VSEPR Theory & Molecular Geometry

The Idea Behind VSEPR

The octet rule tells us which atoms bond, but says nothing about the shape of the molecule. To predict shapes, we use VSEPR theory (Valence Shell Electron Pair Repulsion), proposed by Sidgwick and Powell and developed by Gillespie and Nyholm.

Core idea: The electron pairs in the valence shell of the central atom repel one another, so they arrange themselves as far apart as possible to minimise repulsion. This arrangement decides the molecular shape.

Postulates

1. The shape depends on the number of electron pairs (bonding + lone) around the central atom.
2. Electron pairs arrange to minimise repulsion (maximise separation).
3. Lone pairs occupy more space than bond pairs (they are held by only one nucleus).
4. The repulsion order is: $$lp\text{-}lp > lp\text{-}bp > bp\text{-}bp$$
5. Multiple bonds are treated as a single 'super pair' for geometry, but they repel more strongly than single bonds.

[JEE Tip] Distinguish electron-pair geometry (counts all pairs) from molecular shape (counts only the atoms). $H_2O$ has tetrahedral electron geometry but a bent shape.

The AX$_n$E$_m$ Notation

We label molecules by A (central atom), X (number of bonded atoms), and E (number of lone pairs on the central atom). The sum $n + m$ = steric number = total electron pairs.

Type	Pairs	Shape	Example	Angle
$AX_2$	2	Linear	$BeCl_2$, $CO_2$	$180^\circ$
$AX_3$	3	Trigonal planar	$BF_3$	$120^\circ$
$AX_2E$	3	Bent	$SO_2$	$\approx 119^\circ$
$AX_4$	4	Tetrahedral	$CH_4$	$109.5^\circ$
$AX_3E$	4	Trigonal pyramidal	$NH_3$	$107^\circ$
$AX_2E_2$	4	Bent	$H_2O$	$104.5^\circ$
$AX_5$	5	Trigonal bipyramidal	$PCl_5$	$120^\circ, 90^\circ$
$AX_6$	6	Octahedral	$SF_6$	$90^\circ$
$AX_7$	7	Pentagonal bipyramidal	$IF_7$	$72^\circ, 90^\circ$

Key Point: Count the steric number = (bonded atoms) + (lone pairs). That number fixes the electron geometry; subtracting lone pairs gives the actual molecular shape.

[NEET Important] Learn this table cold — it answers a huge fraction of VSEPR exam questions directly.

How Lone Pairs Distort Shapes

Lone pairs take up more room than bond pairs, so they squeeze the bond angles smaller and choose the least-crowded positions.

The $CH_4 \to NH_3 \to H_2O$ series

All three have 4 electron pairs (tetrahedral electron geometry), but:

• $CH_4$ (0 lp): perfect tetrahedron, $109.5^\circ$.
• $NH_3$ (1 lp): the lone pair pushes the bonds together → pyramidal, $107^\circ$.
• $H_2O$ (2 lp): two lone pairs push harder → bent, $104.5^\circ$.

Lone pairs in trigonal bipyramidal molecules

In $AX_nE_m$ systems based on a trigonal bipyramid, lone pairs usually occupy the roomier equatorial positions (where they have only two $90^\circ$ neighbours instead of three):

• $SF_4$ ($AX_4E$): see-saw shape.
• $ClF_3$ ($AX_3E_2$): T-shape.
• $XeF_2$ ($AX_2E_3$): linear.

In octahedral systems

• $BrF_5$ ($AX_5E$): square pyramidal.
• $XeF_4$ ($AX_4E_2$): square planar (the two lone pairs go opposite each other).

[JEE Tip] In a trigonal bipyramid, lone pairs generally go equatorial. In an octahedron, two lone pairs go trans (opposite) to each other → square planar.

Worked Shape Predictions & Bond-Angle Rules

Let's practise the full method: count valence electrons → find steric number → assign lone pairs → name the shape.

Bond-angle modifiers

1. More lone pairs → smaller bond angle ($CH_4 > NH_3 > H_2O$).
2. More electronegative central atom → larger bond angle ($NH_3\ 107^\circ > PH_3\ 93.5^\circ$, because the bond pairs sit closer to the more electronegative N, increasing bp-bp repulsion).
3. More electronegative surrounding atom → smaller bond angle ($NH_3\ 107^\circ > NF_3\ 102^\circ$, because the electrons are pulled away from N).
4. Multiple bonds repel more than single bonds, slightly opening the angle they make.

Quick examples

• $SO_2$: S has 1 lone pair, 2 bonds → bent, $\approx 119^\circ$.
• $XeF_4$: 6 pairs (4 bonds + 2 lp) → square planar.
• $I_3^-$: central I has 5 pairs (2 bonds + 3 lp) → linear.

Key Point: Bond angle increases with central-atom electronegativity but decreases with surrounding-atom electronegativity and with the number of lone pairs.

[NEET Important] $XeF_4$ is square planar and $XeF_2$ is linear — two of the most-asked noble-gas-compound shapes.

Solved Examples

Example 1: Shape of methane

Predict the shape and bond angle of $CH_4$ using VSEPR.

Solution:

1. Steric number: 4 bond pairs + 0 lone pairs = 4.
2. Electron geometry: tetrahedral.
3. No lone pairs, so molecular shape = tetrahedral.
4. Bond angle: $109.5^\circ$.

Takeaway: $AX_4$ with no lone pairs is a perfect tetrahedron.

Example 2: Shape of water

Why is $H_2O$ bent and not linear?

Solution:

1. Steric number: 2 bond pairs + 2 lone pairs = 4 ($AX_2E_2$).
2. Electron geometry: tetrahedral.
3. Two lone pairs occupy two corners, leaving the two O–H bonds at the other two.
4. Shape: bent, with the angle compressed to $104.5^\circ$ by lone-pair repulsion.

Takeaway: Lone pairs are 'invisible' atoms — they shape the molecule but aren't counted in its name.

Example 3: Shape of ammonia

Predict the shape of $NH_3$.

Solution:

1. Steric number: 3 bond pairs + 1 lone pair = 4 ($AX_3E$).
2. Electron geometry: tetrahedral.
3. One lone pair pushes the three N–H bonds down → trigonal pyramidal.
4. Bond angle: $107^\circ$ (compressed from $109.5^\circ$).

Takeaway: One lone pair turns a tetrahedral arrangement into a pyramid.

Example 4: Shape of PCl₅

Predict the geometry of $PCl_5$.

Solution:

1. Steric number: 5 bond pairs + 0 lone pairs = 5 ($AX_5$).
2. Geometry: trigonal bipyramidal.
3. Bond angles: $120^\circ$ (equatorial-equatorial) and $90^\circ$ (axial-equatorial).
4. Note: axial bonds ($219\,pm$) are longer than equatorial ($204\,pm$).

Takeaway: $AX_5$ = trigonal bipyramidal, with two distinct bond angles and bond lengths.

Example 5: Shape of SF₆

Predict the geometry of $SF_6$.

Solution:

1. Steric number: 6 bond pairs + 0 lone pairs = 6 ($AX_6$).
2. Geometry: octahedral.
3. All bond angles $= 90^\circ$, all S–F bonds equivalent.

Takeaway: $AX_6$ with no lone pairs is a symmetric octahedron, all angles $90^\circ$.

Example 6: Shape of SF₄

Predict the shape of $SF_4$.

Solution:

1. Steric number: 4 bond pairs + 1 lone pair = 5 ($AX_4E$).
2. Electron geometry: trigonal bipyramidal.
3. Lone pair goes equatorial (least repulsion).
4. Shape: see-saw.

Takeaway: In TBP systems, the lone pair generally takes an equatorial site → see-saw shape for $AX_4E$.

Example 7: Shape of XeF₄

Predict the geometry of $XeF_4$.

Solution:

1. Valence electrons on Xe: 8; four go into Xe–F bonds, leaving 4 → 2 lone pairs.
2. Steric number: 4 bonds + 2 lone pairs = 6 ($AX_4E_2$).
3. Electron geometry: octahedral; the two lone pairs sit trans (opposite).
4. Shape: square planar.

Takeaway: $AX_4E_2$ is square planar — the two lone pairs cancel out above and below the plane.

Example 8: Shape of ClF₃

Predict the shape of $ClF_3$.

Solution:

1. Cl valence electrons: 7; three in Cl–F bonds, leaving 4 → 2 lone pairs.
2. Steric number: 3 bonds + 2 lone pairs = 5 ($AX_3E_2$).
3. Electron geometry: trigonal bipyramidal; both lone pairs go equatorial.
4. Shape: T-shaped.

Takeaway: $AX_3E_2$ = T-shaped, with lone pairs occupying two equatorial positions.

Example 9: Bond-angle comparison NH₃ vs NF₃

Why is the bond angle in $NH_3$ ($107^\circ$) larger than in $NF_3$ ($102^\circ$)?

Solution:

1. Both are $AX_3E$ (pyramidal, 1 lone pair).
2. In $NF_3$: F is very electronegative and pulls the bonding electrons away from N, reducing bp-bp repulsion.
3. Smaller bp-bp repulsion → smaller bond angle.
4. Conclusion: $NH_3\ (107^\circ) > NF_3\ (102^\circ)$.

Takeaway: More electronegative surrounding atoms pull electron density away, shrinking the bond angle.

Example 10: Bond-angle comparison NH₃ vs PH₃

Why is the bond angle in $NH_3$ ($107^\circ$) larger than in $PH_3$ ($93.5^\circ$)?

Solution:

1. Both are $AX_3E$.
2. N is more electronegative than P, so the bonding pairs sit closer to N, increasing bp-bp repulsion → larger angle.
3. In $PH_3$, the bonds are further from P (less electronegative), so repulsion is less → smaller angle (closer to pure p-orbital $90^\circ$).

Takeaway: More electronegative central atom → larger bond angle.

Example 11: Shape of the triiodide ion I₃⁻

Predict the shape of $I_3^-$.

Solution:

1. Central I valence electrons: 7, plus 1 for the negative charge = 8; two used in I–I bonds, leaving 6 → 3 lone pairs.
2. Steric number: 2 bonds + 3 lone pairs = 5 ($AX_2E_3$).
3. Electron geometry: trigonal bipyramidal; all three lone pairs go equatorial.
4. Shape: linear.

Takeaway: $AX_2E_3$ = linear (three equatorial lone pairs leave the two bonds axial).

Example 12: Why CO₂ is linear but SO₂ is bent

Both are $AX_2$-type triatomics. Explain the difference.

Solution:

1. $CO_2$: carbon has no lone pair ($AX_2$) → linear, $180^\circ$.
2. $SO_2$: sulphur has one lone pair ($AX_2E$) → bent, $\approx 119^\circ$.
3. Conclusion: the lone pair on S bends the molecule; its absence on C keeps $CO_2$ linear.

Takeaway: A single lone pair on the central atom is enough to bend a triatomic molecule.