: The provided answer (A) is incorrect. The correct answer is B.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect Statement 3.
�� Option C → Omits correct Statement 1 and includes incorrect Statement 3.
�� Option D → Omits correct Statements 2 and 4.

🎯 Strategy

Used

Elimination

Application:

�� Check each statement against the definition of solvate isomerism.

Final Logic:

�� Only statements 1, 2 and 4 are correct.

🧠 Memory Trick

Solvate = Solvent Shift, Not Ion Shift

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

12 The correct unit for expressing molar conductivity, which can be utilized to experimentally distinguish between solvate isomers by evaluating the number of ions produced, is:

A S cm² mol⁻¹

B V m⁻¹

C mol L⁻¹

D S cm⁻¹

💡 Short Explanation

�� Molar conductivity depends on ion concentration. �� It is expressed per mole of electrolyte. �� It helps distinguish ion-producing compounds.

✅ Correct Answer Explanation

Molar conductivity (Λm) is defined as the conductance of all ions produced by one mole of electrolyte. → The commonly used unit is S cm² mol⁻¹. → Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Unit of electric field.
�� Option C → Unit of concentration.
�� Option D → Unit of specific conductance (conductivity), not molar conductivity.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

�� Match the physical quantity with its proper unit.

Final Logic:

�� Molar conductivity is expressed as S cm² mol⁻¹.

🧠 Memory Trick

Λm = S cm² mol⁻¹

📚 Class XII – Chemistry Part I – Chapter: Electrochemistry – Topic: Molar Conductivity

13 Which sub-type of isomerism is critically and exclusively dependent on the presence of non-superimposable mirror images (chirality)?

A Geometrical Isomerism

B Optical Isomerism

C Coordination Isomerism

D Linkage Isomerism

💡 Short Explanation

�� Chirality leads to optical activity. �� Non-superimposable mirror images are enantiomers. �� Optical isomerism arises from chirality.

✅ Correct Answer Explanation

Optical isomerism occurs when molecules exist as non-superimposable mirror images. → Such molecules are called chiral. → The two mirror-image forms are called enantiomers. → Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Depends on ligand arrangement, not chirality.
�� Option C → Involves ligand exchange between complex ions.
�� Option D → Involves ambidentate ligands.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the defining feature of optical isomerism.

Final Logic:

�� Chirality directly leads to optical isomerism.

🧠 Memory Trick

Chiral → Optical

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

14 Match the specific coordination entity with its capacity to demonstrate stereoisomerism.

List I	List II
1. cis-[PtCl₂(en)₂]²⁺	a. Shows geometrical (fac/mer) isomerism
2. trans-[PtCl₂(en)₂]²⁺	b. Shows optical (d/l) isomerism
3. [Co(NH₃)₃(NO₂)₃]	c. Is optically inactive due to symmetry
4. [Ni(CO)₄]	d. Shows absolutely no stereoisomerism

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

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

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

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

💡 Short Explanation

�� Cis form is optically active. �� Trans form has symmetry. �� [Co(NH₃)₃(NO₂)₃] shows fac-mer isomerism.

✅ Correct Answer Explanation

1-b: cis-[PtCl₂(en)₂]²⁺ is optically active and exhibits d/l isomerism. → 2-c: trans-[PtCl₂(en)₂]²⁺ possesses a plane of symmetry and is optically inactive. → 3-a: [Co(NH₃)₃(NO₂)₃] exhibits fac-mer geometrical isomerism. → 4-d: [Ni(CO)₄] is tetrahedral and does not exhibit stereoisomerism. Therefore: 1-b, 2-c, 3-a, 4-d

❓ Why Other Options Are Incorrect

�� Option B → Interchanges optical activity of cis and trans forms.
�� Option C → Incorrectly assigns fac-mer behavior.
�� Option D → Incorrect assignment for cis complex.

🎯 Strategy

Used

Option Grouping

Application:

�� Match each complex with its stereochemical behavior.

Final Logic:

�� Only Option A contains all correct matches.

🧠 Memory Trick

cis → Optical, trans → Symmetric, MA₃B₃ → fac/mer

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

📄 Passage

This type of isomerism also arises in heteroleptic complexes due to different possible geometric arrangements of the ligands. Important examples of this behaviour are found with coordination numbers 4 and 6. In a square planar complex of formula [MX₂L₂] (X and L are unidentate), the two ligands X may be arranged adjacent to each other in a cis isomer, or opposite to each other in a trans isomer. Other square planar complex of the type MABXL shows three isomers—two cis and one trans. Such geometrical isomerism is not possible for a tetrahedral geometry because the relative positions of the unidentate ligands attached to the central metal atom are the exact same with respect to each other.

Question:
According to the passage, why is geometrical isomerism impossible in tetrahedral complexes?

A Because all four attached ligands must inherently be identical.

B Because the relative spatial positions of the unidentate ligands attached to the central metal are exactly the same with respect to each other.

C Because tetrahedral complexes possess chirality and only form optical isomers.

D Because tetrahedral complexes cannot support a coordination number of 4.

💡 Short Explanation

�� All ligand positions are equivalent. �� Cis-trans distinction is impossible. �� Hence no geometrical isomerism occurs.

✅ Correct Answer Explanation

The passage explicitly states that tetrahedral complexes do not exhibit geometrical isomerism because all ligand positions are equivalent relative to one another. → Therefore cis and trans arrangements cannot be distinguished. → Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Ligands need not be identical.
�� Option C → Tetrahedral geometry does not automatically imply optical isomerism.
�� Option D → Tetrahedral complexes commonly have coordination number 4.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Use the exact reasoning given in the passage.

Final Logic:

�� Equivalent ligand positions prevent geometrical isomerism.

🧠 Memory Trick

Tetrahedral = All Positions Equivalent

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

📄 Passage

Question:
Based directly on the passage provided, how many total geometrical isomers are theoretically possible for a square planar complex of the generic type MABXL?

A Two (one cis, one trans)

B Three (two cis, one trans)

C Four (two cis, two trans)

D Zero

💡 Short Explanation

�� MABXL has multiple ligand arrangements. �� Two cis forms are possible. �� One trans form is possible.

✅ Correct Answer Explanation

The passage explicitly states that square planar complexes of type MABXL exhibit three geometrical isomers. → These consist of two cis forms and one trans form. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Underestimates the number of isomers.
�� Option C → More isomers than stated in the passage.
�� Option D → Geometrical isomerism clearly exists.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

�� Extract the exact information from the passage.

Final Logic:

�� MABXL gives three geometrical isomers.

🧠 Memory Trick

MABXL = 2 Cis + 1 Trans

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

17 In an octahedral complex formulated as [MX₂(L-L)₂], where L-L represents a didentate ligand like ethane-1,2-diamine, which geometrical constraint applies regarding its optical activity?

A Only the trans isomer demonstrates optical activity.

B Only the cis isomer demonstrates optical activity.

C Both cis and trans isomers exhibit robust optical activity.

D Neither isomer is capable of optical activity.

💡 Short Explanation

�� Cis form lacks symmetry. �� Trans form possesses symmetry. �� Optical activity requires chirality.

✅ Correct Answer Explanation

In octahedral complexes of the type [MX₂(L-L)₂], the cis isomer is chiral and exists as optical isomers. → The trans isomer generally possesses a plane or center of symmetry and is optically inactive. → Therefore, only the cis isomer exhibits optical activity.

❓ Why Other Options Are Incorrect

�� Option A → Trans is generally optically inactive.
�� Option C → Trans form does not show optical activity.
�� Option D → Cis form is optically active.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall optical activity of cis and trans octahedral complexes.

Final Logic:

�� Chirality exists only in the cis form.

🧠 Memory Trick

cis = Chiral, trans = Symmetric

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Optical Isomerism in Octahedral Complexes

18 Identify the exact category of geometrical isomerism represented when the three equivalent donor atoms of the same ligand occupy positions situated sequentially around the meridian of the octahedron in an [Ma₃b₃] complex.

A fac isomer

B mer isomer

C cis isomer

D trans isomer

💡 Short Explanation

�� Three identical ligands lie along a meridian. �� One pair becomes trans. �� Characteristic of mer isomer.

✅ Correct Answer Explanation

In meridional (mer) isomerism, three identical ligands occupy positions around a meridian plane of the octahedron. → Two identical ligands are trans while the third remains cis to both. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Fac arrangement occupies one triangular face.
�� Option C → Cis refers to adjacent positions only.
�� Option D → Trans refers to only one opposite pair.

🎯 Strategy

Used

Concept Recall

Application:

�� Distinguish fac and mer arrangements.

Final Logic:

�� Meridian arrangement corresponds to mer isomer.

🧠 Memory Trick

Mer = Meridian

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

19 Optical isomerism manifests most frequently and commonly in which structural category of coordination complexes?

A Tetrahedral complexes containing only unidentate ligands

B Square planar complexes containing only unidentate ligands

C Octahedral complexes incorporating didentate (chelating) ligands

D Linear coordination entities

💡 Short Explanation

�� Chelating ligands often create chirality. �� Octahedral geometry favors optical isomerism. �� Enantiomers are commonly observed.

✅ Correct Answer Explanation

Optical isomerism is particularly common in octahedral complexes containing didentate ligands such as ethane-1,2-diamine. → Chelation frequently generates non-superimposable mirror images. → Hence Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Usually do not show common optical isomerism.
�� Option B → Square planar complexes rarely exhibit optical activity.
�� Option D → Linear complexes generally do not show optical isomerism.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the most common class of optically active coordination compounds.

Final Logic:

�� Octahedral chelate complexes most commonly exhibit optical isomerism.

🧠 Memory Trick

Octahedral + Chelate = Optical

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

20 In the context of optical activity in coordination compounds, what definitively characterizes the 'dextro' (d) isomer form?

A It strictly rotates the plane of polarised light to the left.

B It strictly rotates the plane of polarised light to the right.

C It forms a mirror image that is perfectly superimposable.

D It solely forms within square planar ligand geometries.

💡 Short Explanation

�� Dextro means right rotation. �� Laevo means left rotation. �� Measured using a polarimeter.

✅ Correct Answer Explanation

A dextro (d) isomer rotates plane-polarised light in the clockwise (right) direction. → The opposite form is laevo (l), which rotates light to the left. → Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Describes laevo isomer.
�� Option C → Describes an achiral species.
�� Option D → Optical activity is not restricted to square planar complexes.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the definition of dextro and laevo forms.

Final Logic:

�� Dextro means right-handed rotation of plane-polarised light.

🧠 Memory Trick

Dextro = Right, Laevo = Left

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Optical Activity and Enantiomers

CUET UG Chemistry Booster Test – 2 Isomerism

CUET UG Chemistry Booster Test - 2 Isomerism

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