The given reaction sequence involves the following steps:

1. Reaction with Grignard reagent: Compound M reacts with CH₃MgBr (a Grignard reagent) to form an intermediate compound N. In this step, the Grignard reagent adds to the carbonyl group of M, creating a new C–C bond and releasing CH₄ gas as a by-product. The reaction mechanism involves nucleophilic attack by CH₃⁻ on the carbonyl carbon.
2. Hydrolysis: The intermediate N undergoes acid hydrolysis. During this process, the intermediate is converted into CH₃COCH₂COCH₃, which is a β-diketone. Hydrolysis ensures the protonation and stabilization of the newly formed compound.

The electronic configuration of Gd (Gadolinium) is [Xe] 4f⁷ 5d¹ 6s². When Gd loses 3 electrons (one from the 6s orbital and two from the 5d orbital), it becomes Gd³⁺ with the electronic configuration [Xe] 4f⁷. This configuration does not match 4f⁶.

The electronic configuration of Sm (Samarium) is [Xe] 4f⁶ 6s². When Sm loses 2 electrons from the 6s orbital, it becomes Sm²⁺ with the configuration [Xe] 4f⁶. This matches the given 4f⁶ configuration.

The configuration of Sm³⁺ is [Xe] 4f⁵, and that of Tb³⁺ (Terbium) is [Xe] 4f⁸. Neither of these matches the required configuration.

Therefore, the ion with a 4f⁶ configuration is Sm²⁺.

Conductivity in materials depends on the availability of free charge carriers and their mobility.

1. For metallic conductors: Conductivity decreases with an increase in temperature. This is because as temperature rises, the lattice vibrations in the metal increase, leading to more frequent collisions between electrons and the lattice ions. This reduces the mobility of electrons, thereby decreasing conductivity.

2. For semiconductors: Conductivity increases with an increase in temperature. In semiconductors, thermal energy excites more electrons from the valence band to the conduction band, increasing the number of free charge carriers (electrons and holes). This leads to higher conductivity at elevated temperatures.

Thus, the correct answer is that the conductivity of metals decreases while the conductivity of semiconductors increases when heated.

The reaction involves:
I₂ + 2 Na₂S₂O₃ → 2 NaI + Na₂S₄O₆

Here, each molecule of Na₂S₂O₃ donates 1 electron, leading to a total of 2 electrons being transferred per molecule of I₂.

1. Oxidation states of sulfur: In Na₂S₂O₃, the average oxidation state of sulfur is +2. During the reaction, sulfur in Na₂S₂O₃ is partially oxidized to Na₂S₄O₆, where the average oxidation state becomes +2.5.

2. Definition of equivalent weight: The equivalent weight of a substance is given by:
Equivalent Weight = Molecular Weight / n
Here, n represents the number of electrons transferred per molecule in the reaction.

3. Calculation: For Na₂S₂O₃, n = 1, because each molecule transfers 1 electron. Therefore, the equivalent weight of Na₂S₂O₃ is equal to its molecular weight, M.

Thus, the equivalent weight of Na₂S₂O₃ in this reaction is M.

The reactivity order towards an SN1 reaction is determined by the stability of the carbocation intermediate.

Allyl chloride forms an allyl carbocation, which is resonance-stabilized, making it more stable.
Ethyl chloride forms a primary carbocation, which is less stable than the allyl carbocation.
Chlorobenzene, however, does not undergo SN1 reactions readily because the carbocation would be destabilized due to the aromatic ring.

Thus, the reactivity order is I > III > II.

At 25°C, nitration of toluene with a mixture of nitric acid and sulfuric acid (mixed acid) predominantly gives ortho- and para-nitrotoluene as the products. The para isomer is the major product due to steric and electronic factors. The methyl group on toluene is an activating group, which directs the nitro group to the ortho and para positions.

This reaction is a Perkin condensation reaction. Benzaldehyde (PhCHO) reacts with acetic anhydride in the presence of a base (sodium ethoxide) to produce cinnamic acid . The reaction proceeds through the formation of an enolate ion, followed by aldol-like condensation and dehydration to yield the α,β-unsaturated carboxylic acid product.

The reactivity of alkenes towards HBr addition depends on the stability of the carbocation intermediate formed during the reaction:

• MeOCH=CH₂ (III) forms a resonance-stabilized carbocation due to the electron-donating methoxy group, making it the most reactive.
• CH₃–CH=CH₂ (I) forms a secondary carbocation, which is moderately stable.
• CF₃–CH=CH₂ (II) forms a destabilized carbocation due to the electron-withdrawing CF₃ group.
• (IV) forms a sterically hindered carbocation, making it the least reactive.

Ozonolysis of o-xylene involves the cleavage of the double bonds in the aromatic ring, leading to the formation of three carbonyl compounds:

• Compound I: CH₃–C(O)–C(O)–CH₃ (produced once),
• Compound II: CH₃–C(O)–C(O)–H (produced twice),
• Compound III: H–C(O)–C(O)–H (produced thrice).

The reaction sequence involves the formation of a Grignard reagent from Me Br and Mg in dry ether. The Grignard reagent reacts with CO₂ (from dry ice) to form a carboxylate, which upon protonation with H₃O⁺ gives a carboxylic acid. The correct structures of A and B are shown in option (3).

In Lassaigne’s test, compounds containing nitrogen form a characteristic blue or purple color with sodium fusion extract when treated with sodium picrate or other reagents. The compound that does not give a positive test for nitrogen is the one that does not contain nitrogen in its structure.

The given compounds are:

• (A) Nitro phenol (C₆H₄NO₂OH) contains nitrogen and will give a positive test.
• (B) Chlorobenzene (C₆H₅Cl) does not contain nitrogen, and thus does not give a positive test for nitrogen in Lassaigne's test.
• (C) Aniline (C₆H₅NH₂) contains nitrogen and will give a positive test.
• (D) Aminobenzene (C₆H₅NH₂) contains nitrogen and will give a positive test.

Phenol is acidic due to the resonance stabilization of the phenoxide ion. The introduction of an electron-withdrawing group such as -CHO increases acidity by stabilizing the negative charge on the oxygen atom.

• 4-Hydroxybenzaldehyde (II): The -CHO group at the para position exerts the maximum electron-withdrawing effect due to resonance and inductive effects, making II the most acidic.
• 3-Hydroxybenzaldehyde (III): The -CHO group at the meta position provides a lesser electron-withdrawing effect compared to the para position, making III less acidic than II but more acidic than I.
• Phenol (I): Without any substituents, phenol has the lowest acidity among the three compounds.

The given reaction involves a nucleophilic substitution (S_N2) where sodium cyanide (NaCN) is used as the nucleophile and DMF is the solvent. In this reaction, the cyanide ion (CN^-) will attack the electrophilic carbon attached to the halide group (Cl or I) and replace it.

The structure of the reactant is:
C₆H₄ClI
When NaCN is added, the cyanide ion (CN^-) will substitute for the halide group, leading to the formation of a nitrile group (CN).

The major product of this reaction will be:
C₆H₄CN
Thus, the major product is the compound (C) Iodobenzene with one CN group attached to the benzene ring.

The spontaneity of a chemical reaction, including polymerization, is determined by the Gibbs free energy change (ΔG). For a reaction to be spontaneous, ΔG must be negative:
ΔG = ΔH - TΔS
where ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature in Kelvin.

In the case of spontaneous polymerization:

• ΔH < 0: The process is exothermic, releasing energy as bonds are formed between monomers.
• ΔS < 0: Polymerization involves the formation of a more ordered structure from monomers, leading to a decrease in entropy.

The relationship between Gibbs Free Energy change (ΔG°) and the equilibrium constant (K) is given by:
ΔG° = -RT ln K
Here:

• K = 10^-14 (ionic product of water),
• R = 1.987 cal mol^-1 K^-1 (universal gas constant),
• T = 298 K (temperature in Kelvin).

The velocity of an electron in a Bohr orbit is inversely proportional to the principal quantum number (n): \[ v \propto \frac{1}{n} \] For the first Bohr orbit (n = 1), the velocity is v₁. For the third Bohr orbit (n = 3), the velocity is v₃. Using the proportionality: \[ \frac{v_3}{v_1} = \frac{1/3}{1/1} = \frac{1}{3} \] Thus, the ratio v₃ : v₁ is 1:3.

The compressibility factor (Z) is defined as: \[ Z = \frac{PV}{RT} \] For a van der Waals gas, the equation of state is: \[ \left(P + \frac{a}{V^2}\right)(V - b) = RT \] At high pressures, the volume V decreases, making the term b (excluded volume) significant. The term involving a (intermolecular attraction) becomes negligible. Simplifying for high pressure: \[ Z \approx 1 + \frac{Pb}{RT} \] Thus, the compressibility factor is approximately 1 + Pb/RT under high-pressure conditions.

For a spontaneous process at constant temperature and pressure:

• Gibbs Free Energy (ΔG) must be negative: (ΔG_system)_{T, P} < 0.
• The total entropy change (system + surroundings) must be positive: (ΔS_system) + (ΔS_surroundings) > 0.
• Internal energy change (ΔU) can vary depending on the process but is typically negative under adiabatic conditions.

- Option 1: Correct. Viscosity decreases with an increase in temperature because the intermolecular forces are weakened, allowing liquid molecules to flow more easily.
- Option 2: Correct. Surface tension decreases with temperature due to reduced cohesive forces at the liquid surface.
- Option 3: Incorrect. Impurities increase viscosity by disrupting the liquid's intermolecular interactions, making it harder for molecules to flow.
- Option 4: Correct. The effect of impurities on surface tension depends on the type of impurity. For example, surfactants lower surface tension, while some other impurities might increase it.

Therefore, option 3 is incorrect.

The rate constant (k) is governed by the Arrhenius equation: \[ k = A e^{-\frac{E_a}{RT}} \] Here, A is the frequency factor, E_a is the activation energy, R is the gas constant, and T is the temperature. As T increases, the exponential term increases, and hence, k always increases.
The equilibrium constant (K) depends on the reaction's enthalpy change (ΔH): \[ \ln K = -\frac{\Delta H}{RT} + \frac{\Delta S}{R} \] For exothermic reactions (ΔH < 0), K decreases with temperature. For endothermic reactions (ΔH > 0), K increases with temperature.
Thus, the equilibrium constant can either increase or decrease depending on the reaction's nature, but the rate constant always increases with temperature.

The atomic number of ²¹⁴₈₃Bi is 83, and the mass number is 214.
1. Emission of a β-particle: A β-particle emission increases the atomic number by 1 (83 → 84) but keeps the mass number unchanged (214).
2. Emission of an α-particle: An α-particle emission decreases the atomic number by 2 (84 → 82) and the mass number by 4 (214 → 210).

The number of neutrons is calculated as: \[ \text{Number of neutrons} = \text{Mass number} - \text{Atomic number} = 210 - 82 = 128 \]

The radius of the n^th Bohr orbit for a hydrogen-like species is given by: \[ r = \frac{n^2 a_0}{Z} \] where n is the principal quantum number, a₀ is the Bohr radius for hydrogen, and Z is the atomic number. For the 1st Bohr orbit of hydrogen (n=1, Z=1), r = a₀.

We need to find a species where r = a₀. Let us check the options:

• Option 1: n=2, Li²⁺ (Z=3): \[ r = \frac{2^2 a_0}{3} = \frac{4}{3} a_0 \]
• Option 2: n=2, Be³⁺ (Z=4): \[ r = \frac{2^2 a_0}{4} = a_0 \]
• Option 3: n=2, He⁺ (Z=2): \[ r = \frac{2^2 a_0}{2} = 2 a_0 \]
• Option 4: n=3, Li²⁺ (Z=3): \[ r = \frac{3^2 a_0}{3} = 3 a_0 \]

For a first-order reaction, the integrated rate law is: \[ \ln [A] = -kt + \ln [A]_0 \] where [A] is the concentration at time t, k is the rate constant, and [A]₀ is the initial concentration. Taking the logarithm to base 10: \[ \log [A] = -\frac{k}{2.303} t + \log [A]_0 \] This equation is of the form y = mx + c, where:

• y = log [A],
• x = t,
• m = -k/2.303 (slope),
• c = log [A]₀.

- HCl is a monoprotic acid, so 0.1 M HCl provides 0.1 M H⁺ ions.
- H₂SO₄ is a diprotic acid, so 0.2 M H₂SO₄ provides \( 2 \times 0.2 = 0.4 \, \text{M} \, \text{H}^+ \) ions.

When equal volumes of the two solutions are mixed, the total volume doubles, and the concentration of each acid is halved: \[ \text{New concentration of HCl} = \frac{0.1}{2} = 0.05 \, \text{M H}^+ \] \[ \text{New concentration of H}_2\text{SO}_4 = \frac{0.4}{2} = 0.2 \, \text{M H}^+ \] The total H⁺ concentration is: \[ 0.05 \, \text{M} + 0.2 \, \text{M} = 0.25 \, \text{M} \]

The boiling point elevation is a colligative property, which depends on the number of solute particles in the solution:

• KNO₃: Strong electrolyte, dissociates into 2 ions (K⁺ and NO₃^-).
• NaCl: Strong electrolyte, dissociates into 2 ions (Na⁺ and Cl^-).
• CH₃COOH: Weak electrolyte, partially dissociates into CH₃COO^- and H⁺ ions.
• Sucrose: Non-electrolyte, does not dissociate.

The solubility of silver halides in water decreases as the size of the halide ion increases. This is because:

• Larger halide ions (e.g., I^-) have lower lattice energy compared to smaller halide ions (e.g., F^-).
• The lattice energy dominates over the hydration energy, making the solubility lower for larger ions.

Part (i): CuSO₄ added to saturated (NH₄)₂SO₄ solution
When CuSO₄ is added to a saturated solution of (NH₄)₂SO₄, it introduces Cu²⁺ ions into the solution. Cu²⁺ is a Lewis acid and reacts with water, increasing the concentration of H⁺ ions. This reaction results in an increase in acidity: \[ \text{Cu}^{2+} + H_2O \rightarrow \text{[Cu(H$_2$O)$_6$]}^{2+} \rightarrow \text{[Cu(H$_2$O)$_5$(OH)]}^+ + H^+ \] Part (ii): SbF₅ added to anhydrous HF
SbF₅ is a strong Lewis acid and forms the superacid HF-SbF₅ when mixed with HF. This combination greatly increases the acidity of the solution as it stabilizes the fluoride ion (F^-), shifting the equilibrium towards producing more H⁺ ions: \[ \text{HF} + \text{SbF}_5 \rightarrow \text{HSbF}_6 + \text{H}^+ \]
In both cases, the acidity of the system increases. Thus, the correct answer is:
\[ \boxed{\text{Increases, Increases.}} \]

- ClO₃^-: Central Cl atom has 1 lone pair.
- XeF₄: Central Xe atom has 2 lone pairs.
- SF₄: Central S atom has 1 lone pair.
- I₃^-: Central I atom has 3 lone pairs.

Thus, I₃^- has the maximum number of lone pairs (3) on its central atom.

Mohr's salt, chemically known as ammonium iron(II) sulfate, has the formula \((NH_4)_2[Fe(SO_4)_2] \cdot 6H_2O\).
When dissolved in water, it dissociates completely as follows: \[ (NH_4)_2[Fe(SO_4)_2] \cdot 6H_2O \rightarrow 2 \text{NH}_4^+ + \text{Fe}^{2+} + 2 \text{SO}_4^{2-} \] From this dissociation, the ions produced are:

• 2 NH₄⁺ (ammonium ions),
• 1 Fe²⁺ (iron(II) ion),
• 2 SO₄^2- (sulfate ions).

Ligand-to-Metal Charge Transfer (LMCT): LMCT occurs when there is a transfer of electrons from the ligand (oxygen) to the metal ion (Mn or Cr).
Paramagnetism: Paramagnetic species have unpaired electrons in their electronic configuration.

Analysis of the given species:

• MnO₄^-: Mn is in the +7 oxidation state with no unpaired electrons (3d⁰). This species shows LMCT but is diamagnetic.
• MnO₄^2-: Mn is in the +6 oxidation state with one unpaired electron (3d¹). This species exhibits both LMCT and paramagnetism.
• Cr₂O₇^2-: Cr is in the +6 oxidation state with no unpaired electrons (3d⁰). This species is diamagnetic and shows LMCT.
• CrO₄^2-: Cr is also in the +6 oxidation state with no unpaired electrons. This species shows LMCT but is diamagnetic.

The structure of P₄O₁₀ is based on four phosphorus atoms forming a tetrahedron with oxygen atoms bridging between them. There are six P–O–P linkages in the molecule.

The reaction sequence involves:

• Hydroboration-Oxidation: B₂H₆, H₂O₂/OH^- adds water across the double bond with anti-Markovnikov regioselectivity.
• Oxymercuration-Demercuration: Hg(OAc)₂, NaBH₄/OH^- adds water across the double bond with Markovnikov regioselectivity.

The concentration of HCl is 10^-8 M. Since HCl is a strong acid, the [H⁺] from HCl is 10^-8 M. However, the autoionization of water also contributes [H⁺] = 10^-7 M. Therefore, the total [H⁺] is: \[ \text{Total [H$^+$]} = 10^{-8} + 10^{-7} \approx 1.1 \times 10^{-7} \, \text{M}. \] The pH is given by: \[ \text{pH} = -\log [\text{H}^+] \approx -\log (1.1 \times 10^{-7}) \approx 6.96. \] Thus, the pH is greater than 6 and less than 7.

Given:

• k = 1.65 × 10^-4 S cm^-1,
• C = 0.02 M,
• λ^°_H⁺ = 349.1 S cm² mol^-1,
• λ^°_CH3COO^- = 40.9 S cm² mol^-1.

1. Calculate Λ_m: \[ \Lambda_m = \frac{k}{C} = \frac{1.65 \times 10^{-4}}{0.02} = 0.00825 \, \text{S cm$^2$ mol$^{-1}$}. \]
2. Calculate Λ^°: \[ \Lambda^\circ = \lambda^\circ_\text{H$^+$} + \lambda^\circ_\text{CH$_3$COO$^-$} = 349.1 + 40.9 = 390 \, \text{S cm$^2$ mol$^{-1}$}. \]
3. Calculate α: \[ \alpha = \frac{\Lambda_m}{\Lambda^\circ} = \frac{0.00825}{390} \approx 0.021. \]

• H₃PO₃: Contains two -OH groups and one H atom directly bonded to P.
• H₃PO₄: Contains three -OH groups and one P=O bond.

The SN2 reaction mechanism is a bimolecular nucleophilic substitution reaction characterized by:

• A single step where the nucleophile attacks the electrophilic carbon and the leaving group departs simultaneously.
• Statement 1: True. The reaction rate depends on the concentration of both the nucleophile and the substrate, and hence increases with increasing nucleophilicity.
• Statement 2: True. The "2" in SN2 indicates that the reaction is second-order, dependent on two reactants.
• Statement 3: True. Tertiary substrates do not follow the SN2 mechanism due to steric hindrance.
• Statement 4: False. Optical rotation may change, but it is not guaranteed to shift from (+) to (-) or vice versa, as this depends on the specific nucleophile and substrate.

To identify the enantiomer of Y, we need to find structures that are mirror images of Y but non-superimposable.

• Option A: Not a mirror image of Y.
• Option B: Matches the mirror image of Y.
• Option C: Matches the mirror image of Y.
• Option D: Identical to Y, not an enantiomer.

• (1): Correct. The oxidation number of Cr in CrO₅ is +6.
• (2): Correct. For the reaction, ΔH > ΔU because Δn_g > 0.
• (3): Incorrect. The pH of 0.1 N H₂SO₄ is slightly higher than 0.1 N HCl due to incomplete dissociation.
• (4): Incorrect. RT/F = 0.0257 V at 25°C, not 0.0591 V.

• [CoF₆]^3-: Paramagnetic due to high-spin configuration.
• [Co(NH₃)₆]³⁺: Diamagnetic due to low-spin configuration with NH₃.
• [Fe(OH₂)₆]²⁺: Paramagnetic due to unpaired electrons.
• [Fe(CN)₆]^4-: Diamagnetic due to low-spin configuration with CN^-.

• Statement 1: False. Solid I₂ is only slightly soluble in water.
• Statement 2: True. I₂ dissolves in water in the presence of KI due to I₃^- complex formation.
• Statement 3: True. I₂ is freely soluble in non-polar solvents like CCl₄.
• Statement 4: False. Solubility increases in hot water, but I₂ is not freely soluble.