CONCEPT:

Reduction of Aldehydes to Alcohols

• Aldehydes, like benzaldehyde, can be reduced to primary alcohols (such as benzyl alcohol) using specific reducing agents.
• The choice of reducing agent depends on the reactivity and the type of reduction required.

EXPLANATION:

• NaBH₄ (Sodium borohydride) — NaBH₄ is a mild reducing agent commonly used to reduce aldehydes and ketones to their corresponding alcohols. It selectively reduces benzaldehyde to benzyl alcohol.

• LiAlH₄ (Lithium aluminum hydride) — LiAlH₄ is a much stronger reducing agent and can reduce a wider range of compounds, including aldehydes, but it is typically reserved for more reactive functional groups like esters or carboxylic acids.
• Zn/Hg with HCl (Clemmensen reduction) — This reagent is used for the reduction of ketones and aldehydes to alkanes, not alcohols.
• Cu/H₂SO₄ — This reagent is not commonly used for the reduction of aldehydes to alcohols and is more involved in oxidation or other types of reactions.

CONCLUSION:

• NaBH₄ is the reagent used to reduce benzaldehyde to benzyl alcohol.

CONCEPT:

Molecular Orbital Theory (MOT) and Unpaired Electrons in O₂ and O₂⁻

• The molecular orbital theory explains the bonding and electronic structure of diatomic molecules, including the oxygen molecule (O₂).
• Oxygen (O₂) has a total of 16 electrons, and when arranged in molecular orbitals, it has two unpaired electrons in the π* anti-bonding orbitals. This makes O₂ paramagnetic.
• When O₂ gains an extra electron to form O₂⁻ (superoxide ion), this extra electron fills one of the previously unpaired π* orbitals, leaving only one unpaired electron in O₂⁻.

EXPLANATION:

• O₂ (Molecular oxygen) — Oxygen has 2 unpaired electrons in the π* anti-bonding orbitals according to molecular orbital theory, making it paramagnetic.
• O₂⁻ (Superoxide ion) — The addition of one electron to O₂ results in one unpaired electron, as the added electron pairs with one of the previously unpaired electrons.

CONCLUSION:

• O₂ has 2 unpaired electrons, while O₂⁻ has 1 unpaired electron is the correct statement.

Concept:

The shape of a molecule is determined by the number of lone pairs and bond pairs around the central atom, which is predicted using Valence Shell Electron Pair Repulsion (VSEPR) theory.

Here is a table showing the relationship between lone pairs, bond pairs, and the resulting shape:

Explanation: 

Based on the VSEPR theory, the shapes of the molecules can be explained as follows:

• CO₂: Carbon dioxide has two bonding pairs and no lone pairs, giving it a linear shape (180° bond angle).

• SF₆: Sulfur hexafluoride has six bonds around sulfur and adopts an octahedral structure (90° bond angle).

• PF₃: Phosphorus trifluoride has one lone pair and three bonding pairs, resulting in a trigonal pyramidal shape.

• CH₄: Methane has four bonding pairs and no lone pairs, leading to a tetrahedral shape (109.5° bond angle).

Conclusion:

The correct matches for the species and their shapes are as follows:

A-I, B-IV, C-II, D-I

Concept:

Ionization enthalpy is the energy required to remove the most loosely bound electron from a gaseous atom in its ground state. The general trend of ionization enthalpy along the period and group is influenced by several factors:

• Across a Period: Ionization enthalpy increases from left to right across a period due to increasing nuclear charge, which pulls the electrons closer to the nucleus, making them harder to remove.

• Down a Group: Ionization enthalpy decreases as we move down a group due to an increase in atomic size and shielding effect, making it easier to remove electrons from the outermost shell.

• Shielding Effect: The inner electrons shield the outer electrons from the full attraction of the nucleus, making the ionization energy lower.

• Atomic Radius: As the atomic radius increases, the distance between the nucleus and the outermost electron increases, lowering the ionization enthalpy.

• Electronic Configuration: Elements with stable electronic configurations (such as noble gases) have much higher ionization enthalpies compared to elements with an unstable configuration.

Explanation: 

• Sodium (Na) has the lowest ionization enthalpy because it is an alkali metal with a single electron in its outermost shell, making it easy to remove.

• Magnesium (Mg) has a higher ionization enthalpy than Na due to a smaller atomic radius and a higher nuclear charge, making electron removal more difficult.

• Aluminum (Al) follows Mg, but its ionization energy is slightly lower than expected because of its electron configuration (3p electron is easier to remove than a 3s electron).

• Silicon (Si), Phosphorus (P), and Sulfur (S) have progressively higher ionization enthalpies as their nuclear charge increases across the period.

• Chlorine (Cl) has the highest ionization enthalpy because it is a halogen with a high effective nuclear charge and a small atomic radius.

Conclusion:

The correct order of first ionization enthalpy for Na, Mg, Al, Si, P, S, and Cl is: Na < Mg < Al < Si < P < S < Cl

Concept:

Boiling point is the temperature at which the vapor pressure of a liquid equals the external pressure. As external pressure increases, the liquid requires a higher temperature to reach that vapor pressure, which leads to an increase in boiling point. This is why, under higher pressures (like in a pressure cooker), water boils at a temperature above 100°C. Conversely, under lower pressures (like at high altitudes), water boils at a lower temperature.

• Boiling Point and Pressure: The boiling point increases with an increase in external pressure because the liquid needs more energy (in the form of heat) to match the higher external pressure.

• Vapor Pressure: Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. Boiling occurs when vapor pressure equals the surrounding atmospheric pressure.

• External Pressure: As external pressure rises, more heat is needed for the vapor pressure to equal the external pressure, thereby raising the boiling point.

Explanation: 

• The Assertion (A) is correct: The boiling point of water does indeed increase as the external pressure increases.

• The Reason (R) is also correct: The boiling point is the temperature at which the vapor pressure of a liquid becomes equal to the external pressure. This statement correctly describes what happens at the boiling point.

• Since Reason (R) explains why the boiling point increases with external pressure, it serves as the correct explanation of  Assertion (A).

Conclusion:

The correct answer is: Both A and R are correct, and R is the correct explanation of A.

Concept:

The reaction between aluminum chloride (AlCl₃) and sodium hydroxide (NaOH) can proceed in two steps. Initially, aluminum hydroxide (Al(OH)₃) is formed as a precipitate. In excess NaOH, the precipitate dissolves to form sodium aluminate.

• Initial Reaction: In the first step, aluminum chloride reacts with sodium hydroxide to form aluminum hydroxide and sodium chloride.

• Excess NaOH: In the presence of excess NaOH, aluminum hydroxide dissolves to form sodium aluminate NaAl(OH)₄.

Explanation: 

• In the reaction, aluminum chloride first reacts with sodium hydroxide as follows:

  AlCl₃ + 3 NaOH → Al(OH)₃ + 3 NaCl

  In the presence of excess NaOH, aluminum hydroxide dissolves:

  Al(OH)₃ + NaOH → NaAl(OH)₄

• Therefore, the products of the reaction are sodium aluminate (NaAl(OH)₄), sodium chloride (NaCl), and water (H₂O).

Conclusion:

The correct answer is: X: NaAl(OH)₄, Y: NaCl, and Z: H₂O.

Concept:

Ozonolysis is a reaction in which alkenes are cleaved at the double bond by ozone (O₃) to form carbonyl compounds such as aldehydes or carboxylic acids. In the presence of water, aldehydes may be further oxidized to carboxylic acids.

• Ozonolysis of Alkenes: The double bond of the alkene is cleaved, and the two fragments are converted into carbonyl-containing products. The exact products depend on the structure of the alkene.

• Reaction Mechanism: The first step involves the formation of an ozonide intermediate, which decomposes in the presence of water to give aldehydes or carboxylic acids.

Explanation: 

• In the case of 2-pentene, the double bond is between carbon 2 and carbon 3. Ozonolysis will cleave this bond, producing two fragments. One fragment will be propanal (CH₃CH₂CHO) and the other will be ethanal (CH₃CHO).

• Since this reaction does not involve further oxidation, the products are aldehydes rather than carboxylic acids.

Conclusion:

The correct answer is: Propanal and Ethanal.

Concept:

The Kolbe-Schmitt reaction is a carboxylation reaction in which sodium phenoxide reacts with carbon dioxide to form sodium salicylate, which is then converted to salicylic acid upon acidification. This reaction is used in the industrial synthesis of salicylic acid, a precursor to aspirin.

• Electrophile: In this reaction, the electrophile is carbon dioxide (CO₂), which reacts with sodium phenoxide to form the salicylate ion.

• Reaction Conditions: The reaction occurs at high temperature and pressure to drive the carboxylation step.

• Product: The product formed is sodium salicylate, which is converted to salicylic acid upon treatment with an acid (H⁺).

• Reaction Mechanism:

Explanation: 

• Statement 1 is correct: CO₂ acts as the electrophile in the reaction, reacting with sodium phenoxide.

• Statement 2 is correct: The product formed is sodium salicylate.

• Statement 3 is correct: The reaction occurs under high pressure to facilitate the carboxylation of phenoxide.

• Statement 4 is incorrect: Sodium salicylate is converted to salicylic acid only after acidification, not directly in the reaction.

Conclusion:

The correct answer is: The product 'B' formed in the above reaction is salicylic acid upon acidification, not directly after the reaction.

Concept:

The reaction begins with butanoic acid (C₃H₇COOH) as the reactant. The transformation involves the following steps:

• Step 1: Butanoic acid reacts with phosphorus tribromide (PBr₃), which converts the carboxylic acid (-COOH) group into a corresponding acyl bromide (C₃H₇COBr).

• Step 2: The acyl bromide then reacts with a primary amine (R-NH₂) to form an amide (C₃H₇CONH-R) through nucleophilic substitution.

• Step 3: The amide undergoes reduction using lithium aluminum hydride (LiAlH₄), which reduces the amide to a primary amine (C₃H₇CH₂NH-R).

• Step 4: Finally, the reaction is treated with acid (H₃O⁺) for workup, yielding the final product, which is an alkyl amine (N-alkylbutylamine).

Explanation: 

• In the first step, PBr₃ replaces the hydroxyl group of butanoic acid with a bromine atom, forming butanoyl bromide.

• In the second step, an amine (CH₃-NH₂) acts as a nucleophile, attacking the carbonyl carbon of butanoyl bromide and forming an amide.

Concept:

The compound  [PtCl₄​(NH₃​)₂​] is a coordination complex where platinum (Pt) is the central metal atom. The oxidation state of platinum can be calculated by considering the charges on the ligands. The chloride ions (Cl⁻) are monodentate ligands, while the ammonia (NH₃) is a neutral ligand. Square planar geometry is commonly associated with platinum complexes, especially when Pt is in the +4 oxidation state.

• Oxidation State: The oxidation state of Pt can be calculated by knowing that each chloride ion contributes a -1 charge, while ammonia is neutral. In [PtCl₄​(NH₃​)₂​], Pt must have an oxidation state of +4 to balance the 4 negative charges from Cl⁻ ligands.

• Geometry: Platinum in the +4 oxidation state generally adopts a square planar geometry, especially in complexes with coordination number 6, like [PtCl₄​(NH₃​)₂].

Explanation:

• The compound  [PtCl₄​(NH₃​)₂​] has four chloride ions, each contributing a -1 charge, and two neutral ammonia ligands. Hence, the oxidation state of Pt is +4 to balance the negative charge from the Cl⁻ ions. Due to this, the complex adopts a square planar geometry, which is characteristic of platinum complexes with this oxidation state.

• Option (2) is incorrect because the ammonia ligands do not lead to octahedral geometry in this case.

• Option (3) is also incorrect, as chloride ions do not act as bridging ligands here.

• Option (4) is incorrect because the oxidation state of Pt is not +2.

Conclusion:

The correct answer is: (1) The oxidation state of Pt in Pt in [PtCl₄​(NH₃​)₂​]  [PtCl₄(NH₃)₂] is +4, and it forms a square planar complex.