1. General Properties

Elements where the last electron enters the outermost p-orbital. General Config: ns²np^1-6.

Inert Pair Effect:

The reluctance of ns² electrons to participate in bonding due to poor shielding by d and f electrons.
Consequence: Stability of lower oxidation state increases down the group. (e.g., Tl⁺¹ > Tl⁺³).

2. Group 13 (Boron Family)

Config: ns²np¹. Elements: B, Al, Ga, In, Tl.

Important Compounds of Boron

1. Diborane (B₂H₆):

Electron deficient hydride. Contains Banana Bonds (3-center-2-electron bonds).
• Four terminal H-atoms are normal (2c-2e).
• Two bridging H-atoms are banana bonds.
• Hybridization of Boron: sp³.

2. Borax (Na₂B₄O₇·10H₂O):

Used in Borax Bead Test (detects colored cations like Co²⁺, Cu²⁺).
Aqueous solution is alkaline.

3. Orthoboric Acid (H₃BO₃):

Monobasic weak Lewis acid (accepts OH^-). Layered structure due to H-bonding.

3. Group 14 (Carbon Family)

Config: ns²np². Elements: C, Si, Ge, Sn, Pb.

Catenation:

Property of self-linking. Order: C >> Si > Ge ≈ Sn. Carbon forms strong C-C bonds.

Allotropes of Carbon

Diamond: sp³ hybridized. Hardest substance. Insulator. 3D network.
Graphite: sp² hybridized. Layered hexagonal structure. Conductor (free electron). Lubricant.
Fullerenes: C₆₀ (Buckminsterfullerene). Soccer ball shape. sp².

[Image of Diamond vs Graphite structure]

Silicon Compounds

Silicones: Polymers with R₂SiO repeating units. Water repellent, heat resistant.
Silicates: Basic unit SiO₄^4- (Tetrahedral). Examples: Zeolites (Shape selective catalysts).

4. Group 15 (Nitrogen Family)

Config: ns²np³. Elements: N, P, As, Sb, Bi.

Anomalous Behavior of Nitrogen:

Small size, high electronegativity, absence of d-orbitals. Forms pπ-pπ multiple bonds (N≡N is inert gas). P forms pπ-dπ bonds.

Important Compounds

Ammonia (NH₃): Haber's Process. Basic, trigonal pyramidal.
Nitric Acid (HNO₃): Ostwald's Process. Strong oxidizing agent. Makes Iron passive.
Phosphorus:
• White P: P₄ tetrahedron, highly reactive, glow in dark (chemiluminescence).
• Red P: Polymeric, less reactive.
Phosphine (PH₃): Rotten fish smell. Used in Holme's signals.

5. Group 16 (Chalcogens)

Config: ns²np⁴. Elements: O, S, Se, Te, Po.

Ozone (O₃):

Prepared by silent electric discharge on O₂. Strong oxidizing agent (liberates nascent oxygen). Depleted by CFCs and NO.
Tailings of Mercury: Hg loses meniscus in contact with O₃.

Sulphuric Acid (H₂SO₄):

King of Chemicals. Prepared by Contact Process (V₂O₅ catalyst). Strong dehydrating agent (chars sugar to Carbon).

6. Group 17 (Halogens)

Config: ns²np⁵. Elements: F, Cl, Br, I.

Anomalous Fluorine:

Most electronegative. Bond dissociation enthalpy of F₂ is lower than Cl₂ due to e^--e^- repulsion in small F atom. Forms H-bonds (HF is liquid).

Interhalogen Compounds

XX', XX'₃, XX'₅, XX'₇ (where X is larger halogen). More reactive than pure halogens (except F₂) because X-X' bond is weaker.

7. Group 18 (Noble Gases)

Config: ns²np⁶. Elements: He, Ne, Ar, Kr, Xe, Rn.

Xenon Compounds (Neil Bartlett)

Compound	Hybridization	Shape
XeF₂	sp³d	Linear
XeF₄	sp³d²	Square Planar
XeF₆	sp³d³	Distorted Octahedral
XeO₃	sp³	Pyramidal

Numericals & HOTS

Q1. Banana Bonds Calculation

In a molecule of Diborane (B₂H₆), how many 2-center-2-electron (2c-2e) bonds and 3-center-2-electron (3c-2e) bonds are present?

Solution:
Structure: Each Boron is bonded to 2 terminal Hydrogens and 2 bridging Hydrogens.

1. Terminal B-H bonds: Normal covalent bonds. There are 4 such bonds. (Type: 2c-2e).
2. Bridging B-H-B bonds: "Banana bonds". There are 2 such bonds. (Type: 3c-2e).

Ans: 4 (2c-2e) and 2 (3c-2e)

Q2. Structure & Basicity

Determine the basicity of Hypophosphorous acid (H₃PO₂) and Orthophosphorous acid (H₃PO₃) based on their structures.

Solution:
Basicity depends on the number of ionizable P-OH bonds (H bonded to P is not acidic).

1. H₃PO₂: Structure has 2 P-H bonds and 1 P-OH bond.
→ Basicity = 1 (Monobasic).

2. H₃PO₃: Structure has 1 P-H bond and 2 P-OH bonds.
→ Basicity = 2 (Dibasic).

Q3. VSEPR Theory Application

Calculate the number of lone pairs on the central atom and the shape of XeOF₄.

Solution:
Xe (Group 18) has 8 valence electrons.
O forms double bond (uses 2e), 4 F form single bonds (use 4e).
Total electrons used = 2 + 4 = 6.
Remaining electrons = 8 - 6 = 2 (1 Lone Pair).

Electron Pairs (EP) = Sigma bonds + Lone Pairs = 5 (4 F + 1 O) + 1 = 6.
Hybridization: sp³d².
Geometry: Octahedral. Shape (ignoring LP): Square Pyramidal.

Q4. Marshall's Acid

Calculate the oxidation state of Sulphur in H₂S₂O₈ (Peroxodisulphuric acid).

Solution:
Formula method: 2(+1) + 2x + 8(-2) = 0 → 2x = 14 → x = +7. (Wrong, Max for S is +6).

Structure Method:
Structure has a peroxide linkage (-O-O-).
So, 2 Oxygens are -1, remaining 6 are -2.

2(+1) + 2x + 6(-2) [Oxide] + 2(-1) [Peroxide] = 0
2 + 2x - 12 - 2 = 0
2x = 12
x = +6

Q5. Stability Trends

Why does PbCl₄ act as a strong oxidizing agent, whereas SnCl₂ acts as a reducing agent?

Solution:
Due to Inert Pair Effect:
For Pb (Group 14 bottom), +2 state is more stable than +4. So, Pb⁺⁴ tries to gain electrons to become Pb⁺². Hence, PbCl₄ is an Oxidizing Agent.

For Sn, +4 state is more stable than +2. So, Sn⁺² loses electrons to become Sn⁺⁴. Hence, SnCl₂ is a Reducing Agent.

Q6. Drago's Rule (HOTS)

Arrange NH₃, PH₃, and AsH₃ in decreasing order of bond angles and explain the sharp drop.

Solution:
Order: NH₃ (107.8°) > PH₃ (93.6°) > AsH₃ (91.8°).

Reason: NH₃ involves sp³ hybridization. In PH₃ and AsH₃, the central atom is large and less electronegative. According to Drago's Rule, hybridization is negligible, and bonding occurs via almost pure p-orbitals (which are at 90°). Hence, angle is close to 90°.

Q7. Xenon Reaction

1 mole of XeF₆ reacts completely with water. How many moles of HF are produced?

Solution:
Reaction for complete hydrolysis:
XeF₆ + 3H₂O → XeO₃ + 6HF

From stoichiometry:
1 mole XeF₆ produces 6 moles HF.

Q8. ClF₃ Geometry

Predict the shape of Chlorine Trifluoride (ClF₃) using VSEPR theory.

Solution:
Central Atom: Cl (7 valence e^-).
Bond Pairs: 3 (with F).
Lone Pairs: (7 - 3)/2 = 2.

Total Pairs = 5 (sp³d). Geometry is Trigonal Bipyramidal.
To minimize repulsion, 2 Lone Pairs occupy equatorial positions.
Shape: T-shaped (Bent T).

Q9. Pyrosilicate Formula

Two SiO₄^4- tetrahedra share one corner oxygen. What is the formula of the resulting anion?

Solution:
Basic Unit: SiO₄.
When 2 units join sharing 1 oxygen:
Formula = 2 × (SiO₄) - 1 Oxygen (shared) = Si₂O₇.
Charge = 2 × (-4) - (-2 for removed O) = -6.
Formula: Si₂O₇^6- (Pyrosilicate)

Q10. I₃^- Ion

What is the hybridization and shape of the I₃^- ion?

Solution:
Central Atom: I. Valence e^- = 7.
Add 1 e^- for negative charge = 8.
Bond Pairs: 2 (with two terminal I atoms).
Lone Pairs: (8 - 2)/2 = 3.

Total Pairs = 2 + 3 = 5 (sp³d).
Geometry: Trigonal Bipyramidal.
3 Lone Pairs occupy equatorial positions (120° apart) to minimize repulsion.
Shape: Linear

Important Reactions

1. Boron Compounds

Borax Bead Test:

Na₂B₄O₇ · 10H₂O &xrightarrow{\Delta} NaBO₂ + B₂O₃

(Transparent glassy bead)

Heating Boric Acid:

H₃BO₃ &xrightarrow{370K} HBO₂ &xrightarrow{Red Hot} B₂O₃

2. Nitrogen & Phosphorus

Haber's Process (Ammonia):

N₂ + 3H₂ ↔ 2NH₃ ; ΔH = -46.1 kJ/mol

Ostwald's Process (HNO₃):

4NH₃ + 5O₂ &xrightarrow{Pt/Rh} 4NO + 6H₂O

Brown Ring Complex:

[Fe(H₂O)₅(NO)]²⁺

3. Halogens & Xenon

Deacon's Process (Chlorine):

4HCl + O₂ &xrightarrow{CuCl_2} 2Cl₂ + 2H₂O

Hydrolysis of XeF₆:

XeF₆ + 3H₂O → XeO₃ + 6HF

(Complete Hydrolysis - Explosive Solid)

20 Golden Facts (NEET)

1. Inert Pair Effect: The stability of +1 oxidation state increases down Group 13 (Tl⁺¹ > Tl⁺³) and +2 increases down Group 14 (Pb⁺² > Pb⁺⁴) due to poor shielding of d/f electrons holding ns² electrons tight.
2. Inorganic Benzene: Borazine (B₃N₃H₆) is isoelectronic and isostructural with Benzene. However, unlike benzene, it is reactive due to the polarity of B-N bonds.
3. Back Bonding: BF₃ acts as a Lewis acid. Its acidity is less than BCl₃ because of pπ-pπ back bonding between filled p-orbital of F and empty p-orbital of B.
4. Graphite Stability: Thermodynamically, Graphite is the most stable allotrope of Carbon. (Δ_fH° of Graphite = 0, Diamond = 1.90 kJ/mol).
5. Producer vs Water Gas:
Water Gas (Syn Gas): CO + H₂ (High Calorific Value).
Producer Gas: CO + N₂ (Lower Calorific Value).
6. Holme's Signal: A mixture of Calcium Carbide (CaC₂) and Calcium Phosphide (Ca₃P₂). In water, they produce Acetylene (burns) and Phosphine (smoke/spontaneous combustion).
7. Bond Angle Trend: NH₃ (107.8°) > PH₃ (93.6°) > AsH₃ (91.8°) > SbH₃ (91.3°). Drago's Rule explains the sharp drop (almost pure p-orbitals used in PH₃).
8. Boiling Point Anomaly:
Group 15: PH₃ < AsH₃ < NH₃ < SbH₃ < BiH₃.
Group 16: H₂S < H₂Se < H₂Te < H₂O.
(NH₃ and H₂O have high BP due to H-bonding).
9. Nitric Acid Passivity: Concentrated HNO₃ makes Iron and Aluminium passive (non-reactive) due to the formation of a thin impervious oxide layer on the surface.
10. Solid PCl₅: In solid state, PCl₅ exists as an ionic solid: [PCl₄]⁺ (Tetrahedral) and [PCl₆]^- (Octahedral).
11. Oxygen's Magnetism: O₂ is paramagnetic (2 unpaired electrons in π* orbitals). S₂ (in vapor state) is also paramagnetic like O₂.
12. SF₆ Inertness: SF₆ is chemically inert due to steric hindrance; the S atom is protected by 6 F atoms. It is used as a gaseous insulator.
13. Bleaching Actions:
• Cl₂ bleaches by Oxidation (Permanent).
• SO₂ bleaches by Reduction (Temporary).
14. Electron Gain Enthalpy: Unexpectedly, Cl > F and S > O. Fluorine and Oxygen have small sizes, causing inter-electronic repulsion, making it slightly harder to add an electron compared to Cl and S.
15. Glass Etching: HF is the only acid that attacks glass (SiO₂) to form H₂SiF₆ (Fluorosilicic acid). Hence, HF is stored in wax-lined bottles.
16. Neil Bartlett: He prepared the first Noble Gas compound (Xe⁺[PtF₆]^-). He realized O₂ and Xe have similar Ionization Enthalpies (1175 kJ/mol vs 1170 kJ/mol).
17. Aqua Regia: A mixture of Conc. HCl and Conc. HNO₃ in the ratio 3:1. It dissolves Gold (Au) and Platinum (Pt) by forming soluble chloride complexes (HAuCl₄).
18. Hydride Acidity: Acidic strength increases down the group: HF < HCl < HBr < HI. This is because Bond Dissociation Enthalpy decreases as size increases.
19. Phosphinic Acid: H₃PO₂ (Hypophosphorous acid) is monobasic (one P-OH) and a strong reducing agent due to two P-H bonds.
20. ClF₃ Structure: It has a T-shaped structure (sp³d hybridization) with 2 lone pairs at equatorial positions to minimize repulsion.

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