Step 1: Oxidation of the side chains
Alkaline KMnO₄ oxidizes alkyl groups attached to aromatic rings to carboxylic acid groups.
Step 2: Formation of the product
The given reaction produces naphthalene-1,2,4-tricarboxylic acid.

Step 1: Avoid oxidation of Fe²⁺
Prolonged heating can oxidize Fe²⁺ to Fe³⁺, introducing impurities.
Step 2: Reaction equation
4Fe²⁺ + O₂ + 4H⁺ → 4Fe³⁺ + 2H₂O.

Step 1: Evaluate each reaction
(A) SN2: Less substituted carbons are more reactive.
(B) SN1: Stability of the carbocation determines reactivity.
(C) Electrophilic substitution: Electron-donating groups enhance reactivity.
(D) Nucleophilic substitution: Electron-withdrawing groups enhance reactivity.

Step 1: Use Hess's law
Reverse and divide equation (1) by 2, add to equation (2), and solve for ∆H°.
Step 2: Solve for ∆H°
∆H° = (x/2) - y = (x − 2y)/2.

Step 1: Retention factor
B's lower R_f indicates stronger adsorption to the stationary phase.

Step 1: Reaction of lime with water
CaO(s) + H₂O(l) → Ca(OH)₂(aq). 'A' is calcium hydroxide, commonly known as slaked lime.
Step 2: Reaction with CO₂
Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l). 'B' is calcium carbonate (CaCO₃).
Step 3: Further reaction with CO₂
CaCO₃(s) + H₂O(l) + CO₂(g) → Ca(HCO₃)₂(aq), which is soluble.

Step 1: Analyze waste products
(A) Steel plants produce slag, a byproduct of iron and steel making.
(B) Thermal power plants generate fly ash from burning coal.
(C) Fertilizer industries produce gypsum as a byproduct of phosphoric acid production.
(D) Paper mills produce bio-degradable wastes from organic materials.

Step 1: Use Williamson ether synthesis
This reaction involves the reaction of an alkoxide ion (PhO^-) with a primary alkyl halide (MeBr).
Step 2: Reaction mechanism
PhO^-Na⁺ + MeBr → Ph-O-Me + NaBr.
The phenoxide ion acts as a nucleophile, attacking methyl bromide via an SN2 mechanism to form methyl phenyl ether (anisole).

Step 1: Stabilizing forces in protein structures
The secondary and tertiary structures of proteins are stabilized by hydrogen bonding, disulfide bridges (–S–S–), van der Waals forces, and electrostatic attractions. Peroxide linkages (–O–O–) do not play a role in protein stabilization.

Step 1: Analyze compound stability
Sodium superoxide (NaO₂) is highly reactive and tends to disproportionate, making it unstable. Other listed compounds are known and stable.

Step 1: Analyze the two reactions
The first reaction, the water-gas reaction, requires a higher temperature (around 1270 K).
The second reaction, the water-gas shift reaction, occurs at lower temperatures (around 673 K) with a catalyst.

Step 1: Analyze the processes
Adsorption is exothermic (∆H_ads < 0) as energy is released when molecules adhere to a surface.
Micelle formation is endothermic (∆H_mic > 0) due to energy required to overcome hydrophobic repulsion between surfactant molecules.

Step 1: Evaluate each pair
Both CH₂Cl₂ and CHCl₃ have non-zero dipole moments due to differences in electronegativity and asymmetric molecular structures.
Other pairs have at least one compound with zero dipole moment due to symmetry.

Step 1: Role of stabilizers
Cresols are used in froth flotation to stabilize the froth during the concentration of sulphide ores.

Step 1: Evaluate the statements
(A) Incorrect: Reduction potential for Mn is greater than Fe.
(B) Correct: Higher oxidation states are stabilized by oxide ions.
(C) Correct: Cr²⁺ reduces H⁺ to H₂.
(D) Incorrect: Magnetic moment of V²⁺ is approximately 3.87 BM, not 4.4-5.2 BM.

Step 1: Analyze the properties
K₂Cr₂O₇ is more stable in aqueous solution compared to Na₂Cr₂O₇, making it a preferred primary standard.
K₂Cr₂O₇ is less soluble in water than Na₂Cr₂O₇.

Step 1: Analyze the coordination complex
[Co(NH₃)₆]³⁺ has Co³⁺ with an electronic configuration of 3d⁶. NH₃ is a strong field ligand, causing electron pairing in the lower energy t_2g orbitals.
The configuration becomes t⁶_2g, e⁰_g, making it diamagnetic and low spin.

Step 1: Analyze the reactions
(P): Gabriel synthesis produces primary amines.
(Q): Hinsberg’s reagent forms a solid with secondary amines.
(R): The formation of a red dye indicates an aromatic primary amine.

Step 1: Calculate moles
At STP, 1 mole of gas occupies 22.7 L. Moles = Volume/22.7 = 2.8375/22.7 = 0.125 moles.
Step 2: Calculate molecules
Molecules = Moles × Avogadro’s number = 0.125 × 6.022 × 10²³ = 7.527 × 10²².

Step 1: Match polymers with classifications
(A): Nylon-2-Nylon-6 is biodegradable.
(B): Buna-N is synthetic rubber.
(C): Urea-formaldehyde resin is thermosetting.
(D): Dacron is a polyester.

Step 1: Determine the Van’t Hoff Factor
For a weak monobasic acid, the van’t Hoff factor (i) is calculated using the formula:
i = 1 + (n - 1) * α, where n = 2 (for complete dissociation) and α = 0.3.
i = 1 + (2 - 1) * 0.3 = 1.3.
Step 2: Calculate the Percentage Increase
The percentage increase in freezing point depression is given by:
% increase = (^{i - 1}⁄₁) * 100 = (^{1.3 - 1}⁄₁) * 100 = 30%.

Step 1: Determine Products of the Reactions
Reaction 1: Na₂O + H₂O → 2NaOH. Here, X is NaOH, which contains 1 oxygen atom per molecule.
Reaction 2: Cl₂O₇ + H₂O → 2HClO₄. Here, Y is HClO₄, which contains 4 oxygen atoms per molecule.
Step 2: Add the Oxygen Atoms
Each molecule of NaOH contributes 1 oxygen atom, and HClO₄ contributes 4 oxygen atoms.
Total oxygen atoms = 1 (from X) + 4 (from Y) = 5.

Step 1: Calculate Lone Pairs for IF₅
Iodine (I) in IF₅ has 7 valence electrons. Five are used for bonding, leaving 2 electrons (1 lone pair).
Step 2: Calculate Lone Pairs for IF₇
Iodine (I) in IF₇ uses all 7 valence electrons for bonding, leaving no lone pairs.
Step 3: Add the Lone Pairs
Lone pairs on IF₅: 1. Lone pairs on IF₇: 0.
Total = 1 + 0 = 1.

Step 1: Analyze the Molecular Shapes
- N₃⁻: Linear.
- NO₂⁻: Bent (lone pairs on nitrogen).
- I₃⁻: Linear.
- O₃: Bent (lone pairs on central oxygen).
- SO₂: Bent (lone pairs on sulfur).
Step 2: Count the Bent Molecules
Bent molecules: NO₂⁻, O₃, and SO₂.
Total = 3.

Step 1: Evaluate Each Statement
- (1): Correct. Equilibrium requires a closed system to prevent loss of matter.
- (2): Correct. At equilibrium, the forward and reverse processes occur at the same rate.
- (3): Incorrect. The equilibrium values of parameters remain constant but may not be equal.
- (4): Correct. Solubility depends on temperature.
Step 2: Count the Correct Statements
Correct statements: (1), (2), and (4).
Total = 3.

Step 1: Write the Given Data
Initial pressure, P_initial = 940.3 mm Hg.
Initial volume, V_initial = 100 units (assume).
Final volume, V_final = 100 - 40% = 60 units.
Final pressure, P_final = ?
Step 2: Apply Boyle’s Law
Boyle’s Law states: P_initialV_initial = P_finalV_final
Substitute the values:
940.3 × 100 = P_final × 60.
Step 3: Solve for P_final
P_final = (940.3 × 100) / 60 = 1567 mm Hg.
Round to the nearest integer:
P_final = 1567 mm Hg.

Step 1: Combine the Half-Cell Reactions
The reduction steps are:
FeO₄²⁻ + 3e⁻ → Fe³⁺, ΔG₁
Fe³⁺ + e⁻ → Fe²⁺, ΔG₂
FeO₄²⁻ + 4e⁻ → Fe²⁺, ΔG₃
Step 2: Use Gibbs Free Energy Relationships
ΔG₃ = ΔG₁ + ΔG₂.
Substituting ΔG = −nFE°:
−4FE°₃ = −3F(2.2) + (−1F)(0.7).
Step 3: Simplify
4E°₃ = 6.6 + 0.7 = 7.3.
E°₃ = 7.3 / 4 = 1.825 V.
E°₃ = 1.825 × 10⁻³ V.

Step 1: Relationship Between Rate Constants and Effective Half-Life
For independent first-order reactions:
k_eff = k₁ + k₂.
The effective half-life is given by:
ln 2 / t_eff = ln 2 / t₁ + ln 2 / t₂.
Simplify:
1 / t_eff = 1 / t₁ + 1 / t₂.
Step 2: Substitute the Given Half-Lives
t₁ = 12 min, t₂ = 3 min.
1 / t_eff = 1 / 12 + 1 / 3 = 5 / 12.
Step 3: Calculate t_eff
t_eff = 12 / 5 = 2.4 min.
Round to the nearest integer:
t_eff = 2 min.

Step 1: Analyze the Given Statements
(1) Correct: Black bodies emit/absorb electromagnetic radiation.
(2) Correct: Frequency distribution depends on temperature (Planck’s law).
(3) Correct: Intensity vs frequency curve has a maximum value.
(4) Correct: The peak shifts to higher frequency with temperature (Wien’s law).
Step 2: Count the Incorrect Statements
All the statements are correct.
Number of incorrect statements = 0.

Step 1: Analyze the Electronic Configuration of Fe²⁺
Potassium ferrocyanide is K₄[Fe(CN)₆].
In this complex, Fe is in the +2 oxidation state: Fe²⁺.
Electronic configuration: Fe²⁺ = [Ar] 3d⁶.
Step 2: Effect of Strong Field Ligand (CN⁻)
CN⁻ is a strong field ligand, causing pairing of electrons in the d-orbitals.
The t_2g set of orbitals (lower energy) gets fully filled with 6 electrons, forming 3 pairs.
Step 3: Diagrammatic Representation
t_2g⁶ e_g⁰: The t_2g set contains 6 electrons, forming 3 pairs.