Step 1: Set Up the Ratio

From the conditions:

• C = 2B → B = C/2

• C = 3A → A = C/3

Therefore: A : B : C = 2 : 3 : 6

Step 2: Find Three Consecutive Elements Matching Ratio

Ratio: 2 : 3 : 6 ✓ — perfect match!

Step 3: Match Uses

• A = Ti (Titanium) → used in manufacturing artificial joints (biocompatible, strong, light)

• B = V (Vanadium) → vanadium steel used in car spring manufacturing

• C = Cr (Chromium) → used in metal plating (chromium plating gives shiny corrosion-resistant surface)

Step 1: Determine n

Second transition series → n = 5

Configuration: 5s^(5-3) = 5s²; 4d^(2×5) would be 4d¹⁰ (maximum filling of d-sublevel)

Step 2: Identify Element

Configuration: [Kr] 5s² 4d¹⁰ → 48 electrons → Cadmium (Cd, Z = 48)

Step 3: Classification

Since the d-sublevel is completely filled in both the atom (4d¹⁰) and its common Cd²⁺ ion (4d¹⁰), Cd is classified as a non-transition element.

Step 4: Use — Ni-Cd Rechargeable Batteries

Cadmium is famously used in Nickel-Cadmium (Ni-Cd) rechargeable batteries.

Why not (d)? Although CdO is used in rubber/paints, the pairing with Ni in rechargeable batteries is the more characteristic use highlighted in this question context.

Step 1: Identify Element X

X³⁺ has d-sublevel with 2 unpaired electrons → 3d² configuration

Working backward: X atom = X³⁺ + 3 electrons → [Ar] 4s² 3d³

This is Vanadium (V, Z = 23)

Step 2: Identify Oxidation State That Breaks Filled Level

Outer electrons available without breaking [Ar] core:

• 4s² + 3d³ = 5 electrons → maximum normal oxidation = V⁵⁺

To form V⁶⁺, the 6th electron must come from the previous principal level — specifically from the completely filled 3p⁶ in [Ar].

Therefore, V⁶⁺ (X⁶⁺) is the oxidation state produced by breaking down a completely filled principal energy level.

Process X: Crushing

Crushing reduces particle size while conserving mass (purely mechanical). Sintering, in contrast, fuses small particles into larger lumps.

Process Y: Siderite Roasting

Siderite (FeCO₃) contains Fe²⁺. Roasting converts it to Fe₂O₃ containing Fe³⁺.

Roasting siderite increases unpaired electrons from 4 → 5. Hematite already contains Fe³⁺, so roasting wouldn't increase unpaired e⁻.

Step 1: Identify Element X

Filled orbitals: 11 → 22 paired electrons

Half-filled orbitals: 3 → 3 unpaired electrons

Total electrons in X²⁺ = 22 + 3 = 25 electrons

X has 25 + 2 = 27 electrons → Cobalt (Co, Z = 27)

Step 2: Evaluate Each Statement (Element following Co = Ni)

Step 3: The Famous Co-Ni Atomic Mass Anomaly

Although Ni follows Co in atomic number (Z), the atomic mass of Co (58.93) is GREATER than Ni (58.69). This is one of the periodic table's classic anomalies, similar to Ar-K and Te-I.

Statement (a) claims "Co's atomic mass is less than Ni's" — this is FALSE ✓

Other Statements (All True)

(b) ✓ Density of Co (8.90) < Ni (8.91)

(c) ✓ Co compounds containing unpaired d-electrons are paramagnetic (general statement holds for typical Co compounds)

(d) ✓ Effective nuclear charge increases across the period: Co > Fe

Step 1: Identify Salts X and Y

Same cation, but X is soluble in water and Y is not. Both react with HCl to release the same gas Z.

This describes bicarbonate (X) and carbonate (Y) salts where Z = CO₂.

Example: X = Ca(HCO₃)₂ (soluble), Y = CaCO₃ (insoluble in water but dissolves in acids)

Step 2: Test Each Option

Option (b): Heating bicarbonate produces CO₂ and carbonate:

This perfectly matches option (b).

Analysis of Each Salt

Salt A: CaCl₂

CaCl₂ + H₂SO₄ → CaSO₄ + 2HCl↑

HCl is a colorless gas, difficult to oxidize (Cl⁻ is the weakest reducing halide ion) ✓

Salt B: MgI₂

Conc. H₂SO₄ oxidizes HI to I₂ vapors (purple-violet colored vapors):

MgI₂ + H₂SO₄ → MgSO₄ + 2HI

8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O

The resulting iodine solution (I₂ water) can be used to detect first-group anions (like sulfide, sulfite ions) ✓

Salt C: Ag₃PO₄

Ag₃PO₄ + H₂SO₄ → Ag₂SO₄ + H₃PO₄

H₃PO₄ is a non-volatile acid, so no gas is released ✓

Why Not Other Options?

• Option (a) PbSO₄ is incorrect for C? Actually PbSO₄ is insoluble but the question's "no gas" works. However, option (b) is the model answer favored due to the specific behavior of MgI₂ producing detectable iodine vapors.

Concept

For equal masses and equal volumes: M = (mass/M.W.)/V

Higher molar mass → fewer moles → lower concentration

Molar Masses

Ca(OH)₂ has the highest molar mass → lowest concentration.

Solubility Analysis with H₂SO₄

Why Other Options Fail

• Na₂CO₃: precipitates ALL three as carbonates

• NaNO₃: all nitrates are soluble, no separation

• HCl: only Pb²⁺ removed as PbCl₂

Conditions for Law of Mass Action

A reaction must be:

1. Reversible

2. In a closed system

3. Reaches dynamic equilibrium

Analysis

(a) Neutralization → essentially complete (irreversible)

(b) Open container → O₂ escapes, no equilibrium

(c) Precipitation → irreversible (CaCO₃ insoluble)

(d) Reversible decomposition in closed container → reaches equilibrium ✓

Le Chatelier's Principle Analysis

Decreasing volume increases pressure, shifting equilibrium backward (toward fewer gas molecules), reducing C.

Effect of Catalyst at Equilibrium

A catalyst increases both forward and backward rates equally.

As one rate increases, the other increases proportionally → linear relationship with positive slope passing through origin.

The equilibrium position is unchanged, but both rates become higher simultaneously.

Step 1: Find [H⁺]

[H⁺] = 10⁻ᵖᴴ = 10⁻⁴·⁷ ≈ 2 × 10⁻⁵ M

Step 2: Apply Ka Expression for Weak Acid

Ka = [H⁺]² / C

C = [H⁺]² / Ka = (2 × 10⁻⁵)² / (6.2 × 10⁻¹⁰)

C = 4 × 10⁻¹⁰ / 6.2 × 10⁻¹⁰ ≈ 0.645 M

Step 3: Calculate Moles in 300 mL

n = M × V = 0.645 × 0.300 = 0.193 ≈ 0.19 mol ✓

Step 1: Initial Equilibrium

Stoichiometry: 1 N₂H₄ → 1 N₂ + 2 H₂

If [H₂] = 0.2 M → [N₂] = 0.1 M (from 2:1 ratio)

[N₂H₄] remaining = 0.2 − 0.1 = 0.1 M

Step 2: Effect of Temperature Increase

ΔH < 0 → forward reaction is exothermic

Increasing T shifts equilibrium backward (Le Chatelier)

[N₂] must decrease below 0.1 M

Step 3: Choose Answer

Only option < 0.1 M is 0.08 M ✓

Effect of Heating on Water

Water self-ionization: H₂O ⇌ H⁺ + OH⁻ (endothermic)

Raising T → more ionization → both [H⁺] and [OH⁻] increase equally

Key Properties

• Since [H⁺] = [OH⁻], water remains neutral on litmus

• [OH⁻] increases → pOH = −log[OH⁻] decreases ✓

• [H⁺] increases → pH decreases (NOT increases — eliminates options a, d)

• Water is not alkaline (eliminates b)

Electrolysis Rules

• Anode (+): Oxidation occurs; anions discharge

• Cathode (−): Reduction occurs; cations discharge

• Pt electrodes are inert (don't participate)

For CuCl₂ Solution

Zinc Half-Cell Equilibrium

Zn(s) ⇌ Zn²⁺(aq) + 2e⁻

Zinc has high oxidation tendency:

• Zn atoms leave the rod as Zn²⁺ ions → solution contains positive Zn²⁺ ions

• Electrons remain on the rod → rod becomes negatively charged

Step 1: Determine the State of Each Battery

The density of the electrolyte (H₂SO₄) indicates the battery's charge state:

Battery Y (charged) acts as the energy source and charges battery X (discharged).

Step 2: Battery X — Electrolytic Cell (Charging Process)

During charging, reactions are reversed from discharge:

At − electrode (cathode in electrolytic cell):

Pb²⁺ + 2e⁻ → Pb (reduction) ✓

At + electrode (anode in electrolytic cell):

Pb²⁺ → Pb⁴⁺ + 2e⁻ (oxidation, forms PbO₂)

Step 3: Battery Y — Galvanic Cell (Discharging Process)

At + electrode (cathode in galvanic cell):

Pb⁴⁺ + 2e⁻ → Pb²⁺ (reduction) ✓

At − electrode (anode in galvanic cell):

Pb → Pb²⁺ + 2e⁻ (oxidation)

Conclusion

Option (d) correctly describes:

• X: Pb²⁺ → Pb at − electrode (reduction during charging) ✓

• Y: Pb⁴⁺ → Pb²⁺ at + electrode (reduction during discharge) ✓

Faraday's Law: mass ∝ M/n (same charge)

Ratio = 65/32.5 = 2 : 1 ✓ (matches condition)

Both deposit metal at the cathode (X = Au, Y = Zn).

Identifying Electrodes

Iron has higher oxidation potential (+0.409 V) → Fe is anode (oxidized)

Tin: Sn²⁺ + 2e⁻ → Sn (reduction) → Sn is cathode

EMF Calculation

EMF Calculation

EMF = E°(oxidation, anode) + E°(reduction, cathode)

EMF = 0.23 + 0.80 = +1.03 V

Type of Cell

Positive EMF → reaction is spontaneous → Galvanic cell ✓

Reaction 1

CH₂Br-CF₃ + Cl₂ → CHClBr-CF₃ (Halothane)

Halothane is a hydrocarbon derivative (contains halogens) used as a safe anesthetic ✓

Reaction 2

C₂H₂ + H₂O → CH₃CHO (acetaldehyde, Y) — stable tautomer

Oxidation: CH₃CHO → CH₃COOH (acetic acid, Z)

Step 1: Alkylation Products of Toluene

Methyl group (-CH₃) in toluene is ortho/para directing. So methylation produces two main isomers:

• 1,2-Dimethyl benzene (ortho-xylene)

• 1,4-Dimethyl benzene (para-xylene)

Step 2: Identify Isomer X (Used in Artificial Heart Valves)

Artificial heart valves are made from PET (polyethylene terephthalate), a polymer derived from terephthalic acid.

Terephthalic acid is produced by oxidation of para-xylene:

Therefore: X = para-xylene (1,4-dimethyl benzene)

Step 3: Identify Isomer Y

The other isomer is Y = ortho-xylene (1,2-dimethyl benzene)

Oxidation of ortho-xylene produces phthalic acid (1,2-benzene dicarboxylic acid):

Hence statement (c) is correct: oxidation of Y produces phthalic acid ✓

Why Not (d)?

X gives terephthalic acid (not phthalic acid). Phthalic acid is derived from Y, not X.

Dry Distillation Rule

R-COONa + NaOH —(soda lime)→ R-H + Na₂CO₃

The carboxyl carbon is removed; resulting alkane has one fewer carbon.

Analysis of Each Option

Only isobutane is both branched and gaseous (alkanes with ≤4 carbons are gases).

Catalytic Hydration Reaction

3-methyl-1-butyne: (CH₃)₂CH-C≡CH

Hydration (Markovnikov, with Hg²⁺ catalyst):

(CH₃)₂CH-C≡CH + H₂O → (CH₃)₂CH-CO-CH₃ (3-methyl-2-butanone, a ketone)

General Formula

Ketones have general formula CₙH₂ₙO

Reaction with KMnO₄

Ketones do not decolorize acidified KMnO₄ — they resist oxidation (no α-H bonded to C=O that can be easily oxidized in this conditions).

Step 1: Dry Distillation

C₆H₄(C₂H₅)(COONa) + NaOH → C₆H₅-C₂H₅ (ethylbenzene, X) + Na₂CO₃

Step 2: Chlorination with FeCl₃

Ethyl group (-C₂H₅) is an alkyl group → ortho/para directing

Product: mixture of 2-chloro ethylbenzene (ortho) and 4-chloro ethylbenzene (para)

Identify X

Aromatic liquid with bp 80°C → Benzene (C₆H₆)

Reactions

Y: C₆H₆ + 3Cl₂ —UV→ C₆H₆Cl₆ (benzene hexachloride / lindane)

This is an aliphatic derivative (saturated, no aromatic ring) ✓

Z: C₆H₆ + Cl₂ —FeCl₃→ C₆H₅Cl (chlorobenzene)

This is an aromatic derivative (not hydrocarbon — contains Cl).

Identifying the Acids

X: Salt = NaOOC-COONa

→ Acid = HOOC-COOH = Oxalic acid ✓

Y: Salt = C₆H₄(COONa)₂

→ Acid = C₆H₄(COOH)₂ = Phthalic acid ✓

Both are dicarboxylic acids as stated in the problem.

Structural Analysis

Structure: HOOC-C(CH₃)(C₂H₅)-CH₂-C₂H₅ Wait, looking again: HOOC-C(CH₃)(C₂H₅)-CH(C₂H₅)...

From figure: COOH at top, central C has CH₃ and C₂H₅ branches, CH₂ branch leading to another C₂H₅

Longest chain containing COOH: COOH-C-CH₂-CH₂-CH₃ = 5 carbons → pentanoic acid

Substituent Naming

At C2: methyl and ethyl groups → alphabetical: ethyl first

Final name: 2-Ethyl-2-methyl pentanoic acid ✓

Step 1: Find the Simplest Alcohol Containing an Ethyl Branch

To have an ethyl group as a branch in the simplest possible alcohol, the main chain (containing -OH) should be as short as possible while still treating ethyl as a substituent.

The simplest alcohol satisfying this is 2-Ethyl-2-propanol:

Step 2: Derive Compound X (Alkyl Halide)

Alkaline hydrolysis replaces -Cl with -OH:

R-Cl + NaOH(aq) → R-OH + NaCl

To get 2-ethyl-2-propanol, the corresponding alkyl chloride is obtained by replacing -OH with -Cl:

Step 3: Why Not Option (b)?

1-Chloro-2-ethyl butane would give 2-ethyl-1-butanol — this contains 4 carbons in the main chain, which is NOT the simplest alcohol with an ethyl branch.

The simplest is 2-ethyl-2-propanol (only 3 carbons in main chain), so X = 2-Chloro-2-ethyl propane ✓

Reaction Sequence

Step 1 (dehydration): CH₃-CH₂-CH₂-OH → CH₃-CH=CH₂ (propene) + H₂O

Step 2 (hydration, Markovnikov): CH₃-CH=CH₂ + H₂O → CH₃-CHOH-CH₃

X = 2-propanol (isopropanol), a secondary alcohol

Properties of 2-Propanol

• 2° alcohol oxidizes (with KMnO₄) → propanone (acetone) ✓

• 2-propanol decolorizes KMnO₄ as it gets oxidized

Option (c) is correct.

Identifying Compounds

Evaluating Option (b)

Both Fe (A) and Fe (D) react with concentrated H₂SO₄:

Fe + conc. H₂SO₄ → mixture of FeSO₄ and Fe₂(SO₄)₃ ✓

Alloy Analysis

Step 1: Find Molarity of BaCl₂ Solution

n(BaCl₂·2H₂O) = 45 / 244 = 0.1844 mol

M = 0.1844 / 0.500 L = 0.369 M

Step 2: Moles BaCl₂ in 84.2 mL

n(BaCl₂) = 0.369 × 0.0842 = 0.0311 mol

Step 3: Moles of Na₂SO₄ (1:1 ratio with BaCl₂)

n(Na₂SO₄) = 0.0311 mol

Step 4: Find Molar Mass of Hydrated Salt

M(hydrate) = 10 g / 0.0311 mol = 322 g/mol

Step 5: Solve for X

322 = 142 + 18X

18X = 180 → X = 10 ✓

Formula: Na₂SO₄·10H₂O

Testing X₂Y₃

X₂Y₃ → 2X³⁺ + 3Y²⁻

If s = 1×10⁻⁴: [X³⁺] = 2s; [Y²⁻] = 3s

Ksp = (2s)² × (3s)³ = 4s² × 27s³ = 108 s⁵

= 108 × (10⁻⁴)⁵ = 108 × 10⁻²⁰ = 1.08 × 10⁻¹⁸ ✓

Verification

This matches the given Ksp exactly, confirming the formula is X₂Y₃.

Acid Strength Comparison

Y has higher Ka (5.1 × 10⁻⁴ > 1.8 × 10⁻⁵)

→ Y is the stronger acid → more ionization

Implications

• More ionization in Y → more ions → better electrical conductivity

• Higher [H⁺] in Y → lower pH

Cathodic (Sacrificial) Protection Principle

To protect iron from rusting, metal X must be more reactive (more active) than iron — i.e., higher oxidation potential.

Examples: Mg, Zn (more active than Fe)

Reasoning of Option (b)

"X reduces iron ions in solution" → X gives electrons to Fe²⁺/Fe³⁺ converting them back to Fe⁰. This is only possible if X is more active than iron (has higher tendency to lose electrons).

Direction of Electron Flow

Electrons flow from sacrificial metal X TO iron pipe.

According to the figure, this corresponds to direction (1) — from X (above) downward toward the iron pipe.

Why Not (d)?

Statement (d) correctly says X has higher oxidation potential, but incorrectly states electrons flow in direction (2). The arrow (2) points away from iron pipe — wrong direction. So (d) is invalid.

Why Not (c)?

Statement (c) says "X ions are reduced by iron atoms" — this is the opposite (it would protect X, not Fe). Wrong.

Step 1: Mass of Gold Deposited

Au³⁺ + 3e⁻ → Au

Moles Au = (moles e⁻) / 3 = 0.5 / 3 = 0.1667 mol

Mass = 0.1667 × 197 = 32.83 g ≈ 32.8 g ✓

Step 2: Concentration Change

In electroplating, the anode is the same metal (Au) as the coating.

At anode: Au → Au³⁺ + 3e⁻ (replenishes ions)

At cathode: Au³⁺ + 3e⁻ → Au (deposits on spoon)

Ions consumed = ions produced → concentration remains constant ✓

Bifunctional Nature of Glycolic Acid

Glycolic acid has both -OH (alcohol) and -COOH (acid) groups.

Experiment (A): With Methanol

Methanol is an alcohol → reacts with the -COOH group of glycolic acid (esterification)

HO-CH₂-COOH + CH₃OH → HO-CH₂-COOCH₃ + H₂O ✓

Experiment (B): With Acetic Acid

Acetic acid is acid → reacts with the -OH group of glycolic acid (esterification)

HO-CH₂-COOH + CH₃COOH → CH₃COO-CH₂-COOH + H₂O ✓

Identifying Compounds

B: C₂H₄O₂ = CH₃COOH (acetic acid)

C: C₄H₈O₂ — gives amide of B on ammonolysis → ester of acetic acid: CH₃COOC₂H₅ (ethyl acetate)

A: C₃H₆O₃ = CH₃-CHOH-COOH (lactic acid) — has both -OH and -COOH

Why Option (b) is FALSE

Statement says "A does not react with B" — but lactic acid (A) does react with acetic acid (B)!

The -OH of lactic acid can be esterified by the -COOH of acetic acid → ester formation.

Therefore, statement (b) is false (the exception).

Step-by-Step Analysis

Step 1: CH₃COOH + NaOH → CH₃COONa

CH₃COONa + NaOH —(soda lime/Δ)→ CH₄ (Y) + Na₂CO₃

Step 2: CH₄ + Cl₂ —400°C→ CH₃Cl (Z) + HCl

Step 3: Simplest aromatic = benzene (C₆H₆)

C₆H₆ + CH₃Cl —AlCl₃→ C₆H₅-CH₃ = Toluene (W)

Step 4: C₆H₅-CH₃ + [O]/V₂O₅ → C₆H₅-COOH = Benzoic acid (M)

Sequence of Compounds

Compound: Propyl propanoate (CH₃CH₂COOCH₂CH₂CH₃)

Step 1: Alkaline Hydrolysis

CH₃CH₂COOCH₂CH₂CH₃ + NaOH → CH₃CH₂COONa + CH₃CH₂CH₂OH (1-propanol)

Step 2: Dehydration of Propanol

CH₃CH₂CH₂OH —H₂SO₄→ CH₃-CH=CH₂ (propene) + H₂O

Step 3: Catalytic Hydrogenation

CH₃-CH=CH₂ + H₂ —Ni→ CH₃-CH₂-CH₃ (propane) ✓

Analysis

Salt X reacts with HCl to release gas Y. Gas Y passes into ZCl₂ solution → forms black precipitate W.

Black precipitate + ZCl₂ suggests metal sulfides (PbS, CuS are black).

So gas Y = H₂S, and salt X is a sulfide (e.g., Na₂S, FeS).

Answer (1):

(a) Salt ZCl₂: PbCl₂ (or CuCl₂)

(b) Black precipitate W: PbS (or CuS)

Answer (2): Two Other Nitrate Salts

• Pb(NO₃)₂ — Pb²⁺ + H₂S → PbS↓ (black)

• Cu(NO₃)₂ — Cu²⁺ + H₂S → CuS↓ (black)

Part (a): 1 mol compound + 2 mol Br₂ → saturated

Requires a triple bond: D: CH≡CH (acetylene)

CH≡CH + 2Br₂ → CHBr₂-CHBr₂

IUPAC name: 1,1,2,2-Tetrabromoethane

Part (b): HBr + compound (Markovnikov)

Using C: CH₃-CH=CH₂ (propene)

CH₃-CH=CH₂ + HBr → CH₃-CHBr-CH₃ (H adds to C with more H; Br to C with less H)

IUPAC name: 2-Bromopropane

Part (c): 2 mol compound + 2 mol Cl₂ → two mono-chloro isomers

Using F: Toluene with Cl₂/FeCl₃ (substitution on ring, ortho/para directing by -CH₃)

Products (two isomers):

• 2-Chloromethyl benzene (ortho)

• 4-Chloromethyl benzene (para)

Or alternatively named: 2-Chloro toluene and 4-Chloro toluene