Why VSEPR is needed
The Lewis concept of bonding lets us count electrons and draw structures, but it carries no information about the three-dimensional arrangement of atoms in space. It cannot tell us that $\ce{CO2}$ is linear while $\ce{H2O}$ is bent, even though both are triatomic. To bridge this gap, Sidgwick and Powell proposed in 1940 a theory based on the repulsive interactions of electron pairs in the valence shell, later refined by Nyholm and Gillespie in 1957.
The premise is deliberately economical. Electron pairs around a central atom carry negative charge and therefore repel one another; they settle into the spatial arrangement that places them as far apart as possible. Once we know how many electron pairs surround the central atom, the geometry follows almost mechanically — and from there, the visible shape of the molecule.
Postulates of VSEPR theory
The working model rests on a short list of postulates. Each is worth committing to memory because NEET statement-type questions are framed directly around them.
| # | Postulate |
|---|---|
| 1 | The shape of a molecule depends on the number of valence-shell electron pairs — bonded or non-bonded — around the central atom. |
| 2 | Pairs of electrons in the valence shell repel one another because their electron clouds are negatively charged. |
| 3 | These pairs occupy positions in space that minimise repulsion and so maximise the distance between them. |
| 4 | The valence shell is treated as a sphere, with the electron pairs localised on its surface at maximum separation. |
| 5 | A multiple bond is treated as a single super pair: the two or three electron pairs of a double or triple bond count as one. |
| 6 | Where two or more resonance structures represent a molecule, the VSEPR model applies to any one of them. |
The lone-pair repulsion order
Not all repulsions are equal. Nyholm and Gillespie sharpened the original model by distinguishing lone pairs from bonding pairs. A lone pair is held by only one nucleus, so its electron cloud spreads out and occupies more space. A bonding pair is shared between two nuclei and is drawn in more tightly. The consequence is a strict order of repulsive strength.
Lone pair – lone pair $>$ Lone pair – bond pair $>$ Bond pair – bond pair
This single inequality is the engine behind every distortion you will meet. Wherever lone pairs sit on the central atom, they push the bonding pairs together more forcefully than the bonding pairs push back, so real bond angles fall below their idealised values. The order is examined verbatim — NEET 2016 asked candidates to pick exactly this sequence.
Geometry versus shape
The most common conceptual error in this topic is treating geometry and shape as synonyms. They are not.
| Term | What it counts | Example: $\ce{H2O}$ |
|---|---|---|
| Geometry | Arrangement of all electron pairs (bond pairs + lone pairs) around the central atom | Tetrahedral (4 pairs) |
| Shape | Arrangement of only the bonded atoms | Bent / angular |
When the central atom has no lone pairs, geometry and shape coincide: $\ce{CH4}$ has tetrahedral geometry and a tetrahedral shape. When lone pairs are present they occupy space and dictate the geometry, but they are invisible in the shape because they are not atoms. This is why $\ce{NH3}$, with tetrahedral geometry, has a trigonal pyramidal shape, and why $\ce{H2O}$, also with tetrahedral geometry, has a bent shape.
Shapes with no lone pairs
When the central atom carries only bonding pairs, the geometry is the regular polyhedron that places the pairs at maximum separation. NCERT Table 4.6 lists these standard cases for molecules of the type $\ce{AB2}$ through $\ce{AB6}$.
| Type | Bond pairs | Geometry | Bond angle | Example |
|---|---|---|---|---|
| $\ce{AB2}$ | 2 | Linear | 180° | $\ce{BeCl2}$ |
| $\ce{AB3}$ | 3 | Trigonal planar | 120° | $\ce{BF3}$ |
| $\ce{AB4}$ | 4 | Tetrahedral | 109.5° | $\ce{CH4}$ |
| $\ce{AB5}$ | 5 | Trigonal bipyramidal | 120° & 90° | $\ce{PCl5}$ |
| $\ce{AB6}$ | 6 | Octahedral | 90° | $\ce{SF6}$ |
The five-coordinate and six-coordinate cases deserve a closer look because they reappear constantly in NEET matching questions. In $\ce{PCl5}$ the five bonding pairs occupy the corners of a trigonal bipyramid: three equatorial bonds lie in a plane at 120° to each other, and two axial bonds point straight up and down at 90° to the equatorial plane. The axial and equatorial bonds are therefore not equivalent. In $\ce{SF6}$ the six bonding pairs sit at the corners of a regular octahedron, every $\ce{F-S-F}$ angle equal to 90°.
Shapes with lone pairs
Once one or more lone pairs sit on the central atom, the parent geometry is retained but the visible shape changes, and bond angles shrink in proportion to the number of lone pairs. NCERT Tables 4.7 and 4.8 work through the standard cases. The three four-pair examples are the bedrock of the topic.
| Type | Bond pairs | Lone pairs | Shape | Example |
|---|---|---|---|---|
| $\ce{AB2E}$ | 2 | 1 | Bent | $\ce{SO2}$ |
| $\ce{AB3E}$ | 3 | 1 | Trigonal pyramidal | $\ce{NH3}$ |
| $\ce{AB2E2}$ | 2 | 2 | Bent | $\ce{H2O}$ |
| $\ce{AB4E}$ | 4 | 1 | See-saw | $\ce{SF4}$ |
| $\ce{AB3E2}$ | 3 | 2 | T-shape | $\ce{ClF3}$ |
| $\ce{AB2E3}$ | 2 | 3 | Linear | $\ce{XeF2}$ |
VSEPR predicts the shape; hybridisation explains the orbitals behind it. Read the two together to lock the geometry-orbital link.
Master geometry table
The table below is the single sheet to memorise for this topic. It maps the number of electron pairs to the parent geometry, then to the shape after lone pairs are removed, with the corresponding bond angle and a worked example for each row.
| Electron pairs | Geometry | Lone pairs | Shape | Bond angle | Example |
|---|---|---|---|---|---|
| 2 | Linear | 0 | Linear | 180° | $\ce{BeCl2}$ |
| 3 | Trigonal planar | 0 | Trigonal planar | 120° | $\ce{BF3}$ |
| 3 | Trigonal planar | 1 | Bent | <120° | $\ce{SO2}$ |
| 4 | Tetrahedral | 0 | Tetrahedral | 109.5° | $\ce{CH4}$ |
| 4 | Tetrahedral | 1 | Trigonal pyramidal | 107° | $\ce{NH3}$ |
| 4 | Tetrahedral | 2 | Bent | 104.5° | $\ce{H2O}$ |
| 5 | Trigonal bipyramidal | 0 | Trigonal bipyramidal | 120° & 90° | $\ce{PCl5}$ |
| 5 | Trigonal bipyramidal | 1 | See-saw | <90°, <120° | $\ce{SF4}$ |
| 5 | Trigonal bipyramidal | 2 | T-shape | <90° | $\ce{ClF3}$ |
| 5 | Trigonal bipyramidal | 3 | Linear | 180° | $\ce{XeF2}$ |
| 6 | Octahedral | 0 | Octahedral | 90° | $\ce{SF6}$ |
| 6 | Octahedral | 1 | Square pyramidal | <90° | $\ce{BrF5}$ |
The CH4 > NH3 > H2O trend
The most heavily examined consequence of VSEPR is the bond-angle order across $\ce{CH4}$, $\ce{NH3}$ and $\ce{H2O}$. All three have four electron pairs around the central atom, so the parent geometry is tetrahedral in every case. What differs is the number of lone pairs.
| Molecule | Bond pairs | Lone pairs | Shape | Bond angle | Dominant repulsion |
|---|---|---|---|---|---|
| $\ce{CH4}$ | 4 | 0 | Tetrahedral | 109.5° | bp–bp only |
| $\ce{NH3}$ | 3 | 1 | Trigonal pyramidal | 107° | lp–bp |
| $\ce{H2O}$ | 2 | 2 | Bent | 104.5° | lp–lp & lp–bp |
With no lone pair, $\ce{CH4}$ keeps the ideal 109.5°. The single lone pair on $\ce{NH3}$ presses the three $\ce{N-H}$ bonds together through the stronger lp–bp repulsion, cutting the angle to 107°. The two lone pairs on $\ce{H2O}$ add lp–lp repulsion on top of lp–bp, squeezing the $\ce{H-O-H}$ angle further to 104.5°. The trend $\ce{CH4} > \ce{NH3} > \ce{H2O}$ is a direct printout of the repulsion order lp–lp $>$ lp–bp $>$ bp–bp.
Geometry ≠ shape, and bond angle never increases with lone pairs
Two traps fire together here. First, candidates report $\ce{NH3}$ and $\ce{H2O}$ as "tetrahedral" — that is the geometry, but the shapes are pyramidal and bent. Second, the bond-angle order is sometimes inverted. NEET 2016 deliberately offered the false statement "the H–O–H angle in $\ce{H2O}$ is larger than the H–C–H angle in $\ce{CH4}$" as a distractor.
More lone pairs ⇒ stronger compression ⇒ smaller angle. The order is fixed: $\ce{CH4}\,(109.5°) > \ce{NH3}\,(107°) > \ce{H2O}\,(104.5°)$.
ClF3 and XeF2 explained
The trigonal bipyramidal family is where VSEPR earns its keep, because axial and equatorial positions are not equivalent. An equatorial site has only two close neighbours at 90°, while an axial site has three. Since lone pairs demand the most room and 90° repulsions dominate the energy, lone pairs always go equatorial to minimise the number of strong lp–bp contacts.
Predict the shapes of $\ce{SF4}$, $\ce{ClF3}$ and $\ce{XeF2}$ using VSEPR.
Each central atom is $\ce{sp^3d}$-type with five electron pairs, so the parent geometry is trigonal bipyramidal in all three. $\ce{SF4}$ has four bond pairs and one lone pair ($\ce{AB4E}$): the equatorial lone pair gives a see-saw shape. $\ce{ClF3}$ has three bond pairs and two lone pairs ($\ce{AB3E2}$): both lone pairs equatorial give a T-shape. $\ce{XeF2}$ has two bond pairs and three lone pairs ($\ce{AB2E3}$): all three lone pairs equatorial leave the two $\ce{F}$ atoms axial, producing a linear shape.
The lone-pair count also governs which molecule suffers the worst lp–lp repulsion. Among $\ce{IF5}$, $\ce{SF4}$, $\ce{XeF2}$ and $\ce{ClF3}$, the central atom carries one, one, three and two lone pairs respectively, so $\ce{XeF2}$ has the maximum lp–lp repulsion — the answer NEET 2022 was testing.
Counting lone pairs on the central atom
Shape questions stand or fall on the lone-pair count. $\ce{ClF3}$ has two lone pairs on chlorine (NEET 2018), not one; $\ce{XeF2}$ has three. Always derive the count from the total valence electrons minus the bonding electrons, then assign equatorial positions before reading off the shape.
Five electron pairs: 1 lone pair ⇒ see-saw, 2 ⇒ T-shape, 3 ⇒ linear.
VSEPR in one screen
- Shape depends on the number of valence-shell electron pairs around the central atom; pairs arrange to minimise repulsion.
- Repulsion order: lp–lp $>$ lp–bp $>$ bp–bp, because a lone pair occupies more space than a shared bond pair.
- Geometry counts all electron pairs; shape counts only bonded atoms. They coincide only when there are no lone pairs.
- No-lone-pair series: linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral for 2–6 pairs.
- Lone pairs shrink bond angles: $\ce{CH4}\,(109.5°) > \ce{NH3}\,(107°) > \ce{H2O}\,(104.5°)$.
- In trigonal bipyramids lone pairs go equatorial: $\ce{SF4}$ see-saw, $\ce{ClF3}$ T-shape, $\ce{XeF2}$ linear.