Chemical Bonding — vidhyapath 2027

Concept Core

From why atoms bond to predicting molecular geometry — the full picture.

Why Atoms Bond — The Octet Rule

Atoms bond to attain 8 electrons in their outermost shell (2 for H and He — duet rule). Exceptions: H₂ (duet), PCl₅ (expanded octet — 10e⁻), SF₆ (expanded octet — 12e⁻), BF₃ (incomplete octet — 6e⁻). These exceptions are EAPCET favourites.

Types of Chemical Bonds

Ionic Bond: Transfer of electrons. High electronegativity difference (>1.7). Eg: NaCl, MgO. Forms crystal lattice.

Covalent Bond: Sharing of electrons. Similar electronegativities. Eg: H₂, CO₂, CH₄.

Coordinate (Dative) Bond: Both electrons from one atom. Eg: NH₄⁺, H₃O⁺, BF₃·NH₃

Hybridisation — Master Table

Hybrid	Geometry	Angle	Example
`sp`	Linear	180°	BeCl₂, CO₂, C₂H₂
`sp²`	Trigonal planar	120°	BF₃, SO₃, C₂H₄
`sp³`	Tetrahedral	109.5°	CH₄, NH₃, H₂O
`sp³d`	Trigonal bipyramidal	90°,120°	PCl₅
`sp³d²`	Octahedral	90°	SF₆
`sp³d³`	Pentagonal bipyramidal	72°,90°	IF₇

* NH₃ and H₂O are sp³ but distorted — lone pairs reduce bond angle.

VSEPR Theory — Shapes from Electron Pairs

Shape depends on total electron pairs (bonding + lone) around central atom. Lone pairs occupy more space — they compress bond angles.

Bond angle: CH₄ = 109.5° > NH₃ = 107° > H₂O = 104.5°

CH₄

Tetrahedral

109.5°

NH₃

Trigonal pyramidal

107°

H₂O

Angular/bent

104.5°

PCl₅

Trig. bipyramidal

90°/120°

SF₆

Octahedral

90°

XeF₄

Square planar

90°

Bond Polarity & Molecular Polarity

A bond is polar when the two atoms have different electronegativities. But a molecule can be non-polar even with polar bonds — if bond dipoles cancel by symmetry.

CO₂: Two C=O polar bonds, but linear → cancel → non-polar molecule

H₂O: Two O-H bonds, bent shape → don't cancel → polar molecule

BF₃: Three B-F bonds, trigonal planar → cancel → non-polar

NH₃: Three N-H bonds + lone pair → pyramidal → polar

Hydrogen Bonding

Forms between H bonded to F, O, or N (highly electronegative, small) and a lone pair on another F, O, or N.

Intermolecular H-bond: Between molecules. Raises boiling point. Eg: H₂O, HF, NH₃

Intramolecular H-bond: Within same molecule. Lowers boiling point vs intermolecular. Eg: o-nitrophenol < p-nitrophenol BP.

Why HF has lower BP than H₂O? HF has one H-bond per molecule; H₂O has two — stronger H-bond network.

Formal Charge — the Tool for Resonance

Formal Charge = Valence electrons − Non-bonding electrons − ½ × Bonding electrons

FC = V − N − B/2
where V = valence e⁻, N = lone pair e⁻, B = bonding e⁻

The most stable structure has lowest formal charges (ideally zero). In SO₄²⁻, the structure with double bonds has lower formal charge than the all-single-bond structure.

Formula Vault

Rules, formulas, and shortcuts — all in one place.

Hybridisation Number

H = ½(V + M − C + A)

V = valence e⁻, M = monovalent atoms, C = charge (+), A = charge (−). Result = sp, sp², sp³...

Formal Charge

FC = V − N − B/2

V = valence e⁻; N = non-bonding e⁻; B = bonding e⁻

Bond Order (MO Theory)

BO = (N_b − N_a) / 2

N_b = bonding e⁻; N_a = antibonding e⁻

Hybridisation from Bond Angle

sp → 180° | sp² → 120°
sp³ → 109.5°

Lone pairs reduce bond angle below ideal

Electronegativity Order

F > O > N > Cl > Br > C > H

F is most electronegative (3.98 Pauling)

H-Bond Strength Order

F−H···F > O−H···O > N−H···N

Stronger with more electronegative atom

Dipole Moment

μ = q × d

q = charge, d = bond length; units: Debye (D). Zero for symmetric molecules.

Octet Exceptions

Incomplete: BeCl₂, BF₃
Expanded: PCl₅, SF₆, XeF₄

Expanded = d orbitals available (Period 3+)

VSEPR: Lone Pair Effect

lp−lp > lp−bp > bp−bp

Repulsion order. More repulsion = smaller bond angle

Bond Polarity Rule

ΔEN < 0.5: non-polar covalent
0.5–1.7: polar covalent
>1.7: ionic character

Based on Pauling electronegativity difference

Worked Examples

5 problems — from basic shape prediction to EAPCET traps. Click to expand.

EasyFind hybridisation and shape of NH₃▾

Determine the hybridisation, shape, and bond angle of NH₃.

N: valence electrons = 5. Three H atoms each contribute 1e⁻ for bonding.

Total electron pairs around N = 3 (bond pairs) + 1 (lone pair) = 4 pairs → sp³ hybridisation

4 electron pairs → tetrahedral arrangement of electrons. But shape = arrangement of atoms = trigonal pyramidal

1 lone pair reduces angle from 109.5° → 107°

✓ sp³, trigonal pyramidal shape, bond angle = 107°

MediumWhy is the bond angle of H₂O less than NH₃?▾

Both H₂O and NH₃ are sp³ hybridised. H₂O has angle 104.5°, NH₃ has 107°. Explain the difference.

NH₃: 3 bond pairs + 1 lone pair. The 1 lp pushes bond pairs → 107° (less than 109.5°)

H₂O: 2 bond pairs + 2 lone pairs. The 2 lp exert more repulsion → 104.5° (less than NH₃)

Rule: More lone pairs = greater repulsion = smaller bond angle. lp-lp > lp-bp > bp-bp repulsion

✓ H₂O has 2 lone pairs vs NH₃'s 1 → greater compression → 104.5° < 107°

MediumPredict polarity of CO₂ vs SO₂▾

Both CO₂ and SO₂ have polar bonds. Predict which is polar and which is non-polar. Explain.

CO₂: sp hybridisation → linear (O=C=O). Two C=O dipoles point in opposite directions → cancel → μ = 0

SO₂: sp² hybridisation (lone pair on S) → bent/angular shape. Two S=O dipoles do NOT cancel → net dipole ≠ 0

Key difference: CO₂ is linear (symmetric), SO₂ has a lone pair making it bent (asymmetric)

✓ CO₂: non-polar (linear, dipoles cancel) | SO₂: polar (bent, dipoles don't cancel)

EAPCET LevelFind hybridisation of XeF₄ and predict its shape▾

Determine hybridisation and shape of XeF₄. This tests expanded octet and VSEPR together.

Xe: valence electrons = 8. Each F contributes 1e⁻ for bonding. 4 bonds with F.

Remaining non-bonding e⁻ on Xe = 8 − 4 = 4 → 2 lone pairs

Total electron pairs = 4 (bonding) + 2 (lone) = 6 → sp³d² hybridisation

6 electron pairs = octahedral arrangement. 2 lone pairs occupy axial positions (minimise repulsion). 4 F atoms in equatorial plane.

Molecular shape (atoms only) = square planar

✓ XeF₄: sp³d² hybridisation, square planar shape, bond angle = 90°

Trap QuestionArrange HF, HCl, HBr, HI by boiling point — most students get HF wrong▾

Arrange HF, HCl, HBr, HI in order of increasing boiling point. ⚠️ Students assume HF = lowest.

The trap: Students apply only van der Waals forces (larger molecule = higher BP). This gives HI > HBr > HCl > HF. But HF has intermolecular hydrogen bonding!

HF has the strongest H-bond (F is most electronegative). So HF has abnormally high BP despite being smallest.

For HCl, HBr, HI: no H-bonding. Only London dispersion forces. BP increases with molecular mass: HI > HBr > HCl

HF is placed above HCl but below HBr or HI depending on balance. Actual order: HCl (−85°C) < HBr (−67°C) < HF (19°C) < HI (−35°C)

✓ BP order: HCl < HBr < HI < HF (HF is highest due to strong H-bonding)

Mistake DNA

5 errors from distractor analysis — the exact traps EAPCET sets and how to dodge them.

🔷

Confusing Electron Geometry with Molecular Shape

NH₃ has tetrahedral electron geometry but trigonal pyramidal shape. Students write tetrahedral for both.

❌ Wrong

NH₃ shape = tetrahedral
(counting lone pair) ✗

✓ Correct

NH₃ shape = trigonal
pyramidal (atoms only) ✓
e⁻ geometry = tetrahedral

Shape is determined by the positions of ATOMS only, not electron pairs. The lone pair defines the hybridisation but not the shape name.

🌀

Wrong Hybridisation for Expanded Octet Molecules

Students assign sp³ to PCl₅ and SF₆ instead of sp³d and sp³d².

❌ Wrong

PCl₅: sp³ (only 4 pairs) ✗
SF₆: sp³ hybridised ✗

✓ Correct

PCl₅: sp³d (5 pairs) ✓
SF₆: sp³d² (6 pairs) ✓
Use d orbitals (Period 3+)

Only Period 3+ elements (P, S, Cl, Xe) can use d orbitals for expanded octets. First-row elements (N, O) cannot exceed 8 electrons.

💧

Applying H-Bonding to Wrong Molecules

H-bonding requires H bonded to F, O, or N. Students apply it to HCl, CH₄, etc.

❌ Wrong

HCl has H-bonding ✗
CH₄ shows H-bonding ✗
(Cl, C not electronegative
enough)

✓ Correct

H-bonding only in:
HF, H₂O, NH₃ ✓
Alcohols, carboxylic acids ✓
NOT in HCl, CH₄, H₂S

H-bonding needs the small, highly electronegative F, O, or N. Chlorine is too large and not electronegative enough (despite being electronegative in other contexts).

📐

Assuming Polar Bonds = Polar Molecule

CO₂ and CCl₄ have polar bonds but are non-polar molecules because of symmetric cancellation.

❌ Wrong

"CO₂ has polar C=O bonds
→ CO₂ is polar" ✗
"CCl₄ is polar" ✗

✓ Correct

CO₂: linear → dipoles
cancel → non-polar ✓
CCl₄: tetrahedral → non-polar ✓

Test: draw the molecule. If all bond dipoles cancel by symmetry, the molecule is non-polar. Asymmetry (lone pair, different atoms) = polar.

⬆️

Predicting HF as Lowest Boiling Point Among Hydrogen Halides

Applying only van der Waals logic gives the order wrong. HF has anomalously high BP due to hydrogen bonding.

❌ Wrong

BP: HF < HCl < HBr < HI ✗
(only vdW logic)

✓ Correct

BP: HCl < HBr < HI < HF ✓
HF: highest due to
strong H-bonding

HF is an anomaly. Among H₂O, HF, NH₃ — all show H-bonding. H₂O has the highest BP because each molecule forms 2 H-bonds; HF forms only 1.

Chapter Intelligence

EAPCET asks Chemical Bonding questions every year — here is exactly what to prepare.

EAPCET Topic Weightage (2019–2024)

Hybridisation identification

VSEPR — shape prediction

Hydrogen bonding

Molecular polarity

Bond order (MO theory)

Formal charge

High-Yield PYQ Patterns

Find hybridisation of PCl₅, SF₆, XeF₄ Shape of NH₃ vs H₂O Polar vs non-polar molecules H-bond — which molecule? Bond angle order: CH₄>NH₃>H₂O Why CO₂ non-polar but SO₂ polar? Boiling point anomaly: HF, H₂O

Exam Strategy — 5 Marks in 7 Minutes

Hybridisation questions: use the formula H = ½(V + M − C + A). Count, don't memorise each molecule individually. Then look up geometry from the table.
Shape prediction: always ask "how many lone pairs?" first. 0 lp = symmetric shape; each lp compresses angles and changes shape name.
Polarity questions: draw the molecule, mark dipole arrows, check if they cancel by symmetry. Linear and tetrahedral symmetric molecules are non-polar even with polar bonds.
H-bonding questions: the condition is H bonded to F, O, or N only. If a question asks "which has highest BP," look for H-bonding first before molecule size.
Expanded octet molecules (PCl₅, SF₆, XeF₄, ClF₃) appear almost every year. Memorise their hybridisation, shape, and number of lone pairs on central atom.
Bond order from MO theory: O₂ = 2, N₂ = 3, F₂ = 1, Ne₂ = 0 (doesn't exist). Higher bond order = shorter bond = stronger bond.