Chapters

Everything NCERT Class 11 Unit 4 (Chemical Bonding and Molecular Structure) says that can be asked, plus what JEE Main adds beyond the book (marked Beyond NCERT). The short note stays for the last week; this is for a proper revision pass.

Formula sheet

Every formula and counting rule in the chapter on one screen, for the last pass. Details and examples are in the sections below.

Quantity	Formula	Watch
Formal charge	V − (lone-pair e⁻) − ½(bonding e⁻)	Lowest-energy structure has the smallest charges
Electron pairs on central atom	½(V + M − C + A)	M = monovalent atoms; C, A = cation, anion charge
Lone pairs on central atom	pairs − bonded atoms	Double bond counts as one pair
Bond length	R = r(A) + r(B)	Covalent radii
Average bond enthalpy	total dissociation enthalpy / bonds broken	H₂O: (502 + 427)/2 = 464.5 kJ mol⁻¹
Dipole moment	μ = Q × r	1 D = 3.33564 × 10⁻³⁰ C m
Percent ionic character	(μ observed / μ for full charges) × 100	Beyond NCERT
Bond order (MO)	½(N(b) − N(a))	> 0 stable; 0 or < 0 does not exist
Spin-only moment	√(n(n + 2)) BM	n unpaired electrons
s-character	sp 50%, sp² 33%, sp³ 25%	More s → shorter bond, larger angle
Born–Haber	ΔfH = sum of all steps	Lattice formation is the big negative term

0 · What JEE Main asks from this chapter

210 questions in JEE Main 2019–2026, more than 1 per paper; 146 multiple-choice and 64 numerical-answer (mostly "how many of these …"). Grouped by type (counts are approximate):

Type	Share	Typical question
VSEPR shapes, lone pairs, hybridisation (Xe compounds, interhalogens)	~40%	Number of species with two lone pairs on the central atom (2021, 2022)
MO theory: bond order, magnetism, bond length of diatomics	~21%	Increasing bond order of O₂⁺, O₂, O₂⁻, O₂²⁻ (2022)
Lewis structures, formal charge, octet exceptions, resonance, bond parameters, σ/π counts	~17%	Non-bonded electrons in NO₂⁻ (2025)
Dipole moment and polarity	~10%	NH₃ vs NF₃ dipole moment (2024, twice)
Fajans' rules, lattice enthalpy, ionic vs covalent	~6%	Least ionic of BaCl₂, AgCl, KCl, CoCl₂ (2024)
Hydrogen bonding and intermolecular forces	~6%	H-bonds per water molecule in ice (2023)

The counting questions reward one habit: write steric number and lone pairs beside every species before answering.

1 · Kössel–Lewis approach

A chemical bond is the attractive force holding atoms or ions together in a chemical species. Bonding lowers the energy of the system. The theories in order: Kössel–Lewis, VSEPR, valence bond (VB) and molecular orbital (MO).

Lewis (1916): the atom is a positive kernel (nucleus + inner electrons) with an outer shell of up to eight electrons at the corners of a cube. Atoms reach a stable octet by transferring electrons (NaCl) or sharing them (Cl₂).

Lewis symbols: dots for valence electrons. The group valence is the number of dots, or 8 minus it.

Kössel's points:

• Halogens (highly electronegative) and alkali metals (highly electropositive) are separated by the noble gases.
• A halogen gains an electron and an alkali metal loses one, each reaching a noble gas configuration (ns²np⁶; He a duplet).
• The ions are held by electrostatic attraction: the electrovalent bond. Electrovalence = charge on the ion (Ca +2, Cl −1).

Octet rule (Kössel and Lewis, 1916): atoms combine by transferring or sharing valence electrons to get an octet.

Covalent bond (Langmuir, 1919, dropped the cube): a shared electron pair, each atom giving at least one electron. One pair = single bond (Cl–Cl); two = double (CO₂, C₂H₄); three = triple (N₂, C₂H₂).

Writing a Lewis structure:

1. Add the valence electrons of all atoms; add one per negative charge, remove one per positive charge (CO₃²⁻: 24; NH₄⁺: 8).
2. The least electronegative atom goes in the centre (N in NF₃, C in CO₃²⁻); H and F are always terminal.
3. Put single bonds; complete the octets of the outer atoms.
4. Leftover electrons go on the central atom as lone pairs; if it still lacks an octet, make multiple bonds.

Example (NCERT Problems 4.1, 4.2) · CO has 10 valence electrons → :C≡O: (triple bond, one lone pair each). NO₂⁻ has 18 → O=N–O⁻ with a lone pair on N. Its non-bonded electrons: N 2, O(double) 4, O(single) 6 = 12 (2025); electrons around N = 8 (2024).

Formal charge: a bookkeeping charge, assuming each atom owns one electron of each shared pair and both of a lone pair.

V valence electrons of the free atom, L lone-pair electrons, S shared electrons.

• O₃: central O +1, double-bonded O 0, single-bonded O −1.
• Formal charges are not real charges. Use them to choose among Lewis structures: the lowest-energy one has the smallest formal charges.

Example (2026) · HNO₃ (H–O–N(=O)–O⁻): H-bonded O 0, N +1, double-bonded O 0, single O −1.

Limitations of the octet rule:

Exception	Examples
Incomplete octet (fewer than 4 valence e⁻)	LiCl, BeH₂, BCl₃; also AlCl₃, BF₃
Odd-electron molecules	NO, NO₂ (also ClO₂)
Expanded octet (period 3 onward, d orbitals)	PF₅, SF₆, H₂SO₄, many coordination compounds

• Sulphur also obeys the octet in many compounds (SCl₂).
• Other drawbacks: noble gases do form compounds (XeF₂, KrF₂, XeOF₂); it doesn't explain shape; it says nothing about energy or relative stability.

Example (2024) · Exceptions to the octet among CO₂, NO₂, H₂SO₄, BF₃, CH₄, SiF₄, ClO₂, PCl₅, BeF₂, C₂H₆, CHCl₃, CBr₄: NO₂, H₂SO₄, BF₃, ClO₂, PCl₅, BeF₂ = 6.

Example (2022) · "Electron-deficient" among PH₃, B₂H₆, CCl₄, NH₃, LiH, BCl₃: B₂H₆ and BCl₃ = 2.

2 · Ionic bond and lattice enthalpy

An ionic bond forms easily between an element of low ionisation enthalpy and one of highly negative electron gain enthalpy. Most cations come from metals, anions from non-metals; NH₄⁺ is a cation made of non-metals.

Ionic solids are ordered 3-D arrangements of cations and anions (NaCl: rock salt structure). Making the gaseous ions costs energy: for NaCl, ΔᵢH(Na) = 495.8 and ΔegH(Cl) = −348.7, a net +147.1 kJ mol⁻¹. The lattice formation releases 788 kJ mol⁻¹, which pays for it. So stability comes from the lattice enthalpy, not from reaching an octet in the gas.

Lattice enthalpy: energy needed to separate one mole of a solid ionic compound completely into gaseous ions (NaCl: +788 kJ mol⁻¹). It can't be computed from attractions and repulsions alone; crystal geometry enters.

What raises lattice enthalpy · Beyond NCERT: higher charges and smaller ions. MgO ≫ NaCl; LiF > LiCl. Higher lattice enthalpy → higher melting point.

Example (2021) · Melting points: LiF > LiCl (smaller anion) and MgO > NaCl (2+ and 2− charges).

Example (2021) · NaBr from Na(g) and Br(g): 495.8 + (−325.0) + (−728.4) = −557.6 kJ mol⁻¹ (answer 5576 × 10⁻¹).

3 · Bond parameters

Bond length: equilibrium distance between the nuclei of two bonded atoms (measured by spectroscopy, X-ray and electron diffraction).

• Covalent radius: half the distance between two like atoms joined by a covalent bond in one molecule. van der Waals radius: half the distance between two like atoms in neighbouring molecules in the solid. Cl: covalent 99 pm (Cl–Cl 198 pm), van der Waals 180 pm (360 pm apart).
• For a bond A–B, R = r(A) + r(B) (2026: the bond length is r(A) + r(B); the molecule's total length 2(r(A) + r(B))).

Bond	Length (pm)
O–H	96
C–H	107
C–C, C=C, C≡C	154, 133, 120
C–O, C=O	143, 121
C–N, C=N, C≡N	143, 128, 116
N–O, N=O	136, 122

Molecules (pm): H₂ 74, F₂ 144, Cl₂ 199, Br₂ 228, I₂ 267, N₂ 109, O₂ 121, HF 92, HCl 127, HBr 141, HI 160.

Example (2026) · Increasing length: C–H (107) < C≡N (116) < C=O (121) < C–O (143).

Bond angle: the angle between the orbitals holding bonding pairs around the central atom; it shows the shape. H–O–H in water is 104.5°.

Bond enthalpy: energy to break one mole of bonds of a type between gaseous atoms (kJ mol⁻¹). Larger means stronger.

• H–H 435.8; O=O 498; N≡N 946 (one of the highest for a diatomic); H–Cl 431.
• In a polyatomic molecule the same bond differs with its surroundings: in H₂O the first O–H takes 502, the second 427. So use the average (mean) bond enthalpy: 464.5 kJ mol⁻¹.
• Halogens · Beyond NCERT: Cl₂ > Br₂ > F₂ > I₂ (F₂ weakened by lone-pair repulsion; 2026). Carbon–halogen: C–F > C–Cl > C–Br > C–I (2020).

Bond order (Lewis): the number of bonds between two atoms. H₂ 1, O₂ 2, N₂ 3, CO 3.

• Isoelectronic species have equal bond orders: F₂ and O₂²⁻ are 1; N₂, CO and NO⁺ are 3.
• Higher bond order → higher bond enthalpy, shorter bond.

Resonance: when one Lewis structure can't describe a molecule, several canonical structures of similar energy, same nuclear positions, same bonding and non-bonding pairs, are combined into the resonance hybrid (shown with ↔).

• O₃: both O–O bonds are 128 pm, between O–O (148) and O=O (121). Neither structure alone shows that.
• CO₃²⁻: three equal C–O bonds (three canonical forms). CO₂: C–O is 115 pm, between C=O (121) and C≡O (110).
• Resonance stabilises the molecule (the hybrid is lower in energy than any canonical form) and averages bond properties.
• Misconceptions (2023): canonical forms have no real existence; the molecule does not switch between them; there is no equilibrium between them (unlike keto–enol tautomers). The molecule has one structure, the hybrid.

Example (2021) · One π bond and the most canonical forms: CO₃²⁻ (three). SO₂ has two forms; SO₃ has three π bonds.

Polarity and dipole moment. No bond is 100% ionic or covalent. Between like atoms (H₂, O₂, Cl₂) the pair is shared equally: non-polar. In HF the pair shifts to F: polar.

• A vector. Physics draws it from negative to positive; chemistry uses a crossed arrow, cross at the positive end, head at the negative end, showing the shift of electron density (2023 statement: both true).
• A polyatomic molecule's μ is the vector sum of its bond dipoles, so shape decides it.
  • H₂O (bent): 1.85 D = 6.17 × 10⁻³⁰ C m.
  • BeF₂ (linear), BF₃ (trigonal planar), CH₄, CCl₄ (tetrahedral): bond dipoles cancel, μ = 0.
  • NH₃ (1.47 D) > NF₃ (0.23 D): in NH₃ the lone-pair (orbital) dipole points the same way as the N–H resultant; in NF₃ it opposes the N–F resultant.

Arrows point to the negative end. N is more electronegative than H, so the N–H dipoles point up to N, with the lone pair. F is more electronegative than N, so the N–F dipoles point down, against it.

Type	Molecule	μ (D)
AB (linear)	HF, HCl, HBr, HI, H₂	1.78, 1.07, 0.79, 0.38, 0
AB₂	H₂O, H₂S (bent); CO₂ (linear)	1.85, 0.95; 0
AB₃	NH₃, NF₃ (pyramidal); BF₃ (planar)	1.47, 0.23; 0
AB₄	CH₄, CCl₄ (tetrahedral); CHCl₃	0, 0; 1.04

Example (2025) · Increasing μ: NF₃ (0.23) < HBr (0.79) < H₂S (0.95) < CHCl₃ (1.04).

Example (2024) · Zero μ among HF, H₂, H₂S, CO₂, NH₃, BF₃, CH₄, CHCl₃, SiF₄, H₂O, BeF₂: H₂, CO₂, BF₃, CH₄, SiF₄, BeF₂ = 6. Square pyramidal AB₄ (a lone pair below) is polar; square planar and tetrahedral AB₄ are not (2020).

Covalent character of ionic bonds: Fajans' rules. The cation polarises (distorts) the anion's electron cloud, building up charge between them, as in a covalent bond. Covalent character is greater when:

1. the cation is small and the anion large;
2. the cation's charge is high;
3. the cation has an (n − 1)dⁿns⁰ configuration (transition metals) rather than a noble gas one, for the same size and charge.

The factors (2023): polarising power of the cation, polarisability of the anion, extent of distortion of the anion. (Not the anion's polarising power.)

Example (2022) · Covalent character: LiCl > NaCl > KCl > CsCl; CaF₂ < CaCl₂ < CaBr₂ < CaI₂. Lithium halides are somewhat covalent because Li⁺ polarises strongly (2021).

Example (2024) · Least ionic: AgCl (Ag⁺ is 4d¹⁰, a strong polariser). Also SnCl₄ > SnCl₂ and CuCl > NaCl in covalent character (2023).

4 · VSEPR theory

Proposed by Sidgwick and Powell (1940), refined by Nyholm and Gillespie (1957). Lewis can't give shapes; VSEPR can.

Postulates:

• Shape depends on the number of valence-shell electron pairs (bonded and lone) around the central atom.
• Pairs repel, so they sit as far apart as possible on a sphere.
• A multiple bond counts as one (super) pair.
• If resonance structures exist, VSEPR applies to any of them.

A lone pair belongs to one atom, so it spreads out and repels more; that distorts shapes and squeezes bond angles.

Counting pairs · Beyond NCERT shortcut: pairs = ½(V + M − C + A), V = valence electrons of the central atom, M = monovalent atoms (H, halogens), C = positive charge, A = negative charge. Oxygen adds nothing. Lone pairs = pairs − atoms bonded.

Example · XeF₄: ½(8 + 4) = 6 pairs, 4 bonded → 2 lone pairs, square planar. XeO₃: ½(8) = 4 pairs, 3 bonded → 1 lone pair, pyramidal. I₃⁻: ½(7 + 2 + 1) = 5 pairs, 2 bonded → 3 lone pairs, linear.

No lone pairs on the central atom:

Pairs	Geometry	Examples, angle
2	Linear	BeCl₂, CO₂; 180°
3	Trigonal planar	BF₃, BCl₃; 120°
4	Tetrahedral	CH₄, NH₄⁺, CCl₄; 109.5°
5	Trigonal bipyramidal	PCl₅, PF₅; 120° and 90°
6	Octahedral	SF₆; 90°
7	Pentagonal bipyramidal	IF₇ (Beyond NCERT); 72° and 90°

With lone pairs:

Pairs (bp, lp)	Shape	Examples
3 (2, 1)	Bent	SO₂, O₃, NO₂⁻
4 (3, 1)	Trigonal pyramidal	NH₃ (107°), PCl₃, H₃O⁺, ClO₃⁻, XeO₃, SO₃²⁻
4 (2, 2)	Bent	H₂O (104.5°), OF₂, SCl₂, ClO₂⁻
5 (4, 1)	See-saw	SF₄, SeF₄, IF₄⁺, XeO₂F₂
5 (3, 2)	T-shaped	ClF₃, BrF₃, ICl₃
5 (2, 3)	Linear	XeF₂, I₃⁻, ICl₂⁻, IBr₂⁻
6 (5, 1)	Square pyramidal	BrF₅, IF₅, ClF₅, XeOF₄, XeF₅⁺
6 (4, 2)	Square planar	XeF₄, ICl₄⁻, BrF₄⁻
7 (6, 1)	Distorted octahedral	XeF₆ (Beyond NCERT)
7 (5, 2)	Pentagonal planar	XeF₅⁻ (Beyond NCERT)

NCERT's reasons (Table 4.8):

• SO₂: lp–bp > bp–bp, angle 119.5° instead of 120°. NH₃: 107° from 109.5°. H₂O: lp–lp > lp–bp > bp–bp, 104.5°.
• SF₄: in a trigonal bipyramid the lone pair goes equatorial, where it has only two lp–bp repulsions at 90° (axial would have three). Shape: distorted tetrahedron, folded square, see-saw.
• ClF₃: both lone pairs equatorial, the fewest lp–bp repulsions → T-shape. (2025: "the structure with axial lone pairs is most stable" is false.)
• Rule for 5 pairs: lone pairs always go equatorial; for 6 pairs, two lone pairs go opposite (trans).

VSEPR's success and limit: it predicts shapes of p-block compounds well, even when the energy difference between shapes is small. Its theoretical basis is unclear.

Example (2021, 2022) · Two lone pairs on the central atom among SF₄, BF₄⁻, ClF₃, AsF₃, PCl₅, BrF₅, XeF₄, SF₆: ClF₃, XeF₄ = 2. Among SF₄, XeF₄, CF₄, H₂O: XeF₄, H₂O = 2.

Example (2023, 2025) · Linear among SO₂, BeCl₂, CO₂, N₃⁻, NO₂, F₂O, XeF₂, NO₂⁺, I₃⁻, O₃: BeCl₂, CO₂, N₃⁻, XeF₂, NO₂⁺, I₃⁻ = 6. Bent among N₃⁻, NO₂⁻, I₃⁻, O₃, SO₂: 3.

Example (2023) · Square planar among XeF₄, SF₄, SiF₄, BF₄⁻, BrF₄⁻, [Cu(NH₃)₄]²⁺, [FeCl₄]²⁻, [PtCl₄]²⁻: XeF₄, BrF₄⁻, [Cu(NH₃)₄]²⁺, [PtCl₄]²⁻ = 4.

Example (2026) · BrF₃ autoionises to BrF₂⁺ (2 bp, 2 lp: bent) and BrF₄⁻ (4 bp, 2 lp: square planar). SF₄ is isostructural with IF₄⁺ and XeO₂F₂ (see-saw).

Bond angles · Beyond NCERT trends JEE uses:

• More lone pairs, smaller angle: CH₄ 109.5° > NH₃ 107° > H₂O 104.5° (so CH₄ has the largest H–M–H angle; 2020).
• Down a group the hydride angle falls: NH₃ 107° > PH₃ 93.5°; H₂O 104.5° > H₂S 92°. The heavier atom uses almost pure p orbitals.
• More electronegative terminal atoms pull bond pairs away and shrink the angle: OF₂ (103°) < H₂O (104.5°) < Cl₂O (111°, bulky Cl) (2026). NF₃ (102°) < NH₃ (107°).
• BF₃ 120° > PF₃ ~98° > ClF₃ ~87.5° (2024).
• NO₂⁺ 180° > NO₂ 134° > NO₂⁻ 115°.

5 · Valence bond theory

Lewis can't explain why H₂ (435.8 kJ mol⁻¹, 74 pm) and F₂ (155 kJ mol⁻¹, 144 pm) differ, though both have one shared pair; VSEPR gives shapes without a theory. VB theory (Heitler and London, 1927; developed by Pauling) uses atomic orbitals, overlap and hybridisation.

Formation of H₂: as two H atoms approach, new attractions (each nucleus with the other's electron) and repulsions (electron–electron, nucleus–nucleus) arise. Attraction is larger, so energy falls until the two balance at a minimum (74 pm). The energy released is the bond enthalpy; 435.8 kJ mol⁻¹ is needed to break 1 mol of H₂.

Orbital overlap: at the minimum the orbitals partly merge (overlap), pairing electrons of opposite spin. More overlap, stronger bond.

Overlap signs: overlap is positive (same sign/phase and orientation: a bond forms), negative (opposite phase) or zero (orientations that cancel, e.g. s with pₓ when z is the axis). The + and − on lobes are signs of the wave function, not charges. (2022: zero overlap is an out-of-phase overlap arising from orientation: both true, linked.)

Why plain overlap fails: carbon's three p orbitals are at 90°, so CH₄ would have three H–C–H angles of 90° and a fourth bond of no fixed direction. NH₃ and H₂O would have 90° angles, not 107° and 104.5°. Hybridisation fixes this.

σ and π bonds:

	σ (sigma)	π (pi)
Overlap	End to end (head-on, axial) along the internuclear axis	Sideways; axes parallel, perpendicular to the internuclear axis
From	s–s, s–p, p–p	p–p
Strength	Stronger (larger overlap)	Weaker
Electron cloud	Along the axis	Two saucer-shaped clouds above and below the plane

A multiple bond is one σ plus π bonds: double = σ + π; triple = σ + 2π.

Example (2024, 2021, 2025) · Ethene C₂H₄: 5 σ, 1 π. CH₃–CH=CH–C≡CH: 10 σ. Hex-1-en-4-yne (CH₂=CH–CH₂–C≡C–CH₃): 13 σ, 3 π.

6 · Hybridisation

Pauling: atomic orbitals of slightly different energy mix to give a new set of equivalent orbitals (same energy and shape), the hybrid orbitals, which form the bonds.

Features:

1. Number of hybrid orbitals = number of atomic orbitals mixed.
2. Hybrids are equivalent in energy and shape.
3. They form stronger bonds than pure orbitals.
4. They point in preferred directions for minimum repulsion, so the hybridisation gives the geometry.

Conditions: only valence-shell orbitals mix; their energies must be nearly equal; promotion isn't essential first; filled orbitals can also take part.

Type (orbitals mixed)	Geometry, angle	Examples
sp (s + pz)	Linear, 180°; 50% s	BeCl₂, C₂H₂, CO₂
sp² (s + 2p)	Trigonal planar, 120°; 33% s	BCl₃, BF₃, C₂H₄
sp³ (s + 3p)	Tetrahedral, 109.5°; 25% s	CH₄, NH₃, H₂O, C₂H₆
dsp² (dx²−y² + s + 2p)	Square planar	[Ni(CN)₄]²⁻, [PtCl₄]²⁻
sp³d (s + 3p + dz²)	Trigonal bipyramidal	PF₅, PCl₅
sp³d² (s + 3p + 2 outer d)	Octahedral; square pyramidal with 1 lp	SF₆, [CrF₆]³⁻, BrF₅
d²sp³ (2 inner d + s + 3p)	Octahedral	[Co(NH₃)₆]³⁺
sp³d³ · Beyond NCERT	Pentagonal bipyramidal	IF₇; XeF₆ (distorted)

NCERT lists [CrF₆]³⁻ under sp³d² and [Co(NH₃)₆]³⁺ under d²sp³; the two-name distinction is d-block chemistry. (2024: SF₆ and [Co(NH₃)₆]³⁺ are not both sp³d²; PF₅ is sp³d but BrF₅ is sp³d².)

• sp (BeCl₂): Be 2s² → 2s¹2p¹, two sp hybrids at 180°, each overlapping a Cl 3p. Also called diagonal hybridisation.
• sp² (BCl₃): B 2s²2p¹ → 2s¹2p², three sp² at 120°.
• sp³ (CH₄): four sp³ at 109.5°. NH₃: one sp³ holds a lone pair → pyramidal, 107°. H₂O: two lone pairs → bent, 104.5°.
• C₂H₆: each C sp³; C–C is sp³–sp³ σ (154 pm), C–H sp³–s σ (109 pm).
• C₂H₄: each C sp²; C=C is one sp²–sp² σ plus one π from unhybridised p orbitals, perpendicular to the molecular plane (134 pm); C–H 108 pm; H–C–H 117.6°, H–C–C 121°.
• C₂H₂: each C sp; C≡C is one σ and two π; linear.
• d orbitals (period 3 on) have energies close to s and p, so ns, np, nd or (n−1)d, ns, np mix; ns, np, (n−1)d don't (the gap is too big).

PCl₅ (sp³d): trigonal bipyramidal. Three equatorial P–Cl at 120° in a plane; two axial at 90° to it. Axial bonds suffer more repulsion from the equatorial pairs, so they are slightly longer and weaker, which makes PCl₅ reactive. (2022: "PCl₅ has two axial bonds stronger than the equatorial" is the incorrect statement.)

SF₆ (sp³d²): S 3s²3p⁴ → six singly filled orbitals (3s, 3p, 2 × 3d) → six sp³d² at the corners of a regular octahedron.

Example (2021) · N in NO₂⁻ sp², NO₂⁺ sp, NH₄⁺ sp³. SF₄ sp³d, IF₅ sp³d².

Example (2024) · sp³ central atoms among NH₃, SO₂, SiO₂, BeCl₂, CO₂, H₂O, CH₄, BF₃: NH₃, SiO₂, H₂O, CH₄ = 4 (Si in silica is tetrahedral). Benzene has 6 × 3 = 18 sp² hybrid orbitals (2020).

Example (2026) · sp³d among BrF₅, XeF₅⁻, BF₄⁻, ICl₄⁻, XeF₄, SF₄, NH₄⁺, ClF₃, XeF₂, ICl₂⁻: SF₄, ClF₃, XeF₂, ICl₂⁻ = 4. A Lewis-acid EF₃ is sp² (BF₃), a Lewis-base EF₃ is sp³ (NF₃).

Example (2020) · Ideal ML₅: square pyramidal has eight 90° and two 180° angles; trigonal bipyramidal has six 90°, three 120°, one 180°. Total 20.

Back bonding · Beyond NCERT: a filled p orbital of a terminal atom donates into an empty orbital of the central atom.

• BF₃ has the strongest pπ–pπ back donation (2p–2p, same size), so it is the weakest Lewis acid: BF₃ < BCl₃ < BBr₃ < BI₃ (2021).
• (SiH₃)₃N is planar and less basic than pyramidal (CH₃)₃N: the N lone pair delocalises into Si (2019).

7 · Molecular orbital theory

Hund and Mulliken (1932). Features:

1. Electrons in a molecule occupy molecular orbitals (MOs), as in atoms they occupy atomic orbitals.
2. Atomic orbitals of comparable energy and proper symmetry combine.
3. An AO is monocentric; an MO is polycentric (influenced by two or more nuclei).
4. Number of MOs = number of AOs combined: two AOs give one bonding and one antibonding MO.
5. Bonding MO is lower in energy, more stable.
6. An MO gives the electron probability distribution around the nuclei.
7. MOs fill by aufbau, Pauli and Hund.

LCAO (linear combination of atomic orbitals), an approximation because Schrödinger's equation can't be solved for more than one electron:

• Bonding MO: constructive interference; electron density between the nuclei; holds them together; lower energy than either AO.
• Antibonding MO: destructive interference; density away from the region between the nuclei; a nodal plane between them; higher energy.
• The total energy of the two MOs equals that of the two AOs: the antibonding is raised as much as the bonding is lowered.

Conditions for combining (2024): (1) same or nearly the same energy (1s with 1s, not with 2s, for like atoms); (2) same symmetry about the molecular axis (z): 2pz combines with 2pz, not with 2pₓ; (3) maximum overlap.

Example (2025) · With z as the axis, of 2pz + 2pₓ, 2s + 2pₓ, 3dₓᵧ + 3dₓ²₋ᵧ², 2s + 2pz: only 2s + 2pz can combine (same symmetry).

Types: σ MOs are symmetric about the bond axis (σ1s, σ1s from 1s; σ2pz, σ2pz from 2pz). π MOs (from 2pₓ, 2pᵧ) are not: the bonding π has density above and below the axis; the π* has a node between the nuclei (2024: "π bonding MO has lower density above and below the axis" is false).

2s and 2p of two atoms give eight MOs, four of them antibonding (2024).

Energy order (second-period homonuclear diatomics):

The difference: for B₂, C₂, N₂ the σ2pz lies above the π2p pair. Use the second order for 14 or fewer electrons.

(The diagram counts only the n = 2 electrons: 8 bonding, 4 antibonding. With σ1s and σ*1s it is 10 and 6; same bond order.)

What the configuration tells you:

• Stability: N(b) > N(a) stable; N(b) < N(a) unstable.
• Bond order = ½(N(b) − N(a)). Positive → exists; zero or negative → doesn't. 1, 2, 3 mean single, double, triple.
• Bond length falls as bond order rises.
• Magnetism: all MOs doubly filled → diamagnetic (repelled by a magnet); any singly filled → paramagnetic (attracted). O₂ is paramagnetic, which Lewis theory can't explain.

NCERT's molecules:

• H₂: σ1s². BO 1, 435.8 kJ mol⁻¹, 74 pm, diamagnetic.
• He₂: σ1s² σ1s². BO 0: doesn't exist. Be₂ (σ2s² σ2s² after the core) doesn't either.
• Li₂: KK σ2s². BO 1, diamagnetic; exists in the vapour.
• C₂: KK σ2s² σ*2s² π2pₓ² π2pᵧ². BO 2, diamagnetic; its double bond is two π bonds, no σ (unusual).
• O₂: KK σ2s² σ2s² σ2pz² π2pₓ² π2pᵧ² π2pₓ¹ π*2pᵧ¹. BO (10 − 6)/2 = 2, two unpaired electrons, paramagnetic.

Species (electrons)	Bond order	Unpaired e⁻
H₂⁺ (1), He₂⁺ (3)	0.5	1
Li₂ (6)	1	0
B₂ (10)	1	2
C₂ (12)	2	0
N₂⁺, CN (13)	2.5	1
N₂, CO, NO⁺, CN⁻, C₂²⁻ (14)	3	0
N₂⁻, NO, O₂⁺ (15)	2.5	1
O₂, N₂²⁻ (16)	2	2
O₂⁻ (17)	1.5	1
O₂²⁻, F₂ (18)	1	0
Ne₂ (20)	0	0

Heteronuclear species (CO, NO, CN⁻) · Beyond NCERT: count electrons and use the table as if homonuclear.

Bond order order for the oxygen family: O₂⁺ 2.5 > O₂ 2 > O₂⁻ 1.5 > O₂²⁻ 1, so bond length O₂⁺ < O₂ < O₂⁻ < O₂²⁻ (2021, 2022, 2026). π* electrons: O₂ 2, O₂⁺ 1, O₂⁻ 3 → total 6 (2024).

Example (2019, 2023) · NO → NO⁺: bond order 2.5 → 3, paramagnetic → diamagnetic. O₂ → O₂⁺ raises bond order but stays paramagnetic.

Example (2022) · Removing an electron strengthens the bond in NO and O₂ (it leaves a π*); it weakens N₂, C₂ and B₂ (it leaves a bonding MO). C₂ is the one stabilised by gaining an electron (2019).

Example (2019) · C₂²⁻, N₂²⁻, O₂²⁻, O₂: C₂²⁻ is diamagnetic with bond order 3, the shortest bond. N₂⁺ has two π and "half a σ" (σ2pz¹).

Example (2021) · Spin-only μ of B₂⁺ (9 electrons, 1 unpaired) = √3 = 1.73 BM → 173 × 10⁻². A dioxygen species with 1.73 BM is O₂⁻ or O₂⁺ (2020).

Example (2022) · Paramagnetic among B₂, Li₂, C₂, C₂⁻, O₂²⁻, O₂⁺, He₂⁺: B₂, C₂⁻, O₂⁺, He₂⁺ = 4. Diamagnetic among N₂, N₂⁺, N₂⁻, N₂²⁻, O₂, O₂⁺, O₂⁻, O₂²⁻: N₂, O₂²⁻ = 2.

Example (2023) · Unpaired electrons in the HOMO of N₂, N₂⁺, O₂, O₂⁺: 0, 1, 2, 1. The acetylide ion C₂²⁻ matches NO⁺ (BO 3, diamagnetic).

8 · Hydrogen bonding

When H is bonded to a highly electronegative atom (F, O, N), the shared pair shifts away from H, leaving it δ+. This H attracts the electronegative atom of another molecule (or another part of the same one): the hydrogen bond, shown dotted (H–F···H–F). It is much weaker than a covalent bond.

• Definition (NCERT): the attractive force binding the H atom of one molecule with the electronegative atom (F, O or N) of another.
• Strongest in the solid, weakest in the gas. So H-bonding: ice > liquid water > impure water (2023). In ice each water molecule is H-bonded to 4 others.

Types:

1. Intermolecular: between molecules of the same or different compounds (HF, alcohols, water). Raises boiling point and solubility.
2. Intramolecular: H between two electronegative atoms of the same molecule, e.g. o-nitrophenol (between the OH and the NO₂ oxygen). It lowers boiling point and makes the compound more volatile (steam-volatile) than p-nitrophenol · Beyond NCERT.

Effects · Beyond NCERT:

• H₂O (373 K) boils far above H₂S (213 K, below 300 K; 2020): H₂S has no significant H-bonding.
• Strength of intermolecular H-bond: CH₄ (none) < HCN < NH₃ (2022).
• In CuSO₄·5H₂O, four waters coordinate Cu²⁺ and one is held by H-bonds to sulphate (2021).
• Intermolecular forces in decreasing strength: ion–ion > ion–dipole > dipole–dipole (2020). Ethyl acetate (no O–H) has only London and dipole–dipole forces.
• (2021: "dipole–dipole interactions are the only non-covalent interactions giving H-bonds" is false.)

Example (2024) · H-bonding possible among CH₃OH, H₂O, C₂H₆, C₆H₆, o-nitrophenol, HF, NH₃: all but C₂H₆ and C₆H₆ = 5.

9 · Traps

Trap	What goes wrong	Do this
MO order	Using the O₂ order for N₂, C₂, B₂	≤ 14 electrons: π2p below σ2pz
Ion electrons	Forgetting to add or remove electrons	O₂²⁻ has 18; NO⁺ 14; CN⁻ 14
XeF₄, ICl₄⁻	Calling them tetrahedral	2 lone pairs, square planar
SF₄ lone pair	Putting it axial	Equatorial: 2 lp–bp repulsions at 90°
ClF₃	Calling it trigonal planar or pyramidal	T-shaped, both lone pairs equatorial
NF₃ vs NH₃	"F is more electronegative so NF₃ is more polar"	Lone pair opposes in NF₃: 0.23 vs 1.47 D
Zero dipole	Calling a square pyramid non-polar	Only symmetric shapes cancel
PCl₅ bonds	Equal bonds	Axial longer and weaker
Resonance	Forms in equilibrium or real	Only the hybrid exists
C₂	"A double bond is σ + π"	C₂'s double bond is two π bonds
Lewis acid strength	BF₃ strongest (most electronegative F)	Back bonding: BF₃ weakest
Bond-angle trends	Expecting the angle to grow with electronegativity	More electronegative terminal atom → smaller angle (OF₂ < H₂O)
Lone pairs on outer atoms	Counting only the central atom when "total" is asked	O₃ has 6 lone pairs on its oxygens

10 · Numbers worth knowing

• Lattice enthalpy NaCl 788 kJ mol⁻¹; ΔᵢH Na 495.8; ΔegH Cl −348.7.
• Bond enthalpies (kJ mol⁻¹): H–H 435.8, O=O 498, N≡N 946, H–Cl 431, F–F 155; O–H in water 464.5 (average).
• Bond lengths (pm): H₂ 74, N₂ 109, O₂ 121, F₂ 144, Cl₂ 199; C–C 154, C=C 134, C≡C 120; O₃ 128; CO₂ 115.
• Angles: CH₄ 109.5°, NH₃ 107°, H₂O 104.5°, PH₃ 93.5°, H₂S 92°, SO₂ 119.5°, OF₂ 103°.
• Dipoles (D): H₂O 1.85, NH₃ 1.47, HF 1.78, HCl 1.07, CHCl₃ 1.04, H₂S 0.95, HBr 0.79, HI 0.38, NF₃ 0.23.
• 1 D = 3.336 × 10⁻³⁰ C m = 10⁻¹⁸ esu cm.

11 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• G.N. Lewis was an American chemist; Kössel and Lewis gave their theories independently in 1916; Langmuir refined Lewis in 1919.
• Peter Debye (Dutch) won the 1936 Nobel Prize for X-ray diffraction and dipole moments; the debye unit honours him.
• The questions NCERT says bonding theories answer: why atoms combine, why only certain combinations occur, why molecules have definite shapes.