Predict Molecule Geometry with VSEPR

JEE and the medical exams ask about molecule geometry every single year. So you have to be quick at it. The good news? You don't need hybridisation. You need one formula and two short tables.

Let me show you the method I give my students. By the end you'll read the shape of a molecule straight off the page.

First, get one word straight: geometry means two things — structure and shape. They are not the same, and the difference is exactly what trips people up. Hold that thought; we'll come back to it.

Which molecules even have a geometry?

Only covalent ones.

There is no such thing as the shape of an ionic compound. Why? A covalent bond points in a direction — it holds atoms at fixed angles in space. An ionic bond doesn't; it's just opposite charges pulling on each other from every side. No direction, no shape.

So from here on, every molecule we touch is covalent.

What VSEPR is really saying

Three facts. Keep them in your head the whole time.

• Only the valence electrons of the central atom decide the geometry.
• Those electrons sit in two forms — bond pairs and lone pairs.
• Both kinds push the molecule into its structure. But when you name the shape, the lone pairs become invisible.

That last line is the whole trick. The lone pairs are still there, still shoving the bonds around. You just don't draw them when you call out the shape. So a molecule's shape is its structure with the lone pairs taken out.

The one formula

Here is the move that does all the work. For the central atom, find a number we'll call $N$:

$N = \frac{V + M - c}{2}$

Walk through it slowly:

• $V$ = valence electrons of the central atom.
• $M$ = number of atoms bonded to it by a single bond. Atoms that form a double or triple bond do not count here.
• $c$ = the charge. Add 1 for each negative charge, subtract 1 for each positive charge. (In the formula above, $c$ is positive for a positive ion — so it gets subtracted.)
• Divide by 2, because each pair holds two electrons.

The $N$ you get is the total number of electron pairs around the central atom. That number gives you the structure:

$N$	Structure	Bond angle
2	Linear	$180°$
3	Trigonal planar	$120°$
4	Tetrahedral	$109.5°$
5	Trigonal bipyramidal	$120°$ and $90°$
6	Octahedral	$90°$

Then one subtraction gives you the lone pairs:

$\text{lone pairs} = N - M$

Knock those lone pairs out of the structure and you're left with the shape. Done.

See the clean cases first

When there are zero lone pairs, structure and shape are the same molecule. These are the ones to picture before anything else:

Methane: carbon has $V = 4$, four single-bonded H atoms, no charge. $N = (4+4)/2 = 4$, lone pairs $= 4 - 4 = 0$. Tetrahedral, $109.5°$. BF₃: $N = 3$, no lone pairs — trigonal planar, $120°$. CO₂: the oxygens are double-bonded, so $M = 0$, $N = (4+0)/2 = 2$ — linear, $180°$.

Now watch what lone pairs do.

Ammonia: nitrogen has $V = 5$, three single-bonded H, no charge. $N = (5+3)/2 = 4$ → the structure is tetrahedral. But lone pairs $= 4 - 3 = 1$. Take that one lone pair out of a tetrahedron and the shape is pyramidal.

Water: oxygen has $V = 6$, two H, no charge. $N = (6+2)/2 = 4$ → tetrahedral structure again. Lone pairs $= 4 - 2 = 2$. Take out two and the shape is bent.

Same $N$, same structure, three different shapes — and the only thing that changed was the lone-pair count. That's why you keep them separate.

Worked example 1 — PCl₄⁺

What is the shape of the $PCl_4^+$ ion?

• Central atom: P, so $V = 5$.
• Four Cl atoms, each a single bond: $M = 4$.
• Charge is $+1$, so subtract 1.

$N = \frac{5 + 4 - 1}{2} = 4$

Structure is tetrahedral. Lone pairs $= 4 - 4 = 0$. No lone pairs to remove, so the shape is tetrahedral too. $109.5°$.

Worked example 2 — IF₅

Predict the shape of $IF_5$.

• Central atom: I, so $V = 7$.
• Five F atoms, single bonds: $M = 5$.
• No charge.

$N = \frac{7 + 5}{2} = 6$

Structure is octahedral. Lone pairs $= 6 - 5 = 1$. Pull that one lone pair out of an octahedron and the shape is square pyramidal.

Your turn. Find the shape of $SF_4$. Sulphur has 6 valence electrons, four single-bonded fluorines, no charge.

Check: $N = (6+4)/2 = 5$, so the structure is trigonal bipyramidal. Lone pairs $= 5 - 4 = 1$. Remove it and the shape is see-saw.

The whole method in four lines

• Geometry = structure (with lone pairs) + shape (lone pairs removed).
• $N = \dfrac{V + M - c}{2}$. Count only single-bonded atoms in $M$.
• Read $N$ off the table: 2 linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral.
• Lone pairs $= N - M$. Take them out, and what's left is the shape.

That's it. No hybridisation, no orbital diagrams — just one number and one subtraction. Try it on the next ten molecules you meet, and watch how fast it gets.