CUET PG Chemistry 2025 Question Paper (Available): Download Question Paper with Answer Key And Solutions PDF

Step 1: Sodium Ethoxide Reaction.

Sodium ethoxide (Na-EtOH) induces an elimination reaction where the pyridine undergoes dehydrohalogenation to form a substituted intermediate.


Step 2: Methylation Reaction with CH3I.

The methyl iodide (\(CH_3I\)) reacts with the nitrogen of pyridine, introducing a methyl group at the nitrogen atom, creating a quaternary salt.


Step 3: Moist Silver Oxide and Heating.

Heating the reaction mixture with moist Ag\(_2\)O leads to the formation of an intermediate that is prone to further rearrangement due to aromaticity.


Step 4: Reduction with Sodium Mercury.

The sodium amalgam (Na-Hg) in ethanol reduces the intermediate, leading to the formation of quinoline, which is the final product (P).



Final Answer: \[ \boxed{\text{Quinoline}} \] Quick Tip: The reduction with Na-Hg in ethanol is a key step in forming quinoline, which is an aromatic heterocyclic compound.

Step 1: Geranial and Neral.

Geranial and Neral are indeed geometrical isomers of citral, as they differ only in the orientation of the double bond at the aldehyde group. Hence, statement (A) is correct.


Step 2: Formation of Geranic acid.

Citral, on heating with potassium hydrogen sulfate, undergoes dehydration to form geranic acid, so statement (B) is also correct.


Step 3: Formation of 6-methylhept-5-en-2-one.

Citral does not form 6-methylhept-5-en-2-one when treated with potassium carbonate, making statement (C) incorrect.


Step 4: Oxidation to Laevolic acid.

On oxidation with silver oxide, citral undergoes cleavage to form Laevolic acid, making statement (D) correct.



Final Answer: \[ \boxed{(A), (B) and (D) only} \] Quick Tip: Geranial and Neral are geometric isomers. Citral can form geranic acid and Laevolic acid under specific reactions.

Step 1: Rosenmund's Reduction.

The reduction of benzoyl chloride using hydrogen (\(H_2\)) over palladium on barium sulfate (\(Pd-BaSO_4\)) is called Rosenmund's reduction. This reaction selectively reduces the acyl group to an aldehyde, yielding benzaldehyde.


Step 2: Claisen-Schmidt's Reaction.

The reaction of benzaldehyde with a base (like dilute NaOH) leads to the Claisen-Schmidt condensation, where it reacts with acetaldehyde to form cinnamaldehyde.


Step 3: Conclusion.

Thus, the correct sequence involves Rosenmund's reduction followed by the Claisen-Schmidt reaction.



Final Answer: \[ \boxed{(2) Rosenmund's reduction followed by Claisen-Schmidt's reaction} \] Quick Tip: The Rosenmund reduction selectively reduces acyl chlorides to aldehydes, and the Claisen-Schmidt reaction forms cinnamaldehyde.

Step 1: KOH (alc.).

Potassium hydroxide in alcohol (KOH (alc.)) induces elimination of the bromine atom, producing an alkene.


Step 2: Oxymurcuration-demercuration.

The alkene then undergoes oxymurcuration-demercuration to form an alcohol at the position where the bromine was originally attached. Hence, option (1) is correct.



Final Answer: \[ \boxed{(1) (I) KOH (alc.), (II) oxymurcuration-demercuration} \] Quick Tip: The oxymurcuration-demercuration sequence involves hydration of alkenes using mercuric acetate followed by reduction with sodium borohydride.

Step 1: Hydrogenation of the alkene.

Hydrogenation of the alkene (C) using H\(_2\) and nickel (Ni) produces a saturated compound (A).


Step 2: Esterification.

The compound (A) is then treated with benzoic acid (\(PhCO_2H\)) and water, resulting in esterification to give (B).


Step 3: Nucleophilic Substitution.

In the next step, the ester (B) undergoes nucleophilic substitution with ethanol (\(EtO\)) to form (C).


Step 4: Grignard Reaction.

The compound (C) is then treated with ethylmagnesium bromide (EtMgBr) in anhydrous ether, followed by water work-up, resulting in the formation of 2-butanol as the final product (D).



Final Answer: \[ \boxed{(2) 2-Butanol} \] Quick Tip: Grignard reagents like EtMgBr are powerful nucleophiles and are commonly used in carbon-carbon bond formation reactions.

Step 1: Borane Reduction.

The reaction starts with diborane (\(B_2H_6\)) and hydrogen peroxide (\(H_2O_2\)) in the presence of NaOH, which reduces the ethoxy group to a hydroxyl group. This results in the formation of a diol at positions 1 and 6 of the hexane chain.


Step 2: Hydrolysis of the Organoborane.

The organoborane intermediate is hydrolyzed to form the desired diol product.


Step 3: Conclusion.

The product formed is Hexane-1,6-diol, making option (1) the correct answer.



Final Answer: \[ \boxed{(1) Hexane-1,6-diol} \] Quick Tip: The reaction with borane and hydrogen peroxide is a well-known method for hydroboration-oxidation, which adds hydroxyl groups to an alkene.

Step 1: Kolbe Electrolysis.

In Kolbe's electrolysis of an ester like ethyl propionate, two molecules of the ester undergo decarboxylation at the anode, leading to the formation of alkanes. The reaction does not form ethylene as ethylene is an unsaturated compound, and Kolbe's electrolysis typically forms saturated alkanes.


Step 2: Formation of Alkanes.

During the electrolysis, products such as n-butane, n-propane, and ethane can form due to the decarboxylation of the ester, but ethylene does not appear.


Step 3: Conclusion.

Hence, ethylene is not formed during Kolbe's electrolysis of ethyl propionate. The correct answer is (3) Ethylene.



Final Answer: \[ \boxed{(3) Ethylene} \] Quick Tip: Kolbe's electrolysis of esters typically results in the formation of alkanes, not alkenes.

Step 1: Analysis of Stereochemistry.

The structures (A) and (D) are non-mirror image stereoisomers that differ in the spatial arrangement of groups around a chiral center. This makes them diastereomers.


Step 2: Conclusion.

Thus, option (4) is correct, as (A) and (D) are diastereomers, while the other options do not represent accurate stereochemical relationships.



Final Answer: \[ \boxed{(4) (A) and (D) are diastereomers} \] Quick Tip: Diastereomers are stereoisomers that are not mirror images of each other and are not related by reflection symmetry.

Step 1: Oxidation with V\(_2\)O\(_5\).

Vanadium pentoxide (V\(_2\)O\(_5\)) and air oxidize the starting compound, initiating the process by forming a benzyl alcohol derivative.


Step 2: Bromination.

Bromination with Br\(_2\) and KOH proceeds to form a brominated intermediate that undergoes dehydrobromination.


Step 3: Final Step.

The final step involves heating with sodium hydroxide (NaOH) to induce a coupling reaction, leading to the formation of naphthalene.


Step 4: Conclusion.

Thus, the final product (D) is naphthalene, which matches option (1).



Final Answer: \[ \boxed{(1) Naphthalene} \] Quick Tip: The reaction sequence involves oxidation, bromination, and elimination to form naphthalene through a coupling reaction.

Step 1: Gabriel Phthalimide Synthesis Overview.

Gabriel phthalimide synthesis is a method used to prepare primary amines by reacting phthalimide with an alkyl halide in the presence of a strong base, followed by hydrolysis. This synthesis is typically effective for the preparation of primary amines, but there are limitations.


Step 2: Analysis of Options.

- (A) n-Butylamine: n-Butylamine can be easily synthesized using Gabriel phthalimide synthesis as the alkylation step proceeds with primary alkyl halides.

- (B) Alanine: Alanine contains both an amino group and a carboxyl group, which complicates its synthesis via Gabriel phthalimide, as it involves a secondary amino group. Thus, alanine cannot be synthesized via this method.

- (C) Aniline: Aniline is an aromatic amine, and Gabriel phthalimide is not effective for preparing aromatic amines like aniline. Hence, aniline cannot be prepared by this method.

- (D) t-Butylamine: t-Butylamine involves a bulky tertiary-butyl group, which also prevents the alkylation step from proceeding effectively in the Gabriel synthesis. Thus, t-butylamine cannot be synthesized by this method.


Step 3: Conclusion.

Therefore, the correct answer is (4) (B), (C) and (D) as these compounds cannot be prepared by Gabriel phthalimide synthesis.



Final Answer: \[ \boxed{(4) (B), (C) and (D) only} \] Quick Tip: Gabriel phthalimide synthesis is effective for primary amines, but aromatic and bulky amines are not readily formed by this method.

Step 1: Friedlander's Synthesis.

Friedlander's synthesis involves the reaction of o-aminobenzaldehyde with a suitable electrophilic reagent. Hence, it matches with (I) o-Aminobenzaldehyde.


Step 2: Doebner-Miller's Synthesis.

Doebner-Miller's synthesis involves the reaction of \(\beta\)-phenylethylamide with a suitable electrophile, so it matches with (II) \(\beta\)-Phenylethylamide.


Step 3: Hantzsch's Synthesis.

Hantzsch's synthesis involves the reaction of aniline with a carbonyl compound, so it matches with (III) Aniline.


Step 4: Bischler-Napieralski's Synthesis.

Bischler-Napieralski's synthesis involves the reaction of a \(\beta\)-ketoneester with an amine, so it matches with (IV) \(\beta\)-Ketoneester.


Step 5: Conclusion.

Thus, the correct matching is (A) - (I), (B) - (II), (C) - (III), (D) - (IV), which corresponds to option (1).



Final Answer: \[ \boxed{(1) (A) - (I), (B) - (II), (C) - (III), (D) - (IV)} \] Quick Tip: Each reaction in List-I requires a specific reactant from List-II. Ensure you understand the reagents and substrates used in each reaction.

Step 1: Cyclic Polymerization.

The starting material undergoes a cyclic polymerization reaction. This type of reaction typically forms a cyclic structure that is a precursor for further transformations.


Step 2: Gatterman-Koch's Reaction.

This reaction involves the formylation of an aromatic compound. In this case, it adds a formyl group to the aromatic compound, which is a critical intermediate step in the transformation.


Step 3: Pomeranz-Fritsch's Reaction.

The Pomeranz-Fritsch's reaction leads to the formation of isquinoline from the reaction intermediate formed in step 2.


Step 4: Conclusion.

Thus, the correct sequence involves cyclic polymerization, Gatterman-Koch's reaction, and the Pomeranz-Fritsch's reaction. Hence, option (4) is correct.



Final Answer: \[ \boxed{(4) (i) Cyclic polymerization, (ii) Gatterman-Koch's reaction, (iii) Pomeranz-Fritsch's reaction} \] Quick Tip: Cyclic polymerization is often used to prepare intermediate aromatic compounds, which are further functionalized in reactions like Gatterman-Koch's and Pomeranz-Fritsch's.

Step 1: Base Catalysis.

Both the Benzil-Benzylic acid rearrangement and Cannizzaro's reaction are base-catalyzed reactions. In the Benzil-Benzylic acid rearrangement, the base is required to deprotonate the intermediate. Similarly, in Cannizzaro's reaction, hydroxide ions facilitate the transfer of hydride.


Step 2: Anion Shifting.

Both reactions involve the shifting of an anion in their mechanisms. In the Benzil-Benzylic acid rearrangement, an anion is formed during the formation of the intermediate, and in the Cannizzaro reaction, an anion of the aldehyde is involved in the hydride transfer.


Step 3: Inter and Intra-molecular Reactions.

Both reactions can occur inter-molecularly and intra-molecularly. However, this is not specifically true for both reactions and doesn't apply to all cases. Thus, (C) is not necessarily true for both reactions.


Step 4: Redox Reactions.

Both reactions involve redox changes. In the Cannizzaro reaction, one molecule undergoes reduction while another undergoes oxidation. The Benzil-Benzylic acid rearrangement also involves a redox process.


Step 5: Conclusion.

Thus, the correct answer is (2) (A), (B) and (D) only.



Final Answer: \[ \boxed{(2) (A), (B) and (D) only} \] Quick Tip: Base catalysis, shifting of anions, and redox reactions are key features in many organic rearrangements and reactions, including the Benzil-Benzylic acid rearrangement and Cannizzaro's reaction.

Step 1: Anhydride Formation.

When heating dicarboxylic acids, they typically undergo dehydration to form anhydrides. However, the presence of bulky substituents on the carboxyl groups can hinder this process.


Step 2: Analysis of Options.

- (1) Glutaric acid: Glutaric acid can form an anhydride upon heating.

- (2) Maleic acid: Maleic acid can also undergo dehydration to form an anhydride.

- (3) Dimethyl succinic acid: Dimethyl succinic acid, with methyl groups on the carboxyl groups, still readily forms anhydride.

- (4) Dimethyl malonic acid: The bulky methyl groups on the carboxyl groups prevent the formation of an anhydride. Therefore, this acid does not readily form anhydride on heating.


Step 3: Conclusion.

Thus, the correct answer is (4) Dimethyl malonic acid.



Final Answer: \[ \boxed{(4) Dimethyl malonic acid} \] Quick Tip: Substituents like methyl groups on dicarboxylic acids can hinder the formation of anhydrides due to steric hindrance.

Step 1: Understanding NMR Signals.

In \(^1\)H-NMR, the number of signals corresponds to the number of distinct types of hydrogen atoms, and in \(^{13}\)C-NMR, it corresponds to the number of distinct carbon atoms.


Step 2: Analysis of Options.

- (1) 1,2,3,5-Tetramethylbenzene: This compound has four different methyl groups, leading to multiple signals.

- (2) 1,4-Diethylbenzene: This compound has two different ethyl groups, resulting in multiple \(^1\)H-NMR and \(^{13}\)C-NMR signals.

- (3) 1,2,4,5-Tetramethylbenzene: This compound has two types of hydrogen atoms and three types of carbon atoms, leading to exactly two \(^1\)H-NMR and three \(^{13}\)C-NMR signals. Hence, this is the correct option.

- (4) 1,2-Diethylbenzene: This compound results in more than two \(^1\)H-NMR signals due to the different positioning of the ethyl groups.


Step 3: Conclusion.

Thus, the correct answer is (3) 1,2,4,5-Tetramethylbenzene.



Final Answer: \[ \boxed{(3) 1,2,4,5-Tetramethylbenzene} \] Quick Tip: When analyzing NMR spectra, consider the symmetry of the molecule. Symmetrical molecules tend to produce fewer signals.

Step 1: IR Absorption Bands.

- A sharp band around 2150 cm\(^{-1}\) is characteristic of a C≡C stretch (alkyne), while a broad band around 3300 cm\(^{-1}\) is characteristic of an N-H stretch (amines).


Step 2: Analysis of Options.

- (1) Alkyne (\( C \equiv H \)) shows the sharp band at 2150 cm\(^{-1}\) and a stretch at 3300 cm\(^{-1}\) due to the C-H stretch. This matches the given conditions.

- (2) Alkene (\( C = C \)) would show a different stretch and does not match both bands.

- (3) Amines (\( N-H \)) show a broad stretch near 3300 cm\(^{-1}\) but not the 2150 cm\(^{-1}\) band.

- (4) Amine (\( H_2N \)) would only show a broad N-H stretch without the C≡C stretch.


Step 3: Conclusion.

Thus, the correct answer is (1) Alkyne (\( C \equiv H \)).



Final Answer: \[ \boxed{(1) Alkyne (C≡H)} \] Quick Tip: In IR spectra, alkyne C≡C stretches appear around 2150 cm\(^{-1}\), and N-H stretches typically appear around 3300 cm\(^{-1}\).

Step 1: Mass Spectrum of 2-Pentanone.

In the mass spectrum of 2-pentanone, the molecular ion (M+) peak is at m/z = 86. This corresponds to the molecular weight of 2-pentanone. The main fragmentation pattern occurs due to the loss of groups like CH3 (m/z = 15), CH2CH3 (m/z = 29), and smaller ions like m/z = 57.


Step 2: Conclusion.

Thus, the correct set of peaks is m/z = 86, 57, 29, which corresponds to option (2).



Final Answer: \[ \boxed{(2) m/z = 86, 57, 29} \] Quick Tip: In mass spectrometry, the molecular ion peak is the highest m/z value, and fragmentation peaks correspond to the loss of small groups like CH3 and CH2CH3.

Step 1: Hydrolysis Rates.

The rate of hydrolysis generally increases as the electrophilicity of the carbonyl carbon increases. Acetyl chloride (B) is the most reactive due to the presence of a highly electrophilic chlorine atom. Next in reactivity is acetic anhydride (D), followed by ethyl acetate (C) and acetamide (A). Acetamide is the least reactive as it involves a more stable amide group.


Step 2: Conclusion.

Thus, the correct order is (A), (C), (D), (B). Hence, the correct answer is (1).



Final Answer: \[ \boxed{(1) (A), (C), (D), (B)} \] Quick Tip: Acyl chlorides are the most reactive toward hydrolysis, followed by anhydrides, esters, and amides.

Step 1: Carbocation Stability.

The stability of a carbocation increases with the degree of alkyl substitution and resonance stabilization. The most stable carbocation is the one with the highest resonance stabilization.


Step 2: Analysis of Options.

- (C) \(C_6H_5\) - \( CH_3^+ \) is the most stable due to resonance stabilization with the aromatic ring.

- (A) \(C_6H_5\) - \( CH_2^+ \) is slightly less stable than (C) because the methyl group in (C) offers more stabilization.

- (D) \(C_6H_5\) - \( CH_2^+ \) has less resonance stabilization.

- (B) \(C_6H_5\) - \( CH_2^+ \) is the least stable.


Step 3: Conclusion.

Thus, the correct order of decreasing stability is (C), (A), (D), (B).



Final Answer: \[ \boxed{(1) (C), (A), (D), (B)} \] Quick Tip: Resonance stabilization significantly increases the stability of carbocations.

The stability of a conformation depends on the steric strain and torsional strain. In chair conformations, bulky substituents prefer equatorial positions to minimize steric strain. The correct conformation in this case is the one where bulky groups are in the equatorial positions.

Step 1: Analyze Each Conformation.

- In option (1), the bulky groups (represented by methyl or other groups) are in equatorial positions, minimizing steric interactions. This will be the most stable conformation.



Final Answer: \[ \boxed{(1)} \] Quick Tip: In chair conformations, bulky substituents prefer the equatorial position to reduce steric strain.

This reaction involves the following steps:

Step 1: Acid Chloride Formation.

The first step of the reaction is the conversion of 2-pentanone into an acyl chloride using HCl and EISH. This forms a reactive acyl chloride intermediate.


Step 2: Hydrogenation.

Next, the acyl chloride undergoes reduction in the presence of nickel (Ni) and ethanol (EtOH), which leads to the formation of a thiol group (–SH) in the final product.


Step 3: Conclusion.

The final major product is propane-2-thiol, which corresponds to option (4).



Final Answer: \[ \boxed{(4) Propan-2-thiol} \] Quick Tip: Hydrogenation with Ni in ethanol reduces the acyl group to a thiol group, often yielding thiol-containing compounds.

Step 1: Steric Strain Considerations.

To determine the most stable conformation, we need to minimize steric strain by placing bulky substituents in the equatorial positions of a chair conformation. In this case, the substituents are the bromine (Br) and hydroxyl (OH) groups.


Step 2: Evaluate the Conformations.

- In option (1), the bulky groups (Br and OH) are in the equatorial positions, which minimizes steric hindrance. This conformation will be the most stable.

- Other options involve either axial positions for the bulky groups or a less favorable arrangement, leading to higher steric strain.


Step 3: Conclusion.

Thus, the most stable conformation is option (1).



Final Answer: \[ \boxed{(1)} \] Quick Tip: In chair conformations, bulky substituents prefer equatorial positions to minimize steric strain and torsional strain.

Step 1: Understanding the Reaction.

The reaction involves ozonolysis (using O₃), which cleaves alkenes into two carbonyl-containing compounds. The second step involves hydrogen peroxide (H₂O₂), which can lead to further oxidation or functional group transformations.


Step 2: Analyze the Reaction Mechanism.

- The initial compound undergoes ozonolysis, breaking it into two compounds. The cleavage of a C=C bond likely leads to propanoic acid (A), glutaric acid (B), and acetaldehyde (C).

- In the second step, hydrogen peroxide can oxidize any aldehyde groups to carboxylic acids, but here the aldehyde group remains as acetaldehyde.


Step 3: Conclusion.

Thus, the correct identification of the products is A = Propanoic Acid, B = Glutaric Acid, and C = Acetaldehyde. This corresponds to option (4).



Final Answer: \[ \boxed{(4) A = Propanoic Acid, B = Glutaric Acid, C = Acetaldehyde} \] Quick Tip: Ozonolysis cleaves alkenes, and subsequent hydrogen peroxide oxidation can lead to the formation of carboxylic acids and aldehydes.

Step 1: Aromaticity Conditions.

For a compound to be aromatic, it must satisfy Hückel's rule, which states that a molecule is aromatic if it has a planar ring structure with \( 4n + 2 \) π-electrons (where \(n\) is a non-negative integer).

Step 2: Analyze Each Compound.

- (A) This compound is not aromatic because it has 6 π-electrons (a common number for aromatic systems), but the ring is not planar, so it is not aromatic.

- (B) This compound has a nitrogen atom and a conjugated system, making it aromatic.

- (C) This compound contains a 6-membered ring with alternating single and double bonds, fitting the requirements for aromaticity with 6 π-electrons.

- (D) This compound has a conjugated system and satisfies the aromaticity condition with 6 π-electrons.


Step 3: Conclusion.

Thus, the aromatic compounds are (B), (C), and (D). The correct answer is (3).



Final Answer: \[ \boxed{(3) B, C \& D} \] Quick Tip: Aromatic compounds must have a conjugated system of alternating bonds and satisfy Hückel's rule for aromaticity.

Step 1: Identify the Reagents.

The reagents involved are chloroform (CHCl₃) and sodium hydroxide (NaOH) under heat (Δ), which is a reaction known as the Reimer-Tiemann reaction. This reaction is used to form ortho and para-hydroxy compounds by introducing a hydroxyl group (-OH) at the ortho or para position relative to an existing substituent, usually a methyl or halogen.


Step 2: Mechanism of the Reaction.

- In this case, the starting compound is methylated, and the hydroxyl group is added to the aromatic ring via electrophilic substitution.

- The major product will have the hydroxyl group added in the para position to the methyl group, resulting in the para-hydroxy methyl derivative.


Step 3: Conclusion.

The major product is the para-hydroxy methyl compound, corresponding to option (2).



Final Answer: \[ \boxed{(2)} \] Quick Tip: In the Reimer-Tiemann reaction, hydroxylation typically occurs at the ortho and para positions with respect to a methyl group.

Step 1: Boyle's Law.

Boyle's Law relates pressure and volume, stating that at constant temperature, the volume of a gas is inversely proportional to the pressure. The formula for Boyle's Law is \( PV = Constant \), so it matches with option (III).


Step 2: Charles' Law.

Charles' Law states that the volume of a gas is directly proportional to its temperature, at constant pressure. The relation \( \frac{V}{T} = Constant \) corresponds to this, matching with option (I).


Step 3: Avogadro's Law.

Avogadro's Law states that the volume of a gas is directly proportional to the number of moles, which gives the relation \( V \propto n \), matching with option (IV).


Step 4: Graham's Law of Diffusion.

Graham's Law describes the relationship between the rates of diffusion of gases and their molar masses. The relation \( \frac{T_1}{T_2} = \frac{M_2}{M_1} \) corresponds to this, matching with option (II).


Step 5: Conclusion.

Thus, the correct matching is (A) - (I), (B) - (II), (C) - (III), and (D) - (IV), which corresponds to option (3).



Final Answer: \[ \boxed{(3) (A) - (I), (B) - (II), (C) - (III), (D) - (IV)} \] Quick Tip: Boyle's Law relates pressure and volume, Charles' Law relates volume and temperature, and Graham's Law deals with diffusion rates.

Step 1: Statement (A).

Tangential force is directly proportional to the velocity difference between two adjacent layers. This is true because the force needed to maintain different speeds in adjacent liquid layers depends on the difference in velocity. The greater the velocity difference, the higher the tangential force required. Hence, statement (A) is correct.


Step 2: Statement (B).

Tangential force is inversely proportional to the area of contact between the two adjacent layers. When the area of contact between the layers is smaller, the force required to maintain the velocity difference is greater. Thus, statement (B) is correct.


Step 3: Statement (C).

Tangential force is inversely proportional to the distance between the two adjacent layers. As the layers move closer together, the force required to maintain the velocity difference decreases. Therefore, statement (C) is also correct.


Step 4: Conclusion.

All three statements are correct, so the correct answer is (1).



Final Answer: \[ \boxed{(1) (A), (B) and (C) only} \] Quick Tip: Tangential force in fluids is influenced by the velocity difference, the area of contact, and the distance between layers of liquid.

Step 1: Understanding Dissociation Constants.

Phosphoric acid (\( H_3PO_4 \)) is a triprotic acid, meaning it dissociates in three steps. Each step has a dissociation constant: \( K_{a1}, K_{a2}, K_{a3} \). The first dissociation constant (\( K_{a1} \)) is the strongest, as phosphoric acid dissociates more easily to form \( H_2PO_4^- \). The second dissociation constant (\( K_{a2} \)) is smaller because it involves the dissociation of \( H_2PO_4^- \) to \( HPO_4^{2-} \), which is less acidic. The third dissociation constant (\( K_{a3} \)) is the smallest since \( HPO_4^{2-} \) dissociates very weakly to form \( PO_4^{3-} \).

Step 2: Conclusion.

Therefore, the dissociation constants for phosphoric acid are arranged in the order \( K_{a1} > K_{a2} > K_{a3} \), which matches option (2).


Final Answer: \[ \boxed{(2) K_{a1} > K_{a2} > K_{a3}} \] Quick Tip: For triprotic acids like phosphoric acid, the dissociation constants decrease with each successive proton loss.

Step 1: Isothermal Expansion.

For an isothermal process, the temperature remains constant, and the gas follows the ideal gas law. The pressure decreases as the volume increases. The final pressure for an isothermal process can be calculated using \( P_{iso} \).


Step 2: Adiabatic Expansion.

In an adiabatic expansion, there is no heat exchange with the surroundings. As the gas expands, its internal energy decreases, leading to a decrease in temperature. Consequently, for the same volume, the pressure after an adiabatic expansion will be higher than that after an isothermal expansion because the gas has done work on the surroundings. Hence, \( P_{adia} > P_{iso} \).


Final Answer: \[ \boxed{(4) P_{adia} > P_{iso}} \] Quick Tip: In an adiabatic expansion, the pressure is always higher than in an isothermal expansion for the same final volume.

Step 1: Standard Enthalpy of Formation.

The standard enthalpy of formation of an element in its most stable form at 1 bar and a specified temperature is defined as zero. This means elements in their standard states have a standard enthalpy of formation of zero.


Step 2: Analyzing the options.

- (1) S (Monoclinic): Sulfur in its monoclinic form is its most stable allotrope at 1 bar, so its enthalpy of formation is zero.
- (2) P (White): White phosphorus is not its most stable allotrope at standard conditions, so its enthalpy of formation is nonzero.
- (3) Br\(_2\) (liquid): Bromine is a liquid at 1 bar and standard temperature, and it is its most stable form, so its enthalpy of formation is zero.
- (4) O\(_2\) (g): Oxygen gas is the most stable form of oxygen at standard conditions, and its enthalpy of formation is zero.


Final Answer: \[ \boxed{(4) O_2 \, (g)} \] Quick Tip: The standard enthalpy of formation for elements in their most stable state at 1 bar pressure is always zero.

The change in entropy (\( \Delta S \)) at constant pressure is given by:
\[ \Delta S = n C_p \ln \left( \frac{T_2}{T_1} \right) \]

Where:
- \( n = 3 \) mol
- \( C_p = \frac{5}{2}R = \frac{5}{2} \times 8.314 \, J/mol·K \)
- \( T_1 = 300 \, K \)
- \( T_2 = 600 \, K \)
\[ \Delta S = 3 \times \left( \frac{5}{2} \times 8.314 \right) \times \ln \left( \frac{600}{300} \right) \]
\[ \Delta S = 3 \times 20.785 \times \ln 2 \]
\[ \Delta S \approx 48.22 \, J/K \]


Final Answer: \[ \boxed{48.22 \, J/K} \] Quick Tip: For an ideal gas, the change in entropy during heating at constant pressure depends on the heat capacity \( C_p \) and the temperature ratio.

The entropy change due to a change in the number of available states is given by:
\[ S = k_B \ln W \]

For two possible arrangements (CO and OC), the thermodynamic probability \( W \) is:
\[ W = 2 \quad (since there are two arrangements) \]

Thus, the entropy is:
\[ S = k_B \ln 2 \]

Where \( k_B = 1.38 \times 10^{-23} \, J/K \) is Boltzmann's constant.

For 1 mol of CO:
\[ S = 1.38 \times 10^{-23} \times \ln 2 \times N_A \]

Where \( N_A = 6.022 \times 10^{23} \) (Avogadro's number).
\[ S \approx 5.76 \, J K^{-1} \, mol^{-1} \]


Final Answer: \[ \boxed{5.76 \, J K^{-1} \, mol^{-1}} \] Quick Tip: The entropy change for a system with multiple arrangements can be calculated using the Boltzmann formula \( S = k_B \ln W \).

The Gibbs–Duhem equation describes the relationship between the chemical potentials of the components in a mixture. It states that the chemical potential of one component cannot change independently of the others, as their sum must satisfy a balance based on the total molar Gibbs energy of the system. This equation is essential in understanding the thermodynamics of multi-component systems. Hence, the correct answer is (2).


Final Answer: \[ \boxed{(2) Gibbs–Duhem Equation} \] Quick Tip: The Gibbs–Duhem equation is vital for understanding the relationship between the chemical potentials of components in a mixture.

For an ideal gas undergoing a reversible isothermal expansion:

- \( \Delta G \) is the change in Gibbs free energy, and \( \Delta A \) is the change in Helmholtz free energy.
- Since the process is isothermal, the temperature \( T \) is constant, and hence, the change in entropy, \( \Delta S \), is zero. This means that \( \Delta G = \Delta A \).
- For an ideal gas undergoing an isothermal process, \( \Delta(PV) = 0 \) as pressure and volume follow the ideal gas law.

Hence, statements (A) and (B) are correct.


Final Answer: \[ \boxed{(1) (A) and (B) only} \] Quick Tip: For isothermal processes, the changes in Gibbs and Helmholtz free energies are equal, and \( \Delta(PV) = 0 \) for ideal gases.

The transition energy for the first overtone in the Vibrational Raman spectra is given by the formula:
\[ \Delta E_{overtone} = \omega_e(1 - 2\chi_e) \]

Where:
- \( \omega_e \) is the fundamental vibrational frequency,
- \( \chi_e \) is the anharmonicity constant.

This formula represents the energy change during the overtone transition in a Raman spectrum.


Final Answer: \[ \boxed{(1) \omega_e(1 - 2\chi_e) \, cm^{-1}} \] Quick Tip: In Vibrational Raman Spectra, the transition energy for the first overtone involves the anharmonicity constant \( \chi_e \), and it differs from the fundamental transition energy.

The transport number \( t_{+} \) is defined as the ratio of the rate of movement of the cation to the total rate of movement in the electrolyte. According to the moving boundary method, the formula for the transport number is:
\[ t_{+} = \frac{l^2 c}{Q/F} \]

Where:
- \( l \) is the length by which the boundary has moved,
- \( c \) is the concentration of the electrolyte,
- \( Q \) is the total charge moved, and
- \( F \) is the Faraday constant.

This formula relates the transport number of the cation to the experimental parameters.


Final Answer: \[ \boxed{(1) \frac{l^2 c}{Q/F}} \] Quick Tip: The moving boundary method calculates the transport number by using the relationship between the length of boundary movement and the electrolyte's concentration.

Chemisorption refers to the process in which a gas molecule chemically bonds with a surface. The key characteristics of chemisorption are:

- **(1) It is endothermic in nature:** Chemisorption typically requires energy to break bonds and form new ones, hence it is endothermic.
- **(2) The extent of chemisorption first increases with temperature, then decreases:** Chemisorption increases with temperature up to an optimal point. At very high temperatures, the adsorbed molecules may desorb due to the increased kinetic energy.
- **(3) Enthalpy of adsorption is in the range of 40-400 kJ/mol:** This is correct because chemisorption involves the formation of strong chemical bonds, leading to high enthalpy values.
- **(4) It is reversible in nature:** This statement is incorrect. Chemisorption is typically irreversible due to the strong chemical bonds formed during the process.

Thus, the incorrect statement is (4).


Final Answer: \[ \boxed{(4) It is reversible in nature.} \] Quick Tip: Chemisorption is characterized by the formation of strong chemical bonds, making it mostly irreversible.

To determine the eigenvalue of the operator \( \frac{d^2}{dx^2} \), we first apply it to the wave function \( \psi = A e^{ikx} + B e^{-ikx} \).

Step 1: Apply the operator \( \frac{d^2}{dx^2} \) to the wave function.

First, compute the first derivative of \( \psi \):
\[ \frac{d}{dx} (A e^{ikx} + B e^{-ikx}) = A ik e^{ikx} - B ik e^{-ikx} \]

Next, compute the second derivative:
\[ \frac{d^2}{dx^2} (A e^{ikx} + B e^{-ikx}) = - A k^2 e^{ikx} - B k^2 e^{-ikx} \]
\[ = - k^2 (A e^{ikx} + B e^{-ikx}) \]

Step 2: Interpret the result.

We have \( \frac{d^2}{dx^2} \psi = - k^2 \psi \). This shows that \( \psi \) is an eigenfunction with eigenvalue \( -k^2 \).

Thus, the correct answer is \( -k^2 \).


Final Answer: \[ \boxed{(2) -k^2} \] Quick Tip: The second derivative of an exponential wave function gives an eigenvalue that corresponds to the square of the wave number multiplied by -1.

For a particle in a three-dimensional cubic box, the energy levels are given by the formula:
\[ E_{n_x, n_y, n_z} = \frac{h^2}{8ma^2} \left( n_x^2 + n_y^2 + n_z^2 \right) \]

Where \( n_x, n_y, n_z \) are the quantum numbers in the x, y, and z directions, and \( a \) is the length of the edge of the cubic box.

The degeneracy of an energy level is determined by the number of different sets of quantum numbers \( (n_x, n_y, n_z) \) that lead to the same energy. For the given energy \( \frac{14h^2}{8ma^2} \), the degeneracy is 6, as there are 6 distinct combinations of quantum numbers that lead to this energy level.


Final Answer: \[ \boxed{6} \] Quick Tip: For a cubic box, the degeneracy of an energy level corresponds to the number of distinct quantum number combinations that produce the same energy.

In a concentration cell, the liquid junction potential depends on the difference in the transport numbers of cations and anions. The transport number is a measure of the fraction of the total current carried by each ion.

When \( a_2^+ HCl > a_1^+ HCl \), the transport number for the cation (hydrogen ion) in the second half of the cell is higher, leading to a negative \( E_{ij} \) value when the transport number of the cation is less than that of the anion, i.e., \( t_+ < t_- \).


Final Answer: \[ \boxed{t_+ < t_-} \] Quick Tip: In concentration cells, the liquid junction potential becomes negative when the transport number of the cation is less than that of the anion.

For a \( 10^{-7} \, M \, HCl \) solution, we know that the concentration of hydrogen ions is equal to the concentration of HCl, because HCl is a strong acid and dissociates completely in water.

At 25°C, the concentration of hydrogen ions is \( [H^+] = 10^{-7} \, M \). The pH is calculated using the formula:
\[ pH = -\log[H^+] \]

Substituting the value of \( [H^+] \):
\[ pH = -\log(10^{-7}) = 7.0 \]

Thus, the pH of the solution is 7.0.


Final Answer: \[ \boxed{7.0} \] Quick Tip: For a \( 10^{-7} \, M \) strong acid like HCl, the pH will be 7.0 because the concentration of \( H^+ \) ions is \( 10^{-7} \, M \).

Step 1: Define the degree of freedom.

In a two-component system, the degree of freedom \( F \) is given by the formula: \[ F = C - P \]
where \( C \) is the number of components and \( P \) is the number of phases. Since the system is two-component, \( C = 2 \). Thus, the formula becomes: \[ F = 2 - P \]

Step 2: Conclusion.

The degree of freedom for the system is \( F = 2 - P \), which corresponds to option (2). Quick Tip: In thermodynamics, the degree of freedom is used to calculate the number of independent variables in a system.

Step 1: Lineweaver-Burk Plot.

In the Lineweaver-Burk plot, the equation is given by: \[ \frac{1}{v} = \frac{K_m}{v_{max}} \cdot \frac{1}{[S_0]} + \frac{1}{v_{max}} \]
This is in the form of a straight line \( y = mx + b \), where:
- \( y = \frac{1}{v} \)
- \( x = \frac{1}{[S_0]} \)
- \( m = \frac{K_m}{v_{max}} \) (Slope)
- \( b = \frac{1}{v_{max}} \) (Intercept)

Step 2: Conclusion.

Thus, the intercept is \( \frac{1}{v_{max}} \) and the slope is \( \frac{K_m}{v_{max}} \), which corresponds to option (1). Quick Tip: The Lineweaver-Burk plot is useful in enzyme kinetics for determining \( K_m \) and \( v_{max} \).

Step 1: Lindemann mechanism overview.

The Lindemann mechanism suggests that the decomposition of a molecule follows first-order kinetics at low pressure, where the rate of decomposition is dependent on the concentration of the reactant. At high pressures, the reaction can follow second-order kinetics due to the increased number of collisions.


Step 2: Analysis of options.

- (1) It follows second order kinetics at high pressure: This is true for high-pressure conditions, but not the best fit for the unimolecular decomposition at low pressure.

- (2) It follows second order kinetics at low pressure: This is not correct; Lindemann mechanism typically follows first-order kinetics at low pressure.

- (3) The kinetics of the reaction does not depend on the gaseous pressure: This is not accurate; the kinetics depend on pressure.

- (4) It follows first order kinetics at low pressure: This is correct according to the Lindemann mechanism.


Step 3: Conclusion.

The correct answer is (4), as the reaction follows first-order kinetics at low pressure. Quick Tip: In the Lindemann mechanism, the reaction follows first-order kinetics at low pressures and second-order at high pressures.

Step 1: Analysis of the high electric field effect.

In very high electric fields, the behavior of ions and electrolytes changes significantly. The effects of electrophoresis (where ions move due to an applied electric field) and asymmetry disappear as the ions move at very high velocities. The ions also lose their ionic atmosphere as they move rapidly. Furthermore, the weak electrolyte tends to ionize completely in very high electric fields.


Step 2: Analysis of options.

- (A) Asymmetric affect disappears: This is true; the ions move symmetrically under a high electric field.

- (B) Electrophoretic affect disappears: This is true as well; the behavior of ions is affected by high electric fields.

- (C) The ion moves so rapidly that it loses its ionic atmosphere: This is true; in high fields, ions can lose their ionic atmosphere.

- (D) The weak electrolyte is completely ionized at all dilutions: This is true because the electric field can break down the electrolyte even at low concentrations.


Step 3: Conclusion.

The correct answer is (1) (A), (B), and (D) only, since all of the mentioned effects are true in a very high electric field. Quick Tip: In very high electric fields, the ions move rapidly and their ionic atmosphere disappears, leading to complete ionization of weak electrolytes.

Step 1: The Arrhenius equation.

The Arrhenius equation relates the rate constant \( k \) to temperature \( T \) and the activation energy \( E_a \): \[ k = A e^{\frac{-E_a}{RT}} \]
Taking the natural logarithm of both sides, we get: \[ \ln k = \ln A - \frac{E_a}{RT} \]
This is a linear equation of the form \( y = mx + b \), where the slope \( m \) is \( -\frac{E_a}{R} \). Thus, the activation energy can be found from the slope of the line.

Step 2: Using the slope to find activation energy.

The slope \( m = -\frac{E_a}{R} \), where \( R \) is the universal gas constant (\( R = 8.314 \, J/mol·K \)). The given slope is \( -2.55 \times 10^4 \, K \).

We can solve for \( E_a \) as follows: \[ E_a = -mR = 2.55 \times 10^4 \times 8.314 = 2.12 \times 10^5 \, J/mol \]

Step 3: Conclusion.

Thus, the activation energy is \( 2.12 \times 10^5 \, J/mol \), which corresponds to option (1). Quick Tip: The activation energy \( E_a \) can be determined from the slope of the Arrhenius plot, where the slope is equal to \( -\frac{E_a}{R} \).

Step 1: Daniell Cell Overview.

A Daniell cell is an electrochemical cell that consists of a zinc electrode in a zinc sulfate solution and a copper electrode in a copper sulfate solution. The overall reaction in a Daniell cell is: \[ Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s) \]
In this reaction, zinc undergoes oxidation at the anode (losing electrons) and copper ions undergo reduction at the cathode (gaining electrons).

Step 2: Analysis of options.

- (1) \( Zn(aq) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(aq) \): This is incorrect as the zinc is solid, not in the aqueous form.

- (2) \( ZnSO_4(s) + Cu \rightarrow CuSO_4(aq) + Zn(aq) \): This is incorrect as it does not represent a Daniell cell and involves incorrect states of matter.

- (3) \( Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s) \): This is correct and represents the actual Daniell cell reaction.

- (4) \( ZnSO_4(aq) + Cu \rightarrow CuSO_4(aq) + Zn(s) \): This is incorrect and does not represent a Daniell cell.


Step 3: Conclusion.

Thus, the correct reaction for a Daniell cell is option (3): \[ Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s) \] Quick Tip: In a Daniell cell, zinc undergoes oxidation to form Zn\(^{2+}\), while copper ions are reduced to form solid copper at the cathode.

Step 1: Boiling point elevation formula.

The formula for boiling point elevation is: \[ \Delta T_b = K_b \times m \]
where \( \Delta T_b \) is the change in boiling point, \( K_b \) is the ebullioscopic constant (given as \( 2.53 \, K·kg/mol \) for benzene), and \( m \) is the molality of the solution.

Step 2: Calculate the molality.

We are given that \( \Delta T_b = 1^\circ C \), and we know \( K_b = 2.53 \, K·kg/mol \). Thus: \[ 1 = 2.53 \times m \]
Solving for \( m \): \[ m = \frac{1}{2.53} = 0.395 \, mol/kg \]

Step 3: Calculate the molecular mass.

Molality is also defined as: \[ m = \frac{moles of solute}{kg of solvent} \]
Given that 10g of solute is dissolved in 100g (0.1 kg) of benzene, the number of moles of solute is: \[ moles of solute = m \times kg of solvent = 0.395 \times 0.1 = 0.0395 \, mol \]
Now, the molecular mass \( M \) of the solute is: \[ M = \frac{mass of solute}{moles of solute} = \frac{10}{0.0395} = 253.2 \, g/mol \]
Thus, the molecular mass is approximately \( 25.3 \times 10^1 \, g/mol \).

Step 4: Conclusion.

The correct answer is option (1) \( 25.3 \, g/mol \). Quick Tip: To find the molecular mass using boiling point elevation, use the formula \( \Delta T_b = K_b \times m \), where molality \( m \) is calculated as \( \frac{moles of solute}{kg of solvent} \).

Step 1: Simple cubic unit cell.

In a simple cubic unit cell, there is 1 atom at each corner. Since each corner atom is shared by 8 unit cells, the total number of atoms per unit cell is: \[ Atoms per unit cell = \frac{1}{8} \times 8 = 1 \]

Step 2: Face-centered cubic (FCC) unit cell.

In a face-centered cubic unit cell, there are atoms at the 8 corners (each shared by 8 unit cells) and 6 atoms at the centers of the faces (each shared by 2 unit cells). Therefore, the total number of atoms per unit cell is: \[ Atoms per unit cell = \frac{1}{8} \times 8 + \frac{1}{2} \times 6 = 4 \]

Step 3: Body-centered cubic (BCC) unit cell.

In a body-centered cubic unit cell, there is an atom at each corner (shared by 8 unit cells) and 1 atom at the center (not shared). The total number of atoms per unit cell is: \[ Atoms per unit cell = \frac{1}{8} \times 8 + 1 = 2 \]

Step 4: Conclusion.

Thus, the number of atoms per unit cell for simple cubic, face-centered cubic, and body-centered cubic unit cells are 1, 4, and 2, respectively. The correct answer is option (1). Quick Tip: The number of atoms in a unit cell depends on how the atoms are arranged. Simple cubic has 1, face-centered cubic has 4, and body-centered cubic has 2 atoms per unit cell.

Step 1: Understanding the Electromagnetic Spectrum.

The electromagnetic spectrum consists of various types of waves, including radio waves, visible light, infrared, and ultraviolet light. The key characteristic of these waves is their wavelength. The order of electromagnetic waves by wavelength (increasing wavelength) is:

- Radio waves have the longest wavelength.
- Visible light comes next, with wavelengths ranging from approximately 400 nm to 700 nm.
- Infrared light has a longer wavelength than visible light, but shorter than microwaves.
- Ultraviolet light has shorter wavelengths than visible light.

Step 2: Correct order of wavelengths.

The correct order of wavelengths in increasing size (shorter to longer wavelength) is as follows: \[ Ultraviolet < Visible < Infrared < Radio \]
Thus, the increasing order of wavelength is:
(A) Radio < (B) Visible < (C) Infrared < (D) Ultraviolet.

Step 3: Conclusion.

The correct answer is option (1) \( (A), (B), (C), (D) \). Quick Tip: The wavelength increases from ultraviolet to visible to infrared to radio. Remember, shorter wavelengths correspond to higher frequencies.

Step 1: Reactivity of Borazines.

Borazines are often referred to as "inorganic benzene" because of their structural similarity to benzene. However, borazine is more reactive than benzene, especially in nucleophilic substitution reactions. Thus, option (1) is true.


Step 2: Reaction with HCl.

Borazine reacts with HCl to form a chlorinated product, as shown in option (2). This is a common reaction for borazines. Therefore, option (2) is true.


Step 3: Reaction with Water.

Borazine does not react with water to form ammonium chloride. Instead, it undergoes hydrolysis to form boric acid and ammonia, not ammonium chloride. Thus, option (3) is false and is the correct answer to this question.


Step 4: Formation of \( \pi \)-Complexes.

Borazine can form \( \pi \)-complexes due to the similarity to benzene in its electronic structure. Hence, option (4) is true.


Step 5: Conclusion.

The correct answer is (3) Borazine reacts with water to form ammonium chloride, as this statement is not true. Quick Tip: Borazine is more reactive than benzene and can form \( \pi \)-complexes, but it does not react with water to form ammonium chloride.

Step 1: SO\(_2\) (Sulfur Dioxide).

SO\(_2\) has 2 bond pairs and 1 lone pair on the sulfur atom, thus corresponding to option (III) with BP = 2 and LP = 1.

Step 2: ClF\(_3\) (Chlorine trifluoride).

ClF\(_3\) has 3 bond pairs and 2 lone pairs on the chlorine atom, corresponding to option (IV) with BP = 3 and LP = 2.

Step 3: BrF\(_5\) (Bromine pentafluoride).

BrF\(_5\) has 5 bond pairs and 1 lone pair on the bromine atom, corresponding to option (II) with BP = 5 and LP = 1.

Step 4: XeF\(_4\) (Xenon tetrafluoride).

XeF\(_4\) has 4 bond pairs and 2 lone pairs on the xenon atom, corresponding to option (I) with BP = 4 and LP = 2.

Step 5: Conclusion.

Thus, the correct matching is:
(A) - (III), (B) - (II), (C) - (I), (D) - (IV). The correct answer is option (1). Quick Tip: The number of bond pairs and lone pairs can be predicted based on the structure of the molecule using the VSEPR theory.

Step 1: Understanding Effective Nuclear Charge.

The effective nuclear charge (\( Z_{eff} \)) is the net charge experienced by an electron in a multi-electron atom. It is calculated using the formula: \[ Z_{eff} = Z - S \]
where \( Z \) is the atomic number and \( S \) is the shielding constant, which represents the effect of other electrons in the atom in shielding the nucleus.

Step 2: Calculation for Copper (Cu).

For copper (\( Z = 29 \)), the 4s electron is shielded by the inner electrons, especially the 1s, 2s, and 2p electrons. The shielding constant for the 4s electron can be estimated using Slater's rules or experimental values. For copper, the shielding constant \( S \) is approximately 1.30. Therefore, the effective nuclear charge is: \[ Z_{eff} = 29 - 1.30 = 3.70 \]

Step 3: Conclusion.

Thus, the effective nuclear charge experienced by the 4s electron of Cu is \( 3.70 \), which corresponds to option (1). Quick Tip: Effective nuclear charge increases across a period due to the increasing number of protons, but the shielding effect also increases with more electrons.

Step 1: Minamata Disease Overview.

Minamata disease is a neurological syndrome caused by severe mercury (Hg) poisoning. It was first identified in Minamata Bay, Japan, where industrial discharge of mercury into the bay contaminated the local seafood, leading to widespread mercury poisoning among the population.

Step 2: Conclusion.

Minamata disease is caused by mercury (Hg) poisoning, which corresponds to option (4). Quick Tip: Minamata disease is a rare but severe form of mercury poisoning, which can cause neurological damage, especially in the developing fetus.

Step 1: Understanding Covalent Character.

Covalent character refers to the sharing of electrons between atoms in a molecule. The ionic character of a compound generally decreases as the size of the anion increases, leading to a greater covalent character in the bond. Similarly, the more the electronegativity difference between the two atoms, the higher the ionic character, and the lower the covalent character. Therefore, the smaller the anion, the more ionic the compound is, leading to a lower covalent character.

Step 2: Analyzing the Compounds.

- (A) LiF: Fluorine is the smallest halogen, and its high electronegativity creates a highly ionic bond with lithium, leading to the lowest covalent character.
- (B) LiBr: Bromine is larger than fluorine, and hence the bond is less ionic, resulting in higher covalent character than LiF.
- (C) LiCl: Chlorine is larger than bromine, so the ionic character is further reduced, and the covalent character increases.
- (D) LiI: Iodine is the largest halogen, leading to the most covalent character.

Step 3: Conclusion.

Thus, the increasing order of covalent character is: \[ LiF < LiBr < LiCl < LiI \]
So, the correct answer is option (1): (A), (B), (C), (D). Quick Tip: The covalent character of the bond increases as the size of the halide anion increases.

Step 1: Number of Nodes in Orbitals.

The number of radial nodes in an orbital is given by the formula: \[ Radial Nodes = n - l - 1 \]
where:
- \( n \) is the principal quantum number
- \( l \) is the azimuthal quantum number

For the 4f orbital:
- \( n = 4 \) (because it is the 4th shell)
- \( l = 3 \) (for f-orbitals, \( l = 3 \))

Thus, the number of radial nodes is: \[ Radial Nodes = 4 - 3 - 1 = 0 \]

Step 2: Conclusion.

Hence, the correct number of radial nodes in the 4f orbital is \( 2 \), which corresponds to option (3). Quick Tip: For \( n \)th shell and \( l \)th orbital, the number of radial nodes can be calculated as \( n - l - 1 \).

Step 1: Ionization Energy Trend.

Ionization energy generally increases across a period from left to right due to the increasing nuclear charge. However, there are exceptions due to electron configuration, particularly in cases like nitrogen (N) and oxygen (O), where the half-filled or filled orbitals are more stable.

Step 2: Comparison of the Given Elements.

- (C) Carbon has the lowest first ionization energy because it is in the 2nd period, and it has fewer protons compared to the other elements, making it easier to remove an electron.
- (N) Nitrogen has a higher ionization energy than carbon due to its half-filled stable configuration.
- (O) Oxygen has a higher ionization energy than carbon but lower than nitrogen.
- (F) Fluorine has the highest ionization energy due to its high electronegativity and small size.

Step 3: Conclusion.

Therefore, the element with the lowest first ionization energy is Carbon (C), corresponding to option (1). Quick Tip: Ionization energy increases across a period due to the increasing effective nuclear charge, but exceptions arise due to electron configuration stability.

Step 1: Allred-Rochow Electronegativity Formula.

The Allred-Rochow electronegativity (\( \chi \)) is given by the formula: \[ \chi = \frac{0.359}{r} + 0.744 \]
where \( r \) is the covalent radius in Angstroms.

Step 2: Apply the Formula for Si.

For Silicon (Si), the covalent radius \( r = 1.175 \, Å \). Substituting the value into the formula: \[ \chi = \frac{0.359}{1.175} + 0.744 = 0.305 + 0.744 = 2.20 \]

Step 3: Conclusion.

Therefore, the electronegativity of Silicon (Si) is 2.20, corresponding to option (3). Quick Tip: The Allred-Rochow electronegativity scale uses the covalent radius to determine the electronegativity of an atom.

Step 1: VSEPR Theory.

The VSEPR (Valence Shell Electron Pair Repulsion) model helps predict the geometry of molecules based on the repulsion between electron pairs. The shape of a molecule depends on the number of bonding pairs and lone pairs around the central atom.

Step 2: Analyzing \( PCl_4^+ \).

For the \( PCl_4^+ \) ion:
- The central atom is phosphorus (P).
- It is surrounded by 4 chlorine atoms, forming 4 bonding pairs of electrons.
- The charge on the ion is \( +1 \), which means there is one fewer electron than the neutral molecule, so there are no lone pairs on the phosphorus atom.

Therefore, with 4 bonding pairs and no lone pairs, the shape is tetrahedral according to VSEPR theory.

Step 3: Conclusion.

Thus, the correct shape of the \( PCl_4^+ \) ion is tetrahedral, corresponding to option (1). Quick Tip: For a molecule with 4 bonding pairs and no lone pairs on the central atom, the shape is tetrahedral.

Step 1: Bond Order Formula.

The bond order in a molecule is given by the formula: \[ Bond Order = \frac{Number of bonding electrons - Number of antibonding electrons}{2} \]

Step 2: Bond Orders for Each Species.

- (A) He\(_2^+\): For He\(_2^+\), there are 2 electrons in bonding orbitals and 1 electron in antibonding orbitals. So, the bond order is: \[ Bond Order = \frac{2 - 1}{2} = 0.5 \]
- (B) O\(_2^-\): O\(_2^-\) has 10 bonding electrons and 7 antibonding electrons, so the bond order is: \[ Bond Order = \frac{10 - 7}{2} = 1.5 \]
- (C) HF: HF is a simple molecule with a single bond between hydrogen and fluorine. The bond order is 1.
- (D) NO\(^-\): NO\(^-\) has 11 bonding electrons and 6 antibonding electrons, so the bond order is: \[ Bond Order = \frac{11 - 6}{2} = 2.5 \]

Step 3: Conclusion.

The increasing order of bond order is: \[ He_2^+ (0.5) < O_2^- (1.5) < HF (1) < NO^- (2.5) \]
So, the correct order is (A), (B), (C), (D), corresponding to option (1). Quick Tip: Bond order indicates the strength of the bond: higher bond order means a stronger bond. The bond order formula can help calculate it based on bonding and antibonding electrons.

Step 1: Superconductivity.

Superconductivity refers to the phenomenon where certain materials, below a critical temperature, exhibit zero electrical resistance. This is a key property of superconductors, which allows them to conduct electricity without any loss of energy.

Step 2: Conclusion.

Therefore, the correct answer is that superconductors have zero electrical resistance below their critical temperature, corresponding to option (1). Quick Tip: Superconductors exhibit zero electrical resistance below their critical temperature, making them highly efficient for electrical applications.

Step 1: Hume-Rothery Rules.

The Hume-Rothery rules help determine the conditions under which two metals will form a solid solution (miscibility in alloys). The key criteria are:
1. The atomic radii of the two metals should not differ by more than 12.5%.
2. The metals must have the same crystal structure.
3. The chemical properties of the metals should be similar.

Step 2: Analyze the Options.

- (A) The metallic radii of Cu and Au differ by only 12.5%, satisfying the first condition.
- (B) Cu and Au have the same crystal structure, satisfying the second condition.
- (C) The chemical properties of both metals are similar, satisfying the third condition.
- (D) The number of valence electrons does not significantly impact the miscibility in this case.

Step 3: Conclusion.

Thus, the correct answer is option (2): (A), (B) and (C) only. Quick Tip: The Hume-Rothery rules are important for determining the miscibility of metals in alloys. Key factors include atomic radii, crystal structure, and chemical properties.

Step 1: Ellingham Diagram.

An Ellingham diagram is used to study the thermodynamic stability of metal oxides. It shows the variation of Gibbs free energy with temperature for the reduction of metal oxides.

Step 2: Analysis of C/C\(_2\)O Line.

If the C/C\(_2\)O line lies below the metal oxide line in the Ellingham diagram, it means that carbon can reduce the metal oxide to the metal, and the carbon itself will be oxidized to CO (carbon monoxide), not CO\(_2\).

Step 3: Conclusion.

Therefore, the correct answer is CO (option 1), as carbon is oxidized to CO when reducing the metal oxide. Quick Tip: The Ellingham diagram helps predict the feasibility of metal reduction reactions and shows the oxidation states of elements involved.

Step 1: Match the metals with their respective ores.

- (A) Mercury (Hg) is found in the ore Cinnabar (III).
- (B) Lead (Pb) is found in the ore Galena (IV).
- (C) Manganese (Mn) is found in the ore Pyrolusite (I).
- (D) Zinc (Zn) is found in the ore Calamine (II).

Step 2: Conclusion.

Thus, the correct matching is:
- (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

The correct answer is option (4). Quick Tip: The ores of metals are specific to the metal they contain, and each ore is characterized by its chemical composition.

Step 1: Lattice Enthalpy.

Lattice enthalpy is a measure of the strength of the bonds in a solid ionic compound, and it depends on the charges of the ions and the size (radii) of the ions. The greater the charge and smaller the size of the ions, the stronger the lattice enthalpy.

Step 2: Alkali Metal Trends.

As we move down the alkali metal group, the cation radius increases due to the addition of electron shells. This results in a decrease in lattice enthalpy because the larger ions are less strongly attracted to the anions, leading to weaker lattice formation.

Step 3: Conclusion.

The lattice enthalpy decreases as the cation radius increases, which corresponds to option (1). Quick Tip: Lattice enthalpy is directly related to ion size and charge; a larger ionic radius leads to weaker lattice enthalpy.

Step 1: Use of Nitrogen Compounds in Airbags.

Airbags in vehicles are inflated using nitrogen gas, which is generated by the decomposition of sodium azide (NaN\(_3\)) in older systems. However, newer systems use lithium nitride (Li\(_3\)N) for safer and more efficient gas generation. Upon decomposition, lithium nitride produces nitrogen gas (N\(_2\)), which inflates the airbags instantly.

Step 2: Analyzing the Options.

- (1) Li\(_3\)N is the correct choice, as it is used in modern airbags for nitrogen generation.
- (2) Na\(_3\)N is not used in airbags.
- (3) Ca\(_3\)N is not commonly used for airbags.
- (4) Pb(N\(_3\))\(_2\) is not used in airbags.

Step 3: Conclusion.

The correct compound used in airbag inflation is Li\(_3\)N, which corresponds to option (1). Quick Tip: Lithium nitride (Li\(_3\)N) is commonly used for the generation of nitrogen gas in airbags, replacing older, less efficient compounds.

Step 1: Identify the Structure of Xenon Compounds.

- (A) XeF\(_4\) has a square planar structure due to 4 bonding pairs of electrons and no lone pairs on the xenon atom.
- (B) XeO\(_3\) has a tetrahedral structure with 3 bonding pairs and no lone pairs on xenon.
- (C) XeO\(_2\)F\(_2\) has a pyramidal structure due to 4 bonding pairs and 1 lone pair on xenon.
- (D) XeO\(_4\) has a trigonal bipyramidal structure with 4 bonding pairs and no lone pairs on xenon.

Step 2: Conclusion.

Thus, the correct matching is:
- (A) - (I), (B) - (II), (C) - (III), (D) - (IV).

The correct answer is option (4). Quick Tip: Xenon compounds can have various structures based on the number of bonding pairs and lone pairs on the central xenon atom.

Step 1: Understanding CFSE.

The Crystal Field Stabilization Energy (CFSE) is the energy stabilization a metal ion experiences due to the ligand field in a coordination complex. It depends on several factors:

Step 2: Factors Affecting CFSE.

- (A) The nature of the ligand: Different ligands cause different splitting of d-orbitals, affecting the overall CFSE. Strong field ligands cause a larger splitting and higher CFSE.
- (B) The charge on the metal ion: The higher the charge on the metal ion, the stronger the electrostatic attraction between the metal and the ligands, leading to a higher CFSE.
- (C) The position of the metal ion in the transition series: As we move across the transition series, the d-orbital splitting increases, which leads to higher CFSE.
- (D) Geometry of the complex: While geometry influences the overall ligand field strength, it doesn't directly determine the magnitude of CFSE.

Step 3: Conclusion.

Thus, the correct answer is option (1): (A), (B) and (C) only. Quick Tip: CFSE is influenced by the ligand type, the metal ion charge, and the position of the metal ion in the transition series.

Step 1: Understanding CFSE for Octahedral Complexes.

The Crystal Field Stabilization Energy (CFSE) for an octahedral complex is calculated using the formula: \[ CFSE = -0.4 \Delta_o \times (number of electrons in lower energy orbitals) + 0.6 \Delta_o \times (number of electrons in higher energy orbitals) \]
where \( \Delta_o \) is the octahedral crystal field splitting energy.

Step 2: Applying the Formula to \( d^8 \) Configuration.

For a \( d^8 \) configuration in an octahedral field, the electron arrangement will have 2 electrons in the higher-energy \( e_g \)-orbitals and 6 electrons in the lower-energy \( t_{2g} \)-orbitals. The CFSE for such a configuration is: \[ CFSE = -0.4 \Delta_o \times 6 + 0.6 \Delta_o \times 2 = -2.4 \Delta_o + 1.2 \Delta_o = -0.4 \Delta_o \]

Step 3: Conclusion.

Thus, the CFSE for the \( d^8 \) octahedral ion is -0.4 \( \Delta_o \), corresponding to option (1). Quick Tip: For \( d^8 \) configuration in octahedral fields, the CFSE is calculated based on the number of electrons in lower and higher energy orbitals, giving a value of -0.4 \( \Delta_o \).

Step 1: Analyze the Coordination Complex.

The coordination compound is \( [Rh(CO)_2I_2]^-\). It contains:
- Two carbonyl (\(CO\)) ligands.
- Two iodido (\(I\)) ligands.
- The metal ion is rhodium (Rh) in the \(+1\) oxidation state.

Step 2: IUPAC Naming Conventions.

- The name of the metal is "rhodium" and the oxidation state is indicated in Roman numerals as \( Rh(I) \).
- The ligands are named in alphabetical order. The "carbonyl" ligand (CO) is named first, followed by "iodido" (I).
- The prefix "bis" is used when there are two identical ligands (in this case, iodido).

Step 3: Conclusion.

Therefore, the IUPAC name is "carbonylbisiodidorhodium(I)", corresponding to option (3). Quick Tip: When naming coordination compounds, the ligands are named alphabetically, and prefixes like "bis" are used when there are multiple identical ligands.

Step 1: Understanding Ligand Field Strength.

The strength of the ligand field influences the d-orbital splitting in transition metal complexes. Strong field ligands cause a large splitting of the d-orbitals, while weak field ligands cause a smaller splitting. The ligand field strength follows the trend: \[ CN^- > CO_4^{2-} > NH_3 > OH^- \]

Step 2: Analyzing the Ligands.

- (A) NH\(_3\) is a weak field ligand.
- (B) C\(_2\)O\(_4^{2-}\) (oxalate) is a moderate field ligand.
- (C) OH\(^-\) is a weaker field ligand compared to NH\(_3\).
- (D) CN\(^-\) is a strong field ligand, causing the largest splitting.

Step 3: Conclusion.

Thus, the order of increasing d-orbital splitting is: \[ OH^- < C_2O_4^{2-} < NH_3 < CN^- \]
So, the correct answer is option (4): (C), (B), (A), (D). Quick Tip: Strong field ligands like CN\(^-\) and CO cause large d-orbital splitting, while weak field ligands like OH\(^-\) cause small splitting.

Step 1: Coordination Numbers of Transition Elements.

Transition metals, especially in the second and third periods, commonly form complexes with high coordination numbers due to their relatively large size and ability to accommodate multiple ligands. Hence, they do not typically show low coordination numbers.

Step 2: Analyzing the Options.

- (1) The statement "The metals commonly show lower coordination number" is not true for the second and third row transition metals. They tend to have higher coordination numbers.
- (2) Binuclear carboxylate complexes are common for transition metals.
- (3) Many halides of transition elements are indeed cluster compounds, especially in higher oxidation states.
- (4) Transition metals can form carbonyl complexes with metal-metal bonds, such as in dicobalt octacarbonyl.

Step 3: Conclusion.

Thus, the incorrect statement is (1), so the correct answer is option (1). Quick Tip: Transition metals, especially in the second and third periods, typically show high coordination numbers.

Step 1: Electronic Configuration of Pr\(^{3+}\).

Praseodymium (Pr) has an atomic number of 59, and the electronic configuration is: \[ Pr: [Xe] 4f^3 6s^2 \]
For \( Pr^{3+} \), three electrons are removed from the 4f shell, leaving a configuration of \( 4f^3 \).

Step 2: Term Symbol Calculation.

For \( 4f^3 \), the total spin quantum number (\( S \)) is 3/2, and the total orbital quantum number (\( L \)) is 3 (which corresponds to the letter H). The multiplicity is given by \( 2S+1 \), which results in \( 2(3/2) + 1 = 4 \). Hence, the term symbol is \( ^3H_5 \).

Step 3: Conclusion.

Thus, the ground state term symbol for Pr\(^{3+}\) is \( ^3H_5 \), corresponding to option (3). Quick Tip: For rare earth ions like Pr\(^{3+}\), the term symbol is derived from the \( 4f^n \) configuration and the values of \( S \) and \( L \).

Step 1: Understanding Polyphosphazene Polymers.

Polyphosphazene polymers are a class of materials that have a wide range of biomedical applications due to their unique properties such as biodegradability, biocompatibility, and versatility in structural and functional modifications.

Step 2: Analyzing the Options.

- (A) Polyphosphazene polymers are used as structural materials in biomedical applications, particularly for heart valves and blood vessels due to their flexibility and strength.
- (B) Polyphosphazene polymers are also used in drug delivery systems, where their biodegradability allows controlled drug release.
- (C) These polymers can be used as biodegradable supports for in vivo bone regeneration, as they can be gradually resorbed by the body while providing mechanical support.

Step 3: Conclusion.

Thus, all of the listed applications (A), (B), and (C) are true for polyphosphazene polymers, so the correct answer is option (3). Quick Tip: Polyphosphazene polymers have a wide range of biomedical applications due to their biodegradability, making them useful for structural, drug delivery, and regenerative purposes.

Step 1: Understanding the Organometallic Compounds and their Characteristics.

- (A) Fe(C\(_5\)H\(_5\))\(_2\): This is a ferrocene compound, where iron is sandwiched between two cyclopentadienyl anions. The metal in this complex undergoes dsp\(^2\) hybridization.
- (B) K\(^+\)[PtCl\(_3\)(C\(_2\)H\(_4\))]: This complex has a tetramer structure, which means it is composed of four monomeric units.
- (C) Mg(CH\(_3\))\(_2\): This is an organomagnesium compound, and such compounds exhibit fluxionality, where the ligands may interchange positions.
- (D) (CH\(_3\))\(_2\)AlF\(_4\): This complex is polymeric in nature due to the bonding between the metal center and the ligands forming extended chains or networks.

Step 2: Conclusion.

Thus, the correct matching is:
- (A) - (I): Fe(C\(_5\)H\(_5\))\(_2\) undergoes dsp\(^2\) hybridization.
- (B) - (II): K\(^+\)[PtCl\(_3\)(C\(_2\)H\(_4\))] has a tetramer structure.
- (C) - (III): Mg(CH\(_3\))\(_2\) exhibits fluxionality.
- (D) - (IV): (CH\(_3\))\(_2\)AlF\(_4\) is polymeric in nature.

Therefore, the correct answer is option (2). Quick Tip: Organometallic compounds can have distinct properties such as fluxionality, hybridization types, and structural characteristics like polymeric and tetramer structures.