Chemical Bonding & Molecular Structure
Ionic & covalent bonds, VSEPR shapes, hybridisation, MOT & hydrogen bonding
Ionic and Covalent Bonding; VSEPR
Ionic and Covalent Bonds and Lewis StructuresTopic 1
Atoms bond to achieve stable, noble-gas-like electron arrangements — the essence of the octet rule (eight electrons in the valence shell). The Kossel-Lewis approach explains two main ways this happens: by transferring electrons (ionic bonding) or by sharing them (covalent bonding). Which route an element takes depends largely on its ionisation enthalpy and electron gain enthalpy, linking this chapter directly to periodicity.
An ionic (electrovalent) bond forms when a metal transfers electrons to a non-metal, producing oppositely charged ions held together by electrostatic attraction, as in $\text{Na}^+\text{Cl}^-$. The strength of an ionic crystal is measured by its lattice enthalpy, which increases with higher ionic charges and smaller ionic sizes. Fajans' rules add nuance: a small, highly charged cation with a large anion polarises the anion's cloud, giving ionic compounds partial covalent character — a frequent NEET reasoning point.
A covalent bond forms when two atoms share one or more electron pairs, as in $\text{H}_2$, $\text{O}_2$ and $\text{CH}_4$. These are represented by Lewis (electron-dot) structures, drawn by counting total valence electrons, forming bonds, and completing octets (with duplets for hydrogen). Multiple bonds (double, triple) form when atoms share two or three pairs, as in $\text{O}=\text{O}$ and $\text{N}\equiv\text{N}$.
To choose the best Lewis structure, chemists use formal charge, calculated as (valence electrons) − (non-bonding electrons) − ½(bonding electrons). The most stable structure has formal charges closest to zero. The octet rule, while powerful, has clear exceptions NEET likes to test: incomplete octets ($\text{BeCl}_2$, $\text{BF}_3$), odd-electron species ($\text{NO}$), and expanded octets ($\text{PCl}_5$, $\text{SF}_6$) where the central atom holds more than eight electrons.
| Idea | Key point |
|---|---|
| Ionic bond | electron transfer; lattice enthalpy |
| Fajans' rules | small cation + large anion to covalent character |
| Covalent bond | electron sharing; Lewis structures |
| Octet exceptions | $\text{BF}_3$, $\text{NO}$, $\text{PCl}_5$, $\text{SF}_6$ |
Why does $\text{AlCl}_3$ show appreciable covalent character despite being a metal halide?
Show solution
By Fajans' rules, the small, highly charged $\text{Al}^{3+}$ strongly polarises the larger $\text{Cl}^-$ ions, distorting their electron clouds toward sharing — so $\text{AlCl}_3$ has significant covalent character.
Name two molecules that violate the octet rule by having an expanded octet.
Show solution
$\text{PCl}_5$ (10 electrons around P) and $\text{SF}_6$ (12 electrons around S) both expand the octet using available $d$ orbitals.
An ionic bond is formed by:
Lattice enthalpy increases with:
Fajans' rules predict more covalent character for a:
Which molecule has an incomplete octet?
Formal charge is calculated using valence electrons, non-bonding electrons and:
NEET trap: Octet has exceptions — incomplete ($\text{BF}_3$), odd-electron ($\text{NO}$), expanded ($\text{PCl}_5$, $\text{SF}_6$). Use Fajans' rules to judge covalent character of 'ionic' compounds.
VSEPR Theory and Shapes of MoleculesTopic 2
Lewis structures show which atoms are bonded but not the three-dimensional shape of a molecule. The VSEPR (Valence Shell Electron Pair Repulsion) theory fills this gap with one simple idea: electron pairs around a central atom repel one another and arrange themselves as far apart as possible. This single principle predicts the geometry of most NEET molecules quickly.
The key step is to count the electron pairs (bond pairs + lone pairs) on the central atom. Two pairs give a linear shape ($180^{\circ}$); three give trigonal planar ($120^{\circ}$); four give tetrahedral ($109.5^{\circ}$); five give trigonal bipyramidal; and six give octahedral ($90^{\circ}$). These idealised angles apply when all the pairs are bonding.
The crucial refinement is the effect of lone pairs, which occupy more space than bonding pairs and so repel more strongly. The repulsion order is lone pair-lone pair $>$ lone pair-bond pair $>$ bond pair-bond pair. This compresses bond angles when lone pairs are present, giving the classic NEET series: methane $\text{CH}_4$ ($109.5^{\circ}$, no lone pairs) $>$ ammonia $\text{NH}_3$ ($107^{\circ}$, one lone pair) $>$ water $\text{H}_2\text{O}$ ($104.5^{\circ}$, two lone pairs).
Shapes are often described by their actual atom positions, ignoring lone pairs: $\text{NH}_3$ is pyramidal and $\text{H}_2\text{O}$ is bent, even though both are based on a tetrahedral arrangement of electron pairs. Practising the count-pairs-then-adjust-for-lone-pairs routine on common molecules ($\text{CO}_2$, $\text{BF}_3$, $\text{CH}_4$, $\text{NH}_3$, $\text{H}_2\text{O}$, $\text{PCl}_5$, $\text{SF}_6$) covers nearly every shape question NEET asks.
| Electron pairs | Geometry / angle |
|---|---|
| 2 | linear, $180^{\circ}$ |
| 3 | trigonal planar, $120^{\circ}$ |
| 4 | tetrahedral, $109.5^{\circ}$ |
| 5 / 6 | trigonal bipyramidal / octahedral |
Predict the shape and bond angle of ammonia ($\text{NH}_3$).
Show solution
Nitrogen has 3 bond pairs and 1 lone pair (4 electron pairs to tetrahedral arrangement). The lone pair compresses the angle, giving a pyramidal shape with a bond angle of about $107^{\circ}$.
Why is $\text{CO}_2$ linear while $\text{H}_2\text{O}$ is bent?
Show solution
In $\text{CO}_2$ the central carbon has 2 bonding regions and no lone pairs to linear ($180^{\circ}$). In $\text{H}_2\text{O}$ oxygen has 2 bond pairs and 2 lone pairs, whose repulsion bends the molecule to about $104.5^{\circ}$.
VSEPR theory is based on the repulsion between:
A central atom with 4 bond pairs and no lone pairs has the shape:
The bond angle order is:
The shape of $\text{H}_2\text{O}$ is:
The strongest repulsion is between:
NEET tip: Count bond pairs + lone pairs on the central atom for geometry, then shrink angles for lone pairs. Remember $\text{CH}_4\,(109.5^{\circ}) > \text{NH}_3\,(107^{\circ}) > \text{H}_2\text{O}\,(104.5^{\circ})$.
Hybridisation, MOT and Hydrogen Bonding
Valence Bond Theory and HybridisationTopic 3
Valence bond theory (VBT) pictures a covalent bond as the overlap of half-filled atomic orbitals, with greater overlap giving a stronger bond. A head-on overlap forms a strong sigma ($\sigma$) bond, while the sideways overlap of $p$ orbitals forms a weaker pi ($\pi$) bond. A single bond is one $\sigma$; a double bond is one $\sigma$ + one $\pi$; a triple bond is one $\sigma$ + two $\pi$ — a counting rule NEET uses constantly.
Pure atomic orbitals often cannot explain observed shapes (for instance, carbon's four identical bonds in $\text{CH}_4$). The fix is hybridisation — the mixing of atomic orbitals of similar energy to form new, equivalent hybrid orbitals that point toward the bonded atoms. The type of hybridisation fixes the geometry, so identifying it is the master key to molecular shape.
The common hybridisations and their geometries are: $sp$ (2 hybrids, linear, $180^{\circ}$, as in $\text{BeCl}_2$ and $\text{C}_2\text{H}_2$); $sp^2$ (3 hybrids, trigonal planar, $120^{\circ}$, as in $\text{BF}_3$ and $\text{C}_2\text{H}_4$); $sp^3$ (4 hybrids, tetrahedral, $109.5^{\circ}$, as in $\text{CH}_4$, $\text{NH}_3$, $\text{H}_2\text{O}$); $sp^3d$ (5 hybrids, trigonal bipyramidal, as in $\text{PCl}_5$); and $sp^3d^2$ (6 hybrids, octahedral, as in $\text{SF}_6$).
A fast NEET shortcut finds hybridisation by counting the sigma bonds plus lone pairs on the central atom: a total of 2 means $sp$, 3 means $sp^2$, 4 means $sp^3$, 5 means $sp^3d$, and 6 means $sp^3d^2$. Lone pairs occupy hybrid orbitals too, which is why $\text{NH}_3$ and $\text{H}_2\text{O}$ are both $sp^3$ despite being pyramidal and bent rather than tetrahedral. This count-and-classify method handles almost every hybridisation question.
| Hybridisation | Geometry / example |
|---|---|
| $sp$ | linear; $\text{BeCl}_2$, $\text{C}_2\text{H}_2$ |
| $sp^2$ | trigonal planar; $\text{BF}_3$ |
| $sp^3$ | tetrahedral; $\text{CH}_4$, $\text{NH}_3$ |
| $sp^3d$ / $sp^3d^2$ | $\text{PCl}_5$ / $\text{SF}_6$ |
What is the hybridisation of the central atom in $\text{SF}_6$?
Show solution
Sulphur has 6 sigma bonds and 0 lone pairs, so $6 to sp^3d^2$ hybridisation, giving an octahedral shape.
How many sigma and pi bonds are present in ethyne, $\text{C}_2\text{H}_2$ ($\text{H}-\text{C}\equiv\text{C}-\text{H}$)?
Show solution
Two $\text{C}-\text{H}$ single bonds ($2\sigma$) and one $\text{C}\equiv\text{C}$ triple bond ($1\sigma + 2\pi$). Total: $3\sigma$ and $2\pi$ bonds; each carbon is $sp$ hybridised.
A single covalent bond consists of:
The hybridisation of carbon in $\text{CH}_4$ is:
An $sp$ hybridised central atom gives a shape that is:
The hybridisation in $\text{PCl}_5$ is:
A triple bond contains:
NEET tip: Hybridisation = (sigma bonds + lone pairs): 2 $sp$, 3 $sp^2$, 4 $sp^3$, 5 $sp^3d$, 6 $sp^3d^2$. Single = 1$\sigma$; double = 1$\sigma$+1$\pi$; triple = 1$\sigma$+2$\pi$.
Molecular Orbital Theory, Polarity and Hydrogen BondingTopic 4
VBT explains shapes well but fails for some molecules, so molecular orbital theory (MOT) offers a deeper picture. Atomic orbitals combine to form molecular orbitals spread over the whole molecule — a lower-energy bonding orbital and a higher-energy antibonding orbital. Electrons fill these by the same rules as atoms (Aufbau, Pauli, Hund). The key quantity is the bond order $= \tfrac{1}{2}(N_b - N_a)$, where $N_b$ and $N_a$ are the electrons in bonding and antibonding orbitals.
Bond order predicts stability and existence: a bond order of zero means the molecule does not exist, which is why $\text{He}_2$ is not formed. A higher bond order means a stronger, shorter bond. MOT's most celebrated success is explaining the paramagnetism of $\text{O}_2$ — it has two unpaired electrons in its $\pi^*$ antibonding orbitals, a fact the Lewis and valence-bond pictures cannot show. This single example is a NEET favourite.
Whether a molecule is polar depends on its shape and bond dipoles. A bond between atoms of different electronegativity is polar, but the molecule is non-polar if the bond dipoles cancel by symmetry. So $\text{CO}_2$ (linear) is non-polar despite polar bonds, while $\text{H}_2\text{O}$ (bent) is strongly polar — the asymmetry leaves a net dipole moment. Predicting polarity from geometry is a routine NEET skill.
The most biologically important interaction is the hydrogen bond — a strong dipole attraction when hydrogen is bonded to the small, highly electronegative atoms N, O or F. Hydrogen bonding gives water its unusually high boiling point and surface tension, and, crucially for medical aspirants, holds the two strands of the DNA double helix together through base pairing and stabilises the folded shapes of proteins. Intermolecular hydrogen bonds (as in water) raise boiling points, while intramolecular ones (as in ortho-nitrophenol) can lower them — a subtle distinction NEET sometimes probes.
| Concept | Key point |
|---|---|
| Bond order | $\tfrac{1}{2}(N_b - N_a)$; 0 to no molecule |
| $\text{O}_2$ | paramagnetic (2 unpaired $e^-$) |
| Polarity | $\text{H}_2\text{O}$ polar; $\text{CO}_2$ non-polar |
| Hydrogen bond | H with N, O, F; DNA, proteins, water |
Calculate the bond order of $\text{O}_2$ given $N_b = 10$ and $N_a = 6$.
Show solution
Bond order $= \tfrac{1}{2}(N_b - N_a) = \tfrac{1}{2}(10 - 6) = 2$. (MOT also shows two unpaired electrons, so $\text{O}_2$ is paramagnetic.)
Why does water have a much higher boiling point than $\text{H}_2\text{S}$?
Show solution
Water molecules form extensive intermolecular hydrogen bonds (O is small and very electronegative), which require extra energy to break; $\text{H}_2\text{S}$ has only weak van der Waals forces, so it boils much lower.
Bond order in MOT is given by:
$\text{O}_2$ is paramagnetic because it has:
$\text{CO}_2$ is non-polar because it is:
Hydrogen bonding occurs when H is bonded to:
Hydrogen bonds hold together the two strands of:
NEET tip: Bond order $= \tfrac{1}{2}(N_b - N_a)$; zero means no molecule ($\text{He}_2$). MOT explains $\text{O}_2$ paramagnetism. Hydrogen bonding (N/O/F-H) underlies water's properties, DNA and proteins.
Quick Revision — Chemical Bonding & Molecular Structure
- Ionic bond: transfer of electrons (metal + non-metal); strength rises with charge and falls with ion size (lattice energy).
- Fajans' rules: small cation, large anion and high charge to more covalent character.
- Covalent bond: sharing of electron pairs; drawn with Lewis structures; check formal charge and the octet (with exceptions like $\text{BF}_3$, $\text{PCl}_5$, $\text{SF}_6$).
- VSEPR: electron pairs repel and spread out. Lone pairs repel more, reducing bond angles ($\text{CH}_4 = 109.5^{\circ} > \text{NH}_3 = 107^{\circ} > \text{H}_2\text{O} = 104.5^{\circ}$).
- Hybridisation: $sp$ (linear, $180^{\circ}$), $sp^2$ (trigonal planar, $120^{\circ}$), $sp^3$ ($109.5^{\circ}$), $sp^3d$ (trigonal bipyramidal), $sp^3d^2$ (octahedral).
- MOT: bond order $= \tfrac{1}{2}(N_b - N_a)$; explains $\text{O}_2$ paramagnetism and why $\text{He}_2$ does not exist.
- Hydrogen bonding (with N, O, F) gives water its high boiling point and holds the DNA double helix and protein structures together.
- Polarity: a molecule is polar if bond dipoles do not cancel ($\text{H}_2\text{O}$ polar; $\text{CO}_2$ non-polar).
Frequently Asked Questions
Ready to test yourself?
Attempt the full timed mock tests — Main & Advanced level.
Start Mock Test 1 →