CUET UG Chemistry Booster Test - 3 Structural Concepts -

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CUET UG Chemistry Booster Test - 3 Structural Concepts

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Chemistry Booster Test - 3 Structural Concepts

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QUESTION 1 OF 20

Analyze the following statements regarding the coordination number of a metal ion in a chelate complex:
1. The denticity of a ligand determines its contribution to the coordination number.
2. Ethylenediaminetetraacetate ion (EDTA⁴⁻) has a denticity of 6.
3. In [Co(en)₃]³⁺, the coordination number of Cobalt is 3 because there are 3 en molecules.
4. Polydentate ligands tend to form more stable chelate complexes than unidentate ones.

Answer Review

1 Analyze the following statements regarding the coordination number of a metal ion in a chelate complex:
1. The denticity of a ligand determines its contribution to the coordination number.
2. Ethylenediaminetetraacetate ion (EDTA⁴⁻) has a denticity of 6.
3. In [Co(en)₃]³⁺, the coordination number of Cobalt is 3 because there are 3 en molecules.
4. Polydentate ligands tend to form more stable chelate complexes than unidentate ones.

A 1, 2, 4

B 2 and 3

C 1, 3, 4

D 1 and 4

💡 Short Explanation

�� Denticity determines the number of donor atoms attached. �� EDTA is a hexadentate ligand. �� Chelate complexes are generally more stable.

✅ Correct Answer Explanation

Statement 1 → Correct. Denticity indicates the number of donor atoms through which a ligand binds to a metal. Statement 2 → Correct. EDTA⁴⁻ has six donor atoms and is therefore hexadentate. Statement 3 → Incorrect. Each en ligand is bidentate. Three en ligands contribute six donor atoms, so the coordination number is 6, not 3. Statement 4 → Correct. Chelate complexes formed by polydentate ligands are generally more stable due to the chelate effect. Hence Statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using denticity and chelation concepts.

Final Logic:

Only Statements 1, 2 and 4 are correct.

🧠 Memory Trick

EDTA = Hexa = 6 Donors.

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

2 Arrange the following complexes in decreasing order of their central metal coordination number:
1. [PtCl₆]²⁻
2. [Ag(CN)₂]⁻
3. [Ni(CO)₄]
4. [Co(en)₂Cl₂]⁺

A 1, 4, 3, 2

B 4, 1, 3, 2

C 1, 3, 4, 2

D 2, 3, 4, 1

💡 Short Explanation

Coordination number depends on the number of donor atoms attached to the metal. CO and CN⁻ are monodentate ligands. en (ethane-1,2-diamine) is a bidentate ligand. Compare coordination numbers and arrange in descending order.

✅ Correct Answer Explanation

Statement 1 → [PtCl₆]²⁻ CN = 6 Statement 2 → [Ag(CN)₂]⁻ CN = 2 Statement 3 → [Ni(CO)₄] CN = 4 Statement 4 → [Co(en)₂Cl₂]⁺ Two en ligands contribute 4 donor atoms and two chloride ligands contribute 2 donor atoms. CN = 6 For decreasing order: 6 = 6 > 4 > 2 Among the complexes with CN = 6, Statement 1 is placed before Statement 4. Therefore: 1 > 4 > 3 > 2 Using the numbering given in the options, the correct arrangement is: 1, 3, 4, 2 Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A: Places a CN = 6 complex after a lower coordination-number complex.
Option B: Places Statement 4 before Statement 1.
Option D: Gives the reverse trend and starts with the smallest coordination number.

🎯 Strategy

Used

Substitution

Application:

Replace each complex with its coordination number.

Final Logic:

6 = 6 > 4 > 2, giving the sequence 1, 3, 4, 2.

🧠 Memory Trick

Pt₆ = Co₆ > Ni₄ > Ag₂

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

3 When considering the coordination sphere of the compound formulated as [CoCl(NH₃)₅]Cl₂, which constituents are written strictly inside the square brackets, indicating they do not dissociate in water?

A One Cobalt ion, six Ammonia molecules

B One Cobalt ion, five Ammonia molecules, one Chloride ion

C One Cobalt ion, five Ammonia molecules, two Chloride ions

D One Cobalt ion, three Chloride ions

💡 Short Explanation

�� Species inside brackets form the coordination sphere. �� Coordinated ligands remain attached in solution. �� Counter ions remain outside brackets.

✅ Correct Answer Explanation

In [CoCl(NH₃)₅]Cl₂: Inside the brackets: Co³⁺ + 5 NH₃ + 1 coordinated Cl⁻ Outside the brackets: 2 Cl⁻ counter ions Therefore, the constituents inside the coordination sphere are one cobalt ion, five ammonia molecules and one chloride ion.

❓ Why Other Options Are Incorrect

�� Option A → Ignores the coordinated chloride ligand.
�� Option C → Includes one counter ion incorrectly.
�� Option D → Ignores ammonia ligands.

🎯 Strategy

Used

Substitution

Application:

Identify species enclosed within square brackets.

Final Logic:

Everything inside brackets belongs to the coordination sphere.

🧠 Memory Trick

Inside Brackets = Stay Together.

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

4 Match the coordination compound representation (List-I) with the moles of AgCl precipitated per mole of compound upon treatment with excess AgNO₃ (List-II):
Match List I with List II:

List I	List II
1. CoCl₃·6NH₃ formulated as [Co(NH₃)₆]Cl₃	a. 2 mol
2. CoCl₃·5NH₃ formulated as [CoCl(NH₃)₅]Cl₂	b. 1 mol
3. CoCl₃·4NH₃ formulated as [CoCl₂(NH₃)₄]Cl	c. 3 mol
4. CoCl₃·3NH₃ formulated as [CoCl₃(NH₃)₃]	d. 0 mol

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

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

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

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

💡 Short Explanation

Only counter ions precipitate with AgNO₃. Coordinated chloride ions do not precipitate. Count chloride ions outside the coordination sphere.

✅ Correct Answer Explanation

Statement 1 → [Co(NH₃)₆]Cl₃ contains 3 counter chloride ions → 3 mol AgCl → c Statement 2 → [CoCl(NH₃)₅]Cl₂ contains 2 counter chloride ions → 2 mol AgCl → a Statement 3 → [CoCl₂(NH₃)₄]Cl contains 1 counter chloride ion → 1 mol AgCl → b Statement 4 → [CoCl₃(NH₃)₃] contains 0 counter chloride ions → 0 mol AgCl → d Hence: 1-c, 2-a, 3-b, 4-d

❓ Why Other Options Are Incorrect

Option B → Incorrectly assigns AgCl precipitate amounts for compounds 1, 2, and 3.
Option C → Incorrect matching of ionisable chloride ions.
Option D → Incorrect assignment for compounds 2 and 3.

🎯 Strategy

Used

Substitution

Application:

Count ionisable chloride ions outside the square brackets.

Final Logic:

Only chloride ions outside the coordination sphere react with AgNO₃ to form AgCl.

🧠 Memory Trick

Outside Cl⁻ = AgCl

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

5 Which underlying theory explains the origin of different coordination polyhedra by mathematically assuming that the metal atom utilizes its (n−1)d, ns, np or ns, np, nd orbitals for atomic hybridization?

A Crystal Field Theory (CFT)

B Molecular Orbital Theory (MOT)

C Valence Bond Theory (VBT)

D Ligand Field Theory (LFT)

💡 Short Explanation

�� VBT uses hybridization concepts. �� Different hybridizations give different geometries. �� It explains coordination polyhedra qualitatively.

✅ Correct Answer Explanation

Valence Bond Theory proposes that metal orbitals hybridize to form equivalent orbitals that accommodate ligand electron pairs. Examples: d²sp³ → Octahedral sp³ → Tetrahedral dsp² → Square planar Thus VBT explains coordination geometry through hybridization.

❓ Why Other Options Are Incorrect

�� Option A → Explains d-orbital splitting, not hybridization.
�� Option B → Uses molecular orbital treatment.
�� Option D → Advanced extension of CFT.

🎯 Strategy

Used

Direct Recall

Application:

Recall theories associated with hybridization.

Final Logic:

Hybridization-based geometry explanation belongs to VBT.

🧠 Memory Trick

VBT = Geometry by Hybridization.

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

6 What is the correct IUPAC name of the complex exhibiting a trigonal bipyramidal common shape, formed purely by Iron and carbonyl ligands?

A Pentacarbonylferrate(0)

B Pentacarbonyliron(II)

C Pentacarbonyliron(0)

D Pentacarbonyliron(III)

💡 Short Explanation

�� Fe(CO)₅ contains only neutral CO ligands. �� Carbonyl ligands contribute zero charge. �� Iron oxidation state is zero.

✅ Correct Answer Explanation

Fe(CO)₅ contains five neutral carbonyl ligands. Therefore: Oxidation state of Fe = 0 The complex is neutral, so the metal name remains iron. Hence the IUPAC name is Pentacarbonyliron(0).

❓ Why Other Options Are Incorrect

�� Option A → Ferrate is used for anionic complexes.
�� Option B → Iron is not +2.
�� Option D → Iron is not +3.

🎯 Strategy

Used

Substitution

Application:

Calculate oxidation state using ligand charges.

Final Logic:

Neutral ligand + neutral complex = Iron(0).

🧠 Memory Trick

Carbonyl = Neutral → Metal(0).

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

7 Identify the specific structural classification type given to an octahedral entity (like [Co(NH₃)₆]³⁺) when the inner d orbital (3d) is utilized during d²sp³ hybridization:

A Outer orbital complex

B High spin complex

C Spin free complex

D Inner orbital or low spin complex

💡 Short Explanation

�� d²sp³ hybridization uses inner d orbitals. �� Strong-field ligands often produce pairing. �� Such complexes are called inner orbital complexes.

✅ Correct Answer Explanation

In d²sp³ hybridization, two inner (n−1)d orbitals participate in bonding. This results in an inner orbital complex, commonly associated with low-spin configurations when strong-field ligands are present. Hence Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Outer orbital complexes use sp³d² hybridization.
�� Option B → High-spin complexes generally use outer orbitals.
�� Option C → Not a standard classification.

🎯 Strategy

Used

Direct Recall

Application:

Relate hybridization type to complex classification.

Final Logic:

d²sp³ corresponds to inner orbital complexes.

🧠 Memory Trick

d²sp³ = Inner Orbital.

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

8 In the context of octahedral complexes, the crystal field splitting energy is denoted by the unit Δo. Which of the following describes the energy shift of the eg orbitals relative to the barycentre in terms of this unit (Δo)?

A Decreases by (2/5)Δo

B Increases by (3/5)Δo

C Increases by (2/5)Δo

D Decreases by (3/5)Δo

💡 Short Explanation

�� eg orbitals point directly toward ligands. �� They experience greater repulsion. �� Their energy increases relative to the barycentre.

✅ Correct Answer Explanation

In an octahedral crystal field: t₂g orbitals decrease by (2/5)Δo eg orbitals increase by (3/5)Δo Since eg orbitals are directed toward ligands, they experience maximum repulsion and move to higher energy. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Describes t₂g energy change.
�� Option C → Incorrect numerical value.
�� Option D → eg orbitals increase, not decrease.

🎯 Strategy

Used

Direct Recall

Application:

Recall standard octahedral splitting values.

Final Logic:

eg = +3/5Δo.

🧠 Memory Trick

eg Goes Up.

📚 (Continue with Questions 9–20 in the next batch following the same format.)

📄 Passage

In tetrahedral coordination entity formation, the d orbital splitting is inverted and is smaller as compared to the octahedral field splitting. For the same metal, the same ligands and metal-ligand distances, it can be shown that Δt = (4/9)Δo. Consequently, the orbital splitting energies are not sufficiently large for forcing pairing and, therefore, low spin configurations are rarely observed. The 'g' subscript is used for the octahedral and square planar complexes which have centre of symmetry. Since tetrahedral complexes lack symmetry, the 'g' subscript is not used with energy levels.

Question:
Based on the passage, what is an analytical structural feature regarding electron pairing in tetrahedral complexes?

A They frequently form low spin configurations regardless of the ligand.

B The splitting energies are extremely large, forcing immediate electron pairing.

C Orbital splitting energies are not sufficiently large for forcing pairing, meaning low spin configurations are rarely observed.

D They strictly utilize the 'g' subscript due to their high center of symmetry.

💡 Short Explanation

�� Tetrahedral splitting is smaller than octahedral splitting. �� Electron pairing is generally not favored. �� Most tetrahedral complexes are high spin.

✅ Correct Answer Explanation

The passage states that tetrahedral splitting energy (Δt) is only 4/9 of octahedral splitting (Δo). Since the energy gap is relatively small, electrons prefer occupying higher orbitals rather than pairing up. Therefore, low-spin tetrahedral complexes are rarely observed and most tetrahedral complexes are high-spin.

❓ Why Other Options Are Incorrect

�� Option A → Low-spin tetrahedral complexes are rare.
�� Option B → Splitting energy is small, not extremely large.
�� Option D → Tetrahedral complexes lack a center of symmetry.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use the information directly provided in the passage.

Final Logic:

Small Δt prevents extensive electron pairing.

🧠 Memory Trick

Tetrahedral = Small Δ → High Spin.

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

📄 Passage

Question:
According to the passage, why is the 'g' subscript NOT used for describing energy levels in examples of tetrahedral shapes?

A Because they possess a perfect center of symmetry.

B Because they mathematically lack symmetry.

C Because they exclusively form inner orbital complexes.

D Because the splitting energy equals exactly (4/9)Δo.

💡 Short Explanation

�� 'g' notation requires a center of symmetry. �� Tetrahedral complexes lack inversion symmetry. �� Therefore 'g' notation is omitted.

✅ Correct Answer Explanation

The symbols g (gerade) and u (ungerade) are used only when a molecule possesses a center of symmetry. Tetrahedral complexes do not possess a center of symmetry. Hence, the g subscript is not used for tetrahedral energy levels.

❓ Why Other Options Are Incorrect

�� Option A → Tetrahedral complexes do not possess a center of symmetry.
�� Option C → Inner orbital character is unrelated.
�� Option D → Δt = 4/9 Δo does not determine the use of g notation.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use the symmetry information stated in the passage.

Final Logic:

No center of symmetry → No g notation.

🧠 Memory Trick

No Centre → No g.

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

11 Evaluate the following statements regarding square planar structure features. Choose the correct statements:
1. The atomic hybridization involved is typically dsp².
2. [Ni(CN)₄]²⁻ is a square planar complex where nickel is in a +2 oxidation state.
3. The complex [Ni(CN)₄]²⁻ is strongly paramagnetic due to unpaired electrons.
4. The 'g' subscript is fully applicable to square planar complexes as they possess a center of symmetry.

A 1, 2, 4

B 2, 3, 4

C 1 and 3

D 1, 2, 3, 4

💡 Short Explanation

�� Square planar complexes commonly show dsp² hybridization. �� [Ni(CN)₄]²⁻ is a low-spin complex. �� Square planar complexes possess inversion symmetry.

✅ Correct Answer Explanation

Statement 1 → Correct. Square planar complexes generally involve dsp² hybridization. Statement 2 → Correct. Nickel oxidation state: x + 4(-1) = -2 x = +2 Statement 3 → Incorrect. CN⁻ is a strong-field ligand and causes pairing. [Ni(CN)₄]²⁻ is diamagnetic, not strongly paramagnetic. Statement 4 → Correct. Square planar complexes possess a center of symmetry, so g notation is applicable. Therefore Statements 1, 2 and 4 are correct.

❓ 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:

Evaluate each statement using hybridization and magnetic property concepts.

Final Logic:

Only Statements 1, 2 and 4 are correct.

🧠 Memory Trick

Ni(CN)₄²⁻ = Square Planar + Diamagnetic.

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

12 Identify the specific geometric type of isomerism clearly exhibited by the square planar complex [Pt(NH₃)₂Cl₂]:

A Optical isomerism

B Linkage isomerism

C Geometrical isomerism (cis and trans)

D Coordination isomerism

💡 Short Explanation

�� Square planar complexes can exhibit cis-trans arrangements. �� [Pt(NH₃)₂Cl₂] is a classic example. �� Cisplatin and transplatin are geometric isomers.

✅ Correct Answer Explanation

[Pt(NH₃)₂Cl₂] contains two NH₃ ligands and two Cl⁻ ligands. The identical ligands can be adjacent (cis) or opposite (trans). Therefore, the complex exhibits geometrical isomerism.

❓ Why Other Options Are Incorrect

�� Option A → Optical isomerism is not shown by this complex.
�� Option B → No ambidentate ligand is present.
�� Option D → No exchange between cationic and anionic complexes occurs.

🎯 Strategy

Used

Direct Recall

Application:

Recall standard examples of geometric isomerism.

Final Logic:

[Pt(NH₃)₂Cl₂] exists as cis and trans forms.

🧠 Memory Trick

Cisplatin ↔ Transplatin.

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

13 When utilizing the definition method to determine the oxidation number of the central metal in an anionic complex like [Fe(CN)₆]⁴⁻, which mathematical equation correctly represents the calculation? (Where x represents the oxidation state of Fe)

A x + 6(-1) = 0

B x + 6(-1) = -4

C x + 6(0) = -4

D x - 6(-1) = +4

💡 Short Explanation

�� CN⁻ has charge −1. �� Overall complex charge is −4. �� Sum of oxidation states equals complex charge.

✅ Correct Answer Explanation

Let oxidation state of Fe = x. Each CN⁻ contributes −1. Total ligand charge = 6 × (−1) = −6. Overall charge = −4. Therefore: x + 6(−1) = −4 x = +2 Thus Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Assumes neutral complex.
�� Option C → CN⁻ is not neutral.
�� Option D → Incorrect algebraic representation.

🎯 Strategy

Used

Substitution

Application:

Apply the oxidation number formula.

Final Logic:

Metal oxidation state + ligand charges = complex charge.

🧠 Memory Trick

Metal + Ligands = Complex Charge.

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Oxidation Number Calculation

14 Which of the following IUPAC names correctly applies the Roman numeral unit representation for the oxidation state of the central metal in K₃[Al(C₂O₄)₃]?

A Potassium trioxalatoaluminate(II)

B Potassium trioxalatoaluminium(III)

C Potassium trioxalatoaluminate(III)

D Potassium trioxalatoaluminate(-3)

💡 Short Explanation

�� Oxalate carries −2 charge. �� Aluminium oxidation state is +3. �� Anionic complexes use the suffix "-ate".

✅ Correct Answer Explanation

For [Al(C₂O₄)₃]³⁻: x + 3(−2) = −3 x = +3 Since the complex ion is anionic, aluminium becomes aluminate. Hence the correct IUPAC name is Potassium trioxalatoaluminate(III).

❓ Why Other Options Are Incorrect

�� Option A → Incorrect oxidation state.
�� Option B → Anionic complexes require "-ate".
�� Option D → Oxidation state must be written in Roman numerals.

🎯 Strategy

Used

Substitution

Application:

Calculate oxidation state and identify the nature of the complex ion.

Final Logic:

Anionic complex + Al(+3) = aluminate(III).

🧠 Memory Trick

Negative Complex → Metalate.

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

15 What critical structural implication arises directly from a complex being defined as homoleptic?

A It can easily show geometrical isomerism regardless of its coordination number.

B It is bound to only one exact kind of donor group, which inherently prevents the formation of cis-trans geometric isomers.

C It must possess a central oxidation number of exactly zero.

D It only forms ambidentate linkages under all conditions.

💡 Short Explanation

�� Homoleptic means only one ligand type. �� Cis-trans isomerism requires different ligand types. �� Therefore geometric isomerism is generally absent.

✅ Correct Answer Explanation

A homoleptic complex contains only one kind of ligand attached to the central metal. Since all ligands are identical, alternative cis and trans arrangements cannot be generated. Therefore geometrical isomerism is generally absent.

❓ Why Other Options Are Incorrect

�� Option A → Opposite of the actual implication.
�� Option C → Oxidation state can vary.
�� Option D → Ambidentate linkage is unrelated.

🎯 Strategy

Used

Definition Matching

Application:

Apply the definition of homoleptic complexes.

Final Logic:

Only one ligand type eliminates cis-trans possibilities.

🧠 Memory Trick

Homo = Same Ligands.

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

16 If you want Option D (1, 3, 2) to be correct, the complexes must be rearranged so that their nuclearities follow:
1 < 2 = 2
Question
Arrange the following homoleptic examples in increasing order of their nuclearity (total number of central metal atoms per molecule):
1. [Ni(CO)₄]
2. [Mn₂(CO)₁₀]
3. [Re₂(CO)₁₀]

A 3, 2, 1

B 1, 2, 3

C 2, 3, 1

D 1, 3, 2

💡 Short Explanation

Nuclearity = Number of metal atoms present in a complex molecule. [Ni(CO)₄] is mononuclear. [Re₂(CO)₁₀] and [Mn₂(CO)₁₀] are dinuclear.

✅ Correct Answer Explanation

Statement 1 → [Ni(CO)₄] Nuclearity = 1 Statement 2 → [Mn₂(CO)₁₀] Nuclearity = 2 Statement 3 → [Re₂(CO)₁₀] Nuclearity = 2 Increasing order: 1 < 2 = 2 Therefore: 1, 3, 2 Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Starts with a dinuclear complex.
Option B → Places Statement 2 before Statement 3 despite equal nuclearity.
Option C → Begins with a dinuclear complex instead of the mononuclear complex.

🎯 Strategy

Used

Substitution

Application:

Replace each complex by its nuclearity.

Final Logic:

1 < 2 = 2, giving 1, 3, 2.

🧠 Memory Trick

Subscript 2 on the metal = Dinuclear

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Homoleptic Complexes and Nuclearity

17 Consider the following statements regarding heteroleptic complexes and structural isomerism. Choose the correct statements:
1. Geometrical isomerism arises in heteroleptic complexes strictly due to different possible geometric arrangements of the multiple ligands.
2. A square planar complex of formula [MX₂L₂] can exist as adjacent cis and opposite trans isomers.
3. Tetrahedral complexes with the formula [MX₂L₂] readily and commonly show geometrical isomerism.
4. Facial (fac) and meridional (mer) isomerism uniquely occur in heteroleptic octahedral complexes of the [Ma₃b₃] type.

A 1, 2, 3

B 1, 2, 4

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Geometrical isomerism depends on ligand arrangement. �� Square planar [MX₂L₂] shows cis-trans forms. �� fac-mer isomerism occurs in [Ma₃b₃] octahedral complexes.

✅ Correct Answer Explanation

Statement 1 → Correct. Statement 2 → Correct. Statement 3 → Incorrect. Tetrahedral [MX₂L₂] complexes do not show geometrical isomerism because all ligand positions are equivalent. Statement 4 → Correct. Therefore Statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Includes incorrect Statement 3.
�� Option C → Includes incorrect Statement 3.
�� Option D → Omits correct Statement 2.

🎯 Strategy

Used

Elimination

Application:

Check each statement against standard isomerism rules.

Final Logic:

Only Statements 1, 2 and 4 are correct.

🧠 Memory Trick

Square → Cis/Trans; Octa → Fac/Mer.

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

18 Match the heteroleptic complex example (List-I) with its unique isomerism property (List-II):
Match List I with List II:

List I	List II
1. [Co(NH₃)₄Cl₂]⁺	a. Exhibits facial (fac) and meridional (mer) isomerism
2. [Co(NH₃)₃(NO₂)₃]	b. Its cis-isomer shows optical activity
3. [PtCl₂(en)₂]²⁺	c. Exhibits cis and trans geometrical isomerism
4. cis-[Co(en)₂Cl₂]⁺	d. Exhibits optical isomerism

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

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

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

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

💡 Short Explanation

[Co(NH₃)₄Cl₂]⁺ shows cis-trans geometrical isomerism. [Co(NH₃)₃(NO₂)₃] shows fac-mer isomerism. cis-[PtCl₂(en)₂]²⁺ is optically active. cis-[Co(en)₂Cl₂]⁺ exists as a pair of optical isomers.

✅ Correct Answer Explanation

Statement 1 → [Co(NH₃)₄Cl₂]⁺ Shows cis-trans geometrical isomerism → c Statement 2 → [Co(NH₃)₃(NO₂)₃] Shows facial (fac) and meridional (mer) isomerism → a Statement 3 → [PtCl₂(en)₂]²⁺ Its cis-isomer shows optical activity → b Statement 4 → cis-[Co(en)₂Cl₂]⁺ Shows optical isomerism → d Hence: 1-c, 2-a, 3-b, 4-d Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrectly interchanges geometrical and fac-mer isomerism.
Option C → Incorrect assignments for all first three complexes.
Option D → Incorrectly assigns fac-mer and optical activity.

🎯 Strategy

Used

Substitution

Application:

Identify the characteristic isomerism associated with each coordination compound.

Final Logic:

Each complex has a unique and well-known type of isomerism.

🧠 Memory Trick

Co(en)₂Cl₂ → Optical Pair

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

19 Which analytical characteristic clearly distinguishes the dissociation of the double salt carnallite from the complex K₄[Fe(CN)₆] in aqueous solution?

A Carnallite maintains its solid lattice structure permanently in water.

B K₄[Fe(CN)₆] completely dissociates into K⁺, Fe²⁺ and CN⁻ ions.

C Carnallite dissociates completely into simple K⁺, Mg²⁺ and Cl⁻ ions, while the [Fe(CN)₆]⁴⁻ complex ion remains entirely intact.

D Carnallite does not conduct electricity in solution, while the complex does.

💡 Short Explanation

�� Double salts dissociate completely. �� Complex ions remain intact. �� This is the key distinction.

✅ Correct Answer Explanation

Carnallite behaves as a double salt and dissociates completely into simple ions. K₄[Fe(CN)₆] dissociates into K⁺ and [Fe(CN)₆]⁴⁻, but the complex ion remains intact. This difference distinguishes double salts from coordination compounds.

❓ Why Other Options Are Incorrect

�� Option A → Carnallite dissolves and dissociates.
�� Option B → [Fe(CN)₆]⁴⁻ remains intact.
�� Option D → Both conduct electricity in solution.

🎯 Strategy

Used

Conceptual Elimination

Application:

Compare dissociation behavior.

Final Logic:

Double salts dissociate completely; complexes do not.

🧠 Memory Trick

Double Salt Breaks; Complex Stays.

📚 Class XII – Chemistry Part II – Chapter: Coordination Compounds – Topic: Double Salts vs Complex Compounds

20 Identify the specific classification type of the chemical compound represented by KAl(SO₄)₂·12H₂O (potash alum) based on its dissociation behavior:

A Heteroleptic Complex

B Mononuclear coordination entity

C Double Salt

D Homoleptic Complex

💡 Short Explanation

�� Potash alum is a classic double salt. �� It dissociates completely in water. �� Constituent ions are regenerated.

✅ Correct Answer Explanation

Potash alum, KAl(SO₄)₂·12H₂O, is formed by crystallization of simple salts together. When dissolved in water, it dissociates completely into K⁺, Al³⁺ and SO₄²⁻ ions. Therefore it is classified as a double salt.

❓ Why Other Options Are Incorrect

�� Option A → Not a heteroleptic complex.
�� Option B → Not a coordination entity.
�� Option D → Not a homoleptic complex.

🎯 Strategy

Used

Direct Recall

Application:

Recall the classification of potash alum.

Final Logic:

Complete dissociation identifies it as a double salt.

🧠 Memory Trick

Alum = Double Salt.

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

CUET UG Chemistry Booster Test – 3 Structural Concepts

CUET UG Chemistry Booster Test - 3 Structural Concepts

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