🧪 Chemistry · Inorganic 🔥 Day 1

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Inorganic Chemistry — Chemistry

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Periodic Table & Chemical Bonding

Periodicity · VSEPR · Hybridisation · MO theory · Hydrogen bonding

Periodic Table Organisation

s-block (Groups 1–2): alkali and alkaline earth metals — fill ns orbitals
p-block (Groups 13–18): fill np orbitals; includes metalloids, nonmetals, noble gases
d-block (Groups 3–12): transition metals — fill (n−1)d orbitals
f-block (Lanthanides, Actinides): fill (n−2)f orbitals; placed separately below
Period number = number of shells; Group number relates to valence electrons

Periodic Trends

Property	Across Period (→)	Down Group (↓)	Key Reason
Atomic radius	Decreases	Increases	↑ Z_eff across; ↑ shells down
Ionic radius (cation)	Decreases	Increases	Same trend as atomic radius
Ionisation energy (IE)	Increases	Decreases	↑ Z_eff; farther e⁻ easier to remove
Electron affinity (EA)	Increases (general)	Decreases	Compact atoms hold extra e⁻ better
Electronegativity	Increases	Decreases	F most electronegative (3.98)

Key anomalies in IE: Be > B (2s vs 2p); N > O (half-filled 2p stable); Mg > Al; P > S — extra stability of half-filled/filled subshells.

Electron affinity anomalies: N has zero/negative EA (half-filled 2p repels extra e⁻); F < Cl (F is too small, e⁻ repulsion).

Chemical Bonding Types

Bond Type	Formed by	Character	Example
Ionic	Complete e⁻ transfer	Electronegativity diff > 1.7	NaCl, MgO
Covalent (polar)	Shared pairs, unequal	ΔEN = 0.4–1.7	HCl, H₂O
Covalent (non-polar)	Shared pairs, equal	ΔEN < 0.4	H₂, Cl₂, CH₄
Metallic	Sea of electrons	Delocalised	Na, Fe, Cu
Coordinate (dative)	Both e⁻ from one atom	Lewis acid–base	NH₄⁺, BF₃·NH₃

Formal charge = Valence e⁻ − Non-bonding e⁻ − ½(Bonding e⁻)
Lattice energy ∝ (charge product)/(ionic radii sum); more negative = more stable ionic compound
Bond order = (bonding e⁻ − antibonding e⁻) / 2; higher BO → shorter, stronger bond

VSEPR Theory — Molecular Shapes

BP + LP	Bond Pairs	Lone Pairs	Shape	Example
2	2	0	Linear (180°)	BeCl₂, CO₂
3	3	0	Trigonal planar (120°)	BF₃, SO₃
3	2	1	Bent (~120°)	SO₂, SnCl₂
4	4	0	Tetrahedral (109.5°)	CH₄, NH₄⁺
4	3	1	Trigonal pyramidal (~107°)	NH₃, PCl₃
4	2	2	Bent (~104.5°)	H₂O, H₂S
5	5	0	Trigonal bipyramidal	PCl₅
6	6	0	Octahedral (90°)	SF₆, PCl₆⁻
6	4	2	Square planar	XeF₄, ICl₄⁻

Lone pair repulsion > bond pair repulsion → lone pairs compress bond angles (NH₃: 107°, H₂O: 104.5° vs 109.5° tetrahedral).

Hybridisation

Hybridisation	Geometry	Bond angle	Examples
sp	Linear	180°	BeCl₂, C₂H₂, CO₂
sp²	Trigonal planar	120°	BF₃, C₂H₄, SO₃, graphite
sp³	Tetrahedral	109.5°	CH₄, NH₃, H₂O, diamond
sp³d	Trigonal bipyramidal	90°, 120°	PCl₅, SF₄
sp³d²	Octahedral	90°	SF₆, PCl₆⁻, XeF₄

Molecular Orbital Theory

MOs formed by LCAO: bonding MO (lower energy, σ or π), antibonding MO (higher energy, σ* or π*)
Filling order (homonuclear diatomics up to N₂): σ1s, σ*1s, σ2s, σ*2s, π2p (before σ2p), σ2p, π*2p, σ*2p
For O₂, F₂: σ2p fills before π2p (order reverses after N₂)
Bond order = (bonding − antibonding)/2; paramagnetism if unpaired electrons in MOs
O₂: BO = 2, paramagnetic (2 unpaired e⁻ in π*2p) — explains liquid O₂ being attracted to magnets
N₂: BO = 3, diamagnetic; CO: BO = 3, diamagnetic (isoelectronic with N₂)

Hydrogen Bonding

Forms between H bonded to highly electronegative atom (F, O, N) and lone pair of another F/O/N
Intermolecular H-bonding: H₂O, HF, NH₃, carboxylic acids — raises BP and melting point
Intramolecular H-bonding: o-nitrophenol, salicylaldehyde — lowers BP (no intermolecular association)
HF has highest BP among hydrogen halides (strong H-bonding); H₂O has anomalously high BP vs H₂S
Ice is less dense than water (open H-bond network) — maximum density at 4°C

💡 JEE Tip: For hybridisation, count: steric number = bond pairs + lone pairs on central atom. sp = 2, sp² = 3, sp³ = 4, sp³d = 5, sp³d² = 6. XeF₂ has 3 LP + 2 BP = 5 → sp³d → linear shape (3 LPs in equatorial positions).

⚠ Common Error: IE₁ of N > O because N has half-filled 2p (stable). IE₁ of Be > B because 2s (Be) > 2p (B) in stability. These two anomalies are the most frequently tested. Don't just state "increases across period" without noting exceptions.

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Periodic Table & Bonding — Formula Sheet

Periodic Trends (Key Relationships)

$Z_{eff} = Z - \sigma$ (Slater's rules)

Effective nuclear charge; $\sigma$ = shielding constant. Higher $Z_{eff}$ → smaller radius, higher IE

IE order anomaly: Be > B, N > O, Mg > Al, P > S

Half-filled (N, P) and fully-filled (Be, Mg) subshells are extra stable → higher IE than neighbours

Lattice energy $\propto |q^+||q^-|/(r^+ + r^-)$

Higher charge and smaller ions → more negative lattice energy → more stable ionic compound

Lewis Structures & Formal Charge

$FC = V - N - \dfrac{B}{2}$

Formal charge $=$ Valence $e^-$ − Non-bonding $e^-$ − $\tfrac{1}{2}$ Bonding $e^-$; best structure minimises $|FC|$

Total e⁻ in Lewis structure = Σ(valence e⁻) ± charge

Add electrons for negative charge; subtract for positive charge before drawing structure

VSEPR & Hybridisation

Steric number (SN) = BP + LP on central atom

SN=2 → sp (linear); SN=3 → sp² (trigonal); SN=4 → sp³; SN=5 → sp³d; SN=6 → sp³d²

Bond angle reduction: each LP reduces angle by ~2.5°

NH₃ = 109.5° − 2.5° = 107°; H₂O = 109.5° − 5° = 104.5°

Molecular Orbital Theory

Bond order $= \dfrac{N_b - N_a}{2}$

$N_b$ = bonding electrons, $N_a$ = antibonding electrons; $BO > 0$ for stable molecule

O₂: BO=2, paramagnetic | N₂: BO=3, diamagnetic

O₂ has 2 unpaired e⁻ in degenerate π*2p orbitals; N₂ has no unpaired electrons

MO filling order (up to N₂): σ1s σ*1s σ2s σ*2s π2p π2p σ2p …

For O₂, F₂: σ2p comes before π2p. The crossover happens at N₂/O₂ boundary.

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Periodic Table & Bonding — Solved Examples

Arrange the following in increasing order of: (a) First ionisation energy: Na, Mg, Al, Si, P, S, Cl, Ar. (b) Atomic radius: Li, Na, K, Cs. (c) Electronegativity: F, Cl, Br, I.

(a) IE₁ across period 3 (general increase, but anomalies at Mg→Al and P→S):
Na < Mg > Al < Si < P > S < Cl < Ar
So increasing order: Na < Al < Mg < Si < S < P < Cl < Ar. Mg > Al because 2s is more stable than 2p; P > S because half-filled 3p is stable.
(b) Atomic radius down Group 1 — increases with shells: Li < Na < K < Cs ✓
(c) Electronegativity down Group 17 — decreases: I < Br < Cl < F ✓. F is the most electronegative element (Pauling scale: 4.0).

✓ Key pattern: Electronegativity and IE generally increase across a period and decrease down a group. But always check the Mg/Al and N/O (P/S) anomalies for IE.

Determine the shape, bond angle, and hybridisation of: (a) XeF₂, (b) XeF₄, (c) ClF₃, (d) SF₄, (e) PCl₅. Identify which pairs are isostructural.

(a) XeF₂: Xe: 8 valence e⁻. 2 F use 2 BP → 3 LP remaining. SN = 2+3 = 5 → sp³d. 3 LPs occupy equatorial positions → Linear, 180°.
(b) XeF₄: Xe: 8 valence e⁻. 4 F use 4 BP → 2 LP. SN = 4+2 = 6 → sp³d². 2 LPs trans (opposite) → Square planar, 90°.
(c) ClF₃: Cl: 7 valence e⁻. 3 F use 3 BP → 2 LP. SN = 5 → sp³d. 2 LPs equatorial → T-shaped, ~87°.
(d) SF₄: S: 6 valence e⁻. 4 F use 4 BP → 1 LP. SN = 5 → sp³d. 1 LP equatorial → See-saw (seesaw).
(e) PCl₅: P: 5 valence e⁻. 5 Cl use 5 BP → 0 LP. SN = 5 → sp³d → Trigonal bipyramidal, 90° (axial) and 120° (equatorial).

✓ Isostructural pairs: XeF₂ and I₃⁻ (both linear, 3 LPs); XeF₄ and ICl₄⁻ (both square planar, 2 LPs). Isoelectronic = same number of e⁻ and atoms; isostructural = same shape.

Using MO theory, find the bond order and magnetic character of: (a) O₂, (b) O₂⁺, (c) O₂²⁻, (d) N₂, (e) NO. Also predict which species has the shortest bond length.

O₂ MO config (16 e⁻): σ1s² σ*1s² σ2s² σ*2s² σ2p² π2p⁴ π*2p². $N_b = 10, N_a = 6$. $BO = \dfrac{10-6}{2} = $ 2. 2 unpaired e⁻ in π*2p → paramagnetic.
O₂⁺ (15 e⁻): Remove 1 e⁻ from π*2p → π*2p¹. $BO = \dfrac{10-5}{2} = $ 2.5. 1 unpaired → paramagnetic. Shorter bond than O₂ (higher BO).
O₂²⁻ (18 e⁻): Add 2 e⁻ to π*2p → π*2p⁴. $BO = \dfrac{10-8}{2} = $ 1. All paired → diamagnetic. Longest bond (lowest BO).
N₂ (14 e⁻): σ1s² σ*1s² σ2s² σ*2s² π2p⁴ σ2p². $N_b = 10, N_a = 4$. $BO = \dfrac{10-4}{2} = $ 3. All paired → diamagnetic.
NO (15 e⁻): Filling up to π*2p¹. $BO = \dfrac{10-5}{2} = $ 2.5. 1 unpaired → paramagnetic.

✓ Shortest bond → highest BO → N₂ (BO=3). Bond length order: N₂ < NO ≈ O₂⁺ < O₂ < O₂²⁻. Higher BO = shorter, stronger bond — fundamental MO relationship.

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Periodic Table & Bonding — Visual Concepts

Click any diagram to zoom

Periodic Trends — Direction of Change

Periodic trends summary. Atomic radius increases down a group and across a period from right to left (largest = Cs, bottom-left). Ionisation energy and electronegativity increase up a group and across from left to right (highest = F, top-right).

VSEPR — Common Molecular Shapes

VSEPR shapes for common molecules. Lone pairs (purple clouds) occupy more space than bond pairs, compressing bond angles. The molecular geometry describes only the positions of atoms — lone pairs are invisible in the final shape name.

MO Energy Diagram — O₂ (16 electrons)

MO diagram for O₂. The two degenerate π*2p antibonding orbitals each receive one electron (Hund's rule for MOs), leaving two unpaired electrons. This explains O₂'s paramagnetism — it is attracted to magnetic fields, confirmed experimentally by liquid O₂ clinging to magnets.

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Periodic Table & Bonding — Quick Quiz

Question 1

The correct order of first ionisation energies is:

General trend: Na < Mg < Al < Si. But anomaly: Mg (3s²) > Al (3p¹) because 3s is more stable than 3p. So actual order: Na < Al < Mg < Si. Answer: B ✓. IE₁: Na=496, Al=577, Mg=738, Si=786 kJ/mol.

Question 2

The hybridisation of the central atom in SF₄, ClF₃, and XeF₂ respectively is:

SF₄: S has 4 BP + 1 LP = SN 5 → sp³d. ClF₃: Cl has 3 BP + 2 LP = SN 5 → sp³d. XeF₂: Xe has 2 BP + 3 LP = SN 5 → sp³d. All three have SN=5 → all sp³d. Answer: C ✓

Question 3

Among N₂, O₂, F₂, and Ne₂, the species that does NOT exist (bond order = 0) is:

Ne₂: each Ne has 10 e⁻ → 20 e⁻ total. All MOs filled including σ*2p. N_b = 10, N_a = 10. BO = (10−10)/2 = 0 → Ne₂ does not exist. Answer: A ✓. F₂: BO = (10−8)/2 = 1 (exists but weak bond); O₂: BO=2; N₂: BO=3.

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p-Block Elements

Groups 13–18 · Inert pair effect · Allotropes · Halogens · Noble gases

Group 13 — Boron Family (ns²np¹)

Members: B, Al, Ga, In, Tl. Boron is a metalloid; rest are metals.
Inert pair effect: Down the group, ns² electrons become less available for bonding due to poor shielding by d/f electrons → Tl prefers +1 over +3, In shows both +1 and +3.
Boron: BCl₃ — sp², trigonal planar, Lewis acid (vacant p orbital accepts lone pair). BF₃ + NH₃ → BF₃·NH₃ (coordinate bond).
Aluminium: AlCl₃ forms dimer Al₂Cl₆ in vapour (two Cl bridge-bonds). Amphoteric oxide (Al₂O₃ reacts with both acids and bases).
Diagonal relationship: B resembles Si (both form covalent hydrides, acidic oxides, complex anions like [BO₃]³⁻ and [SiO₄]⁴⁻).

Group 14 — Carbon Family (ns²np²)

Carbon allotropes: Diamond (sp³, 3D network, hardest natural substance), Graphite (sp², layered, conductor due to delocalised π electrons), Fullerene C₆₀ (sp², soccer-ball, 20 hexagons + 12 pentagons).
CO vs CO₂: CO is a ligand (strong field via C lone pair), toxic (binds Hb). CO₂ is linear (sp), acidic oxide.
Silicon: SiO₂ is giant covalent solid (each Si tetrahedral, high mp). Silicones are polymers (Si–O backbone).
Inert pair: Pb prefers +2 (PbO, PbCl₂) over +4; Sn shows both +2 and +4 (SnCl₂ reducing agent, SnCl₄ Lewis acid).
C–C multiple bonding readily; Si cannot (d-orbital back bonding instead).

Group 15 — Nitrogen Family (ns²np³)

N₂ is very stable (triple bond, BE = 945 kJ/mol). P₄ (white phosphorus) — tetrahedral, strained 60° angles, highly reactive.
Hydrides: NH₃ > PH₃ > AsH₃ > SbH₃ — basicity decreases down (less lone pair availability); NH₃ has H-bonding, highest bp.
Oxides of nitrogen: N₂O (laughing gas), NO (radical), NO₂ (brown, acidic), N₂O₄ (dimer of NO₂), N₂O₅ (anhydride of HNO₃).
Oxoacids of P: H₃PO₃ (phosphorous acid) — diprotic (one H on P is non-ionisable); H₃PO₄ (phosphoric) — triprotic.
Nitric acid: Dilute HNO₃ + metal → NO; Concentrated HNO₃ + metal → NO₂. Passivation: Fe, Cr, Al with conc. HNO₃.

Group 16 — Chalcogens (ns²np⁴)

Oxygen: Allotropes O₂ and O₃ (ozone — angular, sp², bent). O₃ is a better oxidising agent than O₂.
Sulphur allotropes: Rhombic S₈ (stable <96°C), Monoclinic S₈ (stable 96–119°C), Plastic sulphur (supercooled). All contain puckered S₈ rings.
Acidic character: H₂O < H₂S < H₂Se < H₂Te — increases down (weaker H–X bond, more stable conjugate base anion).
SO₂: V-shape, reducing agent, acidic oxide; forms H₂SO₃ with water. SO₃ (Lewis acid) + H₂O → H₂SO₄.
Oleum: H₂SO₄ + SO₃ → H₂S₂O₇ (pyrosulphuric acid); Contact process for H₂SO₄ uses V₂O₅ catalyst.

Group 17 — Halogens (ns²np⁵)

Property	F	Cl	Br	I
Colour	Pale yellow	Greenish-yellow	Red-brown	Violet (vapour)
State (25°C)	Gas	Gas	Liquid	Solid
Oxidising power	F₂ > Cl₂ > Br₂ > I₂	Decreases down
Bond dissociation E	F₂: 158 kJ/mol (weakest!)	Cl₂: 243 (strongest)	Br₂: 193	I₂: 151
HX acid strength	HF (weak acid)	HCl < HBr < HI	Increases (weaker H–X)

F₂ anomaly: Weakest X–X bond (small size → lone pair repulsion). Only −1 oxidation state (no d orbitals to expand valence).
Interhalogen compounds: XX′ type (ICl, IBr), XX′₃ (ICl₃), XX′₅ (BrF₅ — square pyramidal), XX′₇ (IF₇ — pentagonal bipyramidal).
Bleaching powder: Ca(OCl)Cl — formed by Cl₂ + slaked lime.
Oxoacids: HOCl < HOClO < HOClO₂ < HOClO₃ — acid strength increases with more O atoms (more electronegative, weakens O–H bond).

Group 18 — Noble Gases

All ns²np⁶ configuration (He: 1s²). High ionisation energy, chemically inert.
Xenon compounds: XeF₂ (linear, sp³d), XeF₄ (square planar, sp³d²), XeF₆ (distorted octahedral, sp³d³), XeO₃ (pyramidal), XeOF₄ (square pyramidal).
Kr also forms KrF₂. Rn is radioactive (½-life 3.8 days).
He: lightest noble gas, used in airships (non-flammable), cryogenics (bp 4.2 K). Ar: most abundant noble gas in atmosphere (0.93%).

💡 JEE Tip: F₂ has the WEAKEST bond among halogens (not the strongest!) due to lone-pair repulsion in the small F₂ molecule. Cl₂ has the strongest X–X bond. Also: HF is a weak acid (strong H-bonding), while HCl, HBr, HI are strong acids — HI is the strongest.

⚠ Common Error: H₃PO₃ appears to be triprotic (3 H atoms) but is only diprotic — the H directly bonded to P is non-acidic. Similarly H₃PO₂ is monoprotic. Count only H atoms bonded to O (not to P) for basicity of phosphorus oxoacids.

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p-Block Elements — Formula Sheet

Key Reactions

4HNO₃(conc) + Cu → Cu(NO₃)₂ + 2NO₂ + 2H₂O

Concentrated HNO₃ oxidises Cu to give NO₂ (brown gas)

8HNO₃(dil) + 3Cu → 3Cu(NO₃)₂ + 2NO + 4H₂O

Dilute HNO₃ gives NO (colourless gas, turns brown in air → NO₂)

2SO₂ + O₂ ⇌ 2SO₃ (V₂O₅, 450°C)

Contact process step 2; then SO₃ + H₂SO₄ → H₂S₂O₇ → H₂SO₄ + H₂O

2Cl₂ + 2Ca(OH)₂ → Ca(OCl)Cl + CaCl₂ + H₂O

Formation of bleaching powder; active component is Ca(OCl)Cl

Oxidation States & Trends

Inert pair effect: stability of +2 > +4 for heavier Group 14/15/16

Pb²⁺ more stable than Pb⁴⁺; Tl⁺ more stable than Tl³⁺; Bi³⁺ more stable than Bi⁵⁺

Halogen oxidising power: F₂ > Cl₂ > Br₂ > I₂

F₂ can oxidise Cl⁻ to Cl₂; Cl₂ can oxidise Br⁻ to Br₂ and I⁻ to I₂; Br₂ can only oxidise I⁻

Oxoacid strength: HClO < HClO₂ < HClO₃ < HClO₄

More O atoms → higher oxidation state on Cl → more electron withdrawal → weaker O–H → stronger acid

Bond Dissociation Energies — Halogens

F₂: 158 < I₂: 151 < Br₂: 193 < Cl₂: 243 kJ/mol

F₂ is weakest (lone-pair repulsion in tiny F atom). Cl₂ is strongest dihalogen bond. Order (increasing): F₂ ≈ I₂ < Br₂ < Cl₂

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p-Block Elements — Solved Examples

Compare: (a) diamond vs graphite (properties and hybridisation); (b) O₂ vs O₃ (structure and oxidising ability); (c) rhombic vs monoclinic sulphur.

Diamond: Each C is sp³, tetrahedral network solid. Hardest natural substance, non-conductor (no free e⁻). Graphite: Each C is sp², layered. Unhybridised p electrons form delocalised π system → conducts electricity. Layers held by weak van der Waals → soft, lubricant.
O₂: BO=2, paramagnetic. O₃: Bent (sp²), resonance hybrid BO=1.5. O₃ is stronger oxidising agent (decomposes readily to give nascent O: O₃ → O₂ + [O]). Both are allotropes of oxygen.
Both are S₈ rings (crown-shaped). Rhombic: stable <96°C (transition temperature), orthorhombic crystal. Monoclinic: stable 96–119°C, needle-like crystals. At 119°C both melt (mp of S).

(a) Will Cl₂ displace Br⁻ from NaBr(aq)? Will I₂ displace Cl⁻ from NaCl(aq)? (b) Why is HF a weak acid while HCl is strong? (c) Name the products when excess Cl₂ reacts with NH₃.

Cl₂ + 2NaBr → 2NaCl + Br₂ ✓ (Cl₂ stronger oxidiser than Br₂, displaces Br⁻). I₂ + NaCl → No reaction ✗ (I₂ is weaker than Cl₂, cannot oxidise Cl⁻). Displacement: stronger halogen displaces weaker halide.
HF has a very strong H–F bond (BE = 568 kJ/mol) and extensive H-bonding in solution (F⁻···HF → HF₂⁻). This makes it only partially dissociate: HF ⇌ H⁺ + F⁻, Ka ≈ 6.8×10⁻⁴. HCl bond is much weaker (BE = 432 kJ/mol) and fully dissociates.
Excess Cl₂ + NH₃ → NCl₃ + 3HCl (trichloride). With excess NH₃: 8NH₃ + 3Cl₂ → N₂ + 6NH₄Cl (nitrogen and ammonium chloride). Remember: excess Cl₂ gives NCl₃; excess NH₃ gives N₂.

Determine the basicity (number of ionisable protons) of: H₃PO₄, H₃PO₃, H₃PO₂, H₄P₂O₇. Write the structural formula for each.

Rule: Only H atoms bonded to O (as –OH groups) are ionisable. H atoms bonded directly to P are NOT ionisable.
H₃PO₄ (orthophosphoric acid): Structure: (HO)₃P=O. All 3 H atoms on O → triprotic (basicity 3).
H₃PO₃ (phosphorous acid): Structure: (HO)₂P(=O)H. One H directly on P → only 2 ionisable H → diprotic (basicity 2). P is in +3 state.
H₃PO₂ (hypophosphorous acid): Structure: (HO)P(=O)H₂. Two H directly on P → only 1 ionisable H → monoprotic (basicity 1). P is in +1 state.
H₄P₂O₇ (pyrophosphoric acid): Two PO₄ units sharing one O. 4 –OH groups → tetraprotic (basicity 4). P in +5 state.

✓ Summary: H₃PO₄ (3) > H₄P₂O₇ (4 but 2 P) > H₃PO₃ (2) > H₃PO₂ (1). H₃PO₃ and H₃PO₂ are reducing agents (P–H bonds are oxidised); H₃PO₄ is not a reducing agent.

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p-Block Elements — Visual Concepts

Click any diagram to zoom

Carbon Allotropes — Diamond vs Graphite vs C₆₀

Carbon allotropes. Diamond: sp³ 3D network → hardest, insulator. Graphite: sp² layers → conductor (π electrons), soft lubricant. C₆₀: soccer ball of 20 hexagons (blue) + 12 pentagons (red), discovered 1985 (Kroto, Curl, Smalley).

Halogen Trends — Oxidising Power vs Bond Dissociation Energy

Halogen bond dissociation energies. Note: F₂ has the WEAKEST bond (158 kJ/mol) due to lone-pair repulsion in the small F₂ molecule, yet F₂ is the strongest oxidising agent — oxidising power is driven by electron affinity and hydration energy of F⁻, not bond strength alone.

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p-Block Elements — Quick Quiz

Question 1

The bond dissociation energy of F₂ is lower than that of Cl₂ because:

F is smaller than Cl, so the three lone pairs on each F atom are very close, causing significant repulsion even in the bonded F₂ molecule. This weakens the F–F bond (158 kJ/mol vs Cl–Cl 243 kJ/mol). F has the highest electronegativity (3.98), not lowest. Answer: B ✓

Question 2

H₃PO₃ is diprotic (basicity 2) because:

H₃PO₃ structure: (HO)₂P(=O)H. The H directly attached to P (P–H bond) is not ionisable. Only the two –OH groups give H⁺. So basicity = 2. Answer: C ✓. Similarly H₃PO₂ has 2 P–H bonds → monoprotic.

Question 3

Which of the following noble gas compounds has a square planar structure?

XeF₄: Xe has 4 BP + 2 LP = SN 6 → sp³d². 2 LPs in axial positions (trans) → square planar molecular shape. XeF₂: linear (3 equatorial LPs). XeF₆: distorted octahedral (1 LP). XeO₃: pyramidal (1 LP on Xe). Answer: D ✓

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d & f Block Elements

Transition metals · Variable oxidation state · Colour · Magnetism · Lanthanide contraction

d-Block: Electronic Configuration

General configuration: [noble gas] (n−1)d¹⁻¹⁰ ns⁰⁻². First row: Sc (3d¹4s²) to Zn (3d¹⁰4s²).
Exceptions: Cr: [Ar] 3d⁵4s¹ (half-filled d, more stable). Cu: [Ar] 3d¹⁰4s¹ (fully filled d). Similar for Mo, W, Ag, Au.
When forming ions: ns electrons are lost first (ns is higher in energy for the cation), then nd. Fe²⁺: [Ar] 3d⁶; Fe³⁺: [Ar] 3d⁵.
Zn, Cd, Hg (Group 12) are sometimes excluded as transition metals because they have completely filled d¹⁰ in all oxidation states.

Characteristic Properties of Transition Metals

Property	Explanation	Example
Variable oxidation states	Similar energies of (n−1)d and ns; multiple states accessible	Fe: +2, +3; Mn: +2 to +7; Cr: +2, +3, +6
Coloured compounds	d–d transitions absorb visible light; complementary colour observed	Cu²⁺ (blue), Cr³⁺ (violet), MnO₄⁻ (purple)
Paramagnetism	Unpaired d electrons; $\mu = \sqrt n(n+2) BM$	Fe³⁺: 5 unpaired → μ=5.92 BM (highest)
Catalytic activity	Variable OS + surface adsorption	Fe (Haber), V₂O₅ (Contact), Ni (hydrogenation)
Complex formation	Small size, high charge, vacant d orbitals	[Fe(CN)₆]³⁻, [Cu(NH₃)₄]²⁺
Interstitial compounds	Small atoms (H, C, N) in d-metal lattice gaps	Steel (Fe + C), TiH₂
Alloy formation	Similar atomic radii; mix freely	Bronze (Cu+Sn), Brass (Cu+Zn)

Colour and d–d Transitions

When ligands surround a metal, they split d orbitals into groups of different energy (crystal field splitting).
Electrons absorb photons to jump to higher d level. The frequency absorbed determines the colour seen (complementary).
Sc³⁺, Ti⁴⁺, Cu⁺, Zn²⁺ are colourless — either no d electrons or fully filled d¹⁰ (no d–d transition possible).
MnO₄⁻ is intensely purple due to charge-transfer (not d–d) — LMCT (ligand to metal charge transfer).

Important Oxoanions of Transition Metals

Ion	Colour	Metal OS	Oxidising?
MnO₄⁻ (permanganate)	Purple/violet	Mn +7	Strong (acidic > basic)
Cr₂O₇²⁻ (dichromate)	Orange	Cr +6	Strong (acidic medium)
CrO₄²⁻ (chromate)	Yellow	Cr +6	Moderate
MnO₄²⁻ (manganate)	Green	Mn +6	Moderate

Cr₂O₇²⁻ ⇌ 2CrO₄²⁻ + 2H⁺: Dichromate (orange, acidic) ⇌ Chromate (yellow, basic) — pH-dependent equilibrium.

KMnO₄ in acidic medium: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (Mn: +7 → +2). In basic medium: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ (Mn: +7 → +4).

f-Block: Lanthanides and Actinides

Lanthanides (Ce–Lu): Fill 4f orbitals. Common oxidation state +3 (due to extra stability of [Xe] core). Ce also shows +4; Eu and Yb show +2.
Lanthanide contraction: As 4f electrons are added, nuclear charge increases but 4f electrons are poor shielders → Z_eff increases → ionic radius contracts steadily from La³⁺ to Lu³⁺.
Consequences of lanthanide contraction:
— 4d and 5d elements have nearly identical radii (e.g., Zr ≈ Hf, Nb ≈ Ta) → similar properties, hard to separate.
— Basicity of Ln(OH)₃ decreases from La(OH)₃ to Lu(OH)₃ (smaller ion, more charge density, less hydroxide).
Actinides (Th–Lr): Fill 5f orbitals. Show much wider range of oxidation states (+2 to +6 or +7) because 5f, 6d, and 7s energies are close. Many are radioactive.
Actinide contraction is slightly larger than lanthanide contraction (5f electrons even poorer shielders).

💡 JEE Tip: Magnetic moment formula $\mu = \sqrt{n(n+2)}$ BM where n = number of unpaired electrons. Fe³⁺/Mn²⁺ both have 5 unpaired d electrons → $\mu = \sqrt{35} \approx 5.92$ BM (highest). Zn²⁺ has 0 unpaired → μ = 0 (diamagnetic). Learn: Cr³⁺ (3 unpaired, 3.87 BM), Fe²⁺ (4 unpaired, 4.90 BM).

⚠ Common Error: Cu²⁺ is blue-coloured (NOT Cu⁺ which is colourless — d¹⁰). Similarly, Zn²⁺ is colourless (d¹⁰), Sc³⁺ is colourless (d⁰). Don't confuse Cu metal (reddish) with Cu²⁺ ion (blue in aqueous solution = [Cu(H₂O)₄]²⁺).

📐

d & f Block — Formula Sheet

Magnetic Moment

$\mu = \sqrt{n(n+2)}$ BM (Bohr Magnetons)

n = number of unpaired electrons. Only spin magnetic moment (spin-only formula). Orbital contribution neglected for 3d metals.

$$ n=1: \mu=1.73 | n=2: \mu=2.83 | n=3: \mu=3.87 | n=4: \mu=4.90 | n=5: \mu=5.92 BM $$

Ti³⁺(1), V³⁺(2), Cr³⁺/Co²⁺(3), Fe²⁺/Co³⁺LS(4 or 0), Fe³⁺/Mn²⁺(5)

Redox Half-Reactions

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (acidic)

KMnO₄ in acidic medium: purple → colourless. Gain of 5e⁻. E° = +1.51 V (very strong oxidant)

MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ (basic)

KMnO₄ in neutral/basic medium: purple → brown precipitate MnO₂

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (acidic)

K₂Cr₂O₇: orange → green (Cr³⁺). Gain of 6e⁻. E° = +1.33 V. Only used in acidic medium.

2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O

Chromate (yellow, basic) ⇌ dichromate (orange, acidic). Interconversion via pH.

Electronic Configurations (Key)

Cr: [Ar] 3d⁵ 4s¹ | Cu: [Ar] 3d¹⁰ 4s¹

Half-filled (Cr) and fully filled (Cu) d subshells are extra stable → anomalous 4s¹ instead of 4s²

For ions: remove ns first, then (n−1)d

Fe²⁺: [Ar]3d⁶ (remove 4s² from Fe: [Ar]3d⁶4s²). Fe³⁺: [Ar]3d⁵. Mn²⁺: [Ar]3d⁵.

🧮

d & f Block — Solved Examples

Calculate the spin-only magnetic moment of: (a) Fe³⁺, (b) Cu²⁺, (c) Zn²⁺, (d) Ti³⁺. State whether each is paramagnetic or diamagnetic.

Fe³⁺: Fe is [Ar]3d⁶4s². Remove 4s²4d¹ → wait, Fe³⁺ loses 3e⁻. Fe: [Ar]3d⁶4s² → Fe³⁺: [Ar]3d⁵. Five 3d electrons, all unpaired (d⁵ high spin). n=5 → $\mu = \sqrt{5\times7} = \sqrt{35}$ = 5.92 BM. Paramagnetic.
Cu²⁺: Cu: [Ar]3d¹⁰4s¹ → Cu²⁺: [Ar]3d⁹. One unpaired electron in d⁹. n=1 → $\mu = \sqrt{1\times3} = \sqrt{3}$ = 1.73 BM. Paramagnetic. (Blue colour in solution.)
Zn²⁺: Zn: [Ar]3d¹⁰4s² → Zn²⁺: [Ar]3d¹⁰. All paired. n=0 → μ = 0. Diamagnetic. Colourless compounds.
Ti³⁺: Ti: [Ar]3d²4s² → Ti³⁺: [Ar]3d¹. One unpaired. n=1 → $\mu = \sqrt{3}$ = 1.73 BM. Paramagnetic. (Ti³⁺ is violet — single d electron makes one d–d transition possible.)

10 mL of 0.1 M KMnO₄ (acidified) is used to titrate FeSO₄ solution. What volume of 0.5 M FeSO₄ is required? [MnO₄⁻ + 5Fe²⁺ (acidic) → Mn²⁺ + 5Fe³⁺]

Moles of KMnO₄ = 0.1 × 10/1000 = 10⁻³ mol.
From stoichiometry: 1 mol MnO₄⁻ reacts with 5 mol Fe²⁺ (MnO₄⁻ gains 5e⁻; Fe²⁺→Fe³⁺ loses 1e⁻ each).
Moles of Fe²⁺ required = 5 × 10⁻³ = 5×10⁻³ mol.
Volume of 0.5 M FeSO₄ = (5×10⁻³)/0.5 = 0.01 L = 10 mL.

✓ Method: n-factor of KMnO₄ in acid = 5 (Mn: +7→+2). n-factor of FeSO₄ = 1 (Fe: +2→+3). Milliequivalents must be equal: (0.1×5)×10 = (0.5×1)×V → V = 10 mL ✓.

Explain why: (a) Zr and Hf have almost identical atomic radii despite Hf being in period 6. (b) The 5d metals (W, Re, Os, Ir, Pt, Au) are denser than 4d metals. (c) Au is less reactive than Ag despite being below it in the group.

Going from 4d (Zr, Period 5) to 5d (Hf, Period 6), 14 lanthanide elements are inserted. Their poor 4f shielding causes steady Z_eff increase → ionic/atomic radii contract. By the time we reach Hf, the lanthanide contraction has nearly cancelled the expected increase in radius due to adding a new shell. Result: Zr (160 pm) ≈ Hf (159 pm).
5d metals have similar atomic radii to 4d metals (due to lanthanide contraction) but much higher nuclear masses (more protons + neutrons). Density = mass/volume. Similar radius + greater mass → significantly higher density.
Au (5d) has smaller radius than expected → stronger nucleus-electron attraction → higher ionisation energy and electronegativity. Also, relativistic effects compress 6s orbital in heavy elements like Au, making it harder to lose electrons. Ag has lower IE and is more easily oxidised to Ag⁺. Thus Au is less reactive (nobility of gold).

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d & f Block — Visual Concepts

Click any diagram to zoom

Magnetic Moment vs Unpaired Electrons (3d Metals)

Spin-only magnetic moments. Fe³⁺ and Mn²⁺ (both 3d⁵, 5 unpaired electrons) have the highest magnetic moment (5.92 BM). Zn²⁺ (3d¹⁰) is diamagnetic (μ=0). The curve follows $\mu = \sqrt{n(n+2)}$.

Lanthanide Contraction — Ionic Radii Across 4f Series

Lanthanide contraction: as 14 electrons fill 4f orbitals from La to Lu, poor 4f shielding → steady Z_eff rise → steady radius decrease (~20 pm total). Consequence: 5d metals (Hf, Ta, W…) have radii almost equal to their 4d homologues (Zr, Nb, Mo…), making them harder to separate.

⚡

d & f Block — Quick Quiz

Question 1

The electronic configuration of Cu is [Ar] 3d¹⁰ 4s¹ and not [Ar] 3d⁹ 4s² because:

Fully filled (d¹⁰) and half-filled (d⁵) subshells have extra stability due to symmetrical charge distribution and exchange energy. Cu promotes one 4s electron to 3d to achieve 3d¹⁰ 4s¹ configuration. Similarly Cr achieves 3d⁵ 4s¹. Answer: C ✓

Question 2

Which of the following transition metal ions is COLOURLESS?

Sc³⁺: [Ar] — completely empty d⁰ configuration. No d electrons to undergo d–d transitions → no visible light absorbed → colourless. Cu²⁺ (d⁹) = blue; Fe³⁺ (d⁵) = pale yellow/brown; Mn²⁺ (d⁵) = very pale pink. Answer: B ✓. Also Zn²⁺ (d¹⁰) and Ti⁴⁺ (d⁰) are colourless.

Question 3

Lanthanide contraction is primarily due to:

4f electrons are in inner orbitals with diffuse radial distribution — they shield poorly. As Z increases by 1 and one 4f electron is added across the lanthanides, Z_eff felt by outer electrons increases steadily → ionic radius contracts steadily from La³⁺ (106.1 pm) to Lu³⁺ (86.1 pm). Answer: A ✓

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Coordination Compounds

Werner's theory · IUPAC naming · Isomerism · Crystal Field Theory

Werner's Theory

Primary valence (ionisable): Satisfied by negative ions; equals oxidation state of metal.
Secondary valence (non-ionisable): Satisfied by ligands; equals coordination number (CN).
Secondary valences are directed in space → determines geometry of the complex.
Example: [Co(NH₃)₆]Cl₃ — Co has OS +3 (primary valence = 3), CN = 6 (secondary valence = 6), 3 Cl⁻ are counter-ions.

Key Terminology

Term	Definition	Example
Central metal	Lewis acid (accepts electron pairs)	Fe in [Fe(CN)₆]³⁻
Ligand	Lewis base (donates electron pairs) attached to metal	CN⁻, NH₃, Cl⁻, H₂O
Monodentate	Ligand donates 1 lone pair	Cl⁻, NH₃, H₂O, CN⁻
Bidentate	Donates 2 lone pairs via 2 donor atoms	en (ethylenediamine), ox²⁻, acac⁻
Polydentate	Donates multiple pairs	EDTA⁴⁻ (hexadentate, 6 donor atoms)
Chelate	Complex with ring formed by polydentate ligand	[Cu(en)₂]²⁺
Ambidentate	Can bond through different donor atoms	NO₂⁻ (via N: nitro; via O: nitrito), SCN⁻ (via S or N)

IUPAC Nomenclature Rules

Name the cation first, then the anion (as in ionic nomenclature).
Within the complex ion: ligands are named alphabetically before the metal.
Anionic ligands end in -o: Cl⁻ = chlorido, CN⁻ = cyanido, OH⁻ = hydroxido, NO₂⁻ = nitrito, O²⁻ = oxido.
Neutral ligands retain their names (with exceptions): NH₃ = ammine, H₂O = aqua, CO = carbonyl, NO = nitrosyl.
Multiplying prefixes: di, tri, tetra, penta, hexa (for simple); bis, tris, tetrakis (for ligands containing a number).
Metal oxidation state given in Roman numerals in parentheses after metal name.
If complex is an anion, metal name gets suffix -ate (e.g., ferrate, cuprate, platinate).

Types of Isomerism in Coordination Compounds

Isomerism	Description	Example
Geometric (cis/trans)	Different spatial arrangement of ligands	cis-[Pt(NH₃)₂Cl₂] (anticancer) vs trans-[Pt(NH₃)₂Cl₂]
Optical (Δ/Λ)	Non-superimposable mirror images (chiral)	[Co(en)₃]³⁺ Δ and Λ forms
Linkage	Ambidentate ligand bonded through different atoms	[Co(NH₃)₅(NO₂)]²⁺ vs [Co(NH₃)₅(ONO)]²⁺
Ionisation	Different ions in coordination sphere vs outside	[Co(NH₃)₅Br]SO₄ vs [Co(NH₃)₅SO₄]Br
Hydrate/Solvate	H₂O inside vs outside coordination sphere	[Cr(H₂O)₆]Cl₃ vs [Cr(H₂O)₅Cl]Cl₂·H₂O
Coordination	Exchange of ligands between cation and anion complexes	[Co(NH₃)₆][Cr(CN)₆] vs [Cr(NH₃)₆][Co(CN)₆]

Crystal Field Theory (CFT)

Ligands are treated as point negative charges; they repel metal d electrons, splitting the degenerate d orbitals.
Octahedral field: Splits d orbitals into t₂g (dxy, dyz, dxz — lower energy, 3 orbitals) and eg (dx²−y², dz² — higher energy, 2 orbitals). Splitting energy = Δo. t₂g = −0.4Δo; eg = +0.6Δo.
Tetrahedral field: e (dx²−y², dz²) lower, t₂ (dxy, dyz, dxz) higher. Δt = 4/9 Δo (smaller because only 4 ligands).
Strong vs weak field ligands:
Spectrochemical series (increasing Δo):
I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < SCN⁻ < NH₃ < en < CN⁻ < CO
Low spin (strong field): Large Δo > pairing energy → electrons pair in t₂g first → fewer unpaired e⁻.
High spin (weak field): Small Δo < pairing energy → electrons obey Hund's rule → more unpaired e⁻.
CFSE (Crystal Field Stabilisation Energy) = contribution of t₂g electrons (each −0.4Δo) + eg electrons (each +0.6Δo).

💡 JEE Tip: cis-platin [PtCl₂(NH₃)₂] is the anticancer drug — ONLY the cis form is active (it can cross-link DNA strands within one strand). trans-platin is inactive. Also: CN⁻ and CO are strong field → low spin; Cl⁻, F⁻ are weak field → high spin for most metals.

⚠ Common Error: In IUPAC naming, ligands go in ALPHABETICAL order, not by charge or denticity. "Tetraamminedichloridocobalt(III)" — ammine (a) before chlorido (c) ✓. Also the complex ion is named as one word with no spaces between ligand names and the metal name.

📐

Coordination Compounds — Formula Sheet

Crystal Field Stabilisation Energy (CFSE)

Octahedral CFSE = (−0.4 × n_t₂g + 0.6 × n_eg) × Δo − P × (extra pairs)

n_t₂g = electrons in t₂g; n_eg = electrons in eg. Subtract pairing energy P for each extra pair compared to free ion.

d⁶ (low spin, octahedral): t₂g⁶ eg⁰ → CFSE = 6(−0.4Δo) = −2.4Δo

Maximum CFSE for octahedral. Examples: [Fe(CN)₆]⁴⁻, [Co(NH₃)₆]³⁺. Strongly stabilised.

$$ \Delta t = \frac{4}{9} \times \Delta o $$

Tetrahedral splitting is 4/9 of octahedral (fewer ligands, no direct overlap). Most tetrahedral complexes are high spin.

Oxidation State Calculation

OS of metal = charge of complex − Σ(charges of ligands)

For [Fe(CN)₆]³⁻: OS(Fe) = −3 − 6(−1) = −3 + 6 = +3. For [Cu(NH₃)₄]²⁺: OS(Cu) = +2 − 4(0) = +2.

EAN = Z − OS + 2 × CN (Effective Atomic Number)

Stable complexes often have EAN = noble gas configuration (18-electron rule for transition metals). EAN = atomic number − electrons lost + electrons donated by ligands.

Common Ligands Quick Reference

en: H₂N–CH₂–CH₂–NH₂ (bidentate, chelates via 2 N donors)

Ethylenediamine — most common bidentate amine ligand; forms 5-membered chelate rings.

EDTA⁴⁻: hexadentate (4 O + 2 N donors) — forms 5 rings, very stable chelate

Ethylenediaminetetraacetate; sequesters metal ions; used in analytical chemistry and medicine.

🧮

Coordination Compounds — Solved Examples

Name the following: (a) [Co(NH₃)₄Cl₂]⁺, (b) [Fe(CN)₆]⁴⁻, (c) [Pt(en)₂]²⁺, (d) [CrCl₂(H₂O)₄]⁺.

[Co(NH₃)₄Cl₂]⁺: Ligands alphabetically: chlorido (×2) before ammine (×4)? No — 'a' before 'c' → ammine first! Tetraamminedichloridocobalt(III). [OS: x + 4(0) + 2(−1) = +1 → x = +3.]
[Fe(CN)₆]⁴⁻: Complex anion → metal gets -ate suffix → ferrate. OS: x + 6(−1) = −4 → x = +2. Name: hexacyanidoferrate(II).
[Pt(en)₂]²⁺: en is a bidentate neutral ligand; use 'bis' (not 'di') for ligands containing numerals. OS: x + 2(0) = +2 → x = +2. Name: bis(ethylenediamine)platinum(II).
[CrCl₂(H₂O)₄]⁺: aqua (a) before chlorido (c)? No — 'a'<'c' → aqua comes first. OS: x + 2(−1) + 4(0) = +1 → x = +3. Name: tetraaquadichloridochromium(III).

For a d⁶ octahedral complex, determine: (a) electron configuration, number of unpaired electrons, CFSE for [Fe(H₂O)₆]²⁺ (weak field); (b) same for [Fe(CN)₆]⁴⁻ (strong field). Which has higher magnetic moment?

Fe²⁺ is d⁶. In octahedral field: 6 electrons to distribute between t₂g (lower) and eg (higher).
(a) [Fe(H₂O)₆]²⁺ — H₂O is weak field: Δo < pairing energy → high spin. Fill by Hund's: t₂g³ eg² first, 6th e⁻ pairs in t₂g → t₂g⁴ eg². Unpaired: 4. $\mu = \sqrt{4\times6} = \sqrt{24}$ = 4.90 BM. CFSE = 4(−0.4Δo) + 2(+0.6Δo) = −0.4Δo (low).
(b) [Fe(CN)₆]⁴⁻ — CN⁻ is strong field: Δo > pairing energy → low spin. All 6 electrons pair up in t₂g → t₂g⁶ eg⁰. Unpaired: 0. μ = 0 BM (diamagnetic). CFSE = 6(−0.4Δo) = −2.4Δo (very high stabilisation).

✓ [Fe(H₂O)₆]²⁺ has higher μ (4.90 vs 0 BM). CN⁻ causes larger splitting → all electrons pair → diamagnetic. This is why [Fe(CN)₆]⁴⁻ is yellow (large Δo → absorbs UV edge into violet, transmits yellow), while [Fe(H₂O)₆]²⁺ is pale green.

How many geometric and optical isomers exist for [Co(en)₂Cl₂]⁺? Draw and name them. Which ones are optically active?

[Co(en)₂Cl₂]⁺: Co³⁺, 2 bidentate en ligands, 2 Cl⁻ ligands. Octahedral. The 2 Cl can be cis (90° apart) or trans (180° apart).
trans isomer: 2 Cl directly opposite each other. All en ligands occupy equatorial-ish positions symmetrically. This isomer is superimposable on its mirror image (has C₂ axis + σh plane) → optically inactive. 1 geometric isomer.
cis isomer: 2 Cl adjacent (90° apart). The molecule is chiral (no plane of symmetry). Gives 2 optical isomers: Δ (right-handed, d) and Λ (left-handed, l) forms — non-superimposable mirror images.

✓ Total: 3 isomers — 1 trans (optically inactive) + 2 cis (Δ and Λ, both optically active). All complexes with 3 bidentate ligands [M(AA)₃] are also chiral → 2 optical isomers, e.g., [Co(en)₃]³⁺.

📊

Coordination Compounds — Visual Concepts

Click any diagram to zoom

Crystal Field Splitting — Octahedral vs Tetrahedral

Crystal Field Splitting. Octahedral: 6 ligands split d orbitals into lower t₂g (3 orbitals, −0.4Δo) and upper eg (2 orbitals, +0.6Δo). Tetrahedral: ordering inverts — e is lower, t₂ is higher; Δt = (4/9)Δo. The smaller splitting means tetrahedral complexes are almost always high spin.

Geometric Isomers of [MA₂B₂] — Square Planar (e.g., Cisplatin)

Geometric isomers of cisplatin. cis-[PtCl₂(NH₃)₂]: the two Cl⁻ are 90° apart (adjacent) — this isomer can chelate DNA by forming intrastrand cross-links → anticancer activity. trans-[PtCl₂(NH₃)₂]: Cl⁻ are 180° apart — geometry prevents DNA chelation → biologically inactive.

⚡

Coordination Compounds — Quick Quiz

Question 1

The IUPAC name of [Co(NH₃)₅Cl]²⁺ is:

Ligands in alphabetical order: ammine (a) before chlorido (c). OS: x + 5(0) + (−1) = +2 → x = +3. Name: pentaamminechloridocobalt(III). Note: modern IUPAC uses 'chlorido' not 'chloro'. Answer: C ✓

Question 2

In Crystal Field Theory, which set of d orbitals is lower in energy in an octahedral field?

In an octahedral field, 6 ligands approach along ±x, ±y, ±z axes. The dx²−y² and dz² (eg) orbitals point directly at ligands → greater repulsion → higher energy. The dxy, dyz, dxz (t₂g) point between axes → lower repulsion → lower energy. Answer: B ✓

Question 3

Which of the following pairs represents linkage isomers?

Linkage isomers: same molecular formula but ambidentate ligand (NO₂⁻) bonded through different atoms — N (nitro: Co–NO₂) vs O (nitrito: Co–ONO). A = ionisation isomers; B = coordination isomers; C = geometric isomers. Answer: D ✓

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Metallurgy

Ores · Concentration · Extraction · Refining · Alloys

Occurrence of Metals — Ores

Metal	Important Ores	Formula
Iron	Haematite, Magnetite, Siderite	Fe₂O₃, Fe₃O₄, FeCO₃
Copper	Copper pyrites, Malachite, Cuprite	CuFeS₂, Cu₂(OH)₂CO₃, Cu₂O
Aluminium	Bauxite, Corundum, Cryolite	Al₂O₃·2H₂O, Al₂O₃, Na₃AlF₆
Zinc	Zinc blende, Zincite, Calamine	ZnS, ZnO, ZnCO₃
Lead	Galena, Cerrusite	PbS, PbCO₃
Silver	Argentite, Horn silver	Ag₂S, AgCl
Gold	Native gold, Calaverite	Au, AuTe₂
Mercury	Cinnabar	HgS

Concentration of Ores

Method	Principle	Used for
Gravity separation (levigation)	Difference in specific gravity	Oxide/carbonate ores (haematite)
Froth flotation	Preferential wetting by oil; ore particles attach to froth	Sulphide ores (ZnS, PbS, CuFeS₂)
Magnetic separation	Magnetic properties of ore/gangue	Magnetite, chromite; removes non-magnetic gangue
Leaching (chemical)	Selective dissolution of ore in reagent	Al (NaOH), Au/Ag (NaCN), Ag (dilute H₂SO₄)

Extraction of Metals

Thermodynamic basis — Ellingham diagram: plots ΔG° vs T for metal oxide formation. A metal can reduce another metal's oxide if its ΔG° line lies below the other at that temperature. Carbon and CO are powerful reducing agents at high temperatures.

Method	Principle	Examples
Calcination	Heating ore in absence of air; decomposes carbonate/hydroxide	ZnCO₃→ZnO+CO₂; Al₂O₃·2H₂O→Al₂O₃
Roasting	Heating ore in excess air; converts sulphide to oxide	2ZnS+3O₂→2ZnO+2SO₂
Smelting	Reduction of oxide with coke (C or CO) at high T	Fe₂O₃+3CO→2Fe+3CO₂ (blast furnace)
Self-reduction	Sulphide partially roasted then heated; sulphide reduces oxide	Cu₂S+2Cu₂O→6Cu+SO₂
Electrolytic reduction	Electrolysis of molten ore/solution	Al from molten Al₂O₃ (Hall-Héroult); Na from NaCl
Hydrometallurgy	Leaching + displacement or electrolysis	Cu from CuSO₄ by Fe; Au by NaCN then Zn

Blast Furnace — Iron Extraction

Charge: haematite (Fe₂O₃) + coke (C) + limestone (CaCO₃). Temperature zones (bottom to top): tuyere zone (~1500°C), fusion zone, reduction zone, absorption zone.

CaCO₃ → CaO + CO₂ (flux formation)
C + O₂ → CO₂; CO₂ + C → 2CO (reducing agent)
Fe₂O₃ + 3CO → 2Fe + 3CO₂ (main reduction)
CaO + SiO₂ → CaSiO₃ (slag — removes gangue)
Pig iron (~4% C) tapped from bottom; slag floats above

Hall-Héroult Process — Aluminium

Al₂O₃ dissolved in molten cryolite (Na₃AlF₆) at ~950°C; electrolysed in carbon-lined cell.

Cathode: Al³⁺ + 3e⁻ → Al (liquid, sinks to bottom)
Anode (carbon): 2O²⁻ → O₂ + 4e⁻; O₂ burns anode → CO₂
Anodes must be replaced periodically (consumed)
Cryolite lowers melting point and increases conductivity

Refining of Metals

Method	Principle	Metal
Liquation	Low-mp metal melts and flows away from gangue	Sn, Bi, Pb
Distillation	Volatile metal distilled, collected on cooling	Zn, Hg
Electrolytic refining	Impure metal = anode; pure metal = cathode; same salt solution	Cu, Ag, Au, Al
Zone refining	Impurities more soluble in melt; heater moves along rod	Si, Ge (semiconductors)
Van Arkel method	Volatile iodide formed then thermally decomposed on hot filament	Ti, Zr (ultra-pure)
Mond's process	Ni + 4CO → Ni(CO)₄ (volatile); decomposed at 180°C	Ni

Important Alloys

Alloy	Composition	Use
Steel	Fe + C (0.1–1.5%)	Construction
Stainless steel	Fe + Cr (12–18%) + Ni	Cutlery, utensils
Brass	Cu (70%) + Zn (30%)	Instruments, fittings
Bronze	Cu (90%) + Sn (10%)	Statues, coins
Duralumin	Al + Cu (4%) + Mg + Mn	Aircraft bodies
Solder	Pb (50%) + Sn (50%)	Joining metals
German silver	Cu + Zn + Ni	Utensils, jewellery

💡 JEE Tip: Froth flotation is for sulphide ores — the pine oil makes ore particles hydrophobic so they attach to air bubbles. Gravity separation is for oxide/carbonate ores. Leaching with NaCN is for Au: 4Au + 8NaCN + 2H₂O + O₂ → 4Na[Au(CN)₂] + 4NaOH; gold recovered by: 2Na[Au(CN)₂] + Zn → Na₂[Zn(CN)₄] + 2Au.

⚠ Common Error: Calcination vs Roasting — calcination is in limited/no air (carbonates → oxides); roasting is in excess air (sulphides → oxides + SO₂). Both are pyrometallurgy steps before reduction, not the reduction step itself.

📐

Metallurgy — Key Reactions

Roasting & Reduction

ZnS Roasting

2ZnS + 3O₂ → 2ZnO + 2SO₂

ZnO + C → Zn + CO (reduction)

Blast Furnace (Fe)

Fe₂O₃ + 3CO → 2Fe + 3CO₂

CaO + SiO₂ → CaSiO₃ (slag)

Cu Self-reduction

2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂

2Cu₂O + Cu₂S → 6Cu + SO₂

Gold — Cyanide Process

4Au + 8NaCN + 2H₂O + O₂ → 4Na[Au(CN)₂] + 4NaOH

2Na[Au(CN)₂] + Zn → Na₂[Zn(CN)₄] + 2Au

Mond's Process (Ni)

Ni + 4CO → Ni(CO)₄ (at 60°C)

Ni(CO)₄ → Ni + 4CO (at 180°C)

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Metallurgy — Solved Examples

Example 1: Which method is used to concentrate copper pyrites (CuFeS₂)? Explain the principle.

Froth flotation is used for sulphide ores like copper pyrites.
Principle: Pine oil (collector) is added along with water and the crushed ore. Air is blown through. The ore particles become hydrophobic (water-repelling) due to pine oil, attach to air bubbles, rise to the surface as froth, and are skimmed off. The gangue (rocky material) remains hydrophilic and settles.
Key reagents: Pine oil (collector); frothers (to stabilise froth); depressants (e.g., NaCN to depress ZnS when separating ZnS from PbS).

Example 2: Why is cryolite used in the Hall-Héroult process for extraction of aluminium?

Pure Al₂O₃ has a very high melting point (~2050°C), making direct electrolysis impractical and very costly.
Cryolite (Na₃AlF₆) is added because it: (1) lowers the melting point of the mixture to ~950°C, and (2) increases electrical conductivity of the melt, making the electrolysis efficient. Fluorspar (CaF₂) is also added to further lower the melting point.

Example 3: In electrolytic refining of copper, what happens at the anode and cathode?

Setup: Impure copper = anode; pure copper = cathode; CuSO₄ (aq) = electrolyte.
Anode: Cu → Cu²⁺ + 2e⁻ (anode dissolves). Impurities less reactive than Cu (Au, Ag, Pt) fall as anode mud below the anode.
Cathode: Cu²⁺ + 2e⁻ → Cu (pure copper deposits). More reactive impurities (Zn, Fe) stay in solution as ions.
Net effect: copper transfers from impure anode to pure cathode; precious metals collected as anode mud.

Example 4: Using the Ellingham diagram, explain why aluminium can reduce Cr₂O₃ but not vice versa at 1000°C.

The Ellingham diagram shows ΔG° (formation of metal oxide) vs temperature.
At 1000°C, ΔG° for Al₂O₃ formation ≈ −850 kJ/mol O₂ (more negative than Cr₂O₃ ≈ −500 kJ/mol O₂).
A more negative ΔG° means Al has a stronger affinity for oxygen than Cr. Therefore:
- Al can reduce Cr₂O₃: 2Al + Cr₂O₃ → Al₂O₃ + 2Cr (thermite-type reaction, ΔG° negative → spontaneous)
- Cr cannot reduce Al₂O₃ (would require energy input)
General rule: the metal whose oxide line is lower on the Ellingham diagram can reduce the other metal's oxide.

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Metallurgy — Diagrams

Click any diagram to zoom

Blast furnace: Fe₂O₃ reduced by CO; CaCO₃ flux removes SiO₂ as slag; pig iron (~4% C) tapped from bottom

Five-step metallurgy flow: crushing → concentration → roasting/calcination → reduction → refining

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Metallurgy — Quick Quiz

Question 1

Which concentration method is used for sulphide ores like ZnS and PbS?

Froth flotation exploits the hydrophobic nature of sulphide ore particles — they attach to pine oil-coated air bubbles and rise as froth. Magnetic separation is for magnetic ores (magnetite). Gravity separation is for oxide/carbonate ores. Leaching uses chemical reagents.

Question 2

In Mond's process for nickel purification, nickel tetracarbonyl Ni(CO)₄ is formed at 60°C and decomposed at:

Mond's process: Ni + 4CO → Ni(CO)₄ at 60°C (volatile compound forms). On heating to 180°C: Ni(CO)₄ → Ni + 4CO. Pure nickel deposits on a heated surface; CO is recycled. The CO is toxic — this distinguishes Mond's from other refining methods.

Question 3

Zone refining is used to obtain ultra-pure samples of:

Zone refining is used for semiconductor metals (Si, Ge) that require extremely high purity (parts per billion). A moving heater sweeps impurities along the rod. Copper/silver use electrolytic refining; Ni uses Mond's process; Zn/Hg use distillation.

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Practical Chemistry — Salt Analysis

Cation groups · group reagents · anion tests · flame tests

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1. Qualitative Analysis of Cations

In systematic salt analysis, cations (basic radicals) are precipitated in a fixed order by adding group reagents, so that each group separates out before the next:

Group	Group reagent	Cations
I	dil. HCl	Pb²⁺
II	H₂S in presence of dil. HCl	Cu²⁺, Pb²⁺ (and others)
III	NH₄Cl + NH₄OH	Al³⁺, Fe³⁺
IV	H₂S in presence of NH₄OH	Zn²⁺, Ni²⁺
V	(NH₄)₂CO₃	Ca²⁺, Ba²⁺
VI	no group reagent	Mg²⁺, NH₄⁺

2. Anion Tests & Flame Tests

Anions (acid radicals) are detected with specific reagents — e.g. a carbonate (CO₃²⁻) gives brisk effervescence with dilute acid; a chloride (Cl⁻) gives a white AgCl precipitate with AgNO₃ (soluble in NH₃); a sulphate (SO₄²⁻) gives a white BaSO₄ precipitate insoluble in acid. Some metals impart a characteristic flame colour:

Element	Flame colour
Sodium (Na)	Golden yellow
Potassium (K)	Lilac / violet
Calcium (Ca)	Brick red
Barium (Ba)	Apple green
Copper (Cu)	Bluish green

💡 Cation groups I–VI with reagents dil.HCl → H₂S/HCl → NH₄Cl+NH₄OH → H₂S/NH₄OH → (NH₄)₂CO₃ → (none). Flame: Na golden-yellow, K lilac, Ca brick-red, Ba apple-green, Cu bluish-green.

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Practical Chemistry — Quick Quiz

tap an option to reveal the answer

Question 1

The group reagent used to precipitate Group III cations (Al³⁺, Fe³⁺) is:

Group III cations are precipitated as hydroxides by NH₄OH in the presence of NH₄Cl (which controls the hydroxide-ion concentration so only Group III precipitates).

Question 2

A salt imparts a golden-yellow colour to a flame. The cation is likely:

Sodium gives a persistent golden-yellow flame; potassium is lilac, calcium brick-red and barium apple-green.

Question 3

Brisk effervescence of a gas that turns lime water milky, on adding dilute acid to a salt, indicates the anion:

A carbonate releases CO₂ with dilute acid — brisk effervescence of a gas that turns lime water milky is the classic carbonate test.