Transition Elements

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↑

Step 1: Write Electronic Configurations

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

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

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.

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.

• 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.

• 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
• 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 DuraluminNiAl₃ 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.

• Element (X): Vanadium (V), as V₂O₅ is used as a catalyst for superconducting magnets. Vanadium ([Ar] 3d³ 4s²) has 3 unpaired electrons.
• Element (Y): Contains twice as many unpaired electrons (3 × 2 = 6). This is Chromium (Cr, [Ar] 3d⁵ 4s¹), which has 6 unpaired electrons.
• The maximum oxidation state of Cr (Y) is +6, which is higher than that of V (X), which is +5.

1. Ore Dressing: Roasting the ore to dry and oxidize impurities.
2. Reduction Stage: Using a Midrex furnace (or Blast furnace) to reduce hematite to iron.
3. Steel Production: Using an Oxygen converter (or electric/open-hearth furnace) to produce steel.

Step 1: Identify the Elements

The prompt states that both A²⁺ and B²⁺ ions have 4 unpaired electrons. Let's analyze the 3d configurations:

Step 2: Assign A and B based on Density

The prompt also states that element (A) is less dense than element (B).

Therefore, A = Chromium (Cr) and B = Iron (Fe).

Step 3: Match Uses

Comparing with the options, (c) is the correct match.

Step 1: Identify the Elements

Step 2: Analyze the Stability of Conversions

A conversion to a "lower energy" compound means forming a more stable product. We check the oxidation states.

Step 1: Identify Elements A, B, and C

Step 2: Order by Density

Density generally increases across a period in the d-block. Let's compare the densities of the identified elements.

The correct order of increasing density is:

Step 1: Identify Element B and the Sequence

Given: B³⁺ has the configuration [Ar] 3d³. To find the neutral atom B, we add back 3 electrons. They go into the 4s and 3d orbitals.

Neutral B configuration: [Ar] 4s² 3d⁴. This is the expected configuration for Chromium (Cr), Atomic Number 24 (although its actual configuration is anomalous: 4s¹3d⁵).

The four consecutive elements are:

Step 2: Evaluate Statement (c)

Statement (c) is about element C = Manganese (Mn). It describes losing "the first electron from the (d) sublevel". This typically refers to the oxidation process from a common ion, in this case Mn²⁺ → Mn³⁺.

The number of unpaired electrons decreases from 5 to 4. Since magnetic moment is directly related to the number of unpaired electrons, the magnetic moment decreases. This matches the statement.

Key Concept: Magnetic Moment of Iron

To increase the magnetic moment, we need to increase the number of unpaired electrons. This means oxidizing Fe²⁺ to Fe³⁺.

Analyzing the Options:

(a) FeCl₃ + NH₄OH → Fe(OH)₃ → Fe₂O₃: The iron starts as Fe³⁺ and ends as Fe³⁺. No change.

(b) Fe + H₂SO₄ → FeSO₄; then heat:

• Step 1: Fe + H₂SO₄(dil) → FeSO₄ + H₂. The compound formed is Iron(II) sulfate (Fe²⁺).
• Step 2: 2FeSO₄(s) --(heat)→ Fe₂O₃(s) + SO₂(g) + SO₃(g). The final product is Iron(III) oxide (Fe³⁺).

(c) Limonite (Fe₂O₃·nH₂O) heated: Starts as Fe³⁺ and ends as Fe³⁺. No change.

(d) Siderite (FeCO₃) heated in absence of air: FeCO₃ → FeO + CO₂. The iron starts as Fe²⁺ and ends as Fe²⁺. No change.

1. Identify the precipitates: AgNO₃ forms Ag₃PO₄ (yellow, soluble in NH₄OH) and AgI (yellow, insoluble in NH₄OH).

2. Mass of AgI: Since excess NH₄OH dissolves the Ag₃PO₄, the remaining 8.5g is entirely AgI.
Moles of AgI = 8.5g / (108+127)g/mol = 8.5 / 235 = 0.03617 mol.

3. Find KI mass in mixture: Moles KI = Moles AgI = 0.03617 mol.
Mass KI = 0.03617 × (39+127) = 0.03617 × 166 = 6.00 g.

4. Find Na₃PO₄ mass: Total mixture = 12g. Mass Na₃PO₄ = 12g - 6g = 6.00 g.
Moles Na₃PO₄ = 6.00 / (23×3 + 31 + 16×4) = 6.00 / 164 = 0.03658 mol.

5. Find Ag₃PO₄ mass: Moles Ag₃PO₄ = Moles Na₃PO₄ = 0.03658 mol.
Mass Ag₃PO₄ = 0.03658 × (108×3 + 31 + 16×4) = 0.03658 × 419 = 15.32 g.

6. Total initial yellow precipitate (X): Mass AgI + Mass Ag₃PO₄ = 8.5g + 15.32g = 23.82 g.

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

Alloy Analysis

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

Step 1: Determine Unpaired Electrons in Copper

Step 2: Identify Element X

Double the unpaired electrons = 2 unpaired electrons.

Nickel (Ni): [Ar] 3d⁸ 4s²

Step 3: Analyze the Uses of Nickel

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

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

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

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

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

Step 1: Read the Bar Chart

Step 2: Analyze Stability of Each Process

Step 3: Why W³⁺ → W²⁺ is Easiest

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

Step 1: Identify the Periodic Table Positions

Step 2: Analyze Each Option

Step 3: Conclusion

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

MgCl₂ + H₂SO₄ → MgSO₄ + 2HCl

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

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

m(MgCl₂) = 0.08333 × 95 = 7.916 g

% = (7.916/10) × 100 = 79.16%

The correct option is b) X: Iron, Y: Titanium, Z: Cobalt.

Medical Uses of Transition Metals

• X = Iron (Fe): Stainless steel (Fe + Cr + Ni) is the standard alloy for surgical instruments.
• Y = Titanium (Ti): Biocompatible, lightweight, and corrosion-resistant — ideal for bone implants and artificial joints.
• Z = Cobalt (Co): The radioactive isotope ⁶⁰Co is used in radiotherapy for malignant tumors.

The correct option is a).

Step 1: Identify Elements

• X²⁺ = [Ar]3d⁴ → X = Chromium (Cr, Z=24); Cr = [Ar]3d⁵4s¹
• Y²⁺ = [Ar]3d² → Y = Titanium (Ti, Z=22); Ti = [Ar]3d²4s²

Step 2: Analysis

Chromium has only 6 valence electrons (3d⁵4s¹), so its maximum oxidation state is +6. At X⁶⁺ the configuration is [Ar]3d⁰ — a complete, stable argon core. Removing a 7th electron means breaking into the noble-gas core, which is extremely difficult. ✓

• (b) ✗ Ti⁴⁺ (3d⁰) is its most stable state, so oxidizing Ti²⁺ → Ti⁴⁺ is easy.
• (c) ✗ Cr⁶⁺ is reduced readily to Cr³⁺ (most stable), not Cr²⁺.
• (d) ✗ Ti⁴⁺ (3d⁰) is very stable; it resists gaining an electron, so reduction to Ti³⁺ is not easy.

The correct option is a) Z²⁺ > Y²⁺ > X²⁺.

Identify the Elements (by atomic mass: Y < Z < X)

• Y = Nickel (Ni, 58.7): Ni²⁺ = [Ar]3d⁸ → 2 unpaired e⁻ → μ = 2.83 BM
• Z = Cobalt (Co, 58.9): Co²⁺ = [Ar]3d⁷ → 3 unpaired e⁻ → μ = 3.87 BM
• X = Copper (Cu, 63.5): Cu²⁺ = [Ar]3d⁹ → 1 unpaired e⁻ → μ = 1.73 BM

μ = √[n(n+2)]. Therefore the order is Co²⁺ (3.87) > Ni²⁺ (2.83) > Cu²⁺ (1.73) → Z²⁺ > Y²⁺ > X²⁺ ✓

Note: Ni²⁺ (2 unpaired) has a higher moment than Cu²⁺ (1 unpaired), so Y²⁺ > X²⁺.

The correct option is d).

Process Flow for Stainless Steel

• 1. Sintering: Fine ore dust fused into lumps suitable for blast furnace.
• 2. Roasting: Remove sulfur impurities; convert ore to Fe₂O₃.
• 3. Reduction: Fe₂O₃ + CO → Fe metal in blast furnace.
• 4. Add C + Cr: Controlled carbon + 10–18% Cr → stainless steel.

The correct option is d).

The lowest oxidation state shown in the first transition series is +1, exhibited by Copper (Cu) [as in Cu⁺]. So X = Copper.

• Type of alloy: Copper combines with metals of clearly different atomic radii to form an intermetallic alloy (fixed-ratio compound phases).
• Use: Copper and its alloys are the classic materials for the manufacture of coins. ✓

(c) ✗ refers to Titanium (MiG jets), whose lowest state is +2 — not the lowest in the series.

The correct option is a).

Hydrogen at 800 °C carries out the complete reduction of Fe₂O₃ to metallic iron (Fe); strong heating of iron in air then forms the magnetic oxide Fe₃O₄. ✓

🎯 The Core Chemical Rule: Ground-state transition atoms share identical numbers of unpaired electrons when one has a less-than-half-filled subshell and the other is more-than-half-filled. This points directly to the pairings of Ti / Ni (2 unpaired) or V / Co (3 unpaired).

✅ Identifying the Elements:
• The Electron Decrease: Moving from X²⁺ to X³⁺ decreases the count of unpaired electrons. This only happens in less-than-half-filled subshells where losing an electron lowers the count (e.g., V²⁺ [3d³, 3 unpaired] to V³⁺ [3d², 2 unpaired]). Therefore, Element X is an earlier transition metal (Ti or V).
• Deducing Element Y: Because Element Y shares the same ground-state unpaired electron count but does not match this reduction behavior, it represents the later transition metal counterpart (Ni or Co).

📊 Periodic Trends Evaluation:
• Atomic Mass (Option c): Atomic mass regularly increases across the series from left to right. Since Y sits further right in the series than X, the atomic mass of Y is larger than X.

🎯 The Core Chemical Rule: Industrial alloy design combines different structural types. Interstitial components add tensile strength and rigidity, while substitutional components modify specific chemical and physical traits like elasticity and corrosion resistance.

✅ Structural Anatomy of Vanadium Steel:
• The Interstitial Steel Base: Small carbon atoms fit tightly into the interstitial spaces of the iron crystal structure, blocking metal layers from sliding and reinforcing structural hardness.
• The Substitutional Vanadium Modification: Added vanadium atoms swap into original iron lattice positions because vanadium and iron have closely matching atomic radii and shared lattice geometries, yielding high structural elasticity.

💡 Application Insight: This combination makes vanadium steel ideal for car springs, as it withstands massive mechanical stress and deformation without breaking.

🎯 The Core Chemical Rule: For a compound to reduce Mn⁷⁺ down to Mn²⁺, it must contain a transition element that readily oxidizes by losing electrons to achieve a highly stable, half-filled (3d⁵) or completely filled (3d¹⁰) subshell.

✅ Why Iron(II) Sulfate Succeeds:
• The Reactivity of Iron: In FeSO₄, iron exists as the Fe²⁺ ion (3d⁶). It smoothly oxidizes to Fe³⁺ (3d⁵) by sacrificing one electron. This resulting 3d⁵ system is exceptionally stable because its five 3d orbitals are uniformly half-filled.
• Thermodynamic Driving Force: Because Fe²⁺ is highly driven to transform into stable Fe³⁺, it functions as an excellent reducing agent capable of driving the reduction of Mn⁷⁺.

❌ Why Other Options Fail:
• ScCl₃ & Ti(NO₃)₄: Exist as Sc³⁺ (3d⁰) and Ti⁴⁺ (3d⁰). They are in their highest, fully empty oxidation states and cannot undergo further oxidation.
• ZnCl₂: Features Zn²⁺ (3d¹⁰). Its completely full d-subshell resists losing more electrons.

🎯 The Core Chemical Rule: To determine the true chemical identity of an unknown cation, calculate its atomic number (Z) by adding the positive charge value back to the electron count of its core configuration.

✅ Deciphering the Cations:
• Element X: Given X²⁺ = [Ar] 3d⁶, we add 2 electrons back to the 4s subshell to find the neutral atom configuration: [Ar] 4s² 3d⁶. This corresponds to Iron (Fe, Z=26).
• Element Y: Given Y⁴⁺ = [Ar] 3d⁶, we add 4 total electrons back (2 to 4s and 2 to 3d) to find the neutral atom configuration: [Ar] 4s² 3d⁸. This corresponds to Nickel (Ni, Z=28).

🔍 Row Evaluation:
• Iron (X) is structural and used to build durable reinforced concrete structures.
• Nickel (Y) possesses exactly 5 stable isotopes (unlike cobalt, which features 12 radioactive isotopes). This matches Row (a).

🎯 The Core Chemical Rule: Paramagnetic substances possess at least one unpaired electron in their d-subshell, whereas diamagnetic substances have all their electrons completely paired up (3d⁰ or 3d¹⁰).

✅ Identifying the Three Consecutive Elements:
• Element X (Group 8): Represents **Nickel (₂₈Ni)**. Its common oxidation state is Ni²⁺ ([Ar] 3d⁸), which has 2 unpaired electrons and is paramagnetic. Group 8 elements never form diamagnetic compounds in their standard ionic states, making all its compounds paramagnetic.
• Element Y (Group IB): Represents **Copper (₂₉Cu)**. In the Cu⁺ state ([Ar] 3d¹⁰), it has no unpaired electrons (diamagnetic), while in the Cu²⁺ state ([Ar] 3d⁹), it has 1 unpaired electron (paramagnetic).
• Element Z (Group IIB): Represents **Zinc (₃₀Zn)**. Its only oxidation state is Zn²⁺ ([Ar] 3d¹⁰), where the d-subshell is fully filled. With zero unpaired electrons, all its compounds are strictly diamagnetic.

📊 Conclusion: Since Nickel (Ni), Copper (Cu), and Zinc (Zn) are three consecutive elements with atomic numbers 28, 29, and 30 respectively, their corresponding groups are **Group 8, Group IB, and Group IIB**, which perfectly matches Row (b).