1. General Introduction

d-Block Elements: Elements in which the last electron enters the d-orbital of the penultimate shell (n-1).
Position: Groups 3 to 12.

General Electronic Configuration:

(n-1)d^1-10 ns^1-2

Transition Elements Definition:

Elements which have incompletely filled d-orbitals in their ground state or in any of their stable oxidation states.
Note: Zn, Cd, Hg are d-block elements but NOT transition elements (full d¹⁰ config in ground & stable states).

2. Characteristic Properties

1. Variable Oxidation States

They show variable O.S. because energy difference between (n-1)d and ns orbitals is small.
• Mn shows max range (+2 to +7).
• Os and Ru show highest O.S. (+8).
• Stability depends on d⁰, d⁵, d¹⁰ config.

2. Magnetic Properties

Most are Paramagnetic due to unpaired electrons.
Spin Only Magnetic Moment (μ):

μ = √[n(n+2)] B.M.

(n = number of unpaired electrons)

3. Formation of Colored Ions

Color is due to d-d transition of electrons.
• Unpaired d-electrons absorb light from visible region to jump to higher energy d-orbitals (Crystal Field Splitting).
• Ions with d⁰ (Sc³⁺) or d¹⁰ (Zn²⁺) are Colorless.

4. Catalytic Properties

Due to variable valency (can form unstable intermediates) and large surface area. (e.g., V₂O₅ in Contact Process, Fe in Haber’s Process).

5. Interstitial Compounds & Alloys

Interstitial: Small atoms (H, C, N) trapped in crystal lattice. (Hard, high MP, chemically inert).
Alloys: Formed because transition metals have similar radii, allowing atoms to replace each other in lattice (e.g., Brass, Bronze).

3. Important Compounds

1. Potassium Dichromate (K₂Cr₂O₇):

Prepared from Chromite Ore (FeCr₂O₄).
Chromate-Dichromate Equilibrium:
2CrO₄^2- (Yellow) + 2H⁺ ↔ Cr₂O₇^2- (Orange) + H₂O.
(Yellow in Alkali, Orange in Acid).

2. Potassium Permanganate (KMnO₄):

Prepared from Pyrolusite Ore (MnO₂).
MnO₂ → MnO₄^2- (Green Manganate) → MnO₄^- (Purple Permanganate).
Strong Oxidizing Agent in Acidic, Basic, and Neutral media.

4. f-Block Elements

Last electron enters (n-2)f orbitals. General Config: (n-2)f^1-14 (n-1)d^0-1 ns².

Lanthanoids (4f Series)

Oxidation State: Most common is +3. Ce shows +4 (Noble gas config). Eu shows +2 (Half-filled f⁷).
Chemical Reactivity: Similar to Calcium. React with water to release H₂.

Lanthanoid Contraction:

The steady decrease in atomic and ionic radii from La to Lu.
Cause: Poor shielding effect of 4f electrons.
Consequences:
1. Similarity in size of 4d and 5d series (e.g., Zr ≈ Hf).
2. Difficulty in separating Lanthanides.
3. Decrease in basic strength of hydroxides (La(OH)₃ > Lu(OH)₃).

Actinoids (5f Series)

Contraction: Actinoid contraction is greater than Lanthanoid contraction (5f shielding is poorer than 4f).
Oxidation States: Show wider range (+3, +4, +5, +6, +7) because energy gap between 5f, 6d, and 7s is very small.

5. Uses

Titanium: Structural metal for aerospace (light & strong).
Mischmetal: Alloy of Lanthanides (~95%) + Iron (~5%). Used in lighter flints and bullets.
Ziegler-Natta Catalyst: (TiCl₄ + Al(C₂H₅)₃) for polythene manufacture.
V₂O₅: Catalyst for H₂SO₄ manufacture.

Numericals & HOTS

Q1. Calculating Magnetic Moment

Calculate the spin-only magnetic moment of M²⁺ ion (Z = 27).

Solution:
Element with Z = 27 is Cobalt (Co).
Electronic Config of Co: [Ar] 3d⁷ 4s².
For Co²⁺, remove 2 electrons from 4s: [Ar] 3d⁷.

Unpaired Electrons (n):
d-orbitals: ↑↓ ↑↓ ↑ ↑ ↑ (Hund's Rule).
n = 3.

μ = √[n(n + 2)] = √[3(3 + 2)] = √15.
μ ≈ 3.87 B.M.

Q2. KMnO₄ Stoichiometry

How many moles of acidified KMnO₄ are required to oxidize 1 mole of Ferrous Oxalate (FeC₂O₄)?

Solution:
Oxidation Half:
Fe²⁺ → Fe³⁺ + 1e^-
C₂O₄^2- → 2CO₂ + 2e^-
Total electrons lost by 1 mole FeC₂O₄ = 1 + 2 = 3.

Reduction Half:
MnO₄^- + 5e^- → Mn²⁺ (Acidic medium).
Electrons gained per mole KMnO₄ = 5.

Equating Equivalents:
Moles(KMnO₄) × 5 = Moles(FeC₂O₄) × 3
x × 5 = 1 × 3
x = 0.6 moles

Q3. Cr²⁺ vs Mn³⁺ (HOTS)

Cr²⁺ and Mn³⁺ both have d⁴ configuration. Yet Cr²⁺ is a reducing agent while Mn³⁺ is an oxidizing agent. Why?

Solution:
Cr²⁺ (Reducing): It loses an electron to become Cr³⁺ (d³). In aqueous solution, Cr³⁺ has a half-filled t_2g³ level, which is exceptionally stable in crystal field splitting.

Mn³⁺ (Oxidizing): It gains an electron to become Mn²⁺ (d⁵). The d⁵ configuration is exactly half-filled and symmetric, providing extra stability.
Thus, nature drives them towards stable configurations.

Q4. Identify Compound Formula

The magnetic moment of a chloride of Titanium (Z=22) is 1.73 B.M. Determine the formula of the chloride.

Solution:
μ = 1.73 B.M. implies unpaired electrons (n) = 1 (since √3 ≈ 1.73).

Ti (Z=22): [Ar] 3d² 4s².
To have n=1, the ion must be 3d¹.
This means 2 electrons from 4s and 1 from 3d are removed. Total 3 electrons removed.
Oxidation State = +3.
Formula: TiCl₃

Q5. Basicity Trend

Which is more basic: La(OH)₃ or Lu(OH)₃? Explain using Lanthanoid Contraction.

Solution:
La(OH)₃ is more basic.

Due to Lanthanoid Contraction, size decreases from La³⁺ to Lu³⁺. According to Fajan's rule, as size of cation decreases, the covalent character of the M-OH bond increases.
Therefore, Lu-OH bond is stronger and harder to break (less OH^- release). La-OH is more ionic and releases OH^- easily.

Q6. Copper in Aqueous Solution

Why are Cu⁺ salts unstable in aqueous solution and undergo disproportionation?

Solution:
Reaction: 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s).

Although Cu²⁺ requires high ionization energy to form, it is much more stable in water than Cu⁺ because the Hydration Enthalpy of Cu²⁺ is very high (negative). This energy release compensates for the ionization energy, driving the reaction forward.

Q7. Mechanism of Catalysis

Explain how Iron catalyzes the reaction between Iodide and Persulphate ions.

Solution:
Reaction: 2I^- + S₂O₈^2- → I₂ + 2SO₄^2-.
Iron catalyzes this by cycling between Fe²⁺ and Fe³⁺:

1. 2Fe³⁺ + 2I^- → 2Fe²⁺ + I₂
2. 2Fe²⁺ + S₂O₈^2- → 2Fe³⁺ + 2SO₄^2-
Iron lowers activation energy by offering an alternate path using its variable oxidation states.

Q8. Color Prediction

Out of Ti³⁺, V³⁺, Cu⁺, and Sc³⁺, which ions are colorless in aqueous solution?

Solution:
Color requires d-d transitions (partially filled d-orbitals).
Ti³⁺ (d¹): Colored (Purple).
V³⁺ (d²): Colored (Green).
Cu⁺ (d¹⁰): Full d-orbital. No transition. Colorless.
Sc³⁺ (d⁰): Empty d-orbital. No transition. Colorless.

Ans: Cu⁺ and Sc³⁺

Q9. Zr vs Hf

The atomic radii of Zirconium (Zr, Group 4, Period 5) and Hafnium (Hf, Group 4, Period 6) are almost identical (160 pm vs 159 pm). Why?

Solution:
This is due to Lanthanoid Contraction.
Before Hf, the 4f subshell is filled (Lanthanoids). 4f electrons have poor shielding effect.
The increase in nuclear charge pulls the outer shell inwards more strongly than the shielding provided by 4f electrons can repel it.
This contraction cancels out the expected size increase due to the new shell.

Q10. Zinc's Low Enthalpy

Why does Zinc have the lowest enthalpy of atomization in the 3d series?

Solution:
Enthalpy of atomization depends on the strength of metallic bonding.
Metallic bonding strength increases with the number of unpaired electrons participating in the bond.
Zinc (3d¹⁰ 4s²) has no unpaired electrons. The interatomic bonding is weak, leading to low enthalpy of atomization (and low melting point).

Important Formulae & Reactions

1. Spin Only Magnetic Moment

μ = √[n(n + 2)] B.M.

(n = Number of unpaired electrons)
B.M. = Bohr Magneton

2. K₂Cr₂O₇ Reactions

Chromate-Dichromate Equilibrium:

2CrO₄^2- (Yellow) + 2H⁺ ↔ Cr₂O₇^2- (Orange) + H₂O

Oxidizing Action (Acidic):

Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

(Eq. Wt. = M / 6)

3. KMnO₄ Reactions

Acidic Medium (n=5):

MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O

Neutral/Faintly Alkaline (n=3):

MnO₄^- + 2H₂O + 3e^- → MnO₂ + 4OH^-

Strongly Alkaline (n=1):

MnO₄^- + e^- → MnO₄^2-

20 Golden Facts (NEET)

1. Highest Oxidation State: Osmium (Os) and Ruthenium (Ru) show the highest oxidation state of +8 (e.g., OsO₄). In 3d series, Manganese shows max +7 (KMnO₄).
2. Enthalpy of Atomization: Transition metals have high Δ_aH° because of strong metallic bonding involving both ns and (n-1)d electrons. Max at middle of series (Cr, Mo, W).
3. Lanthanoid Contraction: The atomic radii of 4d and 5d metals in the same group (e.g., Zr and Hf) are almost identical (160 pm and 159 pm). This makes their separation difficult.
4. Colour of Ions:
• Cu⁺ (3d¹⁰): Colourless
• Cu²⁺ (3d⁹): Blue
• Zn²⁺, Sc³⁺, Ti⁴⁺: Colourless (d⁰ or d¹⁰).
5. E° Value of Cu: Copper has a positive E° (Cu²⁺/Cu = +0.34 V). This is because the high hydration enthalpy of Cu²⁺ does not compensate for the high sum of ionization enthalpies (IE₁ + IE₂). Hence, Cu does not liberate H₂ from acids.
6. Interstitial Compounds: They are chemically inert but physical properties change (e.g., steel is harder than pure iron). They retain metallic conductivity.
7. Mischmetal: An alloy of Lanthanoids (~95%), Iron (~5%) and traces of S, C, Ca, Al. Used in magnesium-based alloy for bullets and lighter flints.
8. KMnO₄ Analysis: In volumetric analysis, KMnO₄ acts as a self-indicator. The end point is the appearance of a permanent light pink colour. HCl is NOT used to acidify KMnO₄ because it gets oxidized to Cl₂.
9. Europium & Ytterbium: While most Lanthanoids are +3, Eu (4f⁷) and Yb (4f¹⁴) show +2 oxidation states due to half-filled and full-filled stability. They act as strong Reducing Agents.
10. Spin Only Formula: The magnetic moment of transition metals is determined mainly by spin angular momentum. Orbital contribution is quenched by the crystal field.
11. Cr²⁺ vs Mn³⁺: Both have d⁴ config.
• Cr²⁺ is Reducing (wants to go to stable d³ t_2g³).
• Mn³⁺ is Oxidizing (wants to go to stable d⁵ Mn²⁺).
12. Coinage Metals: Cu, Ag, Au (Group 11). They have d¹⁰s¹ config but are transition metals because Cu²⁺ (d⁹) and Au³⁺ (d⁸) have incomplete d-shells.
13. Actinoid Contraction: It is greater than Lanthanoid contraction from element to element because 5f electrons shield the nuclear charge less effectively than 4f electrons.
14. Density Trend: Density increases from Sc to Cu. Huge jump in density from 3d to 4d/5d series (e.g., Osmium density ~22.6 g/cc).
15. Basicity of Lanthanoids: La(OH)₃ is the most basic and Lu(OH)₃ is the least basic. This is because size decreases (La → Lu), Covalent character increases, and ability to lose OH^- decreases.
16. MnO₄^- Geometry: Permanganate ion has a Tetrahedral geometry involving d³s hybridization of Manganese. The purple colour is due to Charge Transfer (O → Mn), not d-d transition.
17. Photography: AgBr is used in photography. Hypo (Na₂S₂O₃) is used as a fixer to remove unexposed AgBr by forming a soluble complex.
18. Hg is Liquid: Mercury is a liquid metal because its 6s² electrons are very stable (Relativistic effect) and do not participate well in metallic bonding (weak bonding).
19. Disproportionation of Mn: Mn(VI) (Green Manganate, MnO₄^2-) is unstable in acidic medium and disproportionates to Mn(VII) (Purple) and Mn(IV) (Brown MnO₂).
20. Catalytic Activity: Transition metals are good catalysts (e.g., Fe, V₂O₅, Ni) because they provide a surface for adsorption and can change oxidation states to form intermediates.

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