GATE Group Theory — Assigning Point Groups Quickly
Point group assignment is one of the few GATE Chemistry skills that is purely procedural: if you follow the decision tree in the right order you cannot get it wrong, and if you skip a branch you almost certainly will. Everything downstream — reducible representations, SALCs, selection rules, whether a molecule can be chiral or polar — depends on getting the label right first. This guide gives the tree, then applies it to the structures that appear repeatedly.
The five symmetry operations
| Symbol | Operation | Note |
|---|---|---|
| E | Identity | Every molecule has it |
| Cn | Rotation by 360°/n | Highest n = principal axis |
| σ | Reflection | σh ⊥ principal axis; σv contains it; σd bisects two C₂ axes |
| i | Inversion through centre | Equivalent to S₂ |
| Sn | Rotation by 360°/n then reflect ⊥ that axis | S₁ = σ, S₂ = i |
The decision tree — follow it strictly in order
2. Two or more Cn with n > 2? → cubic group: Td, Oh or Ih
3. No Cn at all? → σ only: Cs · i only: Ci · neither: C₁
4. Find the principal Cn. Are there n C₂ axes perpendicular to it?
YES → σh? Dnh · n σd? Dnd · neither? Dn
NO → σh? Cnh · n σv? Cnv · S2n only? S2n · none? Cn
The single most common error is testing for mirror planes before testing for the perpendicular C₂ axes. That step decides between the C family and the D family, and everything after it is conditional on the answer.
Worked assignments — main group
H₂O (bent). Not linear, one C₂ through the oxygen. No C₂ perpendicular to it. Two σv (the molecular plane and the plane bisecting the H–O–H angle). → C2v, order 4.
NH₃ (pyramidal). Principal C₃, no ⊥ C₂ (the lone pair breaks it), three σv each containing one N–H. → C3v, order 6.
BF₃ (trigonal planar). C₃, three ⊥ C₂ along the B–F bonds, plus the molecular plane as σh. → D3h, order 12.
PF₅ (trigonal bipyramidal). Same logic as BF₃ — C₃, three ⊥ C₂ through the equatorial fluorines, σh the equatorial plane. → D3h.
SF₄ (see-saw). The equatorial lone pair leaves only one C₂ and two σv. → C2v. ClF₃ (T-shaped) is also C2v for the same reason.
XeF₄ (square planar). C₄, four ⊥ C₂, σh, and i. → D4h, order 16.
SF₆. Multiple C₄ and C₃ axes → cubic, with i present. → Oh, order 48.
CH₄. Four C₃ axes → cubic, but no i and no C₄. → Td, order 24.
CO₂ linear with i → D∞h. HCN linear without i → C∞v.
Worked assignments — the ones that catch people out
Allene, H₂C=C=CH₂. The two CH₂ groups are mutually perpendicular. There is a C₂ along the C–C–C axis and two more C₂ perpendicular to it, plus two σd that bisect them — but no σh. → D2d. Note there is an S₄ along the long axis, which is exactly why allene, despite looking "unsymmetrical", is achiral.
Ethane, staggered. C₃ along C–C, three ⊥ C₂, three σd, no σh. → D3d. Eclipsed ethane gains a σh and becomes D3h.
Ferrocene. Eclipsed conformer D5h; staggered conformer D5d. Both are examined.
Benzene. C₆, six ⊥ C₂ (three through opposite carbons, three through opposite bond midpoints), σh, i. → D6h, order 24.
trans-N₂F₂. Planar, C₂ perpendicular to the molecular plane through the centre, plus σh (the molecular plane) and i. No ⊥ C₂. → C2h. The cis isomer is C2v.
H₂O₂ (gauche, the actual gas-phase geometry). Only E and one C₂ survive. → C₂ — and because C₂ contains no Sn, this molecule is formally chiral, though rotation about the O–O bond interconverts the enantiomers too fast to resolve.
Coordination complexes
| Complex | Geometry | Point group |
|---|---|---|
| [Co(NH₃)₆]³⁺ | Octahedral, all ligands identical | Oh |
| trans-[CoCl₂(NH₃)₄]⁺ | Octahedral, Cl axial | D4h |
| cis-[CoCl₂(NH₃)₄]⁺ | Octahedral, Cl adjacent | C2v |
| fac-[MX₃Y₃] | Three X on one face | C3v |
| mer-[MX₃Y₃] | Three X meridional | C2v |
| [Ni(CN)₄]²⁻ | Square planar | D4h |
| [NiCl₄]²⁻ | Tetrahedral | Td |
| [Co(en)₃]³⁺ | Tris-chelate propeller | D₃ (chiral) |
Two rules that follow immediately from the label
Chirality. A molecule is chiral if and only if it possesses no improper axis Sn. Since S₁ = σ and S₂ = i, this covers the familiar "no plane, no centre" test but is stricter — it also rules out anything with an S₄, which is how allene and similar cases are caught. In practice: only C₁, Cn and Dn groups can be chiral.
Dipole moment. A permanent dipole must lie along every symmetry element. That is only possible in C₁, Cs, Cn and Cnv. Every D group, every cubic group, Cnh and S2n are necessarily non-polar. So NH₃ (C3v) is polar and BF₃ (D3h) is not, without calculating anything.
Common mistakes
- Assigning symmetry from a Lewis structure rather than the 3D geometry. Lone pairs remove symmetry elements even though they are invisible in the formula — this is what makes NH₃ C3v rather than D3h.
- Skipping the perpendicular-C₂ test. Deciding Cnv versus Dnh on the basis of "it looks symmetrical" fails on SF₄ and ClF₃ every time.
- Confusing σv with σd. Both contain the principal axis; a σd specifically bisects the angle between two perpendicular C₂ axes. This is the Dnh/Dnd decision.
- Forgetting conformation matters. Ethane and ferrocene have different point groups in different rotamers. The question must specify one; if it does not, say so.
- Equating "has a plane" with "achiral" and stopping there. The correct criterion is the absence of any Sn.
- Treating Td as having an inversion centre. It does not; Oh and Ih do.
Order of the common groups
| Group | Order h | Group | Order h |
|---|---|---|---|
| C₁ | 1 | D3h | 12 |
| Cs, Ci, C₂ | 2 | D4h | 16 |
| C2v, C2h | 4 | D6h | 24 |
| C3v | 6 | Td | 24 |
| D3, D3d | 6, 12 | Oh | 48 |
The order is the total number of operations, and it is what you divide by in the reduction formula ni = (1/h) Σ N(R) χ(R) χi(R). Getting h wrong scales every irreducible representation coefficient wrongly, so it is worth checking against this table.
Working through the reduction formula? The summation, the division by the group order and the final integer check are all arithmetic you should verify rather than trust — a non-integer coefficient means an error somewhere in the reducible representation.
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