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VSEPR questions are short, but they are easy to get wrong in a hurry: SF₄ drawn as a tetrahedron, ClF₃ called trigonal planar, XeF₄ given the wrong shape. A fixed four-step method avoids nearly all of these. This post sets out the method, works through the cases JEE and NEET like to ask, and lists the common slips.
What does VSEPR actually say?
The NCERT Class 11 chapter on Chemical Bonding and Molecular Structure gives the main ideas. Electron pairs around a central atom repel each other and spread out to minimise repulsion. A multiple bond is treated as a single domain. Lone pairs are held only by the central atom, so they take up more space, and repulsions fall in the order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.
NCERT credits Nyholm and Gillespie with refining the model; their review, Inorganic Stereochemistry (1957), is the classic statement. The same NCERT section notes that the model's theoretical basis is still debated, even though its predictions work well for p-block compounds. In a later teaching paper, Gillespie (2004) recommends teaching VSEPR at an empirical level first, before valence bond and molecular orbital theories. That is the order to learn it in, too.
Both the JEE (Main) 2026 syllabus and the NEET (UG)-2026 syllabus list "VSEPR theory and shapes of simple molecules".
What are the four steps?
Count domains on the central atom. Domains = atoms bonded to it + lone pairs on it. A quick count that works when the outer atoms are hydrogen, halogens or oxygen: domains = ½(V + M + A − C), where V is the central atom's valence electrons, M the number of hydrogen or halogen atoms bonded to it (oxygen counts zero), A the negative charge and C the positive charge. Lone pairs = domains − bonded atoms.
Choose the electron arrangement. 2 linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral.
Place the lone pairs. In a trigonal bipyramid they go in equatorial positions, where they have fewer 90° neighbours. In an octahedron, two lone pairs sit opposite each other.
Name the shape from the atoms only, then adjust angles: each lone pair squeezes the bond angles a little.
How do the steps work on real molecules?
ClF₃. Domains = ½(7 + 3) = 5. Three are bonds, so two are lone pairs. Five domains make a trigonal bipyramid. Both lone pairs go equatorial, leaving the three fluorines in a T. Shape: T-shaped, with F–Cl–F angles a little under 90°.
SF₄. Domains = ½(6 + 4) = 5, with four bonds and one lone pair, which goes equatorial. NCERT explains why: in an axial position the lone pair would have three repulsions at 90°, in an equatorial position only two. Shape: see-saw.
XeF₄. Domains = ½(8 + 4) = 6: four bonds and two lone pairs. Octahedral arrangement, lone pairs opposite each other. Shape: square planar.
XeF₂ and I₃⁻. Domains = ½(8 + 2) = 5 and ½(7 + 2 + 1) = 5, each with two bonds and three lone pairs, all equatorial. The atoms are left in a straight line: linear.
SO₂. Domains = ½(6 + 0) = 3, since oxygen counts zero. Two are the S=O bonds, each double bond counting once, and one is a lone pair. Trigonal planar arrangement, bent shape, angle below 120°.
Why are the angles in NH₃ and H₂O smaller than 109.5°?
Methane, ammonia and water all have 4 domains, so all start from the tetrahedral angle of 109.5°. Experimental geometries in the NIST Computational Chemistry Comparison and Benchmark Database put the H–N–H angle in ammonia at 106.67° and the H–O–H angle in water at 104.48°. One lone pair squeezes the angle a little; two squeeze it more. The model predicts the order of angles, not their exact values, so use it to rank angles rather than to calculate them.
What do students find hard about shapes?
Furió and Calatayud (1996) analysed what grade 12 and university students need to know and be able to do about molecular geometry and polarity, and diagnosed the reasoning difficulties that block them. A recurring one is skipping the geometry and guessing polarity from the formula. Shape comes first: CO₂ is linear, so its bond dipoles cancel and it is non-polar, while bent H₂O is polar. XeF₄, square planar, is non-polar for the same reason.
Which mistakes cost marks?
Ignoring lone pairs on the central atom, which turns SF₄ into a tetrahedron.
Counting a double bond as two domains, which gives SO₂ the wrong arrangement.
Putting lone pairs axial in a trigonal bipyramid.
Naming the electron arrangement instead of the shape.
Forgetting the charge in ions such as NH₄⁺, where ½(5 + 4 − 1) = 4 domains gives a tetrahedron, or I₃⁻.
Stereochemistry in organic chemistry builds on the same 3D thinking; see our post on stereoisomerism in organic chemistry.
A practice routine
Make a list of 20 molecules and ions: BeCl₂, BF₃, CH₄, NH₃, H₂O, PCl₅, SF₄, ClF₃, XeF₂, SF₆, BrF₅, XeF₄, SO₂, SO₃, NH₄⁺, I₃⁻, NO₂⁻, CO₃²⁻, H₃O⁺, CO₂.
Each day, take five and write the four steps for each, with a sketch.
For each one, add a line on polarity.
On day five, do all twenty from memory against the clock, then check.
Shapes are easier to judge when you can turn a molecule around. Our post on molecule visualisation tools covers options, and the Fermi Concept Library has every concept in chemistry, each with its own lesson and its place in the map, if the idea of electron domains itself is shaky.
Today, run BrF₅ and NO₂⁻ through the four steps and check whether each is polar.
