Solution:

1. Analysis of Statement I:
The structure of Compound A has chlorine (Cl) at position 1 and nitro (NO₂) groups at positions 2 and 4. The correct IUPAC name is 1-chloro-2,4-dinitrobenzene. Therefore, Statement I is incorrect.

2. Analysis of Statement II:
The structure of Compound B has an ethyl group (-C₂H₅) at position 4 and a methyl group (-CH₃) at position 2. The amino group (-NH₂) is given priority. The correct IUPAC name is 4-ethyl-2-methylaniline. Therefore, Statement II is correct.

3. Conclusion:
Statement I is incorrect, and Statement II is correct.

Solution:

1. Understanding S_N2 Reaction Rates:
The rate of S_N2 reactions is influenced by steric hindrance. The reactivity order for alkyl halides is:
Methyl halide > 1° alkyl halide > 2° alkyl halide > 3° alkyl halide.

2. Analysis of the Compounds:
Compound (1) is a tertiary alkyl halide with the highest steric hindrance, resulting in the slowest S_N2 reaction.
Compound (2) is a secondary alkyl halide.
Compound (3) is a primary alkyl halide, undergoing the fastest S_N2 reaction due to minimal steric hindrance.
Compound (4) is also a secondary alkyl halide, slower than primary alkyl halides.

3. Conclusion:
The fastest S_N2 reaction occurs for Compound (3), a primary alkyl halide.

Solution:

1. Combustion Reaction:
The balanced equation for the combustion of glucose is:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O.

2. Moles of Glucose:
Given mass of glucose = 900 g, molar mass of glucose = 180 g mol^-1:
Moles of glucose = Mass / Molar mass = 900 / 180 = 5 mol.

3. Moles of Oxygen Required:
From the balanced reaction:
1 mol of glucose reacts with 6 mol of oxygen.
Therefore, 5 mol of glucose reacts with 5 × 6 = 30 mol of oxygen.

4. Mass of Oxygen:
Molar mass of oxygen (O₂) = 32 g mol^-1:
Mass of oxygen = Moles × Molar mass = 30 × 32 = 960 g.

5. Conclusion:
The amount of oxygen required is 960 g.

Solution:

Step 1: Formation of Compound B (Acetylation):
The reaction of CH₃COCl with alcohol (CH₃OH) in the presence of pyridine results in the acetylation of the hydroxyl group:
CH₃OH + CH₃COCl → CH₃OCOCH₃ (Compound B).

Step 2: Formation of Compound A (Dehydration):
Treatment of Compound B with concentrated H₂SO₄ at elevated temperature leads to an E₁ elimination reaction, resulting in the formation of an alkene:
CH₃OCOCH₃ → CH₂=CH-CH₃ (Compound A).

Conclusion:
Compound A is CH₂=CH-CH₃, and Compound B is CH₃OCOCH₃.

Solution:

1. Understanding Assertion (A):
In group 13 elements (Ga, In, Tl), the stability of the +1 oxidation state increases down the group due to the inert pair effect. The order is Ga < In < Tl, so Assertion (A) is true.

2. Understanding Reason (R):
The inert pair effect refers to the reluctance of the s-electrons to participate in bonding as we move down the group in the periodic table. This effect stabilizes the lower oxidation state (e.g., +1 in group 13 elements). Hence, Reason (R) is true.

3. Link between A and R:
The inert pair effect directly explains why the +1 oxidation state becomes more stable down the group. Therefore, R is the correct explanation of A.

Solution:

1. Matching the Tests with Their Reaction Sequences:

• Borax bead test: Reaction involves formation of MBO₂ complexes. Matches with III.
• Charcoal cavity test: Reaction sequence involves MCO₃ → MCl₂ → M²⁺. Matches with IV.
• Cobalt nitrate test: Reaction involves MCO₃ → MO + Co(NO₃)₂. Matches with I.
• Flame test: Reaction involves MSO₄ + Na₂CO₃ → MCO₃ → MO. Matches with II.

2. Correct Matching:

• A-III
• B-IV
• C-I
• D-II

Solution:

1. Shape Analysis:

• A. NH₃: Ammonia has a trigonal pyramidal shape due to one lone pair on nitrogen. This matches with III.
• B. BrF₅: Bromine pentafluoride has a square pyramidal shape due to five bond pairs and one lone pair on bromine. This matches with I.
• C. PCl₅: Phosphorus pentachloride has a trigonal bipyramidal shape because it has no lone pairs on phosphorus. This matches with IV.
• D. CH₄: Methane has a tetrahedral shape due to four bond pairs and no lone pairs on carbon. This matches with II.

2. Correct Matching:

• A-III
• B-I
• C-IV
• D-II

3. Conclusion: The correct answer is:

3 (A-III, B-I, C-IV, D-II)

Solution:

1. Equilibrium Constants and Reaction Mechanisms: When reactions are added, their equilibrium constants are multiplied to find the overall equilibrium constant.

2. Add the Reactions:

X ↔ Y
Y ↔ Z
Z ↔ W
______________
X ↔ W

3. Multiply the Equilibrium Constants:

K = K₁ × K₂ × K₃
K = 1.0 × 2.0 × 4.0 = 8.0

4. Conclusion: The equilibrium constant for the reaction X ↔ W is 8.0, corresponding to option (3).

Solution:

1. Analyze the Reactions:

• Reaction with Iodine: Iodine reduces to I^-, while thiosulphate is oxidized to S₄O₆^2-.
• Reaction with Bromine: Bromine reduces to Br^-, while thiosulphate is oxidized to SO₄^2-. Bromine oxidizes sulfur to a higher oxidation state than iodine.

2. Identify the Stronger Oxidant: Bromine causes a greater oxidation of sulfur compared to iodine, indicating that bromine is the stronger oxidizing agent.

3. Evaluate the Options:

• Option (1): Incorrect. Both halogens are reduced in these reactions.
• Option (2): Incorrect. Thiosulphate is oxidized in both reactions.
• Option (3): Correct. Bromine is a stronger oxidizing agent than iodine.
• Option (4): Incorrect. The opposite is true.

Solution:

1. Reaction with AgNO₃: The reaction with AgNO₃ produces 2 moles of AgCl, which means there are 2 ionizable Cl^- ions outside the coordination sphere. The remaining chloride ions are part of the coordination sphere.

The complex can be represented as [Co(NH₃)_yCl_n-2]Cl₂, where y is the number of NH₃ ligands.

2. Octahedral Complex: In an octahedral complex, the total number of ligands around the central metal is 6. Therefore:

y + (n - 2) = 6

y = 8 - n

3. Oxidation State of Co:

The oxidation state of Co (x) is determined by balancing the charges. Let the charge on Co be x. NH₃ is neutral, and Cl has a charge of -1:

x + 0 × (8 - n) + (-1) × (n - 2) = +2

x - n + 2 = +2

x = n

4. Calculate x + n:

Since x = n, and there are 2 ionizable Cl^- ions, n = 4. Therefore:

x + n = 4 + 4 = 8

5. Conclusion: The value of x + n is 8.

Solution:

1. Statement (1): Bromine water is a mild oxidizing agent that oxidizes the aldehyde group (-CHO) in glucose to a carboxylic acid (-COOH), forming gluconic acid. However, bromine water cannot oxidize the terminal –CH₂OH group. To form a dicarboxylic acid, a stronger oxidizing agent like HNO₃ is required. Hence, this statement is incorrect.

2. Statement (2): Glucose exists predominantly in its cyclic hemiacetal form, and the open-chain form containing the aldehyde group is present in small amounts. Schiff's test does not detect the hemiacetal form, so this statement is correct.

3. Statement (3): Glucose has 4 chiral carbons in its structure, making this statement correct.

4. Statement (4): Glucose exists in equilibrium between its open-chain form and two cyclic forms (α and β anomers). This statement is correct.

Conclusion: The incorrect statement is (1).

Solution:

1. Types of Carbon Atoms:

• Primary (1°): Carbon bonded to one other carbon atom.
• Secondary (2°): Carbon bonded to two other carbon atoms.
• Tertiary (3°): Carbon bonded to three other carbon atoms.

2. Analyze the Structure:

In the given compound, only one carbon is bonded to two other carbon atoms, making it a secondary carbon (2°).

3. Conclusion: The compound has 1 secondary (2°) carbon atom, corresponding to option (2).

Solution:

1. Criteria for Aromaticity:

• Cyclic: Must form a closed ring.
• Planar: All atoms in the ring must be coplanar.
• Conjugated: Must have a continuous system of overlapping p-orbitals.
• Huckel's Rule: Must have (4n + 2) π electrons, where n is a non-negative integer.

2. Analyze Each Structure:

• A: Naphthalene is aromatic (cyclic, planar, conjugated, 10 π electrons).
• B: Benzene is aromatic (cyclic, planar, conjugated, 6 π electrons).
• C: Biphenyl is non-planar due to steric hindrance between rings. Not aromatic.
• D: Cyclopentadienyl anion is aromatic (cyclic, planar, conjugated, 6 π electrons).

3. Conclusion: B and D are aromatic. The correct option is (1).

Solution:

1. Disproportionation Reaction: A redox reaction where the same element is both oxidized and reduced.

2. Halogens:

• F₂: Cannot disproportionate as it is the most electronegative element and cannot be oxidized further.
• Cl₂, Br₂, I₂: Can exhibit multiple oxidation states and undergo disproportionation. Examples:
  • Cl₂ + 2NaOH → NaCl + NaClO + H₂O
  • Br₂ + 2NaOH → NaBr + NaBrO + H₂O
  • I₂ + 2NaOH → NaI + NaIO + H₂O

3. Conclusion: Cl₂, Br₂, and I₂ can undergo disproportionation. The correct option is (2).

Solution:

1. Statement I:

N(CH₃)₃ and P(CH₃)₃ have lone pairs on nitrogen and phosphorus atoms, respectively, which can coordinate with transition metals to form complexes. Statement I is correct.

2. Statement II:

While N(CH₃)₃ is a σ-donor ligand, P(CH₃)₃ can act as both a σ-donor and a π-acceptor due to the availability of d-orbitals. This makes their bonding nature different. Statement II is incorrect.

3. Conclusion: Statement I is correct, but Statement II is incorrect. The correct option is (3).

Solution:

A. Cl, S: These elements have high negative electron gain enthalpy within their groups. Chlorine has the highest in Group 17. Correct match: IV.

B. Ge, As: These are metalloids and exhibit properties of both metals and non-metals. Correct match: III.

C. Fr, Ra: These are the largest elements in Groups 1 and 2, respectively. Correct match: II.

D. F, O: Fluorine has the highest electronegativity in the periodic table, and oxygen is also highly electronegative. Correct match: I.

Conclusion: The correct matches are A-IV, B-III, C-II, D-I, corresponding to option (3).

Solution:

1. Step 1: Fe³⁺ oxidises iodide (I⁻) to iodine (I₂):

2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂

Step 2: Fe²⁺ reduces persulphate (S₂O₈²⁻) to sulfate (SO₄²⁻):

2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻

Conclusion: A is true because Fe³⁺ oxidises iodide. D is true because Fe²⁺ reduces persulphate. The correct option is (4).

Solution:

A. Fe₄[Fe(CN)₆]₃·xH₂O: This is Prussian Blue. Correct match: III.

B. [Fe(CN)₅NOS]⁴⁻: This is sodium nitroprusside, which forms a violet complex. Correct match: I.

C. [Fe(SCN)]²⁺: This forms a blood red complex. Correct match: II.

D. (NH₄)₃PO₄·12MoO₃: This is ammonium phosphomolybdate, a yellow precipitate. Correct match: IV.

Conclusion: The correct matches are A-III, B-I, C-II, D-IV, corresponding to option (1).

Solution:

1. Crystal Field Theory and t_2g Orbitals:

In octahedral complexes, the d-orbitals split into t_2g (lower energy) and e_g (higher energy) orbitals. Water is a weak field ligand, so electrons follow Hund's rule in high-spin configurations.

2. Electronic Configurations and t_2g Electrons:

• [Fe(H₂O)₆]²⁺: Fe²⁺ is d⁶. In high spin, t_2g contains 4 electrons (even).
• [Co(H₂O)₆]²⁺: Co²⁺ is d⁷. t_2g contains 5 electrons (odd).
• [Co(H₂O)₆]³⁺: Co³⁺ is d⁶. t_2g contains 4 electrons (even).
• [Cu(H₂O)₆]²⁺: Cu²⁺ is d⁹. t_2g contains 6 electrons (even).
• [Cr(H₂O)₆]²⁺: Cr²⁺ is d⁴. t_2g contains 3 electrons (odd).

3. Count Complexes with Even t_2g Electrons:

Three complexes have even t_2g electrons: [Fe(H₂O)₆]²⁺, [Co(H₂O)₆]³⁺, and [Cu(H₂O)₆]²⁺.

Conclusion: The correct answer is (2).

Solution:

1. HVZ (Hell-Volhard-Zelinsky) Reaction:

The HVZ reaction involves halogenation at the α-carbon of carboxylic acids. Red phosphorus (Red P) catalyzes the formation of acyl bromide, which reacts with bromine to undergo α-bromination.

2. Reaction Steps:

• Step 1: Formation of Acyl Bromide:
• Step 2: α-Bromination:
• Step 3: Hydrolysis:

Conclusion: The product is Option (1).

Solution:

1. Wavelength from the Diagram:

The diagram shows one full cycle of the wave as 1.5 pm. Thus, the wavelength (λ) is:

λ = 1.5 × 10^-12 m.

2. Frequency-Wavelength Relation:

The relationship between the frequency (f) and wavelength (λ) is:

c = fλ, where c is the speed of light (3 × 10⁸ m/s).

f = c / λ

3. Calculate the Frequency:

f = (3 × 10⁸) / (1.5 × 10^-12)

f = 2 × 10²⁰ Hz.

However, the diagram indicates that a full cycle includes crest and trough, so λ = 3 pm = 3 × 10^-12 m:

f = (3 × 10⁸) / (3 × 10^-12)

f = 1 × 10²⁰ Hz or x = 5.

4. Conclusion:

The value of x is 5.

Solution:

1. Work for Isothermal Reversible Expansion:

The formula for work in an isothermal process is:

w = -nRT ln(V_f / V_i).

2. Given Values:

n = 1 mol

R = 0.08206 L atm mol^-1 K^-1

T = 18 + 273.15 = 291.15 K

V_i = 10 L, V_f = 100 L

3. Calculate Work:

w = -(1)(0.08206)(291.15) ln(100 / 10)

w = -23.883 × ln(10)

ln(10) ≈ 2.303

w = -23.883 × 2.303 = -55.018 L atm

4. Conclusion:

Work done (x) = 55 L atm.

Solution:

1. Hinsberg's Test:

• Primary amines form sulfonamides soluble in NaOH.
• Secondary amines form insoluble sulfonamides.
• Tertiary amines do not react with Hinsberg's reagent.

2. Analyze the Compounds:

• Aniline (C₆H₅NH₂): Primary amine; soluble in NaOH.
• Cyclohexylamine: Primary amine; soluble in NaOH.
• Methylamine: Primary amine; soluble in NaOH.
• Dimethylamine: Secondary amine; insoluble in NaOH.
• Triethylamine: Tertiary amine; no reaction.

3. Count Soluble Compounds:

Five primary amines are soluble in NaOH after reacting with Hinsberg's reagent.

4. Conclusion:

The number of amines is 5.

Solution:

1. Chiral Centers:

A chiral center is a carbon atom attached to four different groups. By analyzing the given structure, the compound has five chiral centers.

2. Number of Optical Isomers:

The total number of optical isomers for a molecule with n chiral centers is \(2^n\). Since the compound has five chiral centers:

\(2^5 = 32\) optical isomers.

3. Conclusion:

The compound has 32 optical isomers.

Solution:

1. Identify Metal M:

Among the given elements, Zn has the smallest metallic radius due to its position at the end of the 3d transition series (increased nuclear charge).

2. Oxidation State in MO₄^2-:

Oxygen has an oxidation state of -2. Let the oxidation state of Zn be \(x\):

\(x + 4(-2) = -2 \quad \Rightarrow \quad x = +6\).

3. Electronic Configuration of Zn⁶⁺:

Zn's ground-state configuration is [Ar] 3d¹⁰ 4s². After losing six electrons (to form Zn⁶⁺), the configuration becomes [Ar]. There are no unpaired electrons.

4. Magnetic Moment:

The spin-only magnetic moment (\(\mu\)) is given by:

\(\mu = \sqrt{n(n+2)} \, \text{BM}\), where \(n\) is the number of unpaired electrons.

For Zn⁶⁺, \(n = 0\):

\(\mu = \sqrt{0(0+2)} = 0 \, \text{BM}\).

5. Conclusion:

The spin-only magnetic moment of MO₄^2- is 0 BM.

Solution:

1. Octet Rule and Exceptions:

The octet rule states that atoms tend to complete an octet of electrons in their valence shell. Exceptions include:

• Incomplete Octets: Molecules where the central atom has fewer than 8 electrons (e.g., BF₃, BeF₂).
• Expanded Octets: Molecules where the central atom has more than 8 electrons (e.g., PCl₅, H₂SO₄).
• Odd-Electron Molecules: Molecules with an odd number of valence electrons (e.g., NO₂, ClO₂).

2. Analyze the Molecules:

• CO₂: Follows the octet rule.
• NO₂: Odd-electron molecule; exception.
• H₂SO₄: Expanded octet on sulfur; exception.
• BF₃: Incomplete octet on boron; exception.
• CH₄: Follows the octet rule.
• SiF₄: Follows the octet rule.
• ClO₂: Odd-electron molecule; exception.
• PCl₅: Expanded octet on phosphorus; exception.
• BeF₂: Incomplete octet on beryllium; exception.
• C₂H₆: Follows the octet rule.
• CHCl₃: Follows the octet rule.
• CBr₄: Follows the octet rule.

3. Count the Exceptions:

Six molecules are exceptions: NO₂, H₂SO₄, BF₃, ClO₂, PCl₅, and BeF₂.

4. Conclusion:

The number of molecules that are exceptions to the octet rule is 6.

Solution:

1. Reaction:

Aniline reacts with benzenediazonium chloride to form aniline yellow (p-aminoazobenzene):

C6H5NH2 + C6H5N2Cl → C6H5-N=N-C6H4NH2

2. Moles of Aniline:

Molar mass of aniline (C₆H₇N) = 93.13 g/mol

Moles of aniline = mass / molar mass = 279 g / 93.13 g/mol = 3 mol

3. Stoichiometry:

The reaction is 1:1, so 3 moles of aniline will form 3 moles of aniline yellow.

4. Mass of Aniline Yellow:

Molar mass of aniline yellow (C₁₂H₁₁N₃) = 197.24 g/mol

Mass of aniline yellow = moles × molar mass = 3 mol × 197.24 g/mol = 591.72 g

5. Conclusion:

The maximum amount of aniline yellow formed is approximately 591 g (nearest integer).

Solution:

1. Analysis of A:

The time data suggests that the reaction is first-order with respect to A because, for a first-order reaction, the time to reduce concentration by successive halves is proportional (e.g., 1/2 → 1/4).

2. Analysis of B:

The linear decrease in concentration of B with time indicates zero-order behavior with respect to B.

3. Overall Order:

Overall order = (Order with respect to A) + (Order with respect to B)

Overall order = 1 + 0 = 1

4. Conclusion:

The overall order of the reaction is 1.

Solution:

1. Reaction Steps:

Step 1: Oxidation of ethylbenzene to benzoic acid (A) using KMnO₄ and heat.

Step 2: Nitration of benzoic acid to 3-nitrobenzoic acid (B) using HNO₃/H₂SO₄.

2. Structure of B:

The structure of 3-nitrobenzoic acid includes:

• Three π-bonds in the benzene ring
• One π-bond in the carboxyl group (-COOH)
• One π-bond in the nitro group (-NO₂)

3. Total π-bonds:

Total = 3 + 1 + 1 = 5

4. Conclusion:

The major product (B) has 5 π-bonds.

Solution:

1. Boiling Point Elevation:

\(ΔT_b = T_b - T_b^o = 100.52 - 100 = 0.52\) K

2. Molality (m):

Moles of AB₂ = \(10 / 200 = 0.05\) mol

Mass of water = 100 g = 0.1 kg

Molality \(m = 0.05 / 0.1 = 0.5\) mol/kg

3. Van't Hoff Factor (i):

\(ΔT_b = i K_b m\)

\(0.52 = i (0.52)(0.5)\)

\(i = 2\)

\(i = 1 + 2α \quad \Rightarrow \quad 2 = 1 + 2α \quad \Rightarrow \quad α = 0.5\)

4. Degree of Ionization:

\(α = 0.5 = 5 × 10^{-1}\)

5. Conclusion:

The degree of ionization (α) is \(5 × 10^{-1}\).