• X = Manganese (Mn): KMnO₄ is used as a fungicide. Mn is brittle and not used in pure form.
• Y = Iron (Fe): Consecutive to Mn in the first transition series.
• Electron check: Mn²⁺ → [Ar] 3d⁵ → 5 unpaired e⁻ | Fe³⁺ → [Ar] 3d⁵ → 5 unpaired e⁻ ✔
• Mn–Fe alloy is used in railway tracks due to its hardness.

Step 1: Determine Unpaired Electrons in Copper

Step 2: Identify Element X

Double the unpaired electrons = 2 unpaired electrons.

Nickel (Ni): [Ar] 3d⁸ 4s²

Step 3: Analyze the Uses of Nickel

The alloy of Titanium with Aluminium is used in MIG aircraft due to its high strength-to-weight ratio. Nickel does not form this specific aircraft alloy. Therefore, option (d) is NOT a use of Nickel.

Yellow ore = Hydrated iron(III) oxide (Fe₂O₃·nH₂O).

(1) Physical: Surface-tension flotation separates gangue → decreases mass, increases Fe%.

(2) Chemical: Roasting removes water of crystallisation → decreases mass, increases Fe%.

(3): Crushing — only reduces particle size; mass and composition unchanged.

Step 1: Read the Bar Chart

Step 2: Analyze Stability of Each Process

Step 3: Why W³⁺ → W²⁺ is Easiest

The reduction W³⁺(d⁴) → W²⁺(d) is highly favorable because the product achieves the exceptionally stable half-filled d⁵ configuration. This is why Mn³⁺ is a strong oxidizing agent — it readily gains an electron to become Mn²⁺(d⁵).

Step 1: Identify the Periodic Table Positions

Step 2: Analyze Each Option

Step 3: Conclusion

M and C are in the same group of the periodic table. Intermetallic alloys require elements with different electronegativities and atomic sizes (typically from different groups). Therefore, statement (d) is NOT correct.

Both Na₂CO₃ and NaNO₂ dissolve in water with no visible difference, so water cannot distinguish them. KMnO₄/H⁺ is decolorised by NaNO₂ (reducing agent) but not by Na₂CO₃. Dilute acids release CO₂ from carbonate, differentiating them.

NaCl + Pb(NO₃)₂ → PbCl₂↓ (white precipitate, sparingly soluble).

BaCl₂ is fully soluble → no precipitate with Ba²⁺.

Na₂SO₄ and Na₂CO₃ form precipitates with both Pb²⁺ and Ba²⁺, so they cannot be used to distinguish.

Na₂SO₃ + HCl → SO₂↑ (gas) ✔ | Ag₂SO₃ is a white precipitate ✔.

AgBr is pale yellow (not black); AgI is yellow; AgCl is white — so options (a) and (d) are wrong.

Ag₂S is black, so (c) is also wrong — sulfide with HCl gives H₂S, not a typical acid–gas scenario that matches.

NaI + conc. H₂SO₄ → I₂ vapour (violet).

I₂ + Na₂S₂O₃ → 2NaI + Na₂S₄O₆ (colour disappears — standard iodometric reaction).

n(NaOH) = 0.010 L × 1 M = 0.010 mol

V(HCl) needed = n / M = 0.010 / 2 = 0.005 L = 5 mL

Final burette reading = 6.5 + 5.0 = 11.5 mL

Step 1: Identify the Colors Correctly

Important: ICl is dark brown, and ICl₃ is yellow. The equation shows: ICl (dark brown) + Cl₂ ⇌ ICl₃ (yellow).

Step 2: Determine Reaction Type

When heated, the mixture becomes dark brown → more ICl is produced → equilibrium shifts backward (reverse direction).

Heat favors the reverse reaction → reverse reaction is endothermic → forward reaction is exothermic.

Step 3: Determine How to Reduce Brown Color

To reduce brown color (reduce ICl), shift equilibrium forward (toward ICl₃, which is yellow).

Step 4: Conclusion

Forward reaction is exothermic, and decreasing vessel volume (increasing pressure) favors the forward direction (fewer moles of gas) → reduces ICl (dark brown) → reduces brown color.

Answer: Exothermic — decreasing vessel volume → option (c).

In an open container, gases escape, so any equilibrium involving gaseous products (a, b) cannot be maintained. Reaction (d) is irreversible. Only reaction (c) involves all aqueous/dissolved species, so equilibrium can be established in an open vessel.

2AgBr + light energy → 2Ag + Br₂

Ag⁺ is reduced (gains e⁻ from Br⁻) → Ag metal (black image). Br⁻ is oxidised → Br₂.

K_sp = [Ag⁺][Cl⁻] = S² → S = √(1.6×10⁻¹⁰) = 1.265×10⁻⁵ M

Total ions = [Ag⁺] + [Cl⁻] = 2S = 2 × 1.265×10⁻⁵ = 2.53×10⁻⁵ M

pOH = 14 − 12.3 = 1.7 → [OH⁻] = 10^−1.7 = 0.02 M

[Ba(OH)₂] = 0.02/2 = 0.01 M

n = 0.01 × 1.5 = 0.015 mol → m = 0.015 × 171 = 2.565 ≈ 2.56 g

K_p = P(N₂O₄) / [P(NO₂)]²

6.15 = P(N₂O₄) / (0.15)² = P(N₂O₄) / 0.0225

P(N₂O₄) = 6.15 × 0.0225 = 0.1383 atm

Anode (Cu): Cu → Cu²⁺ + 2e⁻ (dissolves, mass decreases, releases Cu²⁺ into solution).

Cathode: Cu²⁺ + 2e⁻ → Cu (deposits, mass increases).

Rate of dissolution = rate of deposition → [Cu²⁺] remains constant. This is the principle of electrorefining.

• NaCl — forms PbCl₂↓ → not suitable.
• Na₂CO₃ — forms PbCO₃↓ and FeCO₃↓ → not suitable.
• Ethyl alcohol — non-electrolyte → not suitable.
• NaNO₃ — all nitrate salts are soluble; does not react with Fe²⁺ or Pb²⁺ → suitable.

Same current → same moles of electrons through each cell. Mass deposited ∝ (M/n):

Highest M/n = 108 (Ag) → greatest cathode mass increase in cell ③.

X is oxidised (anode); Cu²⁺ is reduced (cathode), E°(Cu²⁺/Cu) = +0.34 V.

E_cell = E°_cathode − E°_anode

+1.8 = +0.34 − E°(X) → E°(X) = 0.34 − 1.8 = −1.46 V

During discharge:

• Anode (Pb): Pb + SO₄²⁻ → PbSO₄ + 2e⁻ → anode mass increases.
• Cathode (PbO₂): PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O
• H₂SO₄ is consumed and water is produced → solution density decreases.

A sacrificial anode must be more active (more negative reduction potential) than the protected metal.

B = −1.70 V. Elements more negative: D (−2.52 V) and C (−2.87 V) → both can serve as sacrificial anodes.

A (−0.45 V) is less active than B → cannot protect it.

Propanol → Propene (CH₂=CHCH₃, C₃H₆) = compound X.

Isomer of C₃H₆: Cyclopropane — saturated (single bonds only) but chemically active due to ring strain (opens easily).

C_nH_2n−2BrCl: degree of unsaturation = 1 (one double bond) + Br and Cl substituting 2H. For n = 4: C₃H₄BrCl → 1-bromo-3-chloropropene.

Step 1: Understand the Target Formula

C_nH_2n+2O = general formula for saturated monohydric alcohols (R–OH).

Step 2: Identify Reducible Functional Groups

Functional groups that can be reduced to alcohols:

Step 3: Count Carboxylic Acid Isomers of C₄H₈O₂

Only carboxylic acids reduce to a single alcohol product matching C_nH_2n+2O:

Step 4: Why Esters Don't Count

Esters produce two different alcohols upon reduction (e.g., ethyl ethanoate → ethanol + ethanol, but methyl propanoate → methanol + propanol). The question asks for isomers that reduce to compounds (singular type) with formula C_nH_2n+2O, meaning a single alcohol product.

Step 5: Total Count

Only 2 carboxylic acid isomers reduce to a single alcohol product → Answer: (b) 2.

Dry distillation (with NaOH/CaO) gives alkane with one less carbon. Sodium methanoate (HCOONa) has only 1 carbon, producing H₂ gas, not an alkane.

CaC₂ + H₂O → Acetylene (C₂H₂)

3 C₂H₂ → Benzene (C₆H₆) by trimerisation

C₆H₆ + 3H₂ → Cyclohexane (C₆H₁₂) by hydrogenation (C_nH_2n, 6 CH groups → 6 methyl equivalent positions)

Step 1: Identify the Insecticide

C_nH_nCl_n with n = 6: C₆H₆Cl₆ = Gammexane (BHC — Benzene Hexachloride), a well-known insecticide.

Step 2: Trace the Synthesis from Hexane (C₆H₁)

Step 3: Why UV Light, Not Catalyst?

Step 4: Conclusion

Correct sequence: Catalytic reforming → Halogenation with UV → option (c).

Acid A = Salicylic acid (for aspirin). Oxidation of o-methylphenol (o-cresol) → Salicylic acid.

Acid B = Formic acid (HCOOH). Oxidation of methanol → formic acid. (Formic acid polymer: used in plastics industry context in some curricula.)

X = Glycerol (C₃H₈O₃) | Y = Fatty acid (e.g. hexanoic acid, n≥4)

Glycerol + 3 fatty acids → Triglyceride (fat/oil) = compound Z.

Triglycerides are the raw material for soap manufacturing (saponification).

X = Alcohol (higher BP due to H-bonding) | Y = Ether (isomer, lower BP).

2 R–OH + H₂SO₄ (140°C) → R–O–R + H₂O (intermolecular dehydration → ether).

At 180°C: intramolecular dehydration → alkene (eliminates water), not ether.

• A = Lactic acid (C₃H₆O₃): secondary –OH → oxidised to pyruvic acid (–COOH + C=O).
• B = Citric acid (C₆H₈O₇): tertiary –OH → cannot be oxidised by KMnO₄.
• C = Salicylic acid (C₇H₆O₃): phenolic –OH gives purple/violet colour with FeCl₃.

Step 1: Identify the Target Precipitate

Reddish-brown precipitate = Fe(OH)₃ (iron(III) hydroxide).

Step 2: Analyze Fe₃O₄ Composition

Fe₃O₄ = FeO·Fe₂O₃ = contains both Fe²⁺ and Fe³⁺ ions.

Step 3: Verify Option (a)

Step 4: Why "After a Period" is Key

Step 5: Why Other Options Fail

(b) Oxidation first would work, but the question asks for the simplest correct sequence. Option (a) achieves the same result more directly by allowing air oxidation during the waiting period.

(c) Reduction would convert Fe³⁺ to Fe²⁺, making it harder to get Fe(OH)₃.

(d) Dilute HCl produces FeCl₂ and FeCl₃, but without the waiting period, Fe(OH)₂ (green) would also form.

MgCl₂ + H₂SO₄ → MgSO₄ + 2HCl

MgSO₄ + Na₂CO₃ → MgCO₃↓ (M = 84 g/mol) + Na₂SO₄

n(MgCO₃) = 7/84 = 0.08333 mol = n(MgCl₂)

m(MgCl₂) = 0.08333 × 95 = 7.916 g

% = (7.916/10) × 100 = 79.16%

Mass of FeCl_x = 5.34 × 0.6008 = 3.208 g

Mass of 6H₂O lost = 5.34 − 3.208 = 2.132 g → n(H₂O) = 2.132/18 = 0.11844 mol

n(FeCl_x) = 0.11844/6 = 0.01974 mol

M(FeCl_x) = 3.208/0.01974 = 162.5 g/mol

56 + 35.5x = 162.5 → 35.5x = 106.5 → x = 3

[NO] = 0.15 M, [O₂] = 0.15 M, [NO₂] = 0.25 M

Q = [NO₂] / ([NO][O₂]^½) = 0.25 / (0.15 × √0.15) = 0.25 / 0.0581 ≈ 4.30

Q (4.30) < K_c (6.33) → reaction proceeds forward to increase [NO₂] and reach equilibrium.

Step 1: Set Up the Ionisation Equation

HCN ⇌ H⁺ + CN⁻

Step 2: Calculate x (Ionised Concentration)

x = 0.2 − 0.167 = 0.033 M

[H⁺] = [CN⁻] = 0.033 M

Step 3: Calculate K_a

K_a = [H⁺][CN⁻] / [HCN]

K_a = (0.033)(0.033) / 0.167

K_a = 0.001089 / 0.167

K_a ≈ 5.45 × 10⁻³

Step 4: Verification

If K_a = 5.45 × 10⁻³:

x² / (0.2 − x) = 5.45 × 10⁻³

x² = 5.45 × 10⁻³ × 0.167 = 9.10 × 10⁻⁴

x = √(9.10 × 10⁻⁴) ≈ 0.0302 M ≈ 0.033 M ✓

Step 1: Identify the Original Cell

Higher oxidation potential = more tendency to oxidize = anode.

E_cell = E_ox(anode) − E_ox(cathode) = 0.76 − (−0.34) = 1.10 V

Electron flow: X → Y (external circuit)

Step 2: Test Option (c) — Replace Cathode Y with W

W has E_ox = +2.37 V (very high oxidation potential).

Step 3: Verify Both Conditions

Condition 1: Increased EMF

New E_cell = E_ox(W) − E_ox(X) = 2.37 − 0.76 = 1.61 V

1.61 V > 1.10 V ✔ EMF increased

Condition 2: Reversed Current Direction

Original: electrons flow X → Y (X is anode)

New: electrons flow W → X (W is anode)

The anode has changed from X to W, so the current direction in the external circuit reverses ✔

Step 4: Why Other Options Fail

n(NaCl) = 5.85/58.5 = 0.1 mol; requires 0.1 mol e⁻ = 0.1 F for complete decomposition.

Cathode: Na⁺ + e⁻ → Na → m(Na) = 0.1 × 23 = 2.3 g

Anode: 2Cl⁻ → Cl₂ + 2e⁻ → n(Cl₂) = 0.05 mol → m(Cl₂) = 0.05 × 71 = 3.55 g

0.1 F is exactly enough → complete decomposition.

1-chloropropane + NaOH(aq)/Δ → propan-1-ol (primary alcohol), NOT isopropyl alcohol. Isopropyl alcohol comes from 2-chloropropane. So option (a) is incorrect.

Ethanol (oxidation) → Acetic acid (CH₃COOH)

Benzoic acid (reduction) → Benzyl alcohol (C₆H₅CH₂OH)

Esterification: CH₃COOH + C₆H₅CH₂OH → C₆H₅CH₂–O–CO–CH₃ (benzyl acetate)

Coal mine gas = Methane (CH₄)

CH₄ + Cl₂ (UV) → CH₃Cl (Halogenation)

C₆H₆ + CH₃Cl (AlCl₃) → Toluene (Friedel-Crafts alkylation)

Toluene + 3 HNO₃/H₂SO₄ → 2,4,6-trinitrotoluene (TNT)

Step 1: Analyze the Scheme

The central compound is C_nH_2n (alkene, one double bond).

Step 2: Identify Process (1)

C_nH_2n + H₂O → C_nH_2n+2O

This is catalytic hydration (addition of water across the double bond using acid catalyst).

Product: Monohydroxy alcohol (one –OH group).

Step 3: Identify Process (2)

C_nH_2n + [O] + H₂O → C_nH_2n+2O₂

This is oxidation with cold, dilute KMnO₄ (Baeyer's reagent).

Step 4: Match with Options

Answer: (b) (2): Oxidation → dihydroxy alcohol.

Step 1: Identify the Target Compound

C₇H₆O₂: Degree of unsaturation = (2×7 + 2 − 6)/2 = 5

This corresponds to benzoic acid (C₆H₅COOH): benzene ring (4) + carboxyl group (1) = 5 ✓

Step 2: Identify Coal Tar Product

Fractional distillation of coal tar yields: benzene, toluene, xylene, phenol, naphthalene, etc.

The starting material for this synthesis is benzaldehyde (C₆H₅CHO) or a related compound from coal tar.

Step 3: Verify Sequence (b): Reduction → Alkylation → Oxidation

Step 4: Alternative Interpretation

Starting from phenol (C₆H₅OH) from coal tar:

Step 5: Why Other Options Fail

Answer: (b) Reduction → Alkylation → Oxidation.

(a) Chemical formulas of C and D:

• (A) = Fe₂O₃ (red oxide of iron)
• (B) = Fe (iron metal, from reduction of Fe₂O₃ at 800°C)
• (C) = FeCl₃ (iron reacts with X = Cl₂ gas/Δ → FeCl₃)
• (D) = Fe₂(SO₄)₃ (Fe₂O₃ + conc. H₂SO₄ → Fe₂(SO₄)₃ + H₂O)
• (Z) = NaOH (added to D to give Fe(OH)₃)

(b) Intermetallic alloy for (B) = Fe: Steel (Fe + C) or Stainless steel (Fe + Cr + Ni)

(c) Name of (C) when X_(g) replaced by dilute HCl_(aq):
Fe + 2HCl_(dilute) → FeCl₂ + H₂↑ → Compound = Iron(II) chloride (Ferrous chloride)

Identification:

• (Y) = Simplest tertiary alcohol = 2-methylpropan-2-ol [(CH₃)₃COH]
• (X) = Alkene that gives Y by catalytic hydration = 2-methylpropene [(CH₃)₂C=CH₂]
• (W) = HCl (hydrogen chloride)
• (Z) = Product = 2-chloro-2-methylpropane [(CH₃)₃CCl]

IUPAC Names:

• X: 2-methylpropene (isobutylene)
• Z: 2-chloro-2-methylpropane (tert-butyl chloride)

Structural formulas of saturated isomers of X (C₄H₈):