• Option (b) represents the V⁵⁺ ion.
• Its electronic configuration is [₁₈Ar].
• Vanadium pentoxide (V₂O₅) contains this ion.
• It serves as a catalyst during sulfuric acid production.
• This process is called the contact method.

1. Analysis of the Correct Option (b)

• The Process: The contact method is the industrial process used to manufacture sulfuric acid (H₂SO₄).
• The Catalyst: Vanadium pentoxide (V₂O₅) is the standard catalyst used to speed up the oxidation of sulfur dioxide (SO₂) to sulfur trioxide (SO₃).
• Oxidation State: In V₂O₅, oxygen has a -2 charge. To balance it, Vanadium exists as a V⁵⁺ ion.
• Electron Configuration:
  • Neutral Vanadium (atomic number 23): [₁₈Ar] 4s² 3d³
  • To form V⁵⁺, it loses all 5 valence electrons (2 from 4s and 3 from 3d).
  • The resulting configuration matches the option exactly: [₁₈Ar].

  3. Detailed Rejection of Option (c)

  • The Context: Titanium is biocompatible and extensively used to manufacture artificial joints and bone screws.
  • The Error: It is used in its uncharged, pure metallic elemental form (or as an alloy), not as an ionic compound with an inert [₁₈Ar] noble gas electron shell.

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  4. Detailed Rejection of Option (d)

  • The Context: Vegetable oil hydrogenation requires a catalyst to convert unsaturated fats into saturated fats.
  • The Error: Finely divided Nickel (Ni) metal is the catalyst used for this process. It is used as a neutral element, not as a transition metal ion with a 3d⁶ state.

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2. Detailed Rejection of Option (a)

• The Context: Dry cells (zinc-carbon batteries) use Manganese dioxide (MnO₂). The Mn⁴⁺ ion has a [₁₈Ar] 3d³ configuration.
• The Error: Modern dry car batteries (lithium-ion type) rely on Lithium Cobalt Oxide (LiCoO₂) where Cobalt has a 3d⁶ core, making this statement contradictory and incorrect.

How to Analyze the Graphs

• The Rule: A sudden, massive jump in ionization energy indicates that a stable, completely filled inner electron shell (noble gas core) is being broken into.
• The Deduction: The number of ionization energies before the huge jump equals the number of valence electrons the element possesses.

1. Detailed Analysis of Element X

• Ionization Values: 1^st (578) → 2^nd (1811) → 3^rd (2745) → 4^th (11540)
• The Jump: The energy drastically increases by more than 4 times between the 3^rd and 4^th ionization steps.
• Conclusion: Element X contains exactly 3 valence electrons. Removing the 4^th electron requires breaking its stable octet.
• Chemical Identity: Element X belongs to Group 3A (13) or Group 3B (3). The exact numbers correspond to Aluminium (₁₃Al) or Scandium (₂₁Sc).

2. Detailed Analysis of Element Y

• Ionization Values: 1^st (648) → 2^nd (1364) → 3^rd (2858) → 4^th (4643) → 5^th (6523) → 6^th (12360)
• The Jump: A significant energy surge occurs at the 6^th ionization energy.
• Conclusion: Element Y contains exactly 5 valence electrons. Removing the 6^th electron breaks a stable inner shell.
• Chemical Identity: Element Y belongs to Group 5B (5). These exact values identify it as the transition metal Vanadium (₂₃V).

1. Identifying Element X (Cobalt)

• The Clue: The question states that element X is preceded and followed by elements with similar properties within the same group block.
• Group VIII horizontal trend: In the first transition series, Group VIII contains Iron (₂₆Fe), Cobalt (₂₇Co), and Nickel (₂₈Ni). In this group, the horizontal similarity across the period is exceptionally strong.
• The Middle Element: Since Cobalt (Co) is the middle element—preceded by Iron and followed by Nickel—element X is strictly identified as Cobalt (₂₇Co).

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2. Analysis of the Correct Option (c)

• The Reaction: X³⁺ → X⁴⁺ represents the oxidation of Cobalt(III) to Cobalt(IV).
• Electronic Configuration of Co: [₁₈Ar] 4s² 3d⁷
• Electronic Configuration of Co³⁺: [₁₈Ar] 3d⁶
• Electronic Configuration of Co⁴⁺: [₁₈Ar] 3d⁵
• Why it occurs easily: The 3d⁵ subshell configuration is exactly half-filled. In transition metal chemistry, half-filled d-orbitals are extraordinarily stable due to symmetrical electron distribution and maximum exchange energy. Therefore, Co³⁺ easily loses an electron to reach this highly stable 3d⁵ state.

The correct option is (b).

1. Step 1: Thermal Decomposition of Iron(II) Sulphate

• When Iron(II) sulphate (FeSO₄) is heated, it undergoes thermal decomposition.
• The chemical equation for this reaction is:
  
  
  2FeSO₄(s) →^Δ Fe₂O₃(s) + SO₂(g) + SO₃(g)
  
  
• Result: This step produces Iron(III) oxide (Fe₂O₃), a red solid.

2. Step 2: Reaction with Concentrated Sulphuric Acid

• Iron(III) oxide (Fe₂O₃) is a stable base that only reacts with hot, concentrated acids. It does not react with dilute acids.
• Adding hot, concentrated sulphuric acid (H₂SO₄) to the iron(III) oxide yields the desired product:
  
  
  Fe₂O₃(s) + 3H₂SO₄(conc.) →^Δ Fe₂(SO₄)₃(aq) + 3H₂O(l)
  
  
• Result: This step successfully prepares Iron(III) sulphate (Fe₂(SO₄)₃).

3. Why the Other Options Are Incorrect

• (a): Dilute sulphuric acid cannot react with Iron(III) oxide (Fe₂O₃).
• (c): Adding sodium hydroxide (NaOH) to FeSO₄ forms Iron(II) hydroxide, which would give Iron(II) sulphate upon adding dilute acid, not Iron(III) sulphate.
• (d): Adding sodium hydroxide at the final stage would produce an insoluble iron hydroxide precipitate instead of a soluble sulphate salt.

Detailed Step-by-Step Analysis

1. Understanding the Data

• Before the process: The percentage of iron in the ore is 45%.
• After the process: The mass of the ore decreases to 3.85 Kg, and the percentage of iron increases significantly to 69%.

2. Identifying the Process (Roasting)

Roasting (b) is a chemical process that involves heating the ore strongly in the air.
• The main goals of roasting are to dry the ore, remove moisture, expel volatile impurities as gases, and raise the percentage of iron in the ore to 69% by converting it into Hematite (Fe₂O₃).
• The exact target value of 69% matches the textbook standard for iron ore after the roasting step:


2FeCO₃ (Siderite: 48.5%) + ½O₂ →^Δ Fe₂O₃ (Hematite: 69%) + 2CO₂(g)


2Fe₂O₃·3H₂O (Limonite: 40%) →^Δ 2Fe₂O₃ (Hematite: 69%) + 3H₂O(g)

3. Why the Other Options Are Incorrect

• (a) Crushing & (c) Sintering: These are purely physical processes aimed at changing the particle size of the ore. They do not alter the chemical percentage of iron or cause volatile mass loss to this extent.
• (d) Reduction: Reduction occurs inside the Blast Furnace or Midrex Furnace to produce pure iron metal, not to dress or prepare the ore.

Detailed Step-by-Step Analysis

1. Identifying Salt X (Carbonate or Bicarbonate)

• Observation: A gas evolves that does not change the color of acidified potassium permanganate (KMnO₄).
• Chemical Deduction: Acidified KMnO₄ is a purple oxidizing agent. Gases like SO₂ or H₂S are reducing agents and would decolorize it. However, Carbon dioxide (CO₂) cannot be oxidized further and does not affect KMnO₄.
• Anion Identity: Salt X contains either Carbonate (CO₃^2-) or Bicarbonate (HCO₃^-).
• The Reaction Box:


CO₃^2-(s) + 2H⁺(aq) → H₂O(l) + CO₂(g) ↑



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2. Identifying Salt Y (Thiosulphate)

• Observation: A suspension forms in the solution and the evolved gas affects (decolorizes) the acidified potassium permanganate solution.
• Chemical Deduction: The formation of a yellow suspension of elemental sulfur (S) alongside the evolution of sulfur dioxide gas (SO₂) is the unique signature test for the thiosulphate anion. The SO₂ gas reduces the purple KMnO₄ to a colorless solution.
• Anion Identity: Salt Y contains Thiosulphate (S₂O₃^2-).
• The Reaction Box:


S₂O₃^2-(s) + 2H⁺(aq) → S(s) ↓ (yellow) + SO₂(g) ↑ + H₂O(l)



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3. Identifying Salt Z (Stable Acid Anion)

• Observation: No reaction occurs (no color change, no gas evolved).
• Chemical Deduction: Dilute hydrochloric acid (HCl) can only displace anions derived from less stable acids. A lack of reaction means the anion belongs to a group derived from more stable acids that HCl cannot displace.
• Anion Identity: Salt Z contains an anion from the conc. H₂SO₄ group (Cl^-, Br^-, I^-, NO₃^-) or the BaCl₂ group (SO₄^2-, PO₄^3-).

Detailed Step-by-Step Analysis

1. Identifying Solution W (Sodium sulphide)

• Observation: A black precipitate forms when lead(II) nitrate is added.
• Chemical Deduction: Lead ions (Pb²⁺) react immediately with sulphide ions (S^2-) to form an insoluble black precipitate of lead(II) sulphide (PbS).
• Anion Identity: Solution W must contain Sulphide (S^2-), narrowing our choices down to options B and C.
• The Reaction Box:


Pb²⁺(aq) + S^2-(aq) → PbS(s) ↓ (Black)

2. Identifying Solution Z (Sodium thiosulphate)

• Observation: It decolorizes the distinctive brown color of the iodine solution.
• Chemical Deduction: Iodine solution (I₂) is widely used as a standard volumetric reagent to detect thiosulphate (S₂O₃^2-). Thiosulphate reduces brown iodine molecules into colorless iodide ions (I^-).
• Anion Identity: Solution Z must be Sodium thiosulphate (Na₂S₂O₃), confirming option B as the only correct match.
• The Reaction Box:


2Na₂S₂O₃(aq) + I₂(aq) (Brown) → Na₂S₄O₆(aq) + 2NaI(aq) (Colorless)

3. Verifying Solution X (Ammonium carbonate)

• Observation: A precipitate forms when calcium hydroxide solution is introduced.
• Chemical Deduction: Calcium ions (Ca²⁺) from the hydroxide link up with carbonate ions (CO₃^2-) from the ammonium carbonate to yield a cloudy white precipitate of calcium carbonate (CaCO₃).
• The Reaction Box:


Ca²⁺(aq) + CO₃^2-(aq) → CaCO₃(s) ↓ (White)

4. Verifying Solution Y (Potassium bicarbonate)

• Observation: No precipitate forms upon the initial addition of calcium hydroxide solution.
• Chemical Deduction: All bicarbonate salts (HCO₃^-) are completely soluble in water, meaning no immediate precipitation occurs compared to the direct mixing of a carbonate salt with calcium.

1. Understanding the Two Gases

• Nitric oxide gas (NO): A colorless gas that is easily oxidized because nitrogen is in a +2 oxidation state.
• Hydrogen chloride gas (HCl): A colorless, acidic gas that dissolves readily in water to form hydrochloric acid.

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2. Why Option (a) Cannot Distinguish Between Them

• When passing either NO or HCl gas into a Sodium chloride solution (NaCl), no visible chemical reaction occurs.
• HCl gas will simply dissolve into the aqueous solution without producing any precipitate, color change, or distinct gas evolution.
• Therefore, it is completely ineffective for distinguishing between them.

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3. How the Other Options Successfully Distinguish the Gases

• (b) Exposing each separately to air:
  • NO reacts immediately with oxygen at the mouth of the tube to form reddish-brown fumes of nitrogen dioxide (NO₂).
  • HCl does not form reddish-brown fumes.


2NO(g) + O₂(g) → 2NO₂(g) (Reddish-brown)


• (c) Glass rod wet with ammonia:
  • HCl gas reacts instantly with ammonia gas (NH₃) to form dense white fumes of ammonium chloride (NH₄Cl).
  • NO does not react to form white fumes.


HCl(g) + NH₃(g) → NH₄Cl(s) (Dense White Fumes)


• (d) Acidified potassium permanganate solution:
  • Since NO acts as a reducing agent, it reduces the purple permanganate ions (MnO₄^-) and decolorizes the solution.
  • HCl gas does not show this decolorization effect under standard testing conditions.

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Detailed Step-by-Step Analysis

1. Identifying Acid (1) - Sulphuric Acid

• Observation: Orange vapors evolve when hydrogen bromide gas (HBr) is passed into hot concentrated acid.
• Chemical Deduction: Hot concentrated Sulphuric acid (H₂SO₄) acts as a strong oxidizing agent. It oxidizes colorless hydrogen bromide gas into orange-red bromine vapors (Br₂).
• The Reaction Box:


2HBr(g) + H₂SO₄(conc.) →^Δ SO₂(g) + 2H₂O(l) + Br₂(g) (Orange Vapors)



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2. Identifying Acid (2) - Hydrochloric Acid

• Observation: Dense white clouds form when a glass rod wet with ammonia solution is exposed to the acid vapors.
• Chemical Deduction: Volatile Hydrochloric acid (HCl) releases hydrogen chloride gas, which instantly combines with ammonia gas (NH₃) to form solid ammonium chloride particles suspended as white clouds.
• The Reaction Box:


HCl(g) + NH₃(g) → NH₄Cl(s) (White Clouds)



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3. Identifying Acid (3) - Phosphoric Acid

• Observation: A white precipitate forms upon adding barium nitrate solution.
• Chemical Deduction: Barium nitrate (Ba(NO₃)₂) reacts with phosphate ions from Phosphoric acid (H₃PO₄) to yield an insoluble white precipitate of barium phosphate (Ba₃(PO₄)₂). Nitric acid (Option d) would not react with barium nitrate.
• The Reaction Box:


2H₃PO₄(aq) + 3Ba(NO₃)₂(aq) → Ba₃(PO₄)₂(s) ↓ (White) + 6HNO₃(aq)

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Detailed Step-by-Step Analysis

1. Identifying the Secret Reagent

• The Clue: The question asks for the reagent used in the confirmatory test for one of the cations in the fifth analytical group.
• Fifth Group Cation: The main cation studied in this group is Calcium (Ca²⁺).
• The Confirmatory Test: The confirmatory test for Ca²⁺ involves adding dilute sulphuric acid (H₂SO₄), which forms a distinctive white precipitate of calcium sulphate (CaSO₄).
• Conclusion: The secret reagent we must use to test the salts is dilute H₂SO₄.

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2. Why Option (c) is the Correct Choice

• Anion Detection: Dilute H₂SO₄ is a strong, stable acid that can easily displace less stable acid anions, acting exactly like the dilute HCl group reagent. Therefore, it can detect the Nitrite (NO₂^-) anion in Pb(NO₂)₂ by evolving nitrous fumes.
• Cation Detection: Simultaneously, the sulphate ions (SO₄^2-) react with Lead(II) ions (Pb²⁺) to form a highly insoluble white precipitate. This dual action makes it a perfect detection test for the salt.
• The Combined Reaction Box:


Pb(NO₂)₂(aq) + H₂SO₄(dil.) → PbSO₄(s) ↓ (White) + 2HNO₂(aq)


• The produced nitrous acid (HNO₂) spontaneously decomposes to release nitric oxide gas (NO), which turns into reddish-brown NO₂ gas fumes at the mouth of the tube:


2NO(g) + O₂(g) → 2NO₂(g) ↑ (Reddish-Brown)



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3. Why the Other Options Are Incorrect

• (a) MgCl₂: Dilute acids cannot displace the chloride (Cl^-) ion; concentrated H₂SO₄ with heating is mandatory.
• (b) Fe₂(SO3)₃: This salt is chemically impossible/unstable in reality because Iron(III) (Fe³⁺) is an oxidizing agent that instantly reacts with the reducing sulphite (SO₃^2-) anion via an internal redox pathway.
• (d) Ba₃(PO4)₂: Barium phosphate is a highly insoluble salt that does not dissolve or react with dilute acids.

• (a) N₂ + 3H₂ ⇌ 2NH₃ — Reversible (Haber process). ✓
• (b) CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O — Reversible esterification. ✓
• (c) MgSO₄ + Na₂CO₃ → MgCO₃↓ + Na₂SO₄ — This is an irreversible precipitation reaction (MgCO₃ is insoluble). ✗ Not reversible!
• (d) FeCl₃ + 3NH₄SCN ⇌ Fe(SCN)₃ + 3NH₄Cl — Reversible equilibrium (blood-red colour). ✓

• Both tubes are at the same temperature (50°C) with equal masses of zinc.
• Tube A has HCl at 1M (higher concentration) while Tube B has HCl at 0.5M (lower concentration).
• Higher concentration → higher reaction rate → zinc disappears faster in tube A. ✓
• Upon complete dissolution of equal masses of zinc, the same amount of H₂ is produced in both tubes (since it depends on the mass of zinc, not the acid concentration, assuming excess acid).

• For endothermic reactions: Heat acts as a reactant → Raising temperature shifts equilibrium forward → Kc increases.
• Conversely, lowering temperature shifts equilibrium backward → Kc decreases. ✓ (option c)
• For exothermic reactions: Raising temperature shifts equilibrium backward → Kc decreases. Lowering temperature shifts forward → Kc increases.

• HCl, NaOH, HNO₃ are all strong electrolytes → they fully dissociate (α ≈ 1). ✓
• Boric acid (H₃BO₃) is a very weak acid → it barely dissociates in water (α ≪ 1). ✗

• NaOH neutralizes H⁺ ions → removes H⁺ from the equilibrium.
• By Le Chatelier's principle, equilibrium shifts forward → more CH₃COOH dissociates → dissociation increases.
• The removal of H⁺ (and addition of OH⁻) makes the solution less acidic → pH increases.

K_c = [NO]² × [Cl₂] / [NOCl]²

Let V = volume. Concentrations:

• [NO] = 1.5/V
• [Cl₂] = 3/V
• [NOCl] = 3/V

K_c = (1.5/V)² × (3/V) / (3/V)² = (2.25/V²)(3/V) / (9/V²)

= (6.75/V³) / (9/V²) = 6.75 / (9V) = 0.75/V

0.25 = 0.75/V → V = 0.75/0.25 = 3 L ✓

• Y²⁺ concentration increases → Y is being oxidized (Y → Y²⁺ + 2e⁻) → Y is the anode.
• Since Y is the anode, X is the cathode → X²⁺ ions are being reduced: X²⁺ + 2e⁻ → X
• X²⁺ is the oxidizing agent (it gets reduced). ✓
• Electrons flow from Y (anode) to X (cathode), not X to Y → option (c) is wrong.
• Y has higher oxidation potential (it's oxidized more easily), not X → option (a) is wrong.

• Cd + Zn²⁺ → Non-spontaneous: Cd cannot reduce Zn²⁺ → Zn²⁺ is a weaker oxidizing agent than Cd²⁺. So: Cd²⁺ > Zn²⁺.
• Cd + Cu²⁺ → Spontaneous: Cd can reduce Cu²⁺ → Cu²⁺ is a stronger oxidizing agent than Cd²⁺. So: Cu²⁺ > Cd²⁺.
• Combining: Cu²⁺ > Cd²⁺ > Zn²⁺ ✓

• A sacrificial electrode must be more reactive (higher oxidation potential) than iron → it should be the anode (electrons flow FROM the sacrificial metal TO iron).
• A → Fe: Electrons flow from A to Fe → A is the anode (more reactive than Fe). Cell potential = 1.4 V ✓
• C → Fe: C is also the anode. Cell potential = 0.5 V ✓ (but weaker protection)
• Fe → B and Fe → D: Fe is the anode here → B and D are less reactive than Fe → cannot be sacrificial. ✗
• Between A (1.4V) and C (0.5V), A has higher reactivity difference → A is preferred for best cathodic protection.

Based on the electrochemical system provided in the figure, here is the comprehensive step-by-step chemical breakdown and analysis of the electrolysis of molten silver bromide (AgBr).

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Detailed Step-by-Step Analysis

1. Identifying the Poles of the Cell

• Looking closely at the battery symbol at the top:
  • Electrode B is connected to the longer vertical plate of the battery power supply, representing the positive terminal (+). In an electrolytic cell, the positive pole is the Anode.
  • Electrode A is connected to the shorter vertical plate of the battery power supply, representing the negative terminal (-). In an electrolytic cell, the negative pole is the Cathode.

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2. Half-Cell Reaction at Electrode B (Anode / Positive Pole)

• The Process: Oxidation occurs at the anode. Negatively charged bromide ions (Br^-) from the melt migrate towards the positive Electrode B.
• The Observation: Each bromide ion loses electrons to form reddish-brown bromine gas vapors (Br₂) escaping from the electrode surface.
• The Reaction Box:


2Br^-(l) → Br₂(g) ↑ + 2e^-



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3. Half-Cell Reaction at Electrode A (Cathode / Negative Pole)

• The Process: Reduction occurs at the cathode. Positively charged silver ions (Ag⁺) from the melt migrate towards the negative Electrode A.
• The Observation: Silver ions gain electrons and are reduced, depositing as solid silver metal (Ag) covering the surface of Electrode A.
• The Reaction Box:


Ag⁺(l) + e^- → Ag(s) ↓



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Summary of Expected Exam Questions

This standard curriculum setup typically asks you to match the correct observation/mass change with each electrode. You can directly conclude:

Electrode	Polarity & Role	Chemical Reaction Type	Visible Observation / Consequence
Electrode B	Positive (+) Anode	Oxidation	Reddish-brown bromine gas vapors evolve; Mass stays constant.
Electrode A	Negative (-) Cathode	Reduction	Silver metal is deposited; Mass of the electrode increases.

• In an electrolytic cell: the electrode connected to the negative terminal of the battery is the cathode (reduction), and the one connected to the positive terminal is the anode (oxidation).
• Electrode B = Cathode: Ag⁺ + e⁻ → Ag (silver is deposited). ✓
• Electrode A = Anode: 2Br⁻ → Br₂ + 2e⁻ (oxidation of bromide). ✓

1. Identifying Electrode Polarities

• Looking closely at the battery symbol at the top of the image:
  • The longer vertical line represents the positive terminal (+), which is connected to Electrode X on the right. Therefore, Electrode X is the Anode.
  • The shorter vertical line represents the negative terminal (-), which is connected to Electrode Y on the left. Therefore, Electrode Y is the Cathode.

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2. Electrochemistry Reactions in Aqueous CuCl₂

• At Electrode X (Anode / Positive Pole): Negatively charged chloride ions (Cl⁻) from the solution migrate towards the positive Electrode X. They lose electrons (undergo oxidation) to form chlorine gas bubbles.
  
  
  2Cl⁻(aq) → Cl₂(g) ↑ + 2e⁻
  
  Observation: Pungent chlorine gas (Cl₂) bubbles evolve directly at electrode X. This makes statement (d) the absolute correct choice.
• At Electrode Y (Cathode / Negative Pole): Positively charged copper ions (Cu²⁺) migrate towards the negative Electrode Y. They gain electrons (undergo reduction) and deposit as metallic copper.
  
  
  Cu²⁺(aq) + 2e⁻ → Cu(s) ↓

C₄H₈ isomers include:

Comprehensive Chemical Analysis of C₄H₈ Isomers

The molecular formula C₄H₈ represents an unsaturated hydrocarbon with one double bond (alkene) or a saturated cyclic compound (cycloalkane). There are exactly 5 structural isomers for this formula:

1. Open-Chain Isomers (Alkenes = 3 Isomers)

1. 1-Butene: CH₂=CH-CH₂-CH₃ (Asymmetric alkene)
2. 2-Butene: CH₃-CH=CH-CH₃ (Symmetric alkene)
3. 2-Methylpropene: CH₂=C(CH₃)₂ (Asymmetric alkene)

2. Closed-Chain Isomers (Cycloalkanes = 2 Isomers)

1. Cyclobutane
2. Methylcyclopropane

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Column-by-Column Evaluation

Column X (Isomers obeying Markownikoff's rule)

• Markownikoff's rule applies when adding an asymmetric reagent (like HX or H₂O) to an asymmetric alkene.
• Among the isomers, 1-Butene and 2-Methylpropene are asymmetric because the double-bonded carbons hold unequal counts of hydrogen atoms.
• Count (X) = 2

Column Y (Isomers that decolorize alkaline potassium permanganate)

• Decolorizing purple alkaline KMnO₄ (Baeyer's Test) is a signature oxidation reaction for all carbon-carbon double bonds (alkenes).
• All 3 open-chain alkene isomers contain this double bond and successfully undergo this reaction.
• Count (Y) = 3

Column Z (Isomers that give by catalytic hydration alcohol difficult to be oxidized)

• Catalytic hydration (addition of water) of 2-Methylpropene follows Markownikoff's rule and produces a tertiary alcohol (2-methylpropan-2-ol):


CH₂=C(CH₃)₂ + H₂O →^H⁺ CH₃-C(OH)(CH₃)₂


• Tertiary alcohols lack a hydrogen atom attached directly to the carbinol carbon ($\text{C-OH}$), making them highly resistant to standard chemical oxidation. 2-Methylpropene is the only isomer that gives this result.
• Count (Z) = 1

Column W (Closed chain isomers)

• The ring structural configurations corresponding to the formula $\text{C₄H₈}$ are Cyclobutane and Methylcyclopropane.
• Count (W) = 2

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Detailed Step-by-Step Chemical Conversion

To solve this problem, we must first identify the chemical substances mentioned:

• Simplest Alkyne: Ethyne (Acetylene, C₂H₂).
• Simplest Hydrocarbon: Methane (CH₄).

The correct industrial and laboratory pathway to convert ethyne into methane follows these four consecutive steps:

1. Catalytic Hydration

• Adding water to ethyne (C₂H₂) in the presence of an acid catalyst (H₂SO₄ / HgSO₄ at 60°C) produces an unstable vinyl alcohol intermediate, which quickly rearranges into Acetaldehyde (Ethanal, CH₃CHO).
• The Reaction Box:


C₂H₂ + H₂O →^{H₂SO₄ / HgSO₄} CH₃CHO

2. Complete Oxidation

• Subjecting acetaldehyde to complete chemical oxidation using acidified potassium permanganate or dichromate converts the aldehyde group into a carboxylic acid, yielding Acetic acid (Ethanoic acid, CH₃COOH).
• The Reaction Box:


CH₃CHO + [O] →^{KMnO₄ / H⁺} CH₃COOH

3. Neutralization

• Reacting acetic acid with a strong base like sodium hydroxide (NaOH) neutralizes the acid to form water and the organic salt Sodium acetate (CH₃COONa).
• The Reaction Box:


CH₃COOH + NaOH → CH₃COONa + H₂O

4. Dry Distillation

• Heating solid sodium acetate with soda-lime (NaOH + CaO) removes a molecule of carbon dioxide via decarboxylation, successfully yielding the target simplest hydrocarbon, Methane (CH₄).
• The Reaction Box:


CH₃COONa + NaOH →^{CaO / Δ} CH₄ ↑ + Na₂CO₃

Why the Other Options Are Incorrect

• (a) & (b): They suggest "reduction" instead of neutralization as the third step. Reducing acetic acid would lead back to alcohols or aldehydes rather than setting up the salt needed for methane extraction.
• (c): Suggests "Bayer's oxidation" as the initial step, which is an alkene-specific test utilizing cold alkaline potassium permanganate that cannot process triple-bonded alkynes into aldehydes.

Detailed Step-by-Step Analysis

1. Analyzing Organic Substance X (Violet Color Test)

• The Chemical Rule: Iron(III) chloride solution (FeCl₃) is the standard functional group reagent used to detect the presence of a phenolic OH group (a hydroxyl group directly linked to a benzene ring core).
• Identification: Carbolic acid is the common industrial name for Phenol (C₆H₅OH). When FeCl₃ is added to phenol, it coordinates to form a highly distinctive violet-colored complex solution.

2. Analyzing Organic Substance Y (Catalyst During Preparation)

• The Chemical Rule: Electrophilic aromatic substitution of benzene requires a Lewis acid catalyst to polarize the halogen molecule and generate the strong electrophile necessary to break aromaticity.
• Identification: In the preparation of a Halo benzene (such as chlorobenzene, C₆H₅Cl) via the direct halogenation of benzene, anhydrous Iron(III) chloride (FeCl₃) serves as the mandatory catalyst.
• The Reaction Box:


C₆H₆ (Benzene) + Cl₂ →^{FeCl₃ / UV} C₆H₅Cl (Halo benzene) + HCl

3. Why the Other Options Are Incorrect

• (A): While salicylic acid gives a violet color with FeCl₃, carbolic acid (phenol) is prepared industrially via the hydrolysis of chlorobenzene with NaOH at high temperature and pressure, not using FeCl₃ as a catalyst.
• (C): Sulphuric acid (H₂SO₄) is a strong mineral inorganic acid, not an organic compound, and is not prepared using an iron catalyst.
• (D): This choice completely reverses the roles; halo benzene does not react with FeCl₃ to yield a violet color.

Detailed Step-by-Step Analysis

1. Identifying the Reaction Type (Friedel-Crafts Alkylation)

• The Reactants: The reaction combines Benzene (C₆H₆) with Benzyl chloride (C₆H₅CH₂Cl).
• The Catalyst: Anhydrous Aluminium chloride (AlCl₃) acts as a classic Lewis acid catalyst.
• The Mechanism: This setup triggers a Friedel-Crafts alkylation reaction. The AlCl₃ catalyst polarizes and removes the chloride ion from benzyl chloride, generating a highly reactive carbocation intermediate (C₆H₅CH₂⁺). This carbocation then attacks the aromatic benzene ring.

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2. Determining Products A and B

• When the benzyl carbocation (C₆H₅CH₂⁺) substitutes a hydrogen atom ($\text{H}$) on the benzene ring, the two aromatic groups link together via the central methylene bridge:
• Product A: Diphenylmethane (C₆H₅CH₂C₆H₅).
• The displaced hydrogen atom from the benzene ring immediately combines with the released chloride ion to form the stable inorganic byproduct:
• Product B: Hydrogen chloride gas (HCl).
• The Full Reaction Box:


C₆H₅CH₂Cl + C₆H₆ →^AlCl₃ C₆H₅CH₂C₆H₅ + HCl



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3. Why the Other Options Are Incorrect

• (a) & (b): These options claim that hydrogen gas (H₂) is evolved, ignoring the fact that the chlorine atom from benzyl chloride must form a byproduct.
• (d): Claims that chlorine gas (Cl₂) is released. Free halogens are not produced in alkylation processes; instead, the halogen is eliminated as a halide salt or hydrogen halide.

• Carbolic acid = Phenol (C₆H₅OH)
• Acid (B) = Nitric acid (HNO₃)
• Phenol + conc. HNO₃ in presence of conc. H₂SO₄ → 2,4,6-trinitrophenol = Picric acid

• Propyl formate = HCOOC₃H₇. Acidic hydrolysis gives: Formic acid (HCOOH) + Propanol.
• Compound X = Formic acid (HCOOH) — formula CH₂O₂.
• (a) HCOOH has no structural isomers ✓
• (b) CH₂O₂ ✓
• (c) Formic acid does not react with halogenated acids (HCl, HBr) — that's a property of alcohols, not carboxylic acids ✗
• (d) Formic acid is secreted by ants and some insects ✓

• An alcohol that cannot be prepared by catalytic hydration = Methanol (CH₃OH), because catalytic hydration requires an alkene, and there's no C₁ alkene.
• (a) Can be prepared by the general method (halide + KOH) — technically yes for methyl halide ✓
• (b) Methanol (CH₃OH) has NO structural isomers — it's the simplest alcohol ✗
• (c) Methanol is toxic and causes blindness/death ✓
• (d) Methanol reacts with acids to form esters (e.g., methyl salicylate) ✓

• A = C₂H₆O = Ethanol (C₂H₅OH) — used as a disinfectant/antiseptic in medicine.
• B = C₇H₆O₃ = Salicylic acid — used to make aspirin ,dacron and other medical products.
• Both are used in medical fields. ✓
• Ethanol does not react with NaOH (neutral alcohol), so (b) is wrong.

Detailed Step-by-Step Analysis

The Core Chemical Rule: Acidified potassium permanganate (KMnO₄ / H⁺) is a powerful purple oxidizing agent. It undergoes decolorization (turns from purple to colorless) only when it reacts with substances that can be easily oxidized.

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1. Why Option (c) is the Correct Exception

• Structure Analysis: 2-bromo-2-propanol is a halogenated derivative of a tertiary alcohol (Cyclic or structural formula: (CH₃)₂C(Br)(OH)).
• Oxidation Behavior: The central carbon atom holding the hydroxyl (-OH) group is attached to three other carbon chains and a bromine atom. It contains zero alpha-hydrogen atoms (H-C-OH).
• Conclusion: Because tertiary configurations lack alpha-hydrogens, they are highly stable against chemical oxidation under normal conditions. Therefore, it does not decolorize the purple KMnO₄ solution.

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2. Why the Other Options Do Decolorize KMnO₄

• (a) Hydroxy cyclohexane (Cyclohexanol): This is a secondary alcohol. The carbon bonded to the -OH group possesses one alpha-hydrogen, allowing it to be easily oxidized to cyclohexanone, which decolorizes the solution.
• (b) Isobutyl alcohol (2-methylpropan-1-ol): This is a primary alcohol (R-CH₂OH). It contains two alpha-hydrogens and readily oxidizes first to an aldehyde and then to a carboxylic acid, decolorizing the solution.
• (d) 2-hydroxy propanoic acid (Lactic acid): The alpha-carbon contains a secondary alcohol group (CH₃-CH(OH)-COOH). It has one alpha-hydrogen and is smoothly oxidized into pyruvic acid (CH₃-CO-COOH), decolorizing the solution.

• Two consecutive elements with the same 3d electrons: These are Chromium (Cr, Z=24) [Ar] 3d⁵4s¹ and Manganese (Mn, Z=25) [Ar] 3d⁵4s². Both have 3d⁵.
• Element X is resistant to air → Cr (Chromium) is resistant due to its oxide layer.
• Element Y = Mn (Manganese), which has Z=25 (higher atomic number).
• Higher atomic number → greater effective nuclear charge. ✓

1. Identification of Element X

• The problem specifies that X is a representative element (عنصر ممثل).
• In the Egyptian high school chemistry curriculum for transition metals, Aluminium (Al) is the core representative element that forms essential industrial alloys with transition elements.

2. Role of Elements Y, Z, and L

• Element Y (Nickel - Ni): Aluminium combines with transition metals like Nickel or Copper to form intermetallic alloys known as Duralumin (الديورالومين). These possess definite chemical formulas like NiAl₃ that do not obey the standard laws of valency.
• Element Z (Scandium - Sc): Adding a tiny percentage of Scandium to Aluminium creates an exceptionally hard, lightweight alloy that is ideally suited for manufacturing MiG fighter aircraft.
• Element L (Titanium - Ti): An alloy composed of Aluminium and Titanium is utilized to build spacecraft and high-speed planes because Titanium maintains its mechanical strength and durability at extreme temperatures, whereas the strength of pure aluminium rapidly decreases.

3. Conclusion

Matching these chemical roles to the given options clearly shows that only Option A correctly identifies all four elements: X = Al, Y = Ni, Z = Sc, L = Ti.

Step 1: Molar mass of Na₂CO₃·10H₂O = 2(23) + 12 + 3(16) + 10(18) = 46 + 12 + 48 + 180 = 286 g/mol

Step 2: Moles = 14.3/286 = 0.05 mol in 500 mL → Molarity = 0.05/0.5 = 0.1 M

Step 3: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Moles of Na₂CO₃ in 25 mL = 0.1 × 0.025 = 0.0025 mol

Moles of HCl needed = 2 × 0.0025 = 0.005 mol

Concentration of HCl = 0.005/0.025 = 0.2 M

Initial moles of NaOH = 0.01 × 0.2 = 0.002 mol

Final moles needed = 0.7 × 0.2 = 0.14 mol (assuming volume stays ≈ 200 mL since solid is added)

Additional moles = 0.14 − 0.002 = 0.138 mol

Mass = 0.138 × 40 = 5.52 g ✓

Initial: H₂ = 1 mol, I₂ = 1 mol, HI = 0

At equilibrium: H₂ = 0.3 mol, I₂ = 0.3 mol

Reacted = 1 − 0.3 = 0.7 mol each → HI formed = 2 × 0.7 = 1.4 mol

Concentrations (V = 2L): [H₂] = 0.3/2 = 0.15 M, [I₂] = 0.15 M, [HI] = 1.4/2 = 0.7 M

K_c = [HI]²/([H₂][I₂]) = (0.7)²/(0.15 × 0.15) = 0.49/0.0225 = 21.78 ✓

NH₄OH is a weak base.

α = √(K_b/C) = √(1.8×10⁻⁵/0.1) = √(1.8×10⁻⁴) = 0.0134

[OH⁻] = α × C = 0.0134 × 0.1 = 1.34 × 10⁻³ M

pOH = −log(1.34×10⁻³) = 2.87

pH = 14 − 2.87 = 11.13

Since NH₄OH is a base → pH > 7 ✓ (option B)

1. A (Sodium pentanoate) + NaOH/CaO/Δ → Dry distillation → B = Butane (C₄H₁₀)
2. Butane + Cl₂ → C = 1-chlorobutane (C₄H₉Cl)
3. 1-chlorobutane + KOH (aq) → D = 1-butanol (C₄H₉OH)
4. 1-butanol + Oxidation → E = Butanoic acid (C₃H₇COOH)

Faraday's Law: m = (M × I × t) / (n × F)

Where: M = 23, I = 10 A, t = 10 × 3600 = 36000 s, n = 1 (Na⁺ + e⁻ → Na), F = 96500 C/mol

m = (23 × 10 × 36000) / (1 × 96500) = 8,280,000/96500 = 85.8 g ✓

• Compound A appears in both schemes. From Scheme 2: A + V₂O₅/3O₂ → simplest aromatic acid (benzoic acid, C₆H₅COOH). The substance oxidized by V₂O₅ to give benzoic acid is Toluene (C₆H₅CH₃). So A = Toluene.
• Process Y: CₙH₂ₙ₊₂ → Toluene = Catalytic reforming (converting aliphatic hydrocarbons to aromatic ones). ✓
• From Scheme 1: RX + excess KOH/Δ → Toluene (A). Then Toluene → simplest aliphatic acid (formic acid HCOOH).
• Process X: Toluene → Formic acid. This would be complete oxidation. ✓

🧪 Chemical Explanation

• Compound (A) is Lactic Acid (C₃H₆O₃): Contains a secondary alcohol group (-CHOH-) which is easily oxidizable.
• Compound (B) is Citric Acid (C₆H₈O₇): Contains a tertiary alcohol group (-COH) which is non-oxidizable.
• Compound (C) is Carbolic Acid / Phenol (C₆H₆O): Acts as a weak organic acid but contains zero alcohol groups.

💡 Why Other Options Fail

• Option b: Reverses the correct identities of compounds A and B.
• Option c: Incorrectly lists Phenol (C₆H₆O) as compound A.
• Option d: Compound B here is Salicylic Acid, which does not fit the criteria.

• A = (CH)₆ = C₆H₆ = Benzene
• B = (CH₂)₆ = C₆H₁₂ = Cyclohexane
• (c): Benzene can be produced by reduction of carbolic acid (phenol): C₆H₅OH + Zn dust → C₆H₆ + ZnO ✓
• Cyclohexane is produced by hydrogenation of benzene: C₆H₆ + 3H₂ → C₆H₁₂ ✓

The correct option is a) (A) : Propanoic acid , (B) : Propanol , (C) : Propene.

Detailed Step-by-Step Analysis

1. Structural Logic Breakdown

• B → C (Dehydration): Dehydration of an alcohol removes water to give an alkene (unsaturated hydrocarbon). Thus, B must be an alcohol and C must be an alkene.
• A → B (Reduction): Reduction of compound A gives alcohol B. Since B is an alcohol, A must be a carboxylic acid.
• C → D (Oxidation): Oxidation of an alkene.

2. Verifying Option (a) Pathways

• Reduction Step (A → B): Propanoic acid (CH₃CH₂COOH) is reduced using hydrogen gas over a copper chromate catalyst to produce 1-Propanol (CH₃CH₂CH₂OH).


CH₃CH₂COOH + 2H₂ →^{CuCrO₄ / Δ} CH₃CH₂CH₂OH + H₂O


• Dehydration Step (B → C): Heating 1-Propanol with concentrated sulfuric acid (H₂SO₄) at 180°C eliminates water to form Propene (CH₃CH=CH₂), which is an unsaturated aliphatic hydrocarbon.


CH₃CH₂CH₂OH →^{conc. H₂SO₄ / 180°C} CH₃CH=CH₂ + H₂O

3. Why the Other Options Are Incorrect

• b) & d): These options state that Ethyl alcohol undergoes "reduction" to form aldehydes or acids, which is chemically impossible since primary alcohols can only be oxidized, not reduced, to these forms.
• c): Identifies C as Propane. Propane is a saturated alkane, directly contradicting the explicit condition that C is an unsaturated aliphatic hydrocarbon.

(a) Anode and Cathode of the Lithium Battery:

• The Li⁺ cell is connected to the lead accumulator for charging (it acts as an electrolytic cell).
• Electrode (Y) → connected to negative (−) terminal A → Cathode (Reduction): Li⁺ + e⁻ → Li
• Electrode (X) → connected to positive (+) terminal B → Anode (Oxidation)

(b) Spontaneous Reactions:

• The Lead Accumulator is the source of electric current → spontaneous (galvanic) reactions occur inside it.
• The lithium cell is being charged (non-spontaneous/electrolytic process).

Step 1 — Identify (F): Catalytic hydration of propyne:

F = Acetone (Propanone)

Step 2 — Identify (A): Preservative for frozen fruit = Benzoic acid (C₆H₅COOH)

Step 3 — Identify (B):

B = Sodium benzoate (C₆H₅COONa)

Step 4 — Identify (D): Since (F = Acetone) converts to (D) by process (E), (D) must be Propan-2-ol (Isopropyl alcohol):

D = Propan-2-ol

(1) Structural Formulas:

• (B) Sodium benzoate: C₆H₅–COO⁻Na⁺
• (D) Propan-2-ol: CH₃–CH(OH)–CH₃

(2) Chemical Processes:

• Process (E): Reduction (Catalytic Hydrogenation) — Acetone → Propan-2-ol
• Process (C): Conversion of Sodium benzoate (B) → Propan-2-ol (D) involves multiple steps; overall the process can be classified as dry distillation then halogenation then hydrolysis.