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CONCEPT:

Free Expansion of Gas and Thermodynamic Changes

• When a gas expands into a vacuum (free expansion), there is no external pressure opposing the expansion, i.e., P_ext = 0.
• In free expansion:
• No work is done by the gas (dw = 0) because the external pressure is zero.
• No heat is exchanged with the surroundings (dq = 0) if the system is insulated or thermally isolated.
• For an ideal gas undergoing free expansion, the internal energy (ΔU) remains unchanged since internal energy depends only on temperature and temperature is constant.

EXPLANATION:

• In the problem, two vessels A and B are connected with a stopcock.
• Initially, vessel A contains gas at certain pressure, and vessel B is evacuated (pressure in B is zero).
• Upon opening the stopcock, gas expands freely into vessel B (vacuum), so P_ext = 0.
• Since the assembly is immersed in water and allowed to equilibrate, temperature remains constant.
• Therefore, no work is done (dw = 0), no heat is exchanged (dq = 0), and internal energy change is zero (ΔU = 0).

Hence, the pressure in vessel B before opening the stopcock must be zero.

The correct answer is The pressure in the vessel B before opening the stopcock is zero.

CONCEPT:

Raoult's Law and Vapor Pressure of Solutions

• Raoult's law states that the total vapor pressure of an ideal solution is the sum of the partial vapor pressures of each component.
• The partial vapor pressure of each component is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
• Mathematically, P_total = P_Δ⁰ × X_Δ + P_B⁰ × X_B
• The component with the higher vapor pressure is the more volatile component.
• The least volatile component has the lower vapor pressure.

EXPLANATION:

• Given:
• Moles of volatile liquid Δ = 1
• Moles of volatile liquid B = 3
• Vapor pressure of pure Δ (P_Δ⁰) = 200 mm Hg
• Total vapor pressure of solution (P_solution) = 500 mm Hg
• Mole fractions:
• X_Δ = 1 / (1 + 3) = 1/4 = 0.25
• X_B = 3 / (1 + 3) = 3/4 = 0.75
• Therefore, vapor pressure of pure B is 600 mm Hg.
• Comparing vapor pressures:
• P_Δ⁰ = 200 mm Hg
• P_B⁰ = 600 mm Hg

B has the higher vapor pressure, so B is the more volatile component.

Δ, having the lower vapor pressure, is the least volatile component.

Therefore, the vapor pressure of pure B is 600 mm Hg, and the least volatile component of the solution is A.

CONCEPT:

Stoichiometry and Limiting Reactant in Chemical Reactions

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

• The reaction between calcium carbonate and hydrochloric acid is a typical acid-carbonate reaction:
• Mole ratio between CaCO₃ and HCl is 1:2.
• The limiting reactant determines the amount of product formed.
• Molar masses are used to convert between moles and grams.

EXPLANATION:

• Given:
• Mass of CaCO₃ = 1000 g
• Concentration of HCl = 0.76 M (moles per liter)
• Volume of HCl = 250 mL = 0.250 L
• Molar masses (g/mol): Ca = 40, C = 12, O = 16, H = 1, Cl = 35.5
• Calculate moles of CaCO₃:

Molar mass of CaCO₃ = 40 + 12 + (16 × 3) = 100 g/mol

Moles of CaCO₃ = 1000 g / 100 g/mol = 10 mol

• Calculate moles of HCl:

Moles of HCl = Molarity × Volume = 0.76 mol/L × 0.250 L = 0.19 mol

• Determine limiting reactant:
• Reaction requires 2 moles of HCl per mole of CaCO₃.
• Moles of HCl required for 10 moles CaCO₃ = 10 × 2 = 20 moles.
• Available moles of HCl = 0.19 moles < 20 moles required, so HCl is limiting reactant.
• Calculate moles of CaCl₂ formed (equal to moles of CaCO₃ reacted):

Moles of CaCl₂ formed = moles of HCl / 2 = 0.19 / 2 = 0.095 mol

• Calculate molar mass of CaCl₂:

Ca = 40, Cl = 35.5 × 2 = 71, Total = 40 + 71 = 111 g/mol

• Calculate mass of CaCl₂ produced:

Mass = moles × molar mass = 0.095 × 111 = 10.545 g

Therefore, the mass of CaCl₂ formed is 10.545 g.

When equal volumes are mixed molarity reduce to half.

CONCEPT:

• When two salt solutions AB₂ and XY are mixed, precipitation of AY₂ will occur if the ionic product (Q_sp) exceeds the solubility product constant (K_sp).
• Q_sp is calculated as:

Q_sp = [A⁺][Y⁻]²

• Since equal volumes of solutions are mixed, the concentration of ions is halved due to dilution.

EXPLANATION:

$\text{For precipitation } Q_{SP} = [A^{+2}] [Y^-]^2 > K_{SP} \ (1) , Q_{SP} = (1.8 \times 10^{-3}) \left( \frac{5}{2} \times 10^{-4} \right)^2 < K_{SP} \ (2) , Q_{SP} = (10^{-4}) (0.4 \times 10^{-3})^2 < K_{SP} \ (3) , Q_{SP} = (10^{-2}) (10^{-2})^2 > K_{SP} \ (4) , Q_{SP} = \left( \frac{1.5}{2} \times 10^{-4} \right) \left( \frac{1.5}{2} \times 10^{-3} \right)^2 < K_{SP}$

Conclusion:

Only option 3 gives Q_sp > K_sp, so precipitation of AY₂ occurs.

CONCEPT:

Hybridization, Molecular Geometry, and Dipole Moment

• Hybridization describes the mixing of atomic orbitals on the central atom to form new hybrid orbitals in a molecule.
• The presence of lone pairs affects molecular geometry and the resultant dipole moment.
• Dipole moment depends on the geometry and the difference in electronegativities; molecules with symmetrical geometry or lone pairs arranged symmetrically often have zero dipole moment.
• Hybridizations considered here are sp², sp³, sp³d, and sp³d² corresponding to trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral geometries, respectively.
• Lone pairs on the central atom create asymmetry, often leading to non-zero dipole moments.

EXPLANATION:

• Among molecules with non-zero dipole moment, ClF₃ has the highest number of lone pairs (2) on the central atom.
• Its hybridization is sp³d.

Therefore, the correct answer is: sp³d (Option 4).

CONCEPT:

• Vanillin is an aromatic aldehyde with a phenolic -OH group and a methoxy (-OCH3) group attached to the benzene ring.
• The phenolic group (-OH attached to benzene) is acidic and soluble in NaOH due to phenol's acidic nature.
• Tollen's reagent detects aldehyde groups (-CHO) by oxidizing them to carboxylic acids, giving a silver mirror. Vanillin contains an aldehyde group, so it reacts with Tollen's reagent.
• Self-aldol condensation requires the presence of acidic alpha-hydrogen atoms adjacent to the carbonyl group to form enolate ions under basic conditions.
• Vanillin lacks alpha-hydrogen atoms adjacent to the aldehyde group because the carbonyl carbon is directly bonded to the aromatic ring, thus it does not undergo self-aldol condensation easily.

EXPLANATION:

​

Phenolic group soluble in NaOH   

Benzaldehyde derivative react with Tollen’s reagent.

Vanillin does not give self-aldol reaction due to lack of acidic H for condensation. 

• Statement I: Vanillin will react with NaOH due to the acidic phenolic -OH group which forms phenolate ions soluble in NaOH.
• It also reacts with Tollen's reagent because of the aldehyde (-CHO) group, which is oxidized to carboxylate ions, producing the silver mirror effect.
• Statement II: Vanillin will not undergo self-aldol condensation easily because it lacks acidic alpha-hydrogens required to form the enolate ion necessary for aldol condensation.

Therefore, Statement I is correct but Statement II is incorrect.

CONCEPT:

• Most naturally occurring amino acids contain one chiral center, except glycine, which has two hydrogen atoms attached to the alpha carbon, making it achiral.
• Some amino acids, like isoleucine and threonine, contain two chiral centers.
• Optical activity arises from chirality, so glycine is optically inactive, whereas other amino acids with chiral centers are optically active.
• Glutamic acid and aspartic acid both contain a –COOH group in their side chains (carboxylic acid groups), so glutamic acid is not unique in this feature.
• Cysteine contains a thiol (-SH) group in its side chain, which can form disulfide bonds by dimerization or oxidation, contributing to protein structure stabilization.

EXPLANATION:

• Statement 1: Incorrect. While most amino acids have one chiral center, isoleucine has two chiral centers, and glycine has none.
• Statement 2: Incorrect. Glycine is achiral and thus optically inactive.
• Statement 3: Incorrect. Both glutamic acid and aspartic acid have –COOH groups on their side chains.
• Statement 4: Correct. Cysteine contains a free thiol (-SH) group that readily undergoes dimerization to form disulfide bonds.

Therefore, the correct statement is: Amino acid cysteine easily undergoes dimerization due to the presence of free SH group.

CONCEPT:

• The basicity of amines in aqueous solution depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton (H⁺).
• Alkyl groups like ethyl (CH₃CH₂–) are electron-donating via +I (inductive) effect, which increases electron density on nitrogen and hence increases basicity.
• However, steric hindrance and solvation effects influence basicity. Primary and secondary amines are generally more basic than tertiary amines in aqueous solution due to better solvation of the protonated form.
• Hydrazine (H₂N–NH₂) shows slightly different basicity due to electron interaction between the two nitrogen atoms.

EXPLANATION:

• NH₃ (Ammonia): Least basic among the given due to no alkyl groups.
• H₂N–NH₂ (Hydrazine): Slightly more basic than NH₃ because the lone pair on one nitrogen can be delocalized to some extent.
• CH₃CH₂NH₂ (Ethylamine, primary amine): More basic due to +I effect of ethyl group increasing electron density.
• (CH₃CH₂)₂NH (Diethylamine, secondary amine): Generally most basic as two alkyl groups donate electrons, increasing basicity.
• (CH₃CH₂)₃N (Triethylamine, tertiary amine): Less basic than secondary amine because of steric hindrance and poor solvation.

Therefore, the correct order of basicity is:

NH2–NH2,<NH3<CH3CH2NH2<(CH3CH2)3N<(CH3CH2)2NH

ONCEPT:

• Metallic radius refers to the size of an atom in a metallic state, influenced by electron shielding and effective nuclear charge.
• Gallium (Ga) has filled 3d¹⁰ subshell electrons that poorly shield the nucleus, leading to a stronger effective nuclear charge on the outer electrons, causing contraction in atomic size.
• Due to this d-electron poor shielding, Ga has a smaller metallic radius than Al, despite being below it in the periodic table.
• Ionic radius depends on the number of electrons and nuclear charge after ionization. Both Al³⁺ and Ga³⁺ lose three electrons.
• Since Ga has more protons and poor shielding, its ionic radius is larger than that of Al³⁺.

EXPLANATION:

\implies \text{The metallic radius order of Al & Ga is} \ \mathrm{B} < \mathrm{Ga} < \mathrm{Al} < \mathrm{In} < \mathrm{Tl} \ \underbrace{\mathrm{Ga} < \mathrm{Al}}_{\text{(due to poor shielding of d-subshell electrons)}} \ \implies \text{The ionic radius order of } \mathrm{Al}^{+3} & \mathrm{Ga}^{+3} \text{ is} \ \mathrm{B}^{+3} < \mathrm{Al}^{+3} < \mathrm{Ga}^{+3} < \mathrm{In}^{+3} < \mathrm{Tl}^{+3}

• Statement I: Metallic radius of Al is less than that of Ga — This is incorrect. The order is Ga < Al due to d-electron poor shielding in Ga causing contraction.
• Statement II: Ionic radius of Al³⁺ is less than Ga³⁺ — This is correct. Ga³⁺ has larger ionic radius because the poor shielding effect expands the size compared to Al³⁺.

Therefore, the correct answer is: Statement I is incorrect but Statement II is correct.

CONCEPT:

• In coordination chemistry, Δ_o (octahedral crystal field splitting energy) is the energy difference between the higher and lower sets of d-orbitals in an octahedral complex.
• P refers to the pairing energy, which is the energy required to pair two electrons in the same orbital.
• For octahedral complexes (coordination number 6):
• If Δ_o < P, electrons prefer to occupy higher energy orbitals unpaired → high spin complexes.
• If Δ_o > P, electrons pair in lower energy orbitals → low spin complexes.
• In tetrahedral complexes (coordination number 4):
• The crystal field splitting energy Δ_t is smaller than Δ_o and always less than the pairing energy (Δ_t < P).
• Therefore, tetrahedral complexes are almost always high spin because electrons occupy orbitals singly to minimize pairing.
• Low spin tetrahedral complexes are rare.

Explanation:-

• Statement I: Correct. It accurately describes the relationship between Δ_o and P in octahedral complexes and the formation of high spin or low spin complexes.
• Statement II: Correct. It correctly states that because Δ_o < P in tetrahedral complexes, low spin complexes are rarely formed.

Therefore, both Statement I and Statement II are correct.