📖 Chapter CN04 · NDA Class 11–12 Level🎯 NDA Level : High Priority
Chemical Bonding and States of Matter covers a wide range of NDA questions — from why NaCl is ionic while H₂O is covalent, to why water boils at 100°C (not −80°C), to gas law calculations. Average students who understand the three bond types, memorise VSEPR shapes, know the two gas laws by heart, and can apply PV=nRT can consistently score from this chapter.
📌 What to expect in NDA (based on 2022–2025 pattern): (1) Ionic vs covalent vs metallic bond — basis, examples, properties; (2) VSEPR shapes — linear, trigonal planar, tetrahedral, bent, pyramidal (with bond angles); (3) Hydrogen bonding — why water has unusually high boiling point; H-bond donors vs acceptors; (4) Boyle's law (P₁V₁ = P₂V₂) and Charles's law (V₁/T₁ = V₂/T₂) numerical applications; (5) Ideal gas equation PV = nRT; kinetic theory assumptions; (6) Solution types and concentration terms; colligative properties (elevation of BP, depression of FP).
Topics at a Glance
① Chemical Bonds
Ionic, covalent, metallic
② VSEPR & Geometry
Shapes, bond angles, lone pairs
③ Intermolecular Forces
H-bonding, van der Waals
④ Gas Laws
Boyle, Charles, ideal gas equation
⑤ Solutions
Types, concentration, colligative
1. Chemical Bonding
1.1
Why Do Atoms Bond?
Every atom strives for the lowest energy state — usually a complete outer shell (octet/duplet)
Atoms bond to achieve the electron configuration of the nearest noble gas — a filled valence shell (8 electrons = octet; 2 electrons = duplet for H and Li). The type of bond formed depends on the electronegativity difference (ΔEN) between the bonding atoms.
⚛ Bond Type from Electronegativity Difference
ΔEN = |EN of atom A − EN of atom B|
ΔEN > 1.7 → Ionic bond (electron transfer: metal + non-metal)
ΔEN 0.4–1.7 → Polar covalent (unequal sharing: H–Cl, H–O, H–N)
ΔEN < 0.4 → Non-polar covalent (equal sharing: H₂, Cl₂, C–H)
ΔEN = 0 → Pure covalent (same elements: H–H, O=O, N≡N)
Examples:
NaCl: Na(0.9) vs Cl(3.0) → ΔEN = 2.1 → Ionic ✓
HCl: H(2.1) vs Cl(3.0) → ΔEN = 0.9 → Polar covalent ✓
H₂: H(2.1) vs H(2.1) → ΔEN = 0.0 → Non-polar covalent ✓
Rule of thumb: Metal + Non-metal → Ionic. Non-metal + Non-metal → Covalent. Metal + Metal → Metallic bond.
IONIC BOND
Electron Transfer
Complete transfer of e⁻ from metal to non-metal
Metal loses → forms cation (+)
Non-metal gains → forms anion (−)
Held by electrostatic attraction
Examples: NaCl, KBr, CaF₂, MgO, Al₂O₃
High MP/BP; conduct electricity in molten/solution state; brittle; soluble in water
High MP; lustrous; malleable; excellent conductors
Fig. 1 — Ionic bond (NaCl: electron transfer), Covalent bond (H₂O: electron sharing with lone pairs), Metallic bond (sea of delocalised electrons around M⁺ cations)
📝 TOPIC-WISE PYQ
Chemical Bonding — NDA Pattern Questions
Q1. Which of the following compounds has an ionic bond?
(a) HCl (b) H₂O (c) MgO (d) CO₂
Answer: (c) MgO
Mg (EN=1.2) and O (EN=3.5) → ΔEN = 2.3 > 1.7 → Ionic bond. Mg loses 2 electrons → Mg²⁺; O gains 2 electrons → O²⁻. HCl (ΔEN=0.9) and H₂O (ΔEN=1.4) are polar covalent. CO₂ (ΔEN=1.0 for C–O) is polar covalent. Metal + non-metal → ionic is the quick rule.
Q2. Which property of metals is best explained by the metallic bond (sea of electrons model)?
(a) High melting point (b) Brittleness (c) High electrical conductivity (d) Solubility in water
Answer: (c) High electrical conductivity
The "sea of electrons" consists of freely mobile delocalised electrons. When a potential difference is applied, these electrons flow as electric current — explaining metals' excellent conductivity. Brittleness is a property of ionic crystals (not metals). Metals are ductile and malleable because the sea of electrons lets layers of cations slide past each other without breaking the bond.
Q3. The number of covalent bonds in a molecule of N₂ is:
(a) 1 (b) 2 (c) 3 (d) 4
Answer: (c) 3 — Triple bond
N has 5 valence electrons; needs 3 more for octet. Two N atoms share 3 pairs of electrons → N≡N (triple bond). This consists of 1 sigma (σ) bond and 2 pi (π) bonds. N₂ is extremely stable (ΔH = 945 kJ/mol to break) because of this strong triple bond — relevant to the Haber process for making NH₃ (needs high temperature/catalyst to break N≡N).
2. Molecular Geometry — VSEPR Theory
2.1
VSEPR — Predicting Molecular Shapes
Valence Shell Electron Pair Repulsion — electron pairs repel to maximum separation
VSEPR theory states that electron pairs (bonding pairs and lone pairs) around a central atom arrange themselves to minimise repulsion. Lone pairs repel more than bonding pairs, so they compress bond angles. The shape depends on both the number of bonding pairs (BP) and lone pairs (LP).
📋 Key Rule: Lone pair–lone pair repulsion > Lone pair–bond pair repulsion > Bond pair–bond pair repulsion
Each lone pair reduces bond angle by approximately 2.5°. Starting from tetrahedral (109.5°): add 1 LP → pyramidal (107°); add 2 LP → bent (104.5°).
2 BP + 0 LP
Linear
Bond angle: 180°
BeCl₂, CO₂, HCN, C₂H₂
3 BP + 0 LP
Trigonal Planar
Bond angle: 120°
BF₃, BCl₃, SO₃, NO₃⁻
4 BP + 0 LP
Tetrahedral
Bond angle: 109.5°
CH₄, CCl₄, NH₄⁺, SiH₄
3 BP + 1 LP
Trigonal Pyramidal
Bond angle: ~107°
NH₃, PH₃, PCl₃, AsCl₃
2 BP + 2 LP
Bent / V-Shaped
Bond angle: ~104.5°
H₂O, H₂S, SO₂, O₃
5 BP + 0 LP
Trigonal Bipyramidal
Axial 90°; Equatorial 120°
PCl₅, PF₅
Fig. 2 — VSEPR shapes: Linear (CO₂, 180°), Trigonal Planar (BF₃, 120°), Tetrahedral (CH₄, 109.5°), Bent (H₂O, 104.5° — compressed by 2 lone pairs). Red ellipses = lone pairs.
(a) Trigonal planar (b) Trigonal pyramidal (c) Tetrahedral (d) Linear
Answer: (b) Trigonal pyramidal
N has 5 valence electrons. In NH₃: 3 bond pairs (with H) + 1 lone pair = 4 electron pairs total. Arrangement: tetrahedral electron geometry, but shape (ignoring LP) = trigonal pyramidal. The lone pair compresses H–N–H angle to ~107° (vs 109.5° for perfect tetrahedron). If NH₃ had no lone pair (like CH₄ analogy), it would be tetrahedral.
Q2. The bond angle in H₂O is less than in NH₃. Which statement best explains this?
(a) H₂O has more hydrogen atoms (b) H₂O has two lone pairs vs one in NH₃ — greater repulsion compresses bond angle (c) Oxygen is heavier than nitrogen (d) H₂O has a linear shape
Answer: (b) H₂O has two lone pairs vs one lone pair in NH₃
H₂O: 2 BP + 2 LP → more lone pair repulsion → bond angle 104.5°. NH₃: 3 BP + 1 LP → less LP repulsion → bond angle 107°. Each lone pair exerts greater repulsion than a bond pair, pushing bond pairs closer. So H₂O has a smaller bond angle despite the same tetrahedral electron geometry.
🧠 TRICKY QUESTIONS
Bonding & VSEPR — Conceptual Traps
Q. CO₂ has two polar C=O bonds, yet the molecule is non-polar overall. How?
Answer: The two C=O dipoles are equal and opposite — they cancel out.
CO₂ is linear (O=C=O, 180°). Each C=O bond is polar (O pulls electrons towards itself — higher EN). But since both bonds are identical and directly opposite, their dipole moments are equal in magnitude and point in opposite directions → they cancel completely. Net dipole = 0 → non-polar molecule. Compare to H₂O (bent, 104.5°): the two O–H dipoles do NOT cancel because of the angle → net dipole → polar molecule. This is why H₂O has much higher BP than CO₂.
Q. BF₃ has 3 bond pairs and 0 lone pairs, so its bond angle should be 120°. But NF₃ also has 3 bond pairs — why is its angle only 102°?
Answer: NF₃ has 1 lone pair on N; BF₃ has none on B.
BF₃: B has only 6 valence electrons (incomplete octet) — no lone pair → 3 BP only → perfect trigonal planar (120°).
NF₃: N has 5 valence electrons → 3 bond pairs + 1 lone pair → lone pair compresses bond angle to ~102° (trigonal pyramidal shape). The lone pair on N repels the bonding pairs more than a bond pair would, squeezing the F–N–F angle below 120°.
3. Intermolecular Forces
3.1
Hydrogen Bonding & Van der Waals Forces
Forces between molecules — weaker than covalent bonds but critical for properties
💧 Hydrogen Bond
Forms when H is bonded to a highly electronegative atom (F, O, or N)
The partial positive H (δ+) attracts the lone pair of F/O/N on another molecule
Represented as: X–H ··· Y (dotted line = H-bond; X, Y = F, O, or N)
Intermolecular H-bond: between different molecules (H₂O, NH₃, HF, alcohols)
Intramolecular H-bond: within same molecule (e.g. o-nitrophenol)
Strongest intermolecular force (but much weaker than covalent bonds)
📈 Van der Waals Forces
Weak intermolecular forces present in all molecules
London dispersion forces: temporary dipoles in non-polar molecules; increase with molar mass
Dipole–dipole: between permanent dipoles (polar molecules)
Dipole–induced dipole: polar + non-polar
Increase with: larger molecule (more electrons), greater surface area
Explain why noble gases can be liquefied despite no permanent dipole
📌 Why water (H₂O) has anomalously high boiling point (100°C) — NDA favourite:
H₂S (similar structure) boils at −60°C. H₂O boils at 100°C. The 160°C difference is entirely due to hydrogen bonding. Each water molecule forms up to 4 H-bonds (2 donor, 2 acceptor) creating a strong 3D network. Breaking these H-bonds requires much more energy → higher BP. Similarly: HF (bp 19°C) is much higher than HCl (−85°C) due to H-bonding in HF. NH₃ (bp −33°C) is higher than PH₃ (bp −87°C) for the same reason.
Fig. 3 — Hydrogen bond network in water: each O–H (δ+) points towards a lone pair on O (δ−) of a neighbouring molecule. This network explains water's anomalously high boiling point.
📝 TOPIC-WISE PYQ
Intermolecular Forces — NDA Pattern Questions
Q1. The boiling point of H₂O (100°C) is much higher than H₂S (−60°C) though both have similar molecular structure. The reason is:
(a) H₂O has a higher molecular mass (b) H₂O molecules form strong hydrogen bonds (c) H₂S is a gas at room temperature (d) Sulfur is more electronegative than oxygen
Answer: (b) H₂O molecules form strong hydrogen bonds
Oxygen is strongly electronegative (EN=3.5) and small — H bonded to O becomes highly δ+. These δ+ H atoms attract lone pairs on O of adjacent molecules → strong O–H···O hydrogen bonds. Breaking these requires significant energy → high BP of 100°C. H₂S: S is less electronegative and larger → much weaker H-bonds → BP of −60°C.
Q2. Hydrogen bonding is possible in which of the following molecules?
(a) CH₄ (b) PH₃ (c) HF (d) H₂S
Answer: (c) HF
H-bond requires H bonded to F, O, or N. In HF, H is directly bonded to F (most electronegative element, EN=4.0) → strong H-bond (F–H···F). CH₄: H bonded to C (EN=2.5, not F/O/N) → no H-bond. PH₃: H bonded to P — P not electronegative enough. H₂S: H bonded to S — S not electronegative enough. Only F, O, N qualify.
4. Gas Laws & Kinetic Theory
4.1
Boyle's Law, Charles's Law & Ideal Gas Equation
Three laws + one equation that describes how gases behave — numericals guaranteed in NDA
📌 BOYLE'S LAW (1662) — Constant T
"At constant temperature, pressure of a fixed mass of gas is inversely proportional to its volume"
P ∝ 1/V (T const) → PV = constant → P₁V₁ = P₂V₂
📌 CHARLES'S LAW (1787) — Constant P
"At constant pressure, volume of a fixed mass of gas is directly proportional to its absolute temperature (K)"
V ∝ T (P const) → V/T = constant → V₁/T₁ = V₂/T₂
📌 GAY-LUSSAC'S LAW — Constant V
"At constant volume, pressure of a gas is directly proportional to its absolute temperature"
P ∝ T (V const) → P₁/T₁ = P₂/T₂
📌 AVOGADRO'S LAW — Constant T & P
"Equal volumes of all gases at the same T and P contain equal numbers of molecules"
V ∝ n (T, P const) → V₁/n₁ = V₂/n₂ → 1 mol gas = 22.4 L at STP
⚛ Ideal Gas Equation & Related Formulae
Ideal Gas Equation: PV = nRT
P = pressure (Pa or atm); V = volume (m³ or L); n = moles
R = 8.314 J/mol·K (SI) = 0.0821 L·atm/mol·K
T = temperature in Kelvin (K = °C + 273)
Combined Gas Law: P₁V₁/T₁ = P₂V₂/T₂ (for fixed amount of gas)
Density of ideal gas: d = PM/(RT) where M = molar mass (g/mol)
Standard conditions:
STP (Standard): T = 0°C (273 K), P = 1 atm → 1 mol gas = 22.4 L
SATP: T = 25°C (298 K), P = 1 bar → 1 mol gas = 24.8 L
Dalton's Law of Partial Pressures:
P_total = P₁ + P₂ + P₃ + ... (for mixture of non-reacting gases)
Temperature conversion — critical: Always convert °C to Kelvin before using gas law formulae. Mistake of using °C directly is the most common NDA numerical error.
🔮 Kinetic Theory of Gases — Assumptions
Gas molecules are point masses (negligible volume)
No intermolecular forces between molecules
Molecules in random, continuous motion
Collisions are perfectly elastic (KE conserved)
Average KE of molecules ∝ Absolute temperature (T in K)
Pressure = force per unit area from molecular collisions on walls
📈 Deviations from Ideal Behaviour
Real gases deviate at high pressure and low temperature
At high P: volume of molecules not negligible
At low T: intermolecular forces become significant
van der Waals equation: (P + a/V²)(V − b) = nRT
a = intermolecular attraction correction; b = volume correction
Ideal gas: a=0, b=0 → no interactions, no volume
📝 NDA Numerical Examples — Gas Laws
Boyle's Law: A gas occupies 4 L at 2 atm. What volume at 8 atm (T constant)?
P₁V₁ = P₂V₂ → 2 × 4 = 8 × V₂ → V₂ = 8/8 = 1 L
Charles's Law: Gas volume is 300 mL at 27°C. Volume at 127°C (P constant)?
T₁ = 27 + 273 = 300 K; T₂ = 127 + 273 = 400 K
V₁/T₁ = V₂/T₂ → 300/300 = V₂/400 → V₂ = 400 mL
Ideal Gas: How many moles of gas at 2 atm, 5 L, 300 K? (R = 0.082 L·atm/mol·K)
n = PV/RT = (2 × 5)/(0.082 × 300) = 10/24.6 ≈ 0.41 mol
📝 TOPIC-WISE PYQ
Gas Laws — NDA Pattern Questions
Q1. A gas at 27°C and 1 atm pressure has volume 4 L. At what temperature (in °C) will the volume become 6 L at the same pressure?
(a) 40.5°C (b) 177°C (c) 450°C (d) 18°C
Answer: (b) 177°C
Charles's Law (constant P): V₁/T₁ = V₂/T₂
T₁ = 27 + 273 = 300 K; V₁ = 4 L; V₂ = 6 L
T₂ = T₁ × V₂/V₁ = 300 × 6/4 = 300 × 1.5 = 450 K
T₂ in °C = 450 − 273 = 177°C. Common trap: forgetting to convert back to °C at the end.
Q2. A gas occupies 10 L at STP. What is the number of moles of gas present?
(a) 22.4 mol (b) 0.446 mol (c) 1 mol (d) 2 mol
Answer: (b) 0.446 mol
At STP: 1 mole of any ideal gas = 22.4 L
Moles = Volume / 22.4 = 10 / 22.4 = 0.446 mol
Alternatively using PV = nRT: n = PV/RT = (1 × 10)/(0.0821 × 273) = 10/22.41 ≈ 0.446 mol ✓
Q3. Which of the following is correct regarding an ideal gas? I. Molecules have zero volume. II. There are no intermolecular forces. III. All collisions are elastic.
(a) I only (b) I and II (c) II and III (d) I, II and III
Answer: (d) I, II and III — all three
All three statements are postulates of the kinetic molecular theory (ideal gas assumptions): (I) Gas molecules are point masses with zero volume; (II) No attractive or repulsive forces between molecules; (III) All collisions between molecules (and with walls) are perfectly elastic — kinetic energy is conserved, not converted to heat. Real gases violate all three at high pressure or low temperature.
🧠 TRICKY QUESTIONS
Gas Laws — Classic NDA Calculation Traps
Q. A student uses T = 27°C directly in Charles's law instead of converting to Kelvin. By how much will the answer be wrong?
Answer: Significantly wrong — °C must always be converted to Kelvin.
Charles's law is V ∝ T in Kelvin only. At 0°C, a gas still has substantial volume (molecules still moving). The Kelvin scale starts at absolute zero (−273°C) where motion theoretically stops. Using °C introduces massive error — e.g., V₁/T₁ with T₁=27 instead of T₁=300 is a factor-of-11 error in temperature. Always: K = °C + 273. This single conversion error causes more lost marks in NDA numericals than any other mistake.
Q. If the pressure of a gas is doubled and the temperature is also doubled (in K), what happens to the volume?
Answer: Volume remains unchanged.
Combined gas law: P₁V₁/T₁ = P₂V₂/T₂
P₂ = 2P₁; T₂ = 2T₁ → P₁V₁/T₁ = 2P₁·V₂/2T₁ → P₁V₁/T₁ = P₁V₂/T₁ → V₂ = V₁
Intuition: Doubling T makes gas want to expand (Charles); doubling P compresses it equally → net effect zero. The two changes cancel exactly. This is a frequently tested combined law trap.
5. Solutions
5.1
Types of Solutions & Concentration Terms
A solution is a homogeneous mixture of solute dissolved in solvent
💧 Types by Saturation
Unsaturated: can dissolve more solute at that T; solute dissolves completely
Saturated: maximum solute dissolved at that T; equilibrium with undissolved solute
Supersaturated: more solute than equilibrium allows (unstable); formed by slow cooling
Adding a seed crystal to supersaturated → rapid crystallisation
Solubility increases with T for most solids (NaCl exception: barely)
📈 Concentration Terms
Molarity (M): moles of solute per litre of solution → mol/L
Molality (m): moles of solute per kg of solvent → mol/kg
Mole fraction (χ): moles of component / total moles
Mass fraction (%): mass of solute / total mass × 100
Molality unaffected by temperature (mass doesn't change); Molarity does (volume changes with T)
⚛ Colligative Properties
Depend only on number of solute particles — not their nature
Boiling point elevation: ΔTb = Kb × m (Kb = ebullioscopic constant)
Freezing point depression: ΔTf = Kf × m (Kf = cryoscopic constant)
Osmotic pressure: π = MRT (M = molarity; R = gas constant)
🔮 Solution — Key Formulae
Molarity (M): M = moles of solute / volume of solution (L)
= (mass of solute / molar mass) / V(L)
Molality (m): m = moles of solute / mass of solvent (kg)
Mole fraction: χ_A = n_A / (n_A + n_B) (χ_A + χ_B = 1)
Dilution law: M₁V₁ = M₂V₂ (moles of solute conserved on dilution)
Raoult's Law: P_solution = χ_solvent × P°_solvent
ΔP/P° = χ_solute (relative lowering of vapour pressure)
Boiling point elevation: ΔTb = Kb × m × i
Freezing point depression: ΔTf = Kf × m × i
i = van't Hoff factor (= 1 for non-electrolytes; = number of ions for strong electrolytes)
NaCl: i = 2 (Na⁺ + Cl⁻); CaCl₂: i = 3 (Ca²⁺ + 2Cl⁻)
Osmotic pressure: π = MRT (R = 0.0821 L·atm/mol·K)
Water (H₂O): Kf = 1.86 K·kg/mol (freezing point drops 1.86°C per mole of solute per kg of water). Kb = 0.52 K·kg/mol (BP rises 0.52°C per mole solute). Antifreeze (ethylene glycol) and road salt (NaCl) both work by depression of freezing point.
📝 NDA Numerical Examples — Solutions
Molarity: 4 g of NaOH (molar mass 40 g/mol) dissolved in 500 mL solution.
Moles NaOH = 4/40 = 0.1 mol; V = 0.5 L → M = 0.1/0.5 = 0.2 M
Freezing point depression: 18 g glucose (M=180) in 500 g water. ΔTf = ?
m = (18/180) / 0.5 kg = 0.1/0.5 = 0.2 mol/kg; ΔTf = Kf × m = 1.86 × 0.2 = 0.372°C
New FP of solution = 0 − 0.372 = −0.372°C
van't Hoff factor: 0.1 m NaCl solution: i = 2 → ΔTf = 1.86 × 0.1 × 2 = 0.372°C (same as 0.2 m glucose — NDA tests this coincidence)
Q1. Adding salt (NaCl) to water lowers its freezing point. This is an example of:
(a) Osmosis (b) A colligative property (c) Vapour pressure lowering only (d) Chemical reaction between salt and water
Answer: (b) A colligative property
Freezing point depression (ΔTf = Kf × m × i) depends only on the number of solute particles, not their identity. NaCl dissociates into Na⁺ and Cl⁻ (i=2), doubling the particle count. This is why road salt is so effective — NaCl creates more particles per mole than sugar, giving greater ΔTf. This is a colligative property, alongside BP elevation, vapour pressure lowering, and osmotic pressure.
Q2. The molarity of a solution prepared by dissolving 10 g of NaOH (molar mass = 40 g/mol) in water to make 250 mL of solution is:
(a) 0.25 M (b) 0.5 M (c) 1.0 M (d) 2.0 M
Answer: (c) 1.0 M
Moles of NaOH = 10 / 40 = 0.25 mol
Volume = 250 mL = 0.25 L
Molarity = 0.25 / 0.25 = 1.0 mol/L = 1.0 M
Q3. Which concentration unit is independent of temperature?
(a) Molarity (b) Normality (c) Molality (d) Parts per million
Answer: (c) Molality
Molality = moles of solute / kg of solvent. Mass does not change with temperature. Molarity = moles/L of solution — volume changes with temperature (liquids expand/contract), so molarity changes with temperature. Molality is preferred for colligative property calculations precisely because it is temperature-independent.
🧠 TRICKY QUESTIONS
Solutions — Colligative Property Traps
Q. Which solution has a greater freezing point depression: 0.1 m glucose (C₆H₁₂O₆) or 0.1 m NaCl?
Answer: 0.1 m NaCl has greater ΔTf
Glucose is a non-electrolyte → i = 1 → ΔTf = 1.86 × 0.1 × 1 = 0.186°C
NaCl is a strong electrolyte → i = 2 (Na⁺ + Cl⁻) → ΔTf = 1.86 × 0.1 × 2 = 0.372°C
Same molality but NaCl produces twice as many particles → double the ΔTf. This is the core of the van't Hoff factor — electrolytes are more effective at depressing freezing points per mole. This is why road salt is better than sand for de-icing.
Q. A 1 M solution of NaCl and a 1 M solution of glucose — which has higher osmotic pressure? Why?
Answer: NaCl solution has higher osmotic pressure (π = iMRT)
Osmotic pressure π = iMRT. For glucose (non-electrolyte): i=1; π = 1 × 1 × R × T.
For NaCl: i=2; π = 2 × 1 × R × T — twice as high.
Osmotic pressure, like all colligative properties, depends on particle count — and NaCl provides 2 ions per formula unit. This principle is used in medical saline solutions: normal saline (0.9% NaCl) is isotonic with blood cells; a more concentrated NaCl solution would have higher osmotic pressure than blood → cells shrink (crenation).
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