JEE Main (Engineering) · Chemistry (JEE & NEET)
Chemical Bonding and Molecular Structure
Ionic and covalent bonding, VSEPR geometry, valence bond and molecular orbital theory, hybridisation and hydrogen bonding.
Eight concepts from electron transfer to molecular orbitals. Almost every mark here is a classification with one arithmetic check attached — Fajans, steric number, bond order, percent ionic — so the trap is naming the right rule before you reach for a calculator.
- JEE Main (Engineering)
- Medium level
- 8 concepts
- 5 practice questions
1Ionic, covalent and Fajans' rules
An ionic bond is electron transfer: a metal with low ionisation energy hands an electron to a non-metal with high electron affinity, and the resulting ions pack by Coulomb attraction. A covalent bond is electron sharing between atoms of similar electronegativity. Real bonds sit on a continuum between those two extremes, and the question is how far a supposedly ionic compound has drifted toward covalency.
Fajans' rules name the three levers that pull an ionic bond toward covalent character: a small, highly charged cation (high polarising power), a large, highly charged anion (high polarisability), and a cation with a pseudo-noble-gas or incomplete configuration (more polarising than a noble-gas ion of the same size and charge). AlCl₃ is covalent where NaCl is ionic for exactly those reasons — Al³⁺ is small and highly charged, Cl⁻ is large.
Figure. Fajans' three levers slide a salt toward covalency: a small or highly charged cation and a large, polarisable anion pull electron density onto the cation. The strip is a classification axis, not a bond length.
How it works
- Transfer vs shareLarge ΔEN and a metal–non-metal pair → transfer (ionic). Similar EN → share (covalent).
- Cation polarisesSmall high-charge cations distort the anion's electron cloud; the shared density is covalency.
- Anion yieldsLarge high-charge anions are soft and polarisable — F⁻ resists, I⁻ yields.
| Lever | Raises covalent character when… | Stock example |
|---|---|---|
| Cation size | the cation is smaller | Li⁺ > Na⁺ > K⁺ (same charge) |
| Cation charge | the cation charge is higher | Al³⁺ > Mg²⁺ > Na⁺ |
| Anion size / charge | the anion is larger or more charged | I⁻ > Br⁻ > Cl⁻ > F⁻; O²⁻ > F⁻ |
| Cation configuration | the cation is not noble-gas-like | Cu⁺ more polarising than Na⁺ |
Which pair has more covalent character in the bond?
- NaCl more than AlCl₃, because sodium is more electropositive
- AlCl₃ more than NaCl, because Al³⁺ is small and highly charged
- They are equal — both are metal–chlorine bonds
Al³⁺ is smaller and carries charge +3 against Na⁺ at +1, so it polarises Cl⁻ far more strongly. Fajans predicts AlCl₃ is the covalent one; NaCl stays ionic.
2Formal charge
Formal charge is bookkeeping on a Lewis structure, not the real charge on the atom. For each atom, FC = V - \frac{1}{2}B - L: valence electrons in the free atom, minus half the bonding electrons, minus the lone-pair electrons. The sum of formal charges on every atom equals the charge of the species.
The preferred Lewis structure is the one that keeps formal charges closest to zero and puts any negative formal charge on the more electronegative atom. That is why the carbon-monoxide structure with a triple bond (FC of −1 on C and +1 on O) beats alternatives that load a positive charge onto oxygen.
Figure. Formal charge is bookkeeping on a Lewis count: start from free-atom valence, subtract half the bonding electrons and all lone electrons. It is not the real charge on the atom.
How it works
- Count VValence electrons of the free atom: C = 4, N = 5, O = 6, F = 7.
- Split the bondsEach bond contributes one electron to each atom in the ½B term — a double bond is B = 4.
- Subtract lone pairsEvery lone-pair electron comes off in full. Then sum FCs to recover the species charge.
Formal charges in carbon monoxide
For the triple-bonded Lewis structure of CO, find the formal charge on C and on O.
- C: V = 4, B = 6 (triple bond), L = 2FC = 4 − 3 − 2 = −1
- O: V = 6, B = 6, L = 2FC = 6 − 3 − 2 = +1
- Sum of formal charges−1 + 1 = 0 ✓
- Check against charge of COneutral molecule, sum must be 0
Pro tip. The −1 sits on carbon, the less electronegative atom — awkward, but every alternative Lewis structure for CO is worse. Formal charge is a ranking tool for structures, not a claim about the real charge distribution (which dipole moment measures).
In ozone, one resonance form has a central O double-bonded to one terminal O and single-bonded to the other. The formal charge on the central oxygen is
- 0, because ozone is neutral
- +1, from V = 6, B = 6, L = 2
- −1, because oxygen is electronegative
Central O: V = 6, B = 6 (one double + one single), L = 2 → FC = 6 − 3 − 2 = +1. The singly bound terminal oxygen carries −1, and the three formal charges sum to zero.
3VSEPR geometry
VSEPR says electron pairs around a central atom sit as far apart as they can. The repulsion order is lone-pair–lone-pair > lone-pair–bond-pair > bond-pair–bond-pair, so lone pairs compress bond angles. That is why CH₄ is tetrahedral at 109.5°, NH₃ is pyramidal at about 107°, and H₂O is bent at about 104.5° — same steric number four, successively more lone-pair compression.
Molecular shape is the arrangement of the atoms, not the electron pairs. XeF₄ has steric number six (octahedral electron geometry) but two lone pairs opposite each other, so the molecular shape is square planar. Always name both: the electron geometry from the steric number, then the molecular shape after the lone pairs take their places.
Figure. Lone pairs compress bond angles: methane's tetrahedral 109.5° falls to ~107° in ammonia and ~104.5° in water. The electron geometry stays tetrahedral; the molecular shape loses vertices to lone pairs.
How it works
- Steric numberCount bonded atoms plus lone pairs on the central atom. That sets the electron geometry.
- Place lone pairsLone pairs claim the positions that minimise lp–lp repulsion — axial/trans in octahedral, equatorial in trigonal bipyramidal.
- Read the shapeName the molecular shape from the atoms alone. Angles shrink wherever a lone pair sits beside a bond.
Geometry of XeF₄
Predict the hybridisation and shape of XeF₄.
- Xe valence e⁻ = 8; 4 Xe–F bonds use 42 lone pairs left on Xe
- Steric number = 4 bonded atoms + 2 lone pairs6 → sp³d², octahedral electron geometry
- Place 2 lone pairs to minimise lp–lp repulsionopposite (trans / axial)
- Atoms left in the equatorial planesquare planar molecular shape
Pro tip. In octahedral electron geometry the two lone pairs always go trans. Putting them adjacent would cost an lp–lp right angle; opposite costs only lp–bp. That single placement decision is the whole difference between square planar and a seesaw-like mess.
Why is the H–O–H angle in water smaller than the H–N–H angle in ammonia?
- Oxygen is more electronegative, so the bonds are shorter
- Water has two lone pairs compressing the angle; ammonia has one
- Nitrogen has a higher steric number than oxygen
Both have steric number four. Water's two lone pairs repel more than ammonia's one, so the bond angle falls from ~107° in NH₃ to ~104.5° in H₂O. Electronegativity affects bond polarity, not this VSEPR compression argument.
4Hybridisation from steric number
Hybridisation is the mixing of atomic orbitals on the central atom so that the hybrid set matches the electron geometry VSEPR already demanded. Steric number is the lookup: 2 → sp (linear, 180°), 3 → sp^2 (trigonal planar, 120°), 4 → sp^3 (tetrahedral), 5 → sp^3d (trigonal bipyramidal), 6 → sp^3d^2 (octahedral).
Bent's rule is the refinement once the hybrid type is fixed: more electronegative substituents prefer hybrid orbitals with greater p-character, so the bond to fluorine in PF₃Cl₂ sits in a hybrid richer in p than the bond to chlorine. Use steric number to name the hybrid set; use Bent's rule only when two substituents compete inside that set.
Figure. Steric number is bonded atoms plus lone pairs on the central atom — that integer picks the hybrid set that matches the VSEPR electron geometry. Count first; name the hybrid second.
How it works
- Count SNBonded atoms + lone pairs on the central atom — same count VSEPR used.
- Name the hybridSN 2, 3, 4, 5, 6 → sp, sp², sp³, sp³d, sp³d².
- Bent's ruleInside one hybrid set, the more electronegative substituent takes more p-character.
| SN | Hybrid | Electron geometry | Ideal angle |
|---|---|---|---|
| 2 | sp | Linear | 180° |
| 3 | sp² | Trigonal planar | 120° |
| 4 | sp³ | Tetrahedral | 109.5° |
| 5 | sp³d | Trigonal bipyramidal | 90° / 120° |
| 6 | sp³d² | Octahedral | 90° |
Hybridisation of SF₄
Find the hybridisation and electron geometry of SF₄.
- S valence e⁻ = 6; 4 S–F bonds use 41 lone pair left on S
- Steric number = 4 + 15
- SN 5 → hybridsp³d
- Electron geometry for SN 5trigonal bipyramidal (seesaw molecular shape)
Pro tip. The lone pair in SN 5 takes an equatorial site — that is VSEPR again, not a new hybridisation rule. Hybridisation names the orbital set; VSEPR places the lone pairs inside it.
BeCl₂ has steric number 2. Its hybridisation and shape are
- sp², bent
- sp, linear
- sp³, tetrahedral
SN = 2 bonded atoms + 0 lone pairs = 2 → sp, linear, 180°. Bent would need a lone pair; tetrahedral needs SN 4.
5Bond order and magnetism from MOT
Molecular orbital theory fills bonding and antibonding MOs from the lowest energy up, with Hund's rule in degenerate levels. Bond order is \frac{1}{2}(N_b - N_a): half the difference between electrons in bonding MOs and electrons in antibonding MOs. Higher bond order means a shorter, stronger bond.
O₂ is the set-piece. Its MO configuration leaves two unpaired electrons in the degenerate \pi^* set, so O₂ is paramagnetic — a fact Lewis structures cannot explain. Strip one electron for O₂⁺ and bond order rises to 2.5; add one for O₂⁻ and it falls to 1.5. Both ions stay paramagnetic with one unpaired electron.
Figure. Valence MO ladder for O₂ (O₂/F₂ energy order). Ten electrons sit in bonding levels and six in antibonding ones — bond order 2 — with the two π* electrons unpaired, which is the paramagnetism Lewis structures miss. Levels are schematic in spacing, not to an energy scale.
How it works
- Fill MOsLowest energy first; pair only after each degenerate orbital has one electron.
- Count Nb and NaBonding MOs contribute to Nb; antibonding (starred) MOs to Na.
- Bond order and spinBO = ½(Nb − Na). Unpaired electrons → paramagnetic; all paired → diamagnetic.
Bond order and magnetism of O₂ species
Arrange O₂, O₂⁺ and O₂⁻ in order of bond order and state which are paramagnetic.
- O₂ (16 e⁻): Nb = 10, Na = 6BO = ½(10 − 6) = 2; two unpaired π* e⁻ → paramagnetic
- O₂⁺ (15 e⁻): Nb = 10, Na = 5BO = ½(10 − 5) = 2.5; one unpaired e⁻ → paramagnetic
- O₂⁻ (17 e⁻): Nb = 10, Na = 7BO = ½(10 − 7) = 1.5; one unpaired e⁻ → paramagnetic
- Order of bond order (bond length reverses it)O₂⁺ > O₂ > O₂⁻
Pro tip. Higher bond order means shorter and stronger, so bond length runs O₂⁻ > O₂ > O₂⁺ — the reverse of the bond-order list. Count electrons first; the magnetism and the length order both fall out of the same MO fill.
Which species is diamagnetic?
- O₂
- O₂²⁻
- O₂⁺
O₂²⁻ has 18 valence electrons: Nb = 10, Na = 8, BO = 1, and the π* set is full — all electrons paired. O₂ and O₂⁺ each keep at least one unpaired π* electron.
6MO energy order flips at O₂
The valence MO stacking is not one ladder for every second-period dimer. For B₂, C₂ and N₂ the \pi2p pair sits below \sigma2p_z; for O₂ and F₂ the order flips and \sigma2p_z drops below \pi2p. The 1s core is written KK and ignored; what changes is only the relative order of \sigma2p_z and \pi2p.
The flip matters because it changes which orbitals are the HOMO. N₂ has a filled \sigma2p_z HOMO and is diamagnetic; O₂ has a half-filled \pi^* HOMO and is paramagnetic. Mixing the two orders is the standard way to get the magnetism of N₂ or the bond order of B₂ wrong.
Figure. B₂–N₂ put π2p below σ2p; O₂ and F₂ flip that pair. The swap is the whole exam fork — it decides which levels fill first and why O₂ is paramagnetic with the O₂/F₂ order.
How it works
- Identify the dimerAtoms up to N₂ use the π-below-σ order; O₂ and F₂ use σ-below-π.
- Write the ladderFill from the bottom with the correct order before you count Nb and Na.
- Read HOMOMagnetism and ionisation chemistry follow from which level is only partly filled.
| Molecules | Order (after KK) | HOMO character |
|---|---|---|
| B₂, C₂, N₂ | σ2s < σ*2s < π2p < σ2p_z < π*2p < σ*2p | σ2p_z for N₂ (filled) |
| O₂, F₂ | σ2s < σ*2s < σ2p_z < π2p < π*2p < σ*2p | π*2p for O₂ (half-filled) |
For N₂ the σ2p_z orbital lies
- below the π2p pair, as in O₂
- above the π2p pair — the ≤ N₂ order
- at the same energy as π2p for every second-period dimer
N₂ uses the ≤ N₂ ladder: π2p below σ2p_z. The O₂/F₂ flip puts σ2p_z below π2p; applying that flip to N₂ is the common error.
7Dipole moment and percent ionic character
A bond dipole is \mu = q \times d: partial charge times internuclear separation. The SI unit is C·m; the Debye used in every exam table is 1\,\mathrm{D} = 3.336 \times 10^{-30}\,\mathrm{C\cdot m}. A molecule's net dipole is the vector sum of its bond dipoles, which is why CO₂ is non-polar with two polar bonds — the vectors cancel.
Percent ionic character compares the observed dipole to the dipole the bond would have if a full electron sat at the measured bond length: \%\ \mathrm{ionic} \approx (\mu_{\mathrm{obs}} / \mu_{\mathrm{ionic}}) \times 100 with \mu_{\mathrm{ionic}} = e \times d. HCl at \mu_{\mathrm{obs}} = 1.03\,\mathrm{D} and d = 127\,\mathrm{pm} is only about 17% ionic — a reminder that a "polar covalent" label is doing real quantitative work.
Figure. Observed dipole along H–Cl is 1.03 D; the full-electron ideal at the same bond length is 6.10 D. Their ratio is the percent ionic character (~17%). Arrow length is schematic — not to scale against 6.10 D.
How it works
- Ideal ionic dipoleμ_ionic = e × d, then convert C·m to Debye by dividing by 3.336 × 10⁻³⁰.
- Percent ionicDivide μ_obs by μ_ionic and multiply by 100.
- Molecular netSum bond dipoles as vectors; symmetry can cancel them to zero.
Percent ionic character of HCl
HCl has μ_obs = 1.03 D and bond length 127 pm. Find the percent ionic character. Take 1 D = 3.336 × 10⁻³⁰ C·m and e = 1.602 × 10⁻¹⁹ C.
- d = 127 pm1.27 × 10⁻¹⁰ m
- μ_ionic = e × d = (1.602 × 10⁻¹⁹)(1.27 × 10⁻¹⁰)2.0345 × 10⁻²⁹ C·m
- μ_ionic in D = 2.0345 × 10⁻²⁹ / (3.336 × 10⁻³⁰)6.10 D
- % ionic = (1.03 / 6.10) × 10016.9%
Pro tip. If you forget to convert μ_ionic into Debye and divide 1.03 by 2.03 × 10⁻²⁹, you get nonsense of order 10²⁸. Convert both dipoles into the same unit before you take the ratio — Debye is the unit the data was quoted in.
CO₂ has two polar C=O bonds but a net dipole moment of zero because
- the C=O bonds are actually non-polar
- the two bond dipoles are equal and opposite and cancel
- percent ionic character is defined only for diatomic molecules
Linear O=C=O puts the two bond dipoles on one line in opposite directions. Equal magnitudes cancel; the molecule is non-polar even though each bond is polar.
8Hydrogen bonding
A hydrogen bond is the attraction between a hydrogen bound to a highly electronegative atom (F, O or N) and a lone pair on another F, O or N. Intermolecular hydrogen bonds link neighbouring molecules and raise boiling points — HF, H₂O and NH₃ sit far above their heavier analogues HCl, H₂S and PH₃ for that reason alone.
Intramolecular hydrogen bonds form inside one molecule and do the opposite to the boiling point: they satisfy the H-bond demand without tying molecules together, so the intermolecular network weakens. Ortho-nitrophenol boils lower than para-nitrophenol because the ortho isomer hydrogen-bonds to itself.
Figure. Boiling points of the hydrogen halides. Without hydrogen bonding the series would rise HF → HI with molar mass; HF alone sits above 0 °C because intermolecular H-bonds hold the liquid together. Bar lengths follow the Celsius values from a zero baseline.
How it works
- Need F, O or NH must be bound to F, O or N, and a lone pair on F, O or N must accept it.
- IntermolecularLinks neighbours → higher boiling point, higher viscosity, often higher solubility in water.
- IntramolecularCloses inside one molecule → fewer links between molecules → lower boiling point than the isomer that cannot close.
Ortho-nitrophenol has a lower boiling point than para-nitrophenol primarily because
- the ortho isomer is more ionic
- intramolecular H-bonding in the ortho isomer reduces intermolecular association
- the para isomer has no hydrogen atoms
In the ortho isomer the OH hydrogen bonds to the neighbouring nitro oxygen inside the same molecule. That intramolecular bond replaces intermolecular links, so less energy is needed to separate molecules and the boiling point falls relative to the para isomer.
Notes
- Ionic vs covalent: Ionic bonds form by electron transfer (low IE metal + high electron affinity non-metal); covalent character in ionic bonds is predicted by Fajans' rules (small cation, large anion, high charge).
- VSEPR: Electron pairs orient to minimise repulsion (lp-lp > lp-bp > bp-bp), giving CH_4 tetrahedral 109.5^\circ, NH_3 pyramidal 107^\circ, and H_2O bent 104.5^\circ.
- Hybridisation: mixing atomic orbitals gives sp (linear 180^\circ), sp^2 (trigonal 120^\circ), sp^3 (tetrahedral), sp^3d (trigonal bipyramidal), and sp^3d^2 (octahedral) shapes.
- Molecular Orbital Theory: electrons fill bonding and antibonding MOs; bond order =\frac{1}{2}(N_b-N_a); MOT explains the paramagnetism of O_2 (two unpaired \pi^* electrons).
- Hydrogen bonding: intermolecular H-bonds (HF, H_2O, NH_3) raise boiling points, whereas intramolecular H-bonding (o-nitrophenol) lowers them.
Formulas
- Bond order =\frac{1}{2}(N_b-N_a)
- Formal charge =V-\frac{1}{2}B-L (V = valence e⁻, B = bonding e⁻, L = lone-pair e⁻)
- Dipole moment \mu=q\times d (1\,D=3.336\times10^{-30} C·m)
- % ionic character \approx\frac{\mu_{obs}}{\mu_{ionic}}\times100
- MO order (\leq N_2): \sigma1s<\sigma^*1s<\sigma2s<\sigma^*2s<\pi2p_x=\pi2p_y<\sigma2p_z<\pi^*2p<\sigma^*2p
Exam traps & shortcuts
- Steric number = (bonded atoms + lone pairs): 2→sp, 3→sp^2, 4→sp^3, 5→sp^3d, 6→sp^3d^2.
- For O_2, F_2 the \sigma2p_z lies below the \pi2p; for \leq N_2 the order flips (\pi below \sigma).
- Bent's rule: more electronegative substituents prefer hybrid orbitals with greater p-character.
Reference tables
Five relations this topic actually computes. Steric number and Fajans stay as lookups in their concepts.
| Quantity | Relation | Watch |
|---|---|---|
| Formal charge | FC = V − ½B − L | Sum of FCs = charge of the species |
| Bond order | BO = ½(Nb − Na) | Higher BO → shorter, stronger bond |
| Dipole moment | μ = q × d | 1 D = 3.336 × 10⁻³⁰ C·m |
| % ionic character | (μ_obs / μ_ionic) × 100 | μ_ionic = e × d at the measured bond length |
| MO order (≤ N₂) | σ2s < σ*2s < π2p < σ2p_z < π*2p < σ*2p | O₂, F₂ flip σ2p_z below π2p |
The same count VSEPR and hybridisation share. Molecular shape still needs the lone-pair placement.
| SN | Hybrid | Electron geometry | Example |
|---|---|---|---|
| 2 | sp | Linear | BeCl₂, CO₂ |
| 3 | sp² | Trigonal planar | BF₃, SO₃ |
| 4 | sp³ | Tetrahedral | CH₄; NH₃ pyramidal; H₂O bent |
| 5 | sp³d | Trigonal bipyramidal | PCl₅; SF₄ seesaw |
| 6 | sp³d² | Octahedral | SF₆; XeF₄ square planar |
Recap
Read only this the night before.
- Fajans
- Small high-charge cation + large high-charge anion → more covalent. AlCl₃ covalent, NaCl ionic.
- Formal charge
- FC = V − ½B − L. Prefer structures with FC near zero; put negative FC on the more electronegative atom.
- VSEPR
- lp–lp > lp–bp > bp–bp. CH₄ 109.5°, NH₃ ~107°, H₂O ~104.5°. Shape names atoms; electron geometry names pairs.
- Hybridisation
- SN 2→sp, 3→sp², 4→sp³, 5→sp³d, 6→sp³d². Bent: more electronegative substituent takes more p-character.
- Bond order
- BO = ½(Nb − Na). O₂⁺ (2.5) > O₂ (2) > O₂⁻ (1.5); length runs the other way. O₂ paramagnetic — two unpaired π*.
- MO order
- ≤ N₂ has π below σ2p_z; O₂ and F₂ flip. Wrong ladder → wrong HOMO and wrong magnetism.
- % ionic
- μ = q d; % ionic = μ_obs / (e d) × 100. HCl is only ~17% ionic. CO₂ net μ = 0 by cancellation.
- H-bonds
- Intermolecular raise boiling points (HF, H₂O, NH₃). Intramolecular lower them (o-nitrophenol vs p-).
Practise Chemical Bonding and Molecular Structure
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