313_Chemistry_Eng_Lesson4.pdf). Content covers sections 4.1–4.7.In Lesson 2 you studied atomic structure; in Lesson 3, classification of elements and periodicity. Molecules form when two or more atoms of the same or different elements combine. This lesson answers three fundamental questions: Why do atoms combine? What are the different ways atoms can combine? and What are the shapes of different molecules? The answers underpin all later chemistry — from simple salts to complex biomolecules.
Topics include valence electrons, the octet rule, ionic and covalent bonding, Lewis structures, coordinate bonds, polarity and dipole moment, Born–Haber cycle, Fajan's rules, bond parameters, hydrogen bonding, VSEPR theory, valence bond theory (hybridisation, σ and π bonds, resonance), and molecular orbital theory (MO configurations of H₂, N₂, O₂, F₂).
Electrons in the outermost shell take part in bond formation and determine an atom's combining capacity or valency. The outermost shell is the valence shell; its electrons are valence electrons. Sodium (2,8,1) has one valence electron; chlorine (2,8,7) has seven. Only valence electrons participate in chemical change — inner shells remain essentially unchanged in ordinary reactions.
When two atoms approach each other, the energy of the combined system becomes less than the sum of energies of the separated atoms at large distance. The atoms have combined — a chemical bond has formed. A bond may be visualised as an effect that decreases energy; the resulting molecule has distinct properties unlike its constituent atoms.
How do atoms achieve this energy decrease? Through electronic configuration. Noble gases (He: 2 electrons; others: octet) are chemically inert because their valence shells are stable. Other atoms combine so as to attain the nearest noble gas configuration — by losing, gaining, or sharing electrons. This gives rise to:
The potential energy diagram for H₂ formation (Fig. 4.2) shows energy falling to a minimum at the bond length — maximum orbital overlap, maximum stability. At infinite separation each H atom has its 1s electron under its own nucleus; as atoms approach, attractive nuclear-electron interactions lower energy until repulsion between nuclei and filled orbitals begins to dominate at very short distances. The minimum defines the equilibrium bond length for the molecule.
Chemical bonding is thus fundamentally an electronic phenomenon. Whether a bond is predominantly ionic or covalent depends on how completely electrons are transferred versus shared — a theme developed throughout this lesson.
According to Kossel's theory, atoms attain noble gas configuration by gain or loss of electrons. Sodium (2,8,1) loses one electron (ΔH = +493.8 kJ mol⁻¹) to become Na⁺ (2,8). Chlorine (2,8,7) gains one electron (ΔH = −379.5 kJ mol⁻¹) to become Cl⁻ (2,8,8). The oppositely charged ions are held by electrostatic attraction — the ionic bond. The compound NaCl is an ionic (electrovalent) compound.
Na⁺ formation costs more energy than Cl⁻ formation releases — yet NaCl forms with large overall energy release. The Born–Haber cycle accounts for all steps:
Ionic formation is favoured by: (i) low ionisation energy of metal, (ii) high electron affinity of non-metal, (iii) high lattice energy. Of the five energy terms in the cycle, sublimation and dissociation are relatively small; ionisation energy, electron affinity and lattice energy dominate whether the compound will form. The large negative lattice energy (−754.8 kJ mol⁻¹ for NaCl) reflects the strong electrostatic attraction between oppositely charged ions packed in a crystal — without it, ionic solids would not be stable.
The Born–Haber approach rests on Hess's law (conservation of energy): the enthalpy change for forming NaCl from elements equals the sum of enthalpies of all intermediate steps, regardless of path. This thermochemical analysis explains why students should never judge ionic compound stability from ionisation energy alone.
Kossel's theory works well for Group 1/2 metals with halogens but cannot explain O₂, SO₂, etc. — there is no sensible way for one O atom to lose two electrons while another gains them. The problem was solved by Lewis covalent theory (1916), which remains the most convenient way to represent bonding in simple molecules.
Lewis proposed noble gas configuration is achieved by sharing a pair of electrons. Each atom contributes one electron. In H₂, one shared pair bonds both atoms — a covalent bond; the product is a covalent compound.
Lewis symbols show valence electrons as dots around the element symbol (single dots placed first on each side, then paired). In molecules, dots show bonding and lone pairs. Cl₂: each Cl contributes one electron → shared pair → argon-like octet. O₂: two shared pairs (double bond). N₂: three shared pairs (triple bond).
Second-period elements typically achieve octet (8 valence electrons). Bonding electrons = bond pairs; non-bonding = lone pairs. Single line = single bond; double line = double bond. Lewis structures for ionic compounds show ions in brackets with charges — NaCl as [Na⁺][Cl⁻]. For HF, both H and F achieve stable configurations: H has 2 electrons (like He) and F has octet.
When drawing Lewis structures, place single dots on each side of the symbol before pairing. Different symbols (× and •) may distinguish electrons from different atoms in a bond — purely a drawing convention; all electrons are identical. The octet rule has exceptions (BF₃, PCl₅, SF₆) explained later by expanded octets and hybridisation.
Sometimes both electrons of the shared pair come from one species. BF₃ (electron deficient) accepts a lone pair from NH₃ (electron rich) — a coordinate (dative) bond, shown as B←N. Once formed, it is indistinguishable from a normal covalent bond. Other examples: HNO₃, NH₄⁺.
Covalent compounds have low melting/boiling points, poor electrical conductivity, and dissolve in non-polar solvents. When atoms have different electronegativities, the shared pair shifts toward the more electronegative atom — a polar covalent bond (e.g. HCl: Hδ+—Clδ−). Equal electronegativity gives a pure (non-polar) covalent bond (H₂, Cl₂, O₂, N₂).
Net molecular dipole depends on shape: CO₂ (linear) — bond dipoles cancel, μ = 0. H₂O (bent) — μ = 1.85 D (6.17×10⁻³⁰ C·m). BF₃ (planar) — μ = 0 despite polar B–F bonds. NH₃ (pyramidal + lone pair orbital dipole) — μ = 1.47 D. CCl₄ (tetrahedral) — μ = 0. Dipole moments are vectors and add vectorially; symmetry can cancel individual bond dipoles completely.
H–F (ΔEN large) has μ = 1.90 D; H–Cl 1.04 D; H–Br 0.79 D; H–I 0.38 D — confirming that greater electronegativity difference produces larger charge separation. Terminal exercise note: BF₃ is often mistakenly called polar — it has polar bonds but zero net dipole due to trigonal planar symmetry.
Ionic bonds also have partial covalent character through polarisation of the anion's electron cloud by the cation. Fajan's rules: covalent character increases with (i) small cation, (ii) large anion, (iii) high positive charge on cation, (iv) transition metal cations (n−1)dⁿns⁰ vs ns²np⁶ configuration.
Reference values from the textbook: H–H 436, N≡N 946, O=O 498, C–C 347, C=C 611, C≡C 837 kJ mol⁻¹. Breaking the first O–H in H₂O costs 502 kJ mol⁻¹; the second 427 kJ mol⁻¹ — hence averaging. Bond order in Lewis terms (count of shared pairs) correlates with bond length and enthalpy: triple bonds are shortest and strongest.
A hydrogen bond is attraction between H bonded to N, O, or F and a lone pair on another electronegative atom. Strength: only 4–25 kJ mol⁻¹ (weak vs hundreds for covalent bonds) but explains high boiling points of H₂O and HF, liquid water at room temperature, and low density of ice.
Intermolecular H-bonding: between different molecules (water). Intramolecular: within same molecule (o-nitrophenol, salicylaldehyde) — forms a six-membered ring and reduces boiling point compared to the para isomer. Important in proteins (α-helix, β-sheet) and nucleic acids (DNA base pairing).
In ice, each water molecule forms four H-bonds in a tetrahedral arrangement, creating an open structure — ice is less dense than liquid water. Without hydrogen bonding, water would be a gas at room temperature like H₂S.
VSEPR (Sidgwick & Powell 1940; Nyholm & Gillespie 1957) predicts molecular shape from electron pairs around the central atom.
Postulate 1: Bonding and non-bonding electron pairs arrange to minimise repulsion — farthest apart possible. BeCl₂: 2 bond pairs → linear (180°). BF₃: 3 pairs → trigonal planar (120°). CH₄: 4 pairs → tetrahedral.
Postulate 2: Repulsion order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. Table 4.1 geometries: AX₂ linear (BeCl₂, HgCl₂), AX₃ trigonal planar (BF₃), AX₄ tetrahedral (CH₄, CCl₄), AX₅ trigonal bipyramidal (PCl₅), AX₆ octahedral (SF₆, PF₆⁻).
VSEPR counts all valence-shell electron pairs around the central atom — both shared in bonds and unshared lone pairs. SF₆ has six bond pairs around sulfur with no lone pairs → octahedral (90° and 180° angles). The theory does not require quantum mechanics but successfully predicts geometries that match experimental data and hybridisation models.
Intext 4.2 check: methane has 4 bond pairs, 0 lone pairs → tetrahedral. Electronegativity difference 1.7 → 50% ionic and 50% covalent character in the bond.
VBT (Heitler & London 1927; Pauling) visualises bond formation as overlap of atomic orbitals. Greater overlap → stronger bond. At bond length, overlap is maximum and energy minimum (Fig. 4.2). Explains HF, F₂; for polyatomic molecules needs hybridisation.
Hybridisation mixes atomic orbitals of comparable energy to form equivalent hybrid orbitals (same number as atomic orbitals mixed; all identical shape and energy). Pauling introduced it because ground-state Be (2s²) cannot form two equivalent bonds without excitation and mixing — one 2s electron promotes to 2p, then one 2s and one 2p mix to two sp hybrids pointing 180° apart for BeCl₂ (linear). BCl₃: one 2s + two 2p → three sp² hybrids at 120° in a plane. CH₄: one 2s + three 2p → four sp³ hybrids toward tetrahedron corners (109.5°). PCl₅: sp³d — three equatorial P–Cl at 120°, two axial at 90° to plane. SF₆: sp³d² — regular octahedron.
Hybridisation explains both bond formation and molecular geometry — bridging the gap between Lewis dot localised electrons and the probabilistic orbitals from Lesson 2.
Multiple bonds: C₂H₆ — all σ bonds (sp³–sp³). C₂H₄ (ethene) — sp² hybridisation; one σ (sp²–sp²) + one π (unhybridised p). C₂H₂ (ethyne) — sp hybridisation; one σ + two π bonds (triple bond overall).
When multiple valid Lewis structures exist (canonical structures ↔), the real structure is a resonance hybrid — e.g. O₃ (both O–O bonds equal at 128 pm, between single 148 pm and double 121 pm). Delocalisation of π electrons stabilises the molecule — the resonance hybrid is more stable than any single canonical form. Examples: CO₃²⁻ (three equivalent C–O bonds), SO₂, N₂O, SO₄²⁻, BF₃ (in some representations). Resonance does not mean flipping between structures — bond lengths and energies are averaged. Dashed or broken lines sometimes show delocalised π regions.
Canonical structures must have the same nuclear positions, same number of bonding and non-bonding electrons, and similar energies. Resonance is a limitation of fixed Lewis drawings, not physical oscillation of the molecule.
MOT (Hund & Mulliken 1932): atomic orbitals combine by LCAO to form delocalised molecular orbitals spread over the whole molecule. Combination gives bonding MO (lower energy) and antibonding MO* (higher energy). σ MOs are symmetric about bond axis; π MOs from lateral p overlap.
H₂: (σ1s)², b.o. = 1. He₂: (σ1s)²(σ*1s)², b.o. = 0 — does not exist. N₂ (10 valence e⁻): σ2s² σ*2s² π2px² π2py² σ2pz²; b.o. = ½(8−2) = 3 (triple bond), diamagnetic.
O₂ (12 valence e⁻): … π*2px¹ π*2py¹; b.o. = 2; paramagnetic (unpaired electrons) — key MOT success. F₂: all MOs paired, b.o. = 1, diamagnetic. Energy level diagram Fig. 4.14(a) for O₂/F₂; modified diagram 4.14(b) for B, C, N diatomics (s–p mixing).
For ions: add electrons for negative charge, subtract for positive (O₂⁺ has 11 valence e⁻; O₂²⁻ has 14). Li₂ and Be₂: combining 1s and 2s MOs — Be₂ has b.o. = 0 (like He₂), explaining why Be₂ does not exist (Terminal Exercise 10).
VBT vs MOT: VBT localises bonding between two atoms via orbital overlap; MOT delocalises electrons over the entire molecule. MOT successfully explains O₂ paramagnetism (two unpaired electrons in π* orbitals) which Lewis structures with all paired electrons cannot. Both theories use quantum mechanical principles unlike classical Kossel–Lewis pictures.
F₂ (14 valence electrons) fills all bonding and antibonding MOs through π2p and π*2p levels with no unpaired electrons — diamagnetic, single bond, b.o. = 1. The σ2p_z and σ*2p_z MOs arise from head-on overlap of 2p_z orbitals along the bond axis; π MOs arise from lateral overlap of 2p_x and 2p_y pairs.
Classical theories (Kossel 1916, Lewis 1916) predate wave mechanics; modern VBT and MOT incorporate orbital concepts from Lesson 2. Together they form a layered understanding — Lewis for quick representation, VSEPR for shape, hybridisation for geometry rationale, and MOT for properties like magnetism and bond order in diatomics. Mastering all four levels is essential for NIOS Module 2 exam questions on bonding, geometry, and molecular properties.
Terminal exercise highlights frequently tested skills: drawing Lewis structures for MgCl₂ and N₂O resonance forms; predicting SF₆ as octahedral from VSEPR; explaining CH₄ shape via sp³ hybridisation; computing MO configurations and bond orders for O₂, O₂⁺, O₂⁻, O₂²⁻; explaining why Be₂ does not exist (b.o. = 0); and why AB with ΔEN = 1.7 has 50% ionic character. Intext 4.1 defines electrovalent bond as transfer of electrons held by electrostatic force. Intext 4.3 asks for CO₃²⁻ and SO₂ canonical structures and ammonia shape via sp³ hybridisation with one lone pair compressing H–N–H angles below 109.5°. Intext 4.4 contrasts VBT (localised overlap) with MOT (delocalised MOs) and confirms O₂ paramagnetism.
Atoms combine to lower energy and attain noble gas configurations via ionic transfer (Kossel), covalent sharing (Lewis), or coordinate donation. Ionic compounds have high lattice energies (Born–Haber); covalent compounds show polarity (μ = Q×r), bond parameters, and shapes from VSEPR. VBT explains bonding through orbital overlap and hybridisation (sp, sp², sp³, sp³d, sp³d²) with σ/π bonds and resonance. MOT delocalises electrons in MOs; bond order and paramagnetism (O₂) follow from MO configuration. Hydrogen bonding adds weak but biologically vital intermolecular forces.
Most exam-important points from this chapter:
Combined energy < separated atoms. Atoms attain noble gas config by lose/gain/share electrons — ionic, covalent, or coordinate.
Kossel: electron transfer + lattice energy (Born–Haber). Lewis: electron sharing. High ΔEN → ionic; moderate → polar covalent.
μ = Q×r. Molecular dipole depends on bond polarity AND geometry — CO₂/BF₃/CCl₄ have μ=0; H₂O/NH₃ do not.
Count electron pairs (bond + lone). Repulsion order: LP–LP > LP–BP > BP–BP. VBT: sp/sp²/sp³/sp³d/sp³d² explains observed bond angles.
b.o. = ½(n_b−n_a). H₂ b.o.=1; He₂ b.o.=0. N₂ triple bond; O₂ paramagnetic (π* unpaired) — MOT explains what Lewis cannot.
Extracted from NIOS Chemistry (313) board exam papers in your PDF. Chapter L4 — Chemical Bonding only. Use Model Answer for marking points; Explanation for concept clarity.
9 question(s) · Sources: 313/MAY/205A, 313/MAY/205B, 313/MAY/205C, 313/TUS/105A
PYQ1. In boron trichloride, boron has — (A) sp2 hybridization (B) dsp2 hybridization (C) sp hybridization (D) sp3 hybridization
Model Answer
Answer: (A) sp2 hybridization
Explanation
BCl₃ is trigonal planar — B is sp² hybridised.
Source paper: 313/MAY/205A · Q5 · 1 mark(s) · Chapter L4.
PYQ2. Complete the following choosing from the given options : ( equal, unequal, one, two, zero ) The atomic orbitals of comparable energies give rise to an _____ number of molecular orbitals. He2 is not formed because its bond order is _____
Model Answer
equal; zero. Atomic orbitals of comparable energy give an equal number of MOs. He₂ has bond order zero, so it is not formed.
Explanation
MOT: number of MOs = number of combining AOs. Bond order of He₂ = 0 → unstable.
How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q20 · 2 mark(s) · L4.
PYQ3. In boron trichloride, boron has — (A) sp2 hybridization (B) dsp2 hybridization (C) sp hybridization (D) sp3 hybridization
Model Answer
Answer: (A) sp2 hybridization
Explanation
BCl₃ is trigonal planar — B is sp² hybridised.
Source paper: 313/MAY/205B · Q2 · 1 mark(s) · Chapter L4.
PYQ4. Complete the following choosing from the given options : ( equal, unequal, one, two, zero ) The atomic orbitals of comparable energies give rise to an _____ number of molecular orbitals. He2 is not formed because its bond order is _____
Model Answer
equal; zero. Atomic orbitals of comparable energy give an equal number of MOs. He₂ has bond order zero, so it is not formed.
Explanation
MOT: number of MOs = number of combining AOs. Bond order of He₂ = 0 → unstable.
How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205B · Q28 · 2 mark(s) · L4.
PYQ5. In boron trichloride, boron has — (A) sp2 hybridization (B) dsp2 hybridization (C) sp hybridization (D) sp3 hybridization
Model Answer
Answer: (A) sp2 hybridization
Explanation
BCl₃ is trigonal planar — B is sp² hybridised.
Source paper: 313/MAY/205C · Q11 · 1 mark(s) · Chapter L4.
PYQ6. Complete the following choosing from the given options : ( equal, unequal, one, two, zero ) The atomic orbitals of comparable energies give rise to an _____ number of molecular orbitals. He2 is not formed because its bond order is _____
Model Answer
equal; zero. Atomic orbitals of comparable energy give an equal number of MOs. He₂ has bond order zero, so it is not formed.
Explanation
MOT: number of MOs = number of combining AOs. Bond order of He₂ = 0 → unstable.
How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205C · Q27 · 2 mark(s) · L4.
PYQ7. The linear molecule, which has net dipole moment zero, is — (A) HCl (B) {ZX}e … Bg àíZ-nÌ ‘ (C) HCl
Model Answer
Answer: (B) {ZX}e … Bg àíZ-nÌ ‘|
Explanation
CO₂ is linear O=C=O; bond dipoles cancel → net μ = 0. HCl, H₂O, N₂O are polar.
Source paper: 313/TUS/105A · Q1 · 1 mark(s) · Chapter L4.
PYQ8. Read the passage given below and answer the following questions (out of four attempt any two) : According to VSEPR theory, the electron pairs around the central atom in a molecule arrange themselves in space in such a way that they minimize their mutual repulsion. The lone pair repulsion is much greater than the bond pair repulsion. Name the electron pairs around the central atom in a molecule who arrange themselves in space in such a way that they minimize their mutual repulsion. Which parameter of the molecule is linked to mutual repulsion of electron pairs? When the number of electron pairs around the central atom is five, which geometry is predicted for the molecule? Which electron pair is known as the lone pair of electrons in a molecule?
Model Answer
Answer using key concepts from L4 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.
Explanation
Cross-check with L4 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.
How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/TUS/105A · Q18 · 2 mark(s) · L4.
PYQ9. Predict the shape of methane molecule on the basis of VSEPR theory, specifying the underlying postulate of the theory. dr0 Eg0 B©0 nr0 Ama0 {gÕm§V Ho$ AmYma na ‘oWoZ AUw
Model Answer
Answer using key concepts from L4 (definitions, equations, and one example where useful). Stay within the suggested word range for a 3-mark NIOS question.
Explanation
Cross-check with L4 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.
How to write for NIOS: Use 50–80 words with equation + reason. Open with definition/equation, then reason, end with conclusion. Paper 313/TUS/105A · Q38 · 3 mark(s) · L4.
Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). If the question says draw, a labelled pencil sketch is provided. Explanations open by default.
Draw a sketch of Na and Cl forming NaCl by electron transfer. Why is NaCl an ionic solid?
Final answer: Ionic lattice of Na⁺ and Cl⁻
Ionic bond forms by complete transfer of electrons from electropositive to electronegative atom, followed by Coulomb attraction.
Working formulas: Electron transfer; electrostatic attraction. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.
Sodium gives electron to chlorine; opposite charges stick in a crystal.
Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.
High melting point, conducts when molten/aqueous.
Linked to chapter notes (L4). Remember: Electron transfer; electrostatic attraction. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.
Write Electron transfer; electrostatic attraction before substituting. Keep three significant figures until the end when data allow.
Draw the shape of H₂O and state approximate bond angle. Why is it bent not linear?
Final answer: Bent; ≈104.5°
VSEPR: electron domains around O arrange to minimise repulsion; lone pairs compress H–O–H angle.
Working formulas: Shared pair; H₂O bent ≈104.5°. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.
Oxygen has two lone pairs that push the hydrogens down—like a Mickey Mouse shape.
Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.
Compare with CO₂: linear, no lone pairs on central C.
Linked to chapter notes (L4). Remember: Shared pair; H₂O bent ≈104.5°. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.
Write Shared pair; H₂O bent ≈104.5° before substituting. Keep three significant figures until the end when data allow.
Why is H–Cl polar while Cl–Cl is non-polar?
Final answer: HCl polar; Cl₂ non-polar
Bond polarity arises from electronegativity difference between bonded atoms.
Working formulas: Δχ → partial charges. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.
In HCl the electron pair is tugged toward Cl; in Cl₂ the tug-of-war is fair.
Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.
Molecular polarity also depends on shape (vector sum of bond dipoles).
Linked to chapter notes (L4). Remember: Δχ → partial charges. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.
Write Δχ → partial charges before substituting. Keep three significant figures until the end when data allow.
State hybridisation of C in CH₄ and the geometry.
Final answer: sp³ tetrahedral
Mixing one s and three p orbitals gives four equivalent sp³ hybrids for four σ bonds.
Working formulas: sp³ → tetrahedral 109.5°. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.
Carbon’s four arms point to tetrahedron corners so bonds stay as far apart as possible.
Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.
sp² → trigonal planar; sp → linear.
Linked to chapter notes (L4). Remember: sp³ → tetrahedral 109.5°. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.
Write sp³ → tetrahedral 109.5° before substituting. Keep three significant figures until the end when data allow.
Why does water have abnormally high boiling point for its molar mass?
Final answer: H-bonding raises b.p.
Hydrogen bonds are strong dipole–dipole attractions involving H on electronegative atoms.
Working formulas: H bonded to N,O,F. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.
Water molecules hold hands via H-bonds, so harder to boil.
Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.
Explains ice structure and density anomaly qualitatively.
Linked to chapter notes (L4). Remember: H bonded to N,O,F. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.
Write H bonded to N,O,F before substituting. Keep three significant figures until the end when data allow.
Explain metallic bonding and why metals conduct electricity.
Final answer: Delocalised electrons enable conduction
Metallic bond is attraction between metal cations and mobile valence electrons.
Working formulas: Electron sea model. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.
Electrons roam freely among metal ions—so current can flow and metals are malleable.
Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.
Also explains thermal conductivity and lustre simply.
Linked to chapter notes (L4). Remember: Electron sea model. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.
Write Electron sea model before substituting. Keep three significant figures until the end when data allow.