VSEPR theory — predict shapes of NH₃, H₂O, BF₃, PCl₅, SF₆

Question

Using VSEPR theory, predict the shapes and bond angles of $\text{NH}_3$ , $\text{H}_2\text{O}$ , $\text{BF}_3$ , $\text{PCl}_5$ , and $\text{SF}_6$ .

(NCERT Class 11, Chapter 4 — this is a must-know for boards and entrance exams)

Solution — Step by Step

For each molecule: (1) count total electron pairs around the central atom (bonding + lone pairs), (2) determine the electron pair geometry (how ALL pairs arrange), (3) determine the molecular shape (considering only atoms, not lone pairs).

The key principle: electron pairs repel each other and arrange to maximise the distance between them.

Molecule	Bond pairs	Lone pairs	Electron geometry	Molecular shape	Bond angle
$\text{BF}_3$	3	0	Trigonal planar	Trigonal planar	120°
$\text{NH}_3$	3	1	Tetrahedral	Trigonal pyramidal	107°
$\text{H}_2\text{O}$	2	2	Tetrahedral	Bent (V-shape)	104.5°
$\text{PCl}_5$	5	0	Trigonal bipyramidal	Trigonal bipyramidal	90°, 120°
$\text{SF}_6$	6	0	Octahedral	Octahedral	90°

In $\text{NH}_3$ , the lone pair occupies more space than a bonding pair (it spreads out more since it is held by only one nucleus). This extra repulsion compresses the N-H bonds closer together: bond angle drops from the ideal 109.5° (tetrahedral) to 107°.

In $\text{H}_2\text{O}$ , two lone pairs compress the bond angle even further to 104.5°. The trend: $\text{CH}_4$ (109.5°) > $\text{NH}_3$ (107°) > $\text{H}_2\text{O}$ (104.5°) — each lone pair reduces the angle by about 2-2.5°.

Why This Works

VSEPR theory is based on a simple physical idea: negatively charged electron pairs repel each other and adopt the arrangement that minimises repulsion. The repulsion order is: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.

This explains why lone pairs distort the geometry. In $\text{NH}_3$ , the lone pair pushes the three bonding pairs closer together, giving a pyramidal shape instead of flat trigonal. In $\text{H}_2\text{O}$ , two lone pairs push even harder, giving a bent shape.

Alternative Method

You can also use hybridisation to predict shapes: $sp$ = linear, $sp^2$ = trigonal planar, $sp^3$ = tetrahedral, $sp^3d$ = trigonal bipyramidal, $sp^3d^2$ = octahedral. Both methods give the same results — use whichever feels more natural.

For JEE and NEET, memorise the shapes of these five molecules along with $\text{ClF}_3$ (T-shaped), $\text{XeF}_2$ (linear), $\text{XeF}_4$ (square planar), and $\text{IF}_5$ (square pyramidal). These cover all the common geometries tested. The trick is always: count lone pairs on the central atom.

Common Mistake

Students often confuse electron pair geometry with molecular shape. $\text{NH}_3$ has tetrahedral electron pair geometry but trigonal pyramidal molecular shape. The molecular shape describes the arrangement of atoms only (not lone pairs). When a question asks for “shape,” it means molecular shape. When it asks for “geometry of electron pairs,” include the lone pairs too.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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