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Chemistry — Class 12 — L4: Chemical Bonding

NIOS Code 313 · Module 2 · Atomic Structure and Chemical Bonding

Notes extracted from NIOS Chemistry Course (313), Lesson 4 — Chemical Bonding (313_Chemistry_Eng_Lesson4.pdf). Content covers sections 4.1–4.7.
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Overview — Module 2: Chemical Bonding

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).

Section 1: Valence Electrons and Chemical Bonds (4.1–4.2)

4.1 Valence Electrons

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.

4.2 What Is a Chemical Bond?

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:

  • Ionic (electrovalent) bond — electron transfer
  • Covalent bond — electron sharing
  • Coordinate (dative) covalent bond — both electrons from one species
  • Hydrogen bond — special intermolecular attraction

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.

Section 2: Ionic or Electrovalent Bond (4.3)

4.3 Kossel's Theory and NaCl Formation

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.

Ionic Bond Formation — NaCl Na 2,8,1 e⁻ transfer Na⁺ 2,8 Cl 2,8,7 Cl⁻ 2,8,8 Na⁺ + Cl⁻ → NaCl (lattice: electrostatic attraction)
Fig 4.3 context — Electron transfer gives noble gas configurations; ions form an ionic lattice.

4.3.1 Born–Haber Cycle

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:

  • (a) Sublimation: Na(s) → Na(g); ΔH = +108.7 kJ mol⁻¹
  • (b) Ionisation: Na(g) → Na⁺(g) + e⁻; ΔH = +493.8 kJ mol⁻¹
  • (c) Dissociation: ½Cl₂(g) → Cl(g); ΔH = +120.9 kJ mol⁻¹
  • (d) Electron affinity: Cl(g) + e⁻ → Cl⁻(g); ΔH = −379.5 kJ mol⁻¹
  • (e) Lattice formation: Na⁺(g) + Cl⁻(g) → NaCl(s); ΔH = −754.8 kJ mol⁻¹ (lattice energy)
Na(s) + ½Cl₂(g) → NaCl(s);   ΔHf = −410.9 kJ mol⁻¹
Net exothermic — lattice energy dominates · Favoured when IE is low, EA is high, lattice energy is high

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.

4.3.2 Properties of Ionic Compounds

  • Crystalline solids with regular 3D ionic lattices — hard and brittle
  • High melting and boiling points (strong electrostatic forces)
  • Generally soluble in water; less soluble in non-polar solvents
  • Conduct electricity when molten or in aqueous solution

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.

Section 3: Covalent Bond and Lewis Structures (4.4)

4.4 Covalent Bond

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.

4.4.1 Lewis Structures and Octet Rule

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.

Coordinate Bond — BF₃ + NH₃ → F₃B←NH₃ BF₃ electron deficient lone pair NH₃ electron rich F₃B←NH₃ dative bond Arrow from donor (N) to acceptor (B) — identical to covalent bond once formed
Coordinate covalent bond: NH₃ donates lone pair to electron-deficient BF₃ (also in HNO₃, NH₄⁺).

4.4.2 Coordinate Covalent Bond

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⁺.

4.4.3–4.4.5 Polarity and Dipole Moment

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).

μ = Q × r  |  1 D = 3.336 × 10⁻³⁰ C·m
μ = dipole moment (vector, tail at + centre) · Q = charge · r = separation · Larger ΔEN → larger μ (H–F 1.90 D > H–I 0.38 D)
ΔEN = 1.7 → 50% ionic character
ΔEN < 1.7 → less than 50% ionic · ΔEN > 1.7 → more than 50% ionic · Extreme ΔEN → complete electron transfer (ionic bond)

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.

4.4.6 Fajan's Rules — Covalent Character in Ionic Bonds

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.

4.4.7 Covalent Bond Parameters

  • Bond order: number of bonds between two atoms (1 = single σ, 2 = σ+π, 3 = σ+2π)
  • Bond length: internuclear distance (pm); higher bond order → shorter bond (C–C 154, C=C 134, C≡C 120 pm)
  • Bond angle: angle between bonding orbitals (H₂O 104.5°, NH₃ 107.3°, CH₄ 109.5°)
  • Bond enthalpy (ΔaH): energy to break one mole of bonds; average used when multiple equivalent bonds differ (O–H in H₂O: average 464.5 kJ mol⁻¹)

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.

Section 4: Hydrogen Bonding (4.5)

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.

Section 5: VSEPR Theory (4.6)

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.

VSEPR — Four Electron Pairs (Table 4.2) CH₄ 4 BP, 0 LP 109.5° tetrahedral NH₃ 3 BP, 1 LP LP 107.3° pyramidal H₂O 2 BP, 2 LP 104.5° bent
Table 4.2 — Lone pairs compress bond angles: LP–LP > LP–BP > BP–BP repulsion.

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.

Section 6: Modern Theories — VBT and MOT (4.7)

4.7.1 Valence Bond Theory

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.

4.7.1.1 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.

σ and π Bonds — Ethene vs Ethyne σ C₂H₄: 1σ + 1π C₂H₂: 1σ + 2π σ = head-on overlap along axis · π = sideways p-orbital overlap
Fig 4.6–4.8 — Ethane: all σ (sp³–sp³). Ethene: sp² σ + π. Ethyne: sp σ + two perpendicular π bonds.

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).

4.7.1.3 Resonance

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.

4.7.2 Molecular Orbital Theory

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.

Bond order = ½ (nb − na)
nb = electrons in bonding MOs · na = electrons in antibonding MOs · H₂: ½(2−0)=1 · He₂: ½(2−2)=0 (no bond)

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.

Exam Connections and Intext Checkpoints

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.

Chapter Summary

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.

MCQ Quiz — L4 Chemical Bonding

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Flashcards — L4

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Golden Rules — L4 Chemical Bonding

Most exam-important points from this chapter:

Why atoms bond

Combined energy < separated atoms. Atoms attain noble gas config by lose/gain/share electrons — ionic, covalent, or coordinate.

Ionic vs covalent

Kossel: electron transfer + lattice energy (Born–Haber). Lewis: electron sharing. High ΔEN → ionic; moderate → polar covalent.

Polarity & shape

μ = Q×r. Molecular dipole depends on bond polarity AND geometry — CO₂/BF₃/CCl₄ have μ=0; H₂O/NH₃ do not.

VSEPR & hybridisation

Count electron pairs (bond + lone). Repulsion order: LP–LP > LP–BP > BP–BP. VBT: sp/sp²/sp³/sp³d/sp³d² explains observed bond angles.

MOT essentials

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.

μ = Q × r
ΔHnet = Σ steps (Born–Haber)
Bond order = ½(nb − na)
ΔEN = 1.7 → 50% ionic
σ + π bonds (C₂H₄, C₂H₂)
sp, sp², sp³, sp³d, sp³d²
LP–LP > LP–BP > BP–BP
H-bond: 4–25 kJ mol⁻¹

Section 1: Formulas with Derivations

All key equations from NIOS Chemistry 313, Module 2 — Chemical Bonding (sections 4.1–4.7). Unlock for full derivations, examples, and exam prep.

Dipole Moment — μ = Q × r

Units: Debye (D); 1 D = 3.336 × 10⁻³⁰ C·m

Where

• μ = dipole moment (vector, points negative to positive)
• Q = magnitude of separated charge
• r = distance between charges

When to use

Polar covalent bonds; net molecular dipole (H₂O = 1.85 D; NH₃ = 1.47 D; CO₂ = 0).

Memory aid: "Charge times distance" — more separation or charge → larger μ.

Derivation

Dipole moment measures charge separation. Partial charges δ+ and δ− separated by bond length r give μ = δ × r. Molecular μ is vector sum of bond dipoles — symmetric molecules cancel (CO₂, BF₃).

❌ Assuming polar bonds always give polar molecules.
✓ Geometry matters — CCl₄ has polar C–Cl bonds but μ = 0 (tetrahedral symmetry).

Worked Examples

Basic

Q: Which has higher μ — HCl or Cl₂?

HCl: ΔEN large, no symmetry cancellation

Answer: HCl (μ ≈ 1.03 D); Cl₂ is non-polar

Intermediate

Q: Why is BF₃ non-polar despite polar B–F bonds?

Trigonal planar — three bond dipoles cancel at 120°

Answer: Symmetric geometry cancels dipoles

Exam

Q: Compare μ of NH₃ and NF₃.

NH₃: lone pair adds to dipole. NF₃: lone pair opposes N–F bond dipoles

Answer: NH₃ μ > NF₃ despite higher EN of F

Born–Haber Cycle — NaCl

Na(s) + ½Cl₂(g) → NaCl(s); ΔHf = −410.9 kJ mol⁻¹

Steps (enthalpy, kJ mol⁻¹)

(a) Sublimation Na(s)→Na(g): +108.7
(b) Ionisation Na→Na⁺: +493.8
(c) Dissociation ½Cl₂→Cl: +120.9
(d) Electron affinity Cl+e⁻→Cl⁻: −379.5
(e) Lattice Na⁺+Cl⁻→NaCl(s): −754.8

Favourable ionic formation

Low IE (metal) · High EA (non-metal) · High lattice energy

History: Born and Haber applied Hess's law to explain why endothermic ionisation steps still yield stable ionic solids.

❌ Judging ionic stability from ionisation energy alone.
✓ Large negative lattice energy (−754.8 kJ) drives overall exothermic formation.

Worked Examples

Intermediate

Q: Which step releases most energy in NaCl cycle?

Lattice formation: −754.8 kJ mol⁻¹

Answer: Step (e) lattice energy

Exam

Q: Why does MgO have higher lattice energy than NaCl?

Mg²⁺ and O²⁻ have higher charges; smaller ionic radii → stronger Coulomb attraction

Answer: Higher charge density → stronger lattice

Electronegativity & Bond Character

Pauling scale: ΔEN = |ENA − ENB|

ΔEN = 1.7 → bond is 50% ionic character
ΔEN < 1.7 → <50% ionic (polar covalent)
ΔEN > 1.7 → >50% ionic

When to use

Predict bond polarity; explain why NaCl is ionic but HCl is polar covalent.

Basic

Q: Classify H–H, H–Cl, Na–Cl bonds.

ΔEN: 0, ~0.9, ~2.1

Answer: Non-polar covalent; polar covalent; ionic

MO Bond Order — b.o. = ½(nb − na)

Where

• nb = electrons in bonding MOs
• na = electrons in antibonding MOs

Examples (valence electrons)

H₂: (σ1s)² → b.o. = ½(2−0) = 1
He₂: (σ1s)²(σ*1s)² → b.o. = 0 (does not exist)
N₂: b.o. = 3 (triple bond, diamagnetic)
O₂: b.o. = 2; paramagnetic (π* unpaired e⁻)

Memory aid: "Bonding minus antibonding, halved."

❌ Using Lewis line count for MOT bond order.
✓ MOT bond order counts electrons in bonding vs antibonding MOs only.

❌ Predicting O₂ diamagnetic from Lewis structure.
✓ MOT shows unpaired electrons in π* orbitals → paramagnetic.

Worked Examples

Intermediate

Q: Bond order of F₂?

14 valence e⁻: σ2s² σ*2s² σ2p² π2p⁴ π*2p⁴ → b.o. = ½(8−6)

Answer: 1 (single bond)

Exam

Q: Which is more stable — N₂ or N₂²⁻?

N₂ b.o.=3; N₂²⁻ b.o.=2.5

Answer: N₂ more stable (higher bond order)

Hybridisation & Geometry

sp — 2 orbitals, 180° — BeCl₂, C₂H₂
sp² — 3 orbitals, 120° — BF₃, C₂H₄
sp³ — 4 orbitals, 109.5° — CH₄, H₂O (bent)
sp³d — PCl₅ (trigonal bipyramidal)
sp³d² — SF₆ (octahedral)

σ and π bonds

σ — head-on overlap along axis · π — sideways p overlap · C₂H₄: 1σ C=C + 1π · C₂H₂: 1σ C≡C + 2π

Intermediate

Q: Hybridisation and geometry of NH₃?

4 electron pairs (3 bond + 1 lone) → sp³; lone pair compresses H–N–H to 107.3°

Answer: sp³, pyramidal

Exam

Q: Total σ and π bonds in benzene C₆H₆?

6 C–C σ + 6 C–H σ + 3 delocalised π (resonance)

Answer: 12 σ + 3 π (delocalised)

VSEPR & Hydrogen Bonding

Repulsion order: LP–LP > LP–BP > BP–BP

H-bond: 4–25 kJ mol⁻¹ between H attached to F, O, N and electronegative atom

Bond parameters

Higher bond order → shorter bond: C–C 154 pm, C=C 134, C≡C 120 pm

Section 2: Detailed Definitions

DEFINITION: Ionic (Electrovalent) Bond

Meaning: Electrostatic attraction between oppositely charged ions after electron transfer.
Example: Na⁺ + Cl⁻ → NaCl lattice
Properties: High mp, brittle, conduct when molten/dissolved
Common confusion: No discrete NaCl molecules in solid — extended lattice.

DEFINITION: Covalent Bond

Meaning: Sharing of electron pair between atoms to achieve noble gas configuration.
Example: H₂ — one shared pair
Connects to: Lewis structures, VBT, MOT
Common confusion: Sharing ≠ equal — polar covalent has unequal sharing.

DEFINITION: Coordinate (Dative) Bond

Meaning: Both electrons in shared pair come from one atom.
Example: NH₃ + BF₃ → H₃N→BF₃
Representation: Arrow from donor to acceptor
Common confusion: Once formed, identical to normal covalent bond.

DEFINITION: Resonance

Meaning: Molecule cannot be represented by single Lewis structure; actual structure is hybrid of contributing forms.
Example: Ozone O₃, carbonate CO₃²⁻, benzene
Connects to: Delocalised π electrons
Common confusion: Resonance is not rapid switching — it's simultaneous delocalisation.

DEFINITION: Fajan's Rules

Meaning: Predict covalent character in ionic compounds.
More covalent when: Small/high-charge cation; large/polarisable anion
Example: AgCl more covalent than NaCl
Connects to: Explains low solubility of some "ionic" salts

Section 3: Diagrams & Visuals

Bonding Types Map — L4 Electron transfer Ionic lattice Electron sharing Covalent σ/π VSEPR → geometry MOT: b.o. = ½(n_b − n_a) · H-bond: F/O/N–H···F/O/N

Transfer → ionic · Share → covalent · VSEPR/MOT explain shape and stability

BORN-HABER CYCLE (NaCl) ═══════════════════════════════════════ Na(s) ──sublimation(+108.7)──► Na(g) Na(g) ──IE(+493.8)──► Na⁺(g) + e⁻ ½Cl₂(g) ──dissociation(+120.9)──► Cl(g) Cl(g) + e⁻ ──EA(−379.5)──► Cl⁻(g) Na⁺(g) + Cl⁻(g) ──lattice(−754.8)──► NaCl(s) Net ΔH_f = −410.9 kJ mol⁻¹ (exothermic) ═══════════════════════════════════════

Section 5: Comprehensive Q&A (12 Questions)

Q1: Why do atoms form chemical bonds?

Combined system has lower potential energy than separated atoms. Atoms attain stable noble gas configuration by losing, gaining, or sharing electrons — energy decrease = bond formation.

Q2: Why can't Kossel's theory explain O₂ formation?

No sensible electron transfer — one O cannot lose 2 e⁻ while another gains 2. Covalent sharing (Lewis theory) solves this.

Q3: What is lattice energy and why is it important?

Energy released when gaseous ions form ionic solid. Large negative value stabilises ionic compounds despite endothermic ionisation steps in Born–Haber cycle.

Q4: Difference between σ and π bonds?

σ: head-on overlap along internuclear axis, free rotation. π: sideways p-orbital overlap, electron cloud above/below axis, restricts rotation. Double bond = 1σ + 1π.

Q5: How does VSEPR predict H₂O shape?

4 electron pairs (2 bond + 2 lone) → tetrahedral arrangement → bent shape. LP–LP repulsion > LP–BP → H–O–H = 104.5° (less than 109.5°).

Q6: Why is O₂ paramagnetic?

MOT: two unpaired electrons in π*2p antibonding orbitals. Lewis structure incorrectly predicts all paired electrons.

Q7: Explain hydrogen bonding in water.

H attached to O (high EN) is partially positive; attracts lone pair on neighbouring O. Intermolecular H-bonds → high bp, surface tension, ice less dense than liquid water.

Q8: What is resonance in CO₃²⁻?

Three equivalent Lewis structures with C=O double bond in different positions. Actual structure: three C–O bonds equivalent, bond order 1⅓, delocalised π electrons.

Q9: sp² hybridisation in ethene — how many σ and π?

4 σ (C–C, 4 C–H) + 1 π (C=C sideways p overlap). C=C has 1σ + 1π total between carbons.

Q10: Fajan's rules — why is AgCl less ionic than NaCl?

Ag⁺ smaller and more polarising; Cl⁻ polarisable. Electron cloud distortion gives partial covalent character — AgCl less soluble in water.

Q11: Coordinate bond example and how to draw it?

NH₃ donates lone pair to BF₃: H₃N→BF₃. Arrow from N (donor) to B (acceptor). After formation, bond is normal covalent.

Q12: Exam — Compare bond order and stability of O₂, O₂⁺, O₂⁻.

O₂: b.o.=2. O₂⁺: b.o.=2.5 (more stable). O₂⁻: b.o.=1.5. Higher bond order → shorter bond, stronger, more stable.

Section 6: Tips, Tricks & Exam Hacks

Memory Aids

  • Octet rule: "Gain, lose, or share to eight"
  • VSEPR: "Lone pairs push harder" (LP–LP worst)
  • MOT bond order: "Bonding minus antibonding, halved"
  • H-bond: "FON — Fluorine, Oxygen, Nitrogen with H"
  • 50% ionic: "ΔEN = 1.7 is the dividing line"

Exam Tips

  • Draw Lewis structure before VSEPR geometry
  • Count total electron pairs (bond + lone) for hybridisation
  • MOT: fill σ before π; check paramagnetism via unpaired e⁻
  • Born–Haber: lattice energy is usually largest negative term
  • Resonance: all structures must have same atom connectivity

Section 7: Connections & Relationships

This chapter builds on: L2 atomic structure (valence electrons, quantum numbers), L3 periodicity (EN trends).
This chapter leads to: L7 solutions (intermolecular forces), organic chemistry (hybridisation, resonance), L9 thermodynamics (Born–Haber enthalpies).
Related formulas: Lattice energy links to Hess's law (L9); dipole moment links to solubility (L7).

BONDING THEORY LADDER Lewis (dots, octet) │ VSEPR (shapes from e⁻ pairs) │ VBT (hybridisation, σ/π) │ MOT (MO diagrams, bond order, magnetism) │ Intermolecular: H-bond, dipole-dipole, London

Section 8: Complete Quick Reference

Formulas at a glance:

• μ = Q × r — dipole moment (D)

• Born–Haber: ΔHf = Σ all steps including lattice energy

• b.o. = ½(nb − na) — MOT bond order

• ΔEN = 1.7 → 50% ionic character

Hybridisation: sp (linear) · sp² (trigonal) · sp³ (tetrahedral) · sp³d · sp³d²

VSEPR: LP–LP > LP–BP > BP–BP

Bond lengths (pm): C–C 154 · C=C 134 · C≡C 120 · N≡N 109

H-bond: 4–25 kJ mol⁻¹; F, O, N with H

Decision tree: Metal + non-metal → ionic (Born–Haber). Non-metals → covalent (Lewis). Shape → VSEPR. Stability/magnetism → MOT.

Remember: ✓ He₂ b.o.=0 ✓ O₂ paramagnetic ✓ H₂O bent sp³ ✓ Resonance = delocalisation ✓ Lattice energy drives ionic stability

PYQ — Previous Year Questions

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.

L4 — Chemical Bonding

9 question(s) · Sources: 313/MAY/205A, 313/MAY/205B, 313/MAY/205C, 313/TUS/105A

Section A — MCQ / Objective (from papers)

PYQ1. In boron trichloride, boron has — (A) sp2 hybridization   (B) dsp2 hybridization   (C) sp hybridization   (D) sp3 hybridization

1 mark · Q5 · 313/MAY/205A

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 _____

2 marks · Q20 · 313/MAY/205A

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

1 mark · Q2 · 313/MAY/205B

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 _____

2 marks · Q28 · 313/MAY/205B

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

1 mark · Q11 · 313/MAY/205C

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 _____

2 marks · Q27 · 313/MAY/205C

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

1 mark · Q1 · 313/TUS/105A

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?

2 marks · Q18 · 313/TUS/105A

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.

Section B — Short / Long answer (from papers)

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

3 marks · Q38 · 313/TUS/105A

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.

Problem Solving — L4 Chemical Bonding

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.

Question 1 of 6Ionic

Draw a sketch of Na and Cl forming NaCl by electron transfer. Why is NaCl an ionic solid?

Electron transfer; electrostatic attraction

Pencil sketch (labelled)

Ionic bond idea (NaCl) Na Na → Na⁺ + e⁻ Cl Cl + e⁻ → Cl⁻ e⁻ transfer Na⁺ Cl⁻ electrostatic attraction
Pencil sketch: electron transfer Na → Cl

Solution — step by step with formulas

  1. Na loses one e⁻ → Na⁺; Cl gains e⁻ → Cl⁻.
  2. Ions held by strong electrostatic forces in lattice.

Final answer: Ionic lattice of Na⁺ and Cl⁻

Formulas used in this problem

Electron transfer; electrostatic attraction

Textbook formal language

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.

Easy language (same idea, plain words)

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.

Topic in depth — Ionic bond

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.

Exam tip

Write Electron transfer; electrostatic attraction before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 2 of 6Covalent

Draw the shape of H₂O and state approximate bond angle. Why is it bent not linear?

Shared pair; H₂O bent ≈104.5°

Pencil sketch (labelled)

Water molecule (bent) O H H ≈104.5°
Pencil sketch: H₂O bent shape labelled

Solution — step by step with formulas

  1. Bent shape; angle ≈104.5°.
  2. Two bonding pairs + two lone pairs on O; lone pairs repel more ⇒ bent.

Final answer: Bent; ≈104.5°

Formulas used in this problem

Shared pair; H₂O bent ≈104.5°

Textbook formal language

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.

Easy language (same idea, plain words)

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.

Topic in depth — Covalent bond / VSEPR H₂O

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.

Exam tip

Write Shared pair; H₂O bent ≈104.5° before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 3 of 6Electronegativity

Why is H–Cl polar while Cl–Cl is non-polar?

Δχ → partial charges

Solution — step by step with formulas

  1. Cl more electronegative than H ⇒ shared pair closer to Cl (δ⁺H–Clδ⁻).
  2. Cl–Cl identical atoms ⇒ equal sharing.

Final answer: HCl polar; Cl₂ non-polar

Formulas used in this problem

Δχ → partial charges

Textbook formal language

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.

Easy language (same idea, plain words)

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.

Topic in depth — Polar covalent bond

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.

Exam tip

Write Δχ → partial charges before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 4 of 6Hybridisation

State hybridisation of C in CH₄ and the geometry.

sp³ → tetrahedral 109.5°

Solution — step by step with formulas

  1. sp³; tetrahedral; bond angle 109.5°.

Final answer: sp³ tetrahedral

Formulas used in this problem

sp³ → tetrahedral 109.5°

Textbook formal language

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.

Easy language (same idea, plain words)

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.

Topic in depth — sp³ hybridisation

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.

Exam tip

Write sp³ → tetrahedral 109.5° before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 5 of 6Hydrogen bond

Why does water have abnormally high boiling point for its molar mass?

H bonded to N,O,F

Solution — step by step with formulas

  1. Intermolecular H-bonding between H₂O molecules requires extra energy to separate.

Final answer: H-bonding raises b.p.

Formulas used in this problem

H bonded to N,O,F

Textbook formal language

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.

Easy language (same idea, plain words)

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.

Topic in depth — Hydrogen bonding

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.

Exam tip

Write H bonded to N,O,F before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 6 of 6Metallic

Explain metallic bonding and why metals conduct electricity.

Electron sea model

Solution — step by step with formulas

  1. Positive ions in a sea of delocalised electrons; electrons free to move under potential difference.

Final answer: Delocalised electrons enable conduction

Formulas used in this problem

Electron sea model

Textbook formal language

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.

Easy language (same idea, plain words)

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.

Topic in depth — Metallic bond

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.

Exam tip

Write Electron sea model before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.