Chemical Bonding and Molecular Structure — NEET Chemistry

Bonding is the single most connected topic in chemistry: it explains why NaCl is a hard solid and CO₂ a gas, why water bends and boils high, why O₂ is magnetic. NEET pulls 3–4 questions from here every year — hybridisation, VSEPR shapes, bond order and molecular orbital theory recur without fail. This chapter builds every model in order — Lewis → ionic → covalent → VSEPR → valence bond → molecular orbital — so that by the end you can predict a molecule's shape, polarity and magnetism from its formula alone. Each result is reasoned out, not memorised.

1. Why atoms bond: the octet rule and Lewis structures

Atoms bond to reach a stable noble-gas electron configuration (usually 8 valence electrons — the octet). Two routes:

• Transfer of electrons → ionic bond (metal + non-metal).
• Sharing of electrons → covalent bond (non-metal + non-metal).

Lewis (electron-dot) structures show valence electrons as dots and bonds as shared pairs. To draw one: count total valence electrons, connect atoms with single bonds, complete octets on outer atoms, then place any remainder on the central atom (forming multiple bonds if needed).

Formal charge on an atom $=(\text{valence electrons})-(\text{lone-pair electrons})-\frac{1}{2}(\text{bonding electrons})$. The best Lewis structure minimises formal charges.

Worked example 1.1. Formal charge on each atom in the resonance structure of CO₂, O=C=O? Carbon: $4-0-\frac{1}{2}(8)=0$. Each oxygen: $6-4-\frac{1}{2}(4)=0$. All zero — the structure is favourable.

Limitations of the octet rule (NEET tests these exceptions):

• Incomplete octet: BeCl₂ (4 e⁻ on Be), BF₃ (6 e⁻ on B).
• Expanded octet: PCl₅ (10 e⁻), SF₆ (12 e⁻) — possible from period 3 onward using $d$ orbitals.
• Odd-electron molecules: NO, NO₂ — cannot pair all electrons.

2. The ionic bond and lattice enthalpy

An ionic bond forms by complete transfer of electrons, giving cations and anions held by electrostatic attraction in a giant crystal lattice (not discrete molecules).

Lattice enthalpy — the energy released when gaseous ions form one mole of solid — measures ionic-bond strength. From Coulomb's law it rises with higher ionic charge and smaller ionic size:

$U\propto \frac{q_{+}{q}_{-}}{r_{+}+r_{-}}$

So MgO (2+/2−, small ions) has a far higher lattice enthalpy than NaCl (1+/1−). Ionic compounds are therefore hard, high-melting, brittle, conduct only when molten or dissolved, and are generally water-soluble.

Born–Haber cycle applies Hess's law to find lattice enthalpy indirectly, summing sublimation, ionisation, dissociation, electron gain and lattice steps around a cycle.

Worked example 2.1. Why does MgO melt far higher than NaCl? Lattice enthalpy $\propto q_{+}{q}_{-}/(r_{+}+r_{-})$. MgO has charges $2\times 2=4$ versus $1\times 1=1$ for NaCl, and smaller ions, so its lattice enthalpy is roughly four-plus times larger — hence a much higher melting point.

3. The covalent bond and its parameters

A covalent bond is a shared electron pair. Its key parameters:

• Bond length — the equilibrium internuclear distance; decreases as bond order rises (C≡C < C=C < C–C).
• Bond order — number of bonds between two atoms (1, 2, 3). Higher order → shorter, stronger bond.
• Bond enthalpy — energy to break one mole of bonds; rises with bond order.
• Bond angle — set by geometry and lone pairs (Section 4).

Fajans' rules — covalent character in an "ionic" bond. No bond is purely ionic; the cation distorts (polarises) the anion's electron cloud. Covalent character increases with:

1. Small cation and large anion (more polarising / more polarisable).
2. High charge on either ion.
3. Cation with a non-noble-gas (pseudo-inert, $d$-electron) configuration (e.g. Cu⁺, Ag⁺).

This explains why AlCl₃ is covalent (small, highly charged Al³⁺) while NaCl is ionic, and why AgCl is less soluble/more covalent than expected.

Worked example 3.1. Which is more covalent, AlCl₃ or NaCl? AlCl₃. Al³⁺ is small and triply charged — strongly polarising — so it distorts the chloride cloud heavily, giving AlCl₃ significant covalent character (it even sublimes). Na⁺ is larger and singly charged, so NaCl stays essentially ionic.

4. VSEPR theory — predicting molecular shape

Valence Shell Electron Pair Repulsion theory: electron pairs around the central atom arrange to minimise repulsion, and the lone pairs push harder than bond pairs. Repulsion order:

$\text{lone–lone}>\text{lone–bond}>\text{bond–bond}$

Count the electron pairs (bonding domains + lone pairs) to get the geometry; lone pairs then bend the shape:

The CH₄ → NH₃ → H₂O angle drop (109.5° → 107° → 104.5°) is a classic: replacing bond pairs with lone pairs increases repulsion on the remaining bonds, squeezing the angle. Memorise it.

Worked example 4.1. Predict the shape and angle of H₂O. Oxygen has 2 bond pairs + 2 lone pairs = 4 pairs → tetrahedral electron geometry. The two lone pairs repel strongly, bending the molecule to a bent shape with an angle of 104.5° (below the ideal 109.5°).

Worked example 4.2. Why is CO₂ linear but SO₂ bent, though both are "AO₂"? CO₂'s carbon has no lone pair (2 double bonds, 2 electron domains) → linear, 180°. SO₂'s sulphur has a lone pair (3 domains: 2 bonds + 1 lone pair) → bent, ~119°. The lone pair is the difference.

5. Valence bond theory and hybridisation

Valence bond theory: a covalent bond forms by overlap of half-filled atomic orbitals; greater overlap → stronger bond. Head-on overlap gives a strong σ (sigma) bond; sideways overlap of $p$ orbitals gives a weaker π (pi) bond. A single bond is 1σ; a double bond is 1σ + 1π; a triple bond is 1σ + 2π.

Hybridisation mixes atomic orbitals of similar energy into equivalent hybrid orbitals that point toward the bonded atoms, explaining observed geometry. Count the steric number (σ bonds + lone pairs) on the central atom:

Worked example 5.1. Hybridisation of carbon in ethyne (C₂H₂, H–C≡C–H)? Each carbon has 2 σ bonds (one to H, one to the other C) and 0 lone pairs → steric number 2 → $sp$ hybridised, linear. The triple bond is 1σ ($sp$–$sp$) + 2π (unhybridised $p$–$p$).

Worked example 5.2. Hybridisation of sulphur in SF₆? 6 σ bonds, 0 lone pairs → steric number 6 → $s{p}^3{d}^2$, octahedral. The two extra orbitals come from sulphur's vacant 3d, allowing the expanded octet.

σ vs π count trick: in any molecule, σ bonds = (single bonds) + (one per multiple bond); π bonds = extra bonds in double/triple bonds. Benzene has 6 C–C σ + 6 C–H σ + 3 π (delocalised).

6. Molecular orbital theory (MOT)

Where valence bond theory struggles (why O₂ is paramagnetic), MOT succeeds. Atomic orbitals combine to form molecular orbitals spanning the whole molecule: a lower-energy bonding MO (σ, π) and a higher-energy antibonding MO (σ*, π*). Electrons fill these by Aufbau, Pauli and Hund, exactly as in atoms.

Bond order measures net bonding:

$\text{Bond order}=\frac{N_b-N_a}{2}$

where $N_b$, $N_a$ are electrons in bonding and antibonding MOs. A positive bond order means a stable molecule; zero means it does not exist.

Filling order for second-period diatomics:

• For O₂, F₂, Ne₂ (Z ≥ 8): ${\sigma }_{2s}<{\sigma }_{2s}^{\ast }<{\sigma }_{2p_z}<({\pi }_{2p_x}={\pi }_{2p_y})<({\pi }_{2p_x}^{\ast }={\pi }_{2p_y}^{\ast })<{\sigma }_{2p_z}^{\ast }$.
• For B₂, C₂, N₂ (Z ≤ 7): the ${\pi }_{2p}$ pair lies below ${\sigma }_{2p_z}$ (due to s–p mixing).

Worked example 6.1. Bond order and magnetism of O₂ (16 electrons)? Fill 16 electrons: bonding = 10, antibonding = 6. Bond order $=(10-6)/2=2$ (a double bond). The last two electrons go singly into the two degenerate ${\pi }^{\ast }$ orbitals (Hund) → two unpaired electrons → paramagnetic. This is MOT's great triumph — VBT predicts O₂ diamagnetic, which is wrong.

Worked example 6.2. Why does He₂ not exist? He₂ has 4 electrons: 2 in ${\sigma }_{1s}$ (bonding), 2 in ${\sigma }_{1s}^{\ast }$ (antibonding). Bond order $=(2-2)/2=0$ — no net bond, so He₂ does not form.

Key bond orders to remember: N₂ = 3 (very stable, short bond), O₂ = 2, F₂ = 1. Species like O₂⁺ (bond order 2.5) are stronger than O₂; O₂⁻ (1.5) and O₂²⁻ (1) are weaker.

7. Polarity, dipole moment and resonance

Bond polarity arises from an electronegativity difference: the more electronegative atom carries a partial negative charge. The dipole moment $\mu =q\times d$ (unit: Debye) measures polarity as a vector.

• A molecule can have polar bonds but zero net dipole if the bond vectors cancel by symmetry: CO₂ (linear), BF₃ (trigonal), CH₄, CCl₄ (tetrahedral), SF₆ are all non-polar despite polar bonds.
• H₂O and NH₃ are polar — their lone pairs make them asymmetric, so bond dipoles don't cancel.

Worked example 7.1. Why is CO₂ non-polar but H₂O polar, though both have polar bonds? CO₂ is linear, so its two equal C=O bond dipoles point opposite and cancel → net μ = 0. H₂O is bent (104.5°), so its two O–H dipoles add to a net downward dipole → μ ≈ 1.85 D, polar.

Worked example 7.2. Compare dipole moments of NH₃ and NF₃. Both are pyramidal, but in NH₃ the N–H bond dipoles and the lone-pair dipole point the same way (net large μ ≈ 1.47 D); in NF₃ the N–F dipoles oppose the lone-pair dipole (net small μ ≈ 0.24 D). Same shape, very different polarity — a NEET favourite.

Resonance — when one Lewis structure can't capture the real bonding, the molecule is a hybrid of several structures (e.g. benzene, CO₃²⁻, O₃). Resonance delocalises electrons, lowers energy (resonance stabilisation) and equalises bond lengths (all C–O in CO₃²⁻ are identical, intermediate between single and double).

8. Intermolecular forces and hydrogen bonding

Bonds within molecules are strong; the forces between molecules decide melting/boiling points:

• London dispersion forces — weak, present in all molecules; grow with molecular size/mass (why I₂ is solid, F₂ a gas).
• Dipole–dipole forces — between polar molecules.
• Hydrogen bonding — a strong dipole force when H is bonded to N, O or F; the small, highly electronegative atom leaves H strongly positive.

Hydrogen bonding explains anomalies NEET loves:

• Water's unusually high boiling point and its lower density as ice (open H-bonded lattice).
• HF > HCl in boiling point despite HCl being heavier (HF hydrogen-bonds).
• Ortho-nitrophenol (intramolecular H-bond, lower b.p.) vs para-nitrophenol (intermolecular, higher b.p.).

Worked example 8.1. Why does water boil at 100 °C while H₂S is a gas at room temperature, though both are Group-16 hydrides? Water forms strong hydrogen bonds (O is small and highly electronegative); sulphur is larger and less electronegative, so H₂S has only weak dipole/dispersion forces. The extra energy needed to break water's hydrogen-bond network raises its boiling point dramatically.

9. Common traps NEET sets here

• VSEPR angle drop: CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°) — lone pairs squeeze the angle.
• Shape vs geometry: SF₄ is see-saw (not tetrahedral), XeF₄ is square planar (not octahedral) — lone pairs change the shape.
• CO₂ non-polar, H₂O polar — symmetry cancels dipoles in CO₂; the bent H₂O doesn't cancel.
• O₂ is paramagnetic — only MOT explains it (2 unpaired electrons in π*).
• Bond order ranking: N₂ (3) > O₂ (2) > F₂ (1); O₂⁺ (2.5) is stronger than O₂.
• Fajans: small, highly charged cation → more covalent (AlCl₃ covalent, NaCl ionic).
• NH₃ vs NF₃ dipole — same shape, but the lone-pair/bond dipoles add in NH₃ and oppose in NF₃.
• Hybridisation = σ bonds + lone pairs (steric number), not total bonds — count π separately.

10. Memory aids

• "Steric number = σ + lone pairs → hybridisation" (2→sp, 3→sp², 4→sp³, 5→sp³d, 6→sp³d²).
• "Lone pairs bite the angle" — each lone pair drops the bond angle a few degrees.
• "$(N_b-N_a)/2$" — bond order; positive = exists, zero = doesn't (He₂).
• "O₂ has two lonely electrons" — its π* pair, unpaired → paramagnetic.
• "Small cation, big anion, high charge → covalent" — Fajans in one line.
• "H bonds to N, O, F only" — the hydrogen-bond rule.

11. Exam protocol

1. Draw the Lewis structure; check octets and formal charge; note exceptions (BF₃, PCl₅, NO).
2. Count electron domains for VSEPR; subtract lone pairs to get the shape and angle.
3. Steric number (σ + lone pairs) gives the hybridisation directly.
4. For diatomics, fill MO diagram, compute bond order $(N_b-N_a)/2$, read magnetism from unpaired electrons — remember O₂ is paramagnetic.
5. Judge polarity by symmetry: symmetric shapes (linear, trigonal, tetrahedral, octahedral) cancel dipoles.
6. Use Fajans' rules for covalent character; hydrogen bonding for boiling-point and solubility anomalies.