Chromium (III) Ion & Electron Configuration

Chromium (Cr) forms Chromium (III) oxide (Cr₂O₃), notoriously used as a vibrant green dye in paints.

• 1. Neutral Atom: Chromium has an anomalous electron configuration to maximize stability via a half-filled d-subshell: [Ar] 4s¹ 3d⁵.
• 2. Oxidation State: In Cr₂O₃, Oxygen is -2. To balance the charge, Chromium must be +3.
• 3. Ion Formation: To form Cr³⁺, 3 electrons are removed. Electrons are always lost from the outermost energy level first (4s), then the 3d shell. (Loses 1 from 4s, and 2 from 3d).

Identifying Non-Transition Elements

To identify the non-transition element, trace each ion back to its neutral atomic state by adding the lost electrons back (filling 4s first, then 3d):

• 🔸 (a) X³⁺: [Ar] → Neutral X is [Ar] 4s² 3d¹ (Scandium - Transition).
• 🔸 (b) Y²⁺: [Ar] 3d⁶ → Neutral Y is [Ar] 4s² 3d⁶ (Iron - Transition).
• 🔸 (c) Z²⁺: [Ar] → Neutral Z is [Ar] 4s² 3d⁰ (Calcium - Representative / Non-transition).
• 🔸 (d) W³⁺: [Kr] 4d⁷ → Neutral W is [Kr] 5s² 4d⁸ (Transition).

Oxidation Stability of Transition Ions

A solution must be covered if its ion is unstable and easily oxidized by atmospheric oxygen. Let's analyze the stability based on d-orbital configurations:

• W (V⁵⁺): Has a highly stable d⁰ (empty) configuration.
• Y (Fe³⁺): Has a highly stable d⁵ (half-filled) configuration.
• X (Mn²⁺): Also possesses a stable d⁵ (half-filled) configuration.
• Z (Ti³⁺): Has an unstable d¹ configuration. It rapidly and easily oxidizes in air to the more stable Ti⁴⁺ (d⁰).

Constancy of Atomic Radii

The consecutive elements that share the exact same number of 3d electrons in the first series are:

• Chromium (Cr: 4s¹ 3d⁵) and Manganese (Mn: 4s² 3d⁵)
• Copper (Cu: 4s¹ 3d¹⁰) and Zinc (Zn: 4s² 3d¹⁰)

A defining characteristic of the first transition series is the relative constancy of atomic radii from Chromium to Copper. This occurs due to two opposing forces perfectly balancing each other:

1. An increase in effective nuclear charge (pulling electrons closer).
2. An increase in the shielding effect from the added 3d electrons (pushing electrons away).

Roasting of Siderite Ore

Heating siderite (FeCO₃) strongly in air triggers a two-step chemical process:

1. Thermal Decomposition: FeCO₃ breaks down into FeO and CO₂ gas.
   FeCO₃ → FeO + CO₂↑
2. Oxidation: The FeO is immediately oxidized by air to stable Fe₂O₃.
   4FeO + O₂ → 2Fe₂O₃

Overall Reaction: 4FeCO₃ + O₂ → 2Fe₂O₃ + 4CO₂↑

Properties of Sulphur Dioxide (SO₂)

The reaction between dilute HCl and sodium sulphite (Na₂SO₃) releases sulphur dioxide gas (SO₂).

• Chemical Nature: SO₂ acts as a powerful reducing agent.
• Visual Test: When exposed to a paper wetted with acidified potassium dichromate (K₂Cr₂O₇), it reduces the orange dichromate ion (Cr₂O₇²⁻) into the green chromium (III) ion (Cr³⁺).

Formation of Insoluble Sulphides

Let's analyze the products of each reaction:

• (a) H₂S + Lead (II) acetate: Produces Lead (II) sulphide (PbS), which is highly insoluble and forms a distinct black precipitate.
• (b), (c), and (d): These are all purely redox reactions that yield soluble aqueous products. They are characterized by color changes (e.g., removing iodine's color, turning dichromate green, or decolorizing permanganate), but no solid precipitates are formed.

Halide Reactions with Conc. H₂SO₄

Hot, concentrated sulfuric acid acts as both an acid and an oxidizing agent. Its reaction depends on the halide:

• Salt X (NaCl): Releases HCl gas. HCl is highly stable and difficult for sulfuric acid to oxidize further.
• Salt Y (NaI or NaBr): Releases HI or HBr gas. These are stronger reducing agents and are partially oxidized by the hot H₂SO₄ (e.g., HI rapidly oxidizes to release striking violet iodine vapors).

Identifying Anions by Precipitation Profiles

Let's evaluate the given chemical tests to find the anions:

• Test for X (Barium Chloride): BaCl₂ precipitates both phosphates and sulphates as white solids. However, Barium phosphate dissolves in dilute HCl, whereas Barium sulphate does not. Since X dissolves, X is Phosphate.
• Test for Y (Lead (II) Acetate): Lead acetate produces a white precipitate of Lead sulphate (PbSO₄) when added to sulphates. (If it were a sulphide, it would be black PbS). Thus, Y is Sulphate.

Identifying the Calcium Cation

We are looking for a metal cation that forms a white precipitate with BOTH sulfates and carbonates.

• With Sulfuric Acid: Calcium (Ca²⁺) reacts to form Calcium sulfate (CaSO₄), a white precipitate. (Note: Mg²⁺, Fe²⁺, and Cu²⁺ form soluble sulfates, meaning no precipitate).
• With Sodium Carbonate: Calcium reacts to form Calcium carbonate (CaCO₃), another classic white precipitate.

Interpreting Equilibrium Graphs

The key to these graphs lies in observing where the curves level off (reach equilibrium):

• Graph (X): The product curve (rising from zero) finishes at a higher concentration than the reactant curve. Since [Products] > [Reactants], the math tells us Kc > 1. This means the forward reaction is predominant.
• Graph (Y): The reactants remain higher than the products at equilibrium, yielding a Kc < 1 (backward reaction predominant).

Le Chatelier's Principle & Heterogeneous Equilibrium

In this system, Ca(OH)₂ and CaO are solids, while H₂O is a vapor (gas). Solids do not affect the equilibrium position. We only care about the vapor!

• Withdrawing Water Vapor: According to Le Chatelier's Principle, removing a product causes the system to shift forward (to the right) to replace the lost water vapor.
• The Result: This forward shift naturally generates more products, leading to a direct increase in the mass of solid CaO.
• Note: Altering the amounts of solids (choices a and c) has zero effect on the equilibrium position.

Ostwald's Dilution Law

This is a two-step calculation: first find the new diluted concentration, then calculate the degree of dissociation (α).

1. Step 1: Dilution Formula (M₁V₁ = M₂V₂)
   V₁ = 50 mL, M₁ = 0.2 M.
   V₂ = 50 + 450 = 500 mL.
   M₂ = (0.2 × 50) / 500 = 0.02 M.
2. Step 2: Calculate α
   Using the formula: α = √(K_b / C)
   α = √(1.8 × 10⁻⁵ / 0.02) = √(9 × 10⁻⁴) = 0.03.

Common Ion Effect vs. Acidic Dissolution

Let's evaluate the chemical additions in each tube based on Le Chatelier's Principle:

• Tube A (Adding HCl): The H⁺ ions from the acid react directly with the CO₃²⁻ ions to form CO₂ gas and water. This removes carbonate from the equilibrium, shifting it to the right. The solid CaCO₃ dissolves, causing the precipitate to decrease.
• Tube B (Adding CaCl₂): This introduces extra Ca²⁺ ions (a common ion). The increased concentration of products pushes the equilibrium back to the left, forming more solid CaCO₃. The precipitate increases.

Understanding pH, pOH, and Basicity

Pure distilled water is neutral with a pOH of 7 (and pH of 7).

• The Shift: A decrease in pOH means the concentration of OH⁻ ions increased (since pOH = -log[OH⁻]). Lower pOH means the solution became more basic. Therefore, solution X must be a base.
• Evaluating Choices: We need a basic solution.
  Choice A says "Base pOH = 8", which means pH = 6 (acidic - contradiction).
  Choice C says "Base pH = 8", which means pOH = 6 (basic). Adding a pOH=6 base to pOH=7 water will indeed lower the overall pOH below 7.

Calculating Kp from Partial Pressures

The total pressure involves only the gases in the system. Solid carbon C_(s) is entirely ignored in gas pressure calculations.

1. Find Partial Pressures:
   P_Total = P(CO₂) + P(CO)
   40 atm = P(CO₂) + 31.6 atm
   P(CO₂) = 40 - 31.6 = 8.4 atm.
2. Apply Kp Formula:
   Kp = (P(CO))² / P(CO₂)
   Kp = (31.6)² / 8.4
   Kp = 998.56 / 8.4 ≈ 118.876.

Chemical Interference in a Daniel Cell

The cathode half-cell of a standard Daniel cell contains aqueous Cu²⁺ ions waiting to be reduced.

• The Reaction: Adding Na₂S introduces Sulphide ions (S²⁻), which aggressively bind to Copper to form Copper(II) sulphide (CuS), a highly insoluble black precipitate.
• The Consequence: This massively depletes the [Cu²⁺] concentration available for reduction. Because the active ions are removed chemically rather than being converted to electrical energy, the functional lifespan (consumption time) of the cell decreases drastically.

Calculating Cell EMF (E_cell)

To find the EMF, we must properly align the standard potentials with the actual reaction:

1. Anode (Oxidation): Cl⁻ is oxidized to Cl₂. The provided potential (+1.36 V) is for reduction. Thus, the oxidation potential is -1.36 V.
2. Cathode (Reduction): MnO₄⁻ (Mn⁷⁺) is reduced to Mn²⁺. The provided potential (-1.52 V) is an oxidation potential (from 2+ to 7+). Thus, its reduction potential is +1.52 V.
3. EMF Calculation: E_cell = E_ox(anode) + E_red(cathode) = -1.36 V + 1.52 V = +0.16 V.

Combining Standard Cell Potentials

Using the general formula: E_cell = E_ox(Anode) + E_red(Cathode)

1. Analyze the Given Cells:
   For Cell 1: E_ox(X) + E_red(Ag) = 0.80 V.
   For Cell 2: E_ox(Y) + E_red(Ag) = 1.56 V.
2. Find the Difference: Subtracting Cell 1 from Cell 2 cancels out the Silver (Ag) component:
   E_ox(Y) - E_ox(X) = 1.56 - 0.80 = 0.76 V.
3. Construct New Cell (X, Y): Since Y has a higher oxidation potential, Y will act as the anode, and X will be forced to act as the cathode. EMF = E_ox(Y) + E_red(X) = 0.76 V.

Determining the Type of Metal Protection

The activity of a metal is inversely related to its reduction potential. A lower (more negative) reduction potential means a more active metal.

• Activity Series: A (-0.44) > B (+0.34) > C (+0.80). Metal C is the least active.
• Protection Type: When a metal is coated with a less active metal, the coating acts as a physical barrier. If scratched, the underlying (more active) metal becomes the anode and rusts faster. This setup (coating is less active) is strictly defined as Cathodic Protection.

Electrolysis of Bauxite (Hall-Héroult Process)

During the industrial electrolysis of molten Bauxite (Al₂O₃ dissolved in molten cryolite), there is no water present, meaning no hydrogen gas is produced.

• At the Cathode (Negative): Aluminum ions (Al³⁺) gain electrons (reduction) and pool at the bottom as molten Aluminum metal.
• At the Anode (Positive): Oxide ions (O²⁻) lose electrons (oxidation) and form Oxygen gas (O₂). (Note: This oxygen often reacts with the carbon anodes to form CO₂, requiring regular replacement).

Electrolytic Purification of Metals

In electrolytic purification, the impure metal block is made the anode. Based on the activity series (A > B > C), here is what happens:

• Highly Active A: Because A is more active than B, it easily oxidizes and dissolves into the solution as ions.
• Target Metal B: B also oxidizes, travels through the solution, and safely reduces onto the pure cathode.
• Less Active C: Because C is less active than B, the applied voltage isn't high enough to oxidize it. It remains unreacted solid metal and simply falls down to the bottom below the anode as "anode mud".

Complete Hydrogenation of Hydrocarbons

Let's decipher the identity of each hydrocarbon based on the clues:

• Compound (A): Kept in a smaller ratio in hot countries to manage LPG cylinder pressure. This is Propane (C₃H₈).
• Compound (B): Has 3 carbons but 2 fewer H atoms. This is Propene (C₃H₆).
• Compound (C): Has 3 carbons but 2 fewer H atoms than B. This is Propyne (C₃H₄).

The Chemistry: Completely hydrogenating any 3-carbon unsaturated hydrocarbon maxes out its hydrogen bonds, turning it back into the fully saturated alkane. Thus, all three turn into/remain Propane.

Synthesis of Methane

First, identify the compounds (all have 2 carbons): A is Ethane, B is Ethanol (1 OH group), C is Ethylene glycol (2 OH groups). The target is a saturated hydrocarbon with a lower boiling point than Ethane, which must be Methane (CH₄).

Synthesis Pathway from Ethanol (B):

1. Complete Oxidation: Ethanol is oxidized to Acetic acid (CH₃COOH).
2. Neutralization (Add NaOH): Acetic acid reacts with NaOH to form Sodium acetate (CH₃COONa).
3. Dry Distillation: Heating Sodium acetate with soda lime cleaves the molecule, producing Methane gas.

Ester Isomers

Let's deduce the reactants and the product:

• The Acid: An acid where the number of carbons equals the number of carboxyl groups is Formic acid (HCOOH) (1 C, 1 COOH).
• The Alcohol: The alcohol referenced with a low freezing point around -110°C is Ethanol (C₂H₅OH).
• The Ester: Esterification produces Ethyl formate (HCOOC₂H₅), with a total molecular formula of C₃H₆O₂.

Finding Isomers: How many other esters share the formula C₃H₆O₂? There is only exactly one: Methyl acetate (CH₃COOCH₃).

Reactions of Lactic Acid

The clue "causes muscle contraction/fatigue" immediately identifies acid (A) as Lactic acid (CH₃CH(OH)COOH).

1. Step 1: Reacting lactic acid with NaOH neutralizes it to form Sodium lactate (B).
2. Step 2 (Process C): Heating sodium lactate with soda lime (Dry Distillation) cleaves the carboxylate group, converting the remaining structure into Ethanol (D). (C₂H₆O).
3. Step 3: Complete oxidation of ethanol yields Ethanoic/Acetic acid (E). (C₂H₄O₂).

Preparation of Ethylene Glycol (Antifreeze)

The target antifreeze compound is Ethylene glycol. The journey from cane sugar involves four distinct steps:

1. Hydrolysis: Cane sugar (Sucrose) is broken down into simple sugars (glucose/fructose).
2. Alcoholic Fermentation: Yeast enzymes convert glucose into Ethanol.
3. Dehydration (180°C): Heating ethanol with concentrated H₂SO₄ at exactly 180°C eliminates water to form Ethene.
4. Oxidation (H₂O₂ / Bayer's): Oxidizing ethene adds two hydroxyl (-OH) groups, resulting in Ethylene glycol.

Identifying Non-Carboxylic Organic Acids

Let's map out the identities of these specific compounds:

• Acid A: A weak solid acid with a distinct hospital smell is Phenol (also historically called carbolic acid).
• Acid B: Benzene + mineral acid = Benzenesulfonic acid (sulfonation).
• Acid C: Used to treat burns (and as an explosive) is Picric acid (2,4,6-trinitrophenol).

To prepare B from A: You must first reduce Phenol (A) back to Benzene by heating it with Zinc dust. Then, perform a substitution reaction by adding concentrated sulfuric acid to get Benzenesulfonic acid (B).

Combustion Stoichiometry

Let's work backward from the combustion products to find the molecular formula:

• 3 mol CO₂: Contains exactly 3 Carbon atoms.
• 3 mol H₂O: Contains exactly 6 Hydrogen atoms (3 × 2).

This means the original hydrocarbon must have the formula C₃H₆. Among the choices, Propane is C₃H₈, and Propanol is C₃H₈O. Only Cyclopropane (and propene) fits the formula C₃H₆.

Roles of Core Reagents

Both NaOH and H₂SO₄ play definitive roles in these organic preparations:

• Paraffins (Alkanes): Prepared via dry distillation of sodium salts. The role of NaOH (in soda lime) is to cleave and remove the carboxylate group (-COONa) as sodium carbonate (Na₂CO₃).
• Olefins (Alkenes): Prepared by dehydrating alcohols. Concentrated H₂SO₄ acts as a powerful dehydrating agent, eliminating water from the molecule to form the double bond. (Note: "illuminating" is a known translational typo for eliminating/removing).

Functional Group Analysis

By visually breaking down the molecule, we find three key functional groups that govern its reactivity:

• 1. Carboxyl Group (-COOH): This acidic group reacts strongly with NaHCO₃, releasing CO₂ gas, causing visible effervescence that clouds lime water.
• 2. Phenolic Group (-OH on ring): Phenols famously react with Iron (III) chloride (FeCl₃) to produce a highly characteristic purple complex.
• 3. Ester Linkage (-O-CO-CH₂-NH₂): Acidic hydrolysis severs this ester bond, separating the phenol ring and isolating the amino acid Glycine (NH₂-CH₂-COOH).

Preparation of Carbon Black

The simplest alkyne is ethyne (acetylene). Its catalytic hydration yields Acetaldehyde. The target substance (used heavily in printing inks) is Carbon black.

The Multi-Step Synthesis:

1. Oxidation: Acetaldehyde oxidizes to form Acetic acid.
2. Neutralization: Adding NaOH to acetic acid forms Sodium acetate.
3. Dry Distillation: Heating sodium acetate with soda lime produces Methane gas.
4. Thermal Cracking: Heating methane to 1000°C in the absolute absence of air decomposes it into Hydrogen gas and solid Carbon black.

Selective Precipitation & Amphoterism

To precipitate exactly one cation, we need a reagent that has distinct, highly specific interactions with the metals:

• Aluminum (Al³⁺): It initially precipitates as Al(OH)₃, but because it is amphoteric, adding an abundance of NaOH causes it to re-dissolve completely into soluble Sodium aluminate.
• Calcium (Ca²⁺): Calcium hydroxide Ca(OH)₂ is moderately soluble and does not precipitate effectively at these typical testing concentrations.
• Iron (Fe²⁺): Reacts to form Iron (II) hydroxide, Fe(OH)₂, a stable, insoluble precipitate that does not dissolve in excess base.

Magnetic Properties of Transition Ions

First, identify elements X and Y based on the clues:

• Element X: Has the highest oxidation state (+7) in the series. This is Manganese (Mn).
• Element Y: The consecutive element following Mn is Iron (Fe).

Analyzing the Y²⁺ ion: The neutral Iron atom has a configuration of 3d⁶ 4s². The Fe²⁺ ion loses the two 4s electrons, leaving exactly 3d⁶. According to Hund's rule, filling 5 orbitals with 6 electrons results in 1 pair and 4 unpaired electrons, making the ion strongly paramagnetic.

Limiting Reactant & Excess Concentration

Calculate the molar amounts to determine what is left over after the reaction completes.

1. Initial Moles:
   Moles H₂SO₄ = 0.2 L × 0.2 M = 0.04 mol.
   Moles Ca(OH)₂ = 0.3 L × 0.2 M = 0.06 mol.
2. Reaction & Excess: The balanced equation shows a 1:1 reacting ratio. Thus, 0.04 mol of acid reacts with exactly 0.04 mol of base. The unreacted base left over is: 0.06 - 0.04 = 0.02 mol Ca(OH)₂.
3. New Concentration: The final mixture volume is 200 mL + 300 mL = 500 mL (0.5 L).
   New Molarity = Moles / Volume = 0.02 mol / 0.5 L = 0.04 M.

Equilibrium Constant Expressions

A fundamental rule of writing Equilibrium Constants (Kc) is that pure solids (s) and pure liquids (l) are entirely excluded from the expression because their concentrations remain strictly constant.

• Reaction (b) Analysis: Both CaCO₃ (the reactant) and CaO (one of the products) are solids.
• Writing the Formula: Eliminating the solids leaves only the gaseous product. Therefore, the expression simplifies drastically to: Kc = [CO₂].

Solubility Product (Ksp) at Different Temperatures

This is a reverse-engineering calculation to find Ksp at a higher temperature.

1. Base Solubility at 25°C:
   Ksp = 10^-21. Solubility (S) = √Ksp = 3.16 × 10^-11 mol/L.
2. Concentration that Precipitated:
   Mass = 1.53 × 10^-5 g. Moles = (1.53 × 10^-5) / 97 = 1.577 × 10^-7 mol.
   Precipitated Molarity = Moles / 5 L ≈ 3.15 × 10^-8 mol/L.
3. Total Solubility at 60°C:
   Solubility at 60°C = Solubility at 25°C + Concentration precipitated.
   Since 10^-11 is mathematically negligible compared to 10^-8, the total solubility is essentially 3.15 × 10^-8 mol/L.
4. Calculate Ksp at 60°C:
   Ksp = S² = (3.15 × 10^-8)² ≈ 9.9 × 10^-16, which closely rounds to 1 × 10^-15.

Faraday's Law of Electrolysis

Using Faraday's main formula: Mass = (Current × Time × Equivalent Mass) / 96500

1. Determine Equivalent Mass: For Silver (Ag⁺), valency is 1. Equivalent mass = 108 / 1 = 108 g.
2. Set Up the Equation:
   26.25 = (25 × Time × 108) / 96500.
3. Solve for Time (in seconds):
   Time = (26.25 × 96500) / (25 × 108) = 2533125 / 2700 = 938.19 seconds.
4. Convert to Minutes:
   938.19 / 60 ≈ 15.63 minutes.

Comparing Electrode Activities

In any galvanic cell, electrons spontaneously flow from the Anode (the more active metal) to the Cathode (the less active metal).

• Cell 1 Analysis: The green arrow shows electrons flowing from electrode A to B. Therefore, Activity A > B.
• Cell 2 Analysis: The green arrow shows electrons flowing from electrode C to B. Therefore, Activity C > B.

In both setups, metal B is bullied into acting as the cathode because it is less active than its counterpart.

Distinguishing Acids and Esters

The general formula C_nH_2nO₂ indicates the compound is either a carboxylic acid or an ester.

• Compound A (Reacts cold): Carboxylic acids react readily with cold NaOH (simple neutralization). Thus, A must be an acid, such as Butyric acid.
• Compound B (No cold reaction): Esters do not react with cold NaOH; they require heating (saponification) to break the ester bond. Thus, B must be an ester, such as Ethyl ethanoate.

Multi-Step Synthesis (Backward Analysis)

To find starting material X, we trace the reactions backwards from the final product:

1. Complete Hydrogenation: 1,2-dimethylcyclohexane is formed by fully hydrogenating the aromatic ring of 1,2-dimethylbenzene (o-xylene).
2. Catalytic Reforming: O-xylene (an 8-carbon aromatic) comes from reforming an 8-carbon aliphatic chain with a 6-carbon straight core: 3,4-dimethylhexane.
3. Dry Distillation: This process drops one carbon atom. So, the precursor was a 9-carbon salt: Sodium 3,4-dimethylheptanoate.
4. Neutralization & Oxidation: The salt came from neutralizing an acid, which was produced by completely oxidizing a primary alcohol. This alcohol must be 3,4-dimethyl-1-heptanol.

Preparation of Sodium Benzoate

The renowned organic food preservative that inhibits fungal growth is Sodium benzoate.

The Industrial Pathway:

1. Catalytic Reforming: Normal hexane is heated with a Pt catalyst to form Benzene.
2. Alkylation: Benzene reacts with methyl chloride (Friedel-Crafts) to form Toluene.
3. Oxidation: Toluene is oxidized using V₂O₅ at 400°C to yield Benzoic acid.
4. Neutralization: Benzoic acid is neutralized with NaOH to produce the water-soluble salt Sodium benzoate.

Oxidation of Butanol Isomers

Let's map the isomers of C₄H₁₀O based on their structural methyl count and oxidize them:

• (A) Two methyl groups: This is isobutanol (2-methyl-1-propanol). Complete oxidation yields 2-methylpropanoic acid.
• (B) Three methyl groups: This is tert-butanol (2-methyl-2-propanol). Because the carbon bonded to the OH group has no attached hydrogen atoms, it strongly resists oxidation under normal conditions (No reaction).
• (C) One methyl group: This is the straight-chain 1-butanol. Oxidation seamlessly yields Butanoic acid.

Oxidation and Dehydration Constraints

Let's filter the choices based on the two distinct chemical requirements:

• Constraint 1 (Must Oxidize): The compound must be a primary or secondary alcohol. Tertiary alcohols do not oxidize. This instantly eliminates choices (a) and (c).
• Constraint 2 (Does NOT give 2-methyl-1-butene):
  Choice (d) directly dehydrates to form 2-methyl-1-butene.
  Choice (b) 2,2-Dimethyl-1-propanol (Neopentyl alcohol) is a primary alcohol that oxidizes, but it lacks beta-hydrogens (the adjacent carbon is bonded only to methyl groups). Thus, it cannot undergo standard E2 dehydration to yield simple alkenes without highly complex carbocation rearrangements.

Iron Reaction Pathway

By tracking the reactions visually:

• Compound (B) is a black oxide that reacts with dilute H₂SO₄. It must be FeO (Iron II Oxide).
• Reacting FeO with dilute H₂SO₄ gives Compound (C), which is FeSO₄.
• Heating FeSO₄ strongly yields Compound (D), which is Fe₂O₃.
• Element (A) is elemental Iron (Fe), which also reacts with dil. H₂SO₄ to give FeSO₄.

(a) Process Names:
- Process (1) converts FeO to Fe₂O₃. This is an Oxidation reaction.
- Process (2) converts Fe₂O₃ down to elemental Fe. This is a Reduction reaction.

(b) Chemical Formulas:
- Compound (C): FeSO₄
- Compound (D): Fe₂O₃

Synthesis of Oxalic Acid

Let's map out this sequence from the end backward:

• Compound (F) is a dibasic acid produced from oxidation. It is Oxalic acid.
• It is prepared by the complete oxidation of (E), which must be Ethylene glycol.
• Ethylene glycol is prepared by oxidizing (D) with H₂O₂. Thus, (D) is Ethene.
• Ethene is prepared by dehydrating (C), so (C) is Ethanol.
• Ethanol is produced by the alcoholic fermentation of (B), which is Glucose.
• Glucose comes from the hydrolysis of (A), which is Sucrose.