Step 1: Write Electronic Configurations

Magnetic attraction is directly proportional to the number of unpaired electrons in the d-sublevel.

Ion	Atomic Number (Z)	Configuration	Unpaired Electrons
Mn2+	25	[Ar] 3d5	5 unpaired electrons
Fe2+	26	[Ar] 3d6	4 unpaired electrons
Ti4+	22	[Ar] 3d0	0 unpaired electrons (diamagnetic)

Step 2: Order by Magnetic Attraction

Since 5 > 4 > 0, the correct order of decreasing magnetic attraction is:

Analysis of Ore Dressing

Chemical dressing of iron ore mainly involves roasting (heating the ore strongly in air). This achieves several objectives:

• Moisture is evaporated, reducing the total mass of the ore.
• Volatile impurities like sulfur and phosphorus are oxidized into gases (SO₂, P₂O₅), thus decreasing the percentage of impurities.
• Because impurities are removed, the concentration and percentage of iron in the remaining ore increases.

The Midrex Furnace

The Midrex furnace utilizes water gas (a mixture of CO and H₂) as its reducing agent. This gas is synthesized from natural gas (methane) via catalytic reforming with carbon dioxide and water vapor:

This matches option (d) perfectly.

Step-by-Step Chemical Reactions

Step	Process	Chemical Equation
1	Addition of Alkaline Solution	FeCl3(aq) + 3NH4OH(aq) → Fe(OH)3(s)↓ + 3NH4Cl(aq)
2	Thermal Decomposition (>200°C)	2Fe(OH)3(s) → Fe2O3(s) + 3H2O(g)
3	Reduction (at 230°C - 300°C)	3Fe2O3(s) + CO(g) → 2Fe3O4(s) (Black Oxide) + CO2(g)

Industrial Steelmaking Sequence

The processing pathway from raw limonite ore to steel consists of:

1. Roasting: Dehydrates limonite (2Fe₂O₃·3H₂O) to hematite (Fe₂O₃).
2. Reduction: Reduces Fe₂O₃ in the blast or Midrex furnace to produce molten iron (pig iron).
3. Addition of carbon: Done during the manufacturing stage (oxygen converter) to achieve the carbon composition necessary for high-strength steel.

Reactivity with Dilute HCl

• With Na₂SO₃: HCl reacts to release sulfur dioxide gas (SO₂), which turns acidified potassium dichromate paper green:
  Na2SO3 + 2HCl → 2NaCl + H2O + SO2↑
• With Na₃PO₄: HCl cannot displace phosphate because phosphoric acid is more stable than hydrochloric acid (no observable gas is evolved).

No reaction occurs with pairs (b) or (c). Both salts in pair (d) produce CO₂ gas, making them indistinguishable with HCl alone.

Analysis of Distinguishing Factors

Method	Calcium Carbonate (CaCO3)	Silver Phosphate (Ag3PO4)	Distinguishable?
Color	White	Yellow	Yes
H2O + CO2	Dissolves (forms Ca(HCO3)2)	Insoluble	Yes
NH4OH	Insoluble	Dissolves (forms soluble complex)	Yes
Acidified KMnO4	No reaction / color persists	No reaction / color persists	No

Gas Removal Chemistry

To selectively trap and eliminate H₂S and SO₂ gases:

1. H₂S removal: Reacts with Lead (II) acetate to precipitate solid, black Lead sulphide (PbS):
   (CH3COO)2Pb + H2S → PbS(s) + 2CH3COOH
2. SO₂ removal: Reacts with acidified potassium dichromate, which oxidizes SO₂ and reduces Cr(VI) to Cr(III), turning green:
   K2Cr2O7 + 3SO2 + H2SO4 → K2SO4 + Cr2(SO4)3 + H2O

Precipitation Reactions

Let's check the reactions for Calcium chloride (CaCl₂):

• Reaction with AgNO₃:
  CaCl2(aq) + 2AgNO3(aq) → 2AgCl(s)↓ (white ppt) + Ca(NO3)2(aq)
• Reaction with (NH₄)₂CO₃:
  CaCl2(aq) + (NH4)2CO3(aq) → CaCO3(s)↓ (white ppt) + 2NH4Cl(aq)

Both conditions are satisfied.

Step-by-Step Reaction Mechanism

When excess sodium hydroxide (NaOH) is added to the mixture containing Fe³⁺ and Al³⁺:

1. Iron(III) reaction: Fe³⁺ reacts with OH^- to form a reddish-brown precipitate of Iron(III) hydroxide, which is insoluble in excess NaOH:
   Fe3+(aq) + 3OH-(aq) → Fe(OH)3(s)↓
2. Aluminium reaction: Al³⁺ initially precipitates as Al(OH)₃, but being amphoteric, it dissolves in excess NaOH to form soluble sodium meta-aluminate:
   Al(OH)3(s) + OH-(aq) → AlO2-(aq) + 2H2O(l)
3. Spectators: Na⁺ and Cl^- ions remain in solution unchanged.

Understanding Dynamic Equilibrium

For a system to achieve dynamic equilibrium (Rate_forward = Rate_backward), the reaction must be reversible and occur in a closed system where no products escape:

• (a) is a precipitation reaction that goes to completion (irreversible).
• (b) represents strong acid ionization, which is 100% complete.
• (c) is in an open vessel, so gases escape, preventing the backward reaction from establishing a true equilibrium.
• (d) involves a weak acid (CH₃COOH) in solution, establishing a reversible dynamic equilibrium.

Le Chatelier's Principle & Common Ion Effect

HCl is a strong acid that dissociates completely:

This increases the concentration of H₃O⁺ (common ion) in the system. According to Le Chatelier's principle:

• The equilibrium shifts in the backward direction to consume the excess H₃O⁺.
• As the reaction shifts left, less HCN dissociates, meaning the degree of dissociation (α) decreases.

Thermodynamic and Shift Analysis

The equation is written with "- heat" on the product side, which is equivalent to:

Let's analyze the effects:

• Adding reactant Y₂ shifts the equilibrium in the forward direction.
• Since the forward reaction is endothermic, shifting forward leads to more energy being absorbed.

Factors Affecting Reaction Rate

The rate of a chemical reaction is influenced by:

1. Surface Area: Powdered magnesium reacts much faster than a solid ribbon.
2. Concentration: Higher molarity (0.2 M) of H₂SO₄ results in a faster rate than 0.1 M.
3. Temperature: Elevated temperature (35°C) yields a faster rate than room temperature.

Option (d) optimizes all three rate-enhancing factors.

pH Calculation

Given [H+] = 1.0 × 10-9 M:

Since pH > 7, the solution is basic (a base) with pH = 9.

Step 1: Calculate Partial Pressure of N₂

The total pressure is the sum of all partial pressures:

Step 2: Calculate K_p

Anode vs. Cathode Designation

• Since electrons flow from X to Y, X is the anode (oxidation) and Y is the cathode (reduction).
• In a spontaneous galvanic cell, the anode (X) has a higher oxidation potential and a lower reduction potential than the cathode (Y).

Step 1: Determine Anode and Cathode

Both equations represent standard oxidation reactions relative to the standard hydrogen electrode (SHE). Thus:

• E°_ox(X) = +0.76 V
• E°_ox(Y) = +1.66 V

Since Y has a higher oxidation potential, Y acts as the anode and X acts as the cathode.

Step 2: Calculate Cell EMF

Electrode Reactions

• At the Anode (+) (Copper): Copper metal is oxidized to copper ions, dissolving into solution:
  Cu(s) → Cu2+(aq) + 2e-
• At the Cathode (-) (Iron): Cu²⁺ ions from the solution are reduced to copper metal, plating the iron electrode:
  Cu2+(aq) + 2e- → Cu(s)

Because the rate of oxidation at the anode equals the rate of reduction at the cathode, the concentration of copper ions in solution remains constant, and overall charge neutrality is maintained.

Faraday's Law Calculations

Since the cells are connected in series, the same charge Q passes through all of them. Let's look at the mole of electrons needed for 1 mole of each gas:

1. Oxygen: 2O2- → O2 + 4e- (Requires 4 Faradays per mole) → Moles = Q / 4
2. Nitrogen: 2N3- → N2 + 6e- (Requires 6 Faradays per mole) → Moles = Q / 6
3. Chlorine: 2Cl- → Cl2 + 2e- (Requires 2 Faradays per mole) → Moles = Q / 2

Let's find the ratio of emitted gas volumes (proportional to moles):

Mercury Cell Chemistry

In a Mercury Cell, the reduction half-reaction at the cathode produces liquid mercury:

The mercury cell is compact and steady, making it ideal for portable devices like hearing aids (ear phones).

Lithium-Ion Discharge

During discharge (spontaneous power delivery):

• Anode (Negative): LiC₆ is oxidized:
  LiC6 → C6 + Li+ + e-
• Cathode (Positive): Li⁺ ions are inserted into cobalt oxide:
  CoO2 + Li+ + e- → LiCoO2

Hence, both lithium ions (Li⁺) and electrons move toward the positive electrode (cathode) during discharge.

Structural Formula Analysis

a) \(\text{C}_n\text{H}_n\text{Cl}_{2n+1}\) (Saturated: was used as an anesthetic):Verification: Total monovalent atoms = \(n + (2n + 1) = 3n + 1\). If \(n = 1\), the formula gives \(\text{CHCl}_{3}\) (Chloroform), which was historically used worldwide as a surgical anesthetic. b) \(\text{C}_n\text{H}_{n+1}\text{Cl}_{n+1}\) (Saturated: Used in dry cleaning):Verification: Total monovalent atoms = \((n + 1) + (n + 1) = 2n + 2\), confirming it is saturated. If \(n = 2\), the formula gives \(\text{C}_2\text{H}_3\text{Cl}_3\) (1,1,1-Trichloroethane), which is a major solvent used in the dry cleaning industry. c) \(\text{C}_n\text{H}_{n+1}\text{Cl}_{n-1}\) ( Used in carpet manufacturing) \(\rightarrow \) TrueTotal monovalent atoms = \((n + 1) + (n - 1) = \{2n}\) (confirms it is unsaturated with a double bond).If \(n = 2\), substituting it into the formula gives: \(\text{C}_2\text{H}_3\text{Cl}\) (Chloroethene / Vinyl chloride). d) \(\text{C}_n\text{H}_{n-1}\text{Cl}_{n-1}\) (Unsaturated: Used in carpet manufacturing):Verification: Total monovalent atoms = \((n - 1) + (n - 1) = 2n - 2\), confirming it is unsaturated. If \(n = 6\), the formula gives \(\text{C}_6\text{H}_5\text{Cl}\) (Chlorobenzene). In industrial organic synthesis, Chlorobenzene undergoes alkaline hydrolysis (reacting with \(\text{NaOH}\))

IUPAC Rule Verification

The correct IUPAC name for d is 4,4-dimethyl-1-pentyne

Drawing the Isomers

To have an ethyl branch (–CH₂CH₃), the longest chain must be at least 5 carbons (pentene) to accommodate the branch at a non-terminal carbon:

1. 1-pentene:
   CH2=CH–CH2–CH2–CH3
2. 2-pentene:
   CH3–CH=CH–CH2–CH3
3. 2-methyl-1-butene:
   CH2=C(CH3)–CH2–CH3

These are the only 3 possible alkene structures containing an ethyl branch.

Nucleophilic Substitution Reaction

The starting material is CH3–CH(CH3)–CH(Br)–CH3. Alkaline hydrolysis substitutes the halogen (Br) with a hydroxyl group (OH):

Numbering from the right gives the -OH group the lowest locant (2), so the IUPAC name is 3-methyl-2-butanol.

Markovnikov's Addition Analysis

• For X (2-methyl-1-butene): H adds to CH₂, and Br adds to the tertiary carbon:
  CH3–CH2–C(CH3)=CH2 + HBr → CH3–CH2–C(Br)(CH3)–CH3 (2-bromo-2-methylbutane).
• For Y (2-methyl-2-butene): H adds to CH, and Br adds to the tertiary carbon:
  CH3–CH=C(CH3)2 + HBr → CH3–CH2–C(Br)(CH3)2 (2-bromo-2-methylbutane).

Synthesis Route

Goal: Convert Methane (CH₄) to Benzoic acid (C₆H₅COOH):

1. Strong heating and rapid cooling: Methane is converted to ethyne (C₂H₂) at 1500°C.
2. Polymerization: Ethyne undergoes cyclic polymerization to form Benzene (C₆H₆).
3. Alkylation: Benzene reacts with CH₃Cl to form Toluene (C₆H₅CH₃).
4. Oxidation: Toluene is oxidized using V₂O₅ catalyst to Benzoic acid.

• Starting compound: C₇H₆O₂ is phenyl formate.
• Target compound: C₆H₆O is phenol.
• Reaction: Acid hydrolysis splits the ester.
• Products: It yields phenol directly.

---

💡 Why Other Options Fail

• Basic hydrolysis: Forms a phenoxide salt.
• Dehydration: Removes water instead.
• Neutralization: Acid-base reaction only.

Step-by-Step Synthesis

Starting material is Phenol (C₆H₆O):

1. Heating with Zinc: Reduces phenol to benzene.
2. Alkylation: Friedel-Crafts alkylation to form toluene.
3. Friedel-Crafts reaction: A second alkylation yields 1,2-dimethylbenzene (o-xylene).
4. Hydrogenation: Saturates the benzene ring into 1,2-dimethylcyclohexane.

Dacron Synthesis

Process 1 :

(Oxidation of Ethylene):Ethylene is oxidized (using an alkaline \(\text{KMnO}_{4}\) solution, known as Bayer's reaction) to directly form Ethylene glycol.\(\text{C}_{2}\text{H}_{4}+[\text{O}]+\text{H}_{2}\text{O}\longrightarrow \text{HO-CH}_{2}\text{-CH}_{2}\text{-OH}\)

Process 2 :

(Oxidation of para-methyl toluene):The two methyl side chains of para-methyl toluene (\(p\)-xylene) are oxidized to form Terephthalic acid.\(\text{CH}_{3}\text{-C}_{6}\text{H}_{4}\text{-CH}_{3}+3\text{O}_{2}\longrightarrow \text{HOOC-C}_{6}\text{H}_{4}\text{-COOH}+2\text{H}_{2}\text{O}\)

Process 3 :

(Condensation Polymerization):The resulting Ethylene glycol and Terephthalic acid undergo condensation polymerization to form the final medical polymer, Dacron.

Step 1: Identify Compound A & B

• Compound A: Phenol (carbolic acid) — reacts with HNO₃ (mineral acid) to form picric acid, but not with HCl.
• Compound B: Formaldehyde (HCHO) — oxidizes to formic acid, used in perfumes.

Step 2: Polymerization Product

Phenol reacts with formaldehyde in an acidic or basic medium to form Bakelite, a thermosetting polymer widely used as an insulator in electrical equipment.

Alloy Classification

• Intermetallic Alloys: Formed by metals chemically combining to produce compounds that do not follow normal chemical valency rules (e.g., Cementite Fe₃C, Gold-lead Au₂Pb, Duralumin Al-Cu).
• Substitution Alloys: Standard physical mixtures where atoms substitute for each other in the crystal lattice. Brass (Cu + Zn) is a substitution alloy.

Comparing Coke in equattion and water gas

In the Midrex furnace, water gas acts as reducing agent (CO + H₂). Similarly, in the given equation, to give iron.

Step 1: Calculate KOH Molarity

Step 2: Titration Math

Reaction: H2SO4 + 2KOH → K2SO4 + 2H2O

Gravimetric Analysis Steps

ZnSO₄·XH₂O + BaCl₂ → BaSO₄ + ZnCl₂ + XH₂O From the equation, 1 mole of ZnSO4·XH2O produces 1 mole of BaSO4.

1. Moles of BaSO₄:
   Molar Mass of BaSO4 = 137.3 + 32 + 64 = 233.3 g/mol
Moles = 1.165 g / 233.3 g/mol = 0.005 mol
2. Mass of Anhydrous ZnSO₄:
   Molar Mass of ZnSO4 = 65.4 + 32 + 64 = 161.4 g/mol
Mass of ZnSO4 = 0.005 mol × 161.4 g/mol = 0.807 g
3. Mass of Water:
   Mass of H2O = 1.437 g - 0.807 g = 0.630 g
Moles of H2O = 0.630 g / 18 g/mol = 0.035 mol
4. Determine X:
   X = Moles of H2O / Moles of ZnSO4 = 0.035 / 0.005 = 7

another solution

Balanced Chemical Equation:
ZnSO4·XH2O + BaCl2 → BaSO4 + ZnCl2 + XH2O

From the equation, 1 mole of ZnSO4·XH2O produces 1 mole of BaSO4.

Molar Mass of BaSO4 = 137.3 + 32 + (16 × 4) = 233.3 g/mol
Molar Mass of ZnSO4 = 65.4 + 32 + (16 × 4) = 161.4 g/mol
Molar Mass of H2O = (1 × 2) + 16 = 18 g/mol

Moles of BaSO4 precipitated = 1.165 g / 233.3 g/mol = 0.005 mol
Since the mole ratio is 1:1, Moles of ZnSO4·XH2O = 0.005 mol

Molar Mass of ZnSO4·XH2O = 1.437 g / 0.005 mol = 287.4 g/mol

Mass of XH2O = 287.4 g/mol - 161.4 g/mol = 126 g/mol
X = 126 g/mol / 18 g/mol = 7

Molecular Formula = ZnSO4·7H2O

Step 1: Calculate Solubility (S)

Step 2: Dissociation and K_sp Calculation

Mg(OH)2 ⇌ Mg2+ + 2OH-

Standard Potential Analysis

From Cell 1 (A/C) and Cell 2 (C/B):

• Since electrons flow from left to right in standard galvanic configurations, we find:
  E°ox(A) > E°ox(C) with difference of 1.2 V.
  E°ox(C) > E°ox(B) with difference of 0.8 V.
• Thus, the activity/oxidation potential order is: A > C > B.
• In a cell combining A and B: A is the natural anode and B is the natural cathode with spontaneous EMF = +2.0 V.
• If forced to run with B as anode and A as cathode, the reaction is reversed and becomes non-spontaneous with EMF = -2.0 V.

Electrochemical Activity Series

• Since nickel is more active than copper (Ni > Cu), nickel is oxidised and reduces copper ions. Thus, a nickel container would react and dissolve.
• Because copper is less active than nickel, it cannot displace Ni²⁺ from solution. Thus, nickel salts can be safely stored in copper containers.

Reagent Verification

Reagent	Ethyl Alcohol	Phenol	Distinguishable?
Bromine water	No reaction	White ppt	Yes
Sodium (Na)	H2 gas evolved	H2 gas evolved	No
FeCl3	No reaction	Violet color	Yes

Identification and Analysis

• Compound 1: Aliphatic Alcohol (e.g., Ethanol) — formula C_nH_2n+2O.
• Compound 2: Carboxylic Acid — formula C_nH_2nO₂ (reacts with both alcohols and NaOH).
• Compound 3: Phenol — acidic but cannot be prepared by simply oxidizing an aliphatic alcohol.

Step 1: Identify the Alcohol

Alcohols resistant to oxidation are tertiary alcohols (such as 2-methyl-2-propanol).

Step 2: Conversion Pathway

Reaction Analysis

1. X (Acetylene / Ethyne): H-C≡C-H (undergoes addition in 2 steps).
2. Y (Acetaldehyde / Ethanal): CH3CHO formed via catalytic hydration of X.
3. Z (Ethanol): CH3CH2OH formed by reduction of Y (primary alcohol).
4. W (Ethene): formed by dehydration of Z.

Esterification of Lactic Acid

The bifunctional molecule is lactic acid [CH₃CH(OH)COOH]:

• Reaction 1 (Acid function): reacts at –COOH with methanol to yield CH₃CH(OH)COOCH₃.
• Reaction 2 (Alcohol function): reacts at –OH with acetic acid to yield CH₃CH(OCOCH₃)COOH.

(1) Highest pH Value

Solution C (0.02M). In a weak acid, dilution decreases the hydronium ion concentration [H₃O⁺], which consequently increases the pH value.

(2) Highest Fluoride Ion Concentration

Solution A (0.4M). Higher initial analytical concentration results in a larger total concentration of dissociated F^- ions at equilibrium, because [F-] = √(Ka × Ca).

(3) K_a Value of Solution B

6.7 × 10^-4. The acid dissociation constant (K_a) is temperature-dependent and remains constant for the same acid at room temperature.

(4) Finding Solution with Dissociation Percentage = 4.9%

This matches Solution B.

Step-by-Step Pathway

• A: Formic acid (CH₂O₂)
• B: Methanol (CH₃OH)
• C: Methyl formate (HCOOCH₃)
• D: Formamide (HCONH₂)

Answers:

1- Common Name for Compound (A): Formic acid

2- Structural Formula of the ester formed (Methyl salicylate / Oil of Wintergreen):

3- IUPAC Name for Compound (C): Methyl methanoate

4- Structural Formula of Compound (D):