Point groups and molecular symmetry

12 min read Updated 2026-09-19

Every molecule has a symmetry. That symmetry is a set of operations you can perform on the molecule (rotations, reflections, inversions) that leave every atom in an indistinguishable position. The complete set of such operations is called the point group of the molecule, and knowing it tells you three practical things without solving any equations: whether the molecule has a dipole moment, whether it is chiral, and how many vibrational modes will appear in its IR and Raman spectra.

This guide walks the five symmetry elements, the decision tree for assigning a point group, and three worked examples (H₂O, NH₃, XeF₄) that cover the common cases.

Prerequisites: 3D molecular structure. If MO theory for diatomic molecules is comfortable, symmetry is the tool that generalises the same MO-diagram reasoning to any molecule.

The five symmetry elements

Every point-group operation is one of these five (Cotton 3e Ch. 3):

E — the identity. Do nothing; every atom stays put. Every molecule has E.
Cₙ — proper rotation by 360°/n about an axis. Water has C₂ (rotate 180°). Ammonia has C₃ (rotate 120°). Benzene has C₆ (rotate 60°).
σ — reflection through a plane. Water has two mirror planes (both containing the C₂ axis).
i — inversion through a point. Every atom (x, y, z) maps to (−x, −y, −z). Benzene has i (through the ring centre); methane does not.
Sₙ — improper rotation by 360°/n followed by reflection through a plane perpendicular to the axis. Ethane in eclipsed form has S₃; methane has S₄.

There is also a special “principal axis” — the Cₙ with the largest n. Water’s principal axis is C₂; ammonia’s is C₃; benzene’s is C₆. Mirror planes are named relative to the principal axis:

σₕ (horizontal) — perpendicular to the principal axis.
σᵥ (vertical) — contains the principal axis.
σd (dihedral) — bisects two C₂ axes perpendicular to the principal axis (a special kind of vertical mirror).

The decision tree

Assigning a molecule to its point group is a five-step decision (Miessler 5e Fig. 4-2):

Step 1: Is the molecule linear? If yes, it is D∞h (has a mirror plane perpendicular to the axis: H₂, CO₂, N₂) or C∞v (no such mirror: HCl, HCN). Stop.

Step 2: Does it have multiple high-order rotation axes (>C₂)? If yes, it is one of the “special” groups: Td (tetrahedron: CH₄, CCl₄), Oh (octahedron: SF₆, PtCl₆²⁻), Ih (icosahedron: C₆₀, B₁₂H₁₂²⁻). Stop.

Step 3: Is there a principal Cₙ (n ≥ 2)? If no, check for a single mirror (Cs), inversion only (Ci), or nothing except E (C₁). Stop.

Step 4: With the principal Cₙ identified, are there n C₂ axes perpendicular to it? If yes, the point group is a “D” group. If no, it is a “C” or “S” group.

Step 5: Check for σₕ, σᵥ, and Sₙ. The exact combination names the point group:

C_n: only Cₙ, no mirrors.
Cnv: Cₙ plus n σᵥ planes.
Cnh: Cₙ plus σₕ.
Dn: Cₙ plus n C₂ axes, no σ.
Dnh: Cₙ plus n C₂ plus σₕ (and consequently σᵥ).
Dnd: Cₙ plus n C₂ plus n σd (dihedral mirrors), no σₕ.
S₂n: Sₙ but no Cₙ (rare).

Worked example — water (H₂O)

Point group of water

Formula
$Non-linear \to no multiple high-order axes \to C_{2} present$
Substitute
$No perpendicular C_{2} axes \to C group; 2 σ_{v} planes \to C_{2 v}$
Note
Water is the paradigm C₂ᵥ molecule. The C₂ axis passes through the oxygen and bisects the HOH angle. Two σᵥ planes contain that axis: one is the molecular plane (containing both H atoms), the other is perpendicular. No other symmetry elements — no σₕ, no i. Point group C₂ᵥ. Consequences: water has a dipole moment (aligned with the C₂ axis), is not chiral, and has 3 vibrational modes, all IR-active. This is why the H₂O IR spectrum shows exactly three fundamentals: symmetric stretch at 3657, antisymmetric stretch at 3756, bend at 1595 cm⁻¹.
Result
$H_{2} O \in C_{2 v}$

Worked example — ammonia (NH₃)

Point group of ammonia

Formula
$Non-linear \to no multiple high-order axes \to C_{3} present$
Substitute
$No perpendicular C_{2} axes \to C group; 3 σ_{v} \to C_{3 v}$
Note
Same reasoning as water, one rung up: three-fold rotation axis, three vertical mirror planes containing it, and no σₕ because there is no way to reflect through the horizontal plane that would give back the same molecule (the nitrogen sticks 'up' out of the plane of the three H atoms). Point group C₃ᵥ. Consequences: NH₃ has a dipole moment along C₃, is not chiral (achiral because of the σᵥ planes), and has 4 vibrational modes (2 fundamental frequencies × doubly-degenerate is 4 physical modes but 2 sets), all IR- and Raman-active.
Result
$N H_{3} \in C_{3 v}$

Worked example — XeF₄

Point group of square planar XeF₄

Formula
$Non-linear \to no multiple high-order axes \to C_{4} principal$
Substitute
$4 perpendicular C_{2} axes \to D group; σ_{h} present \to D_{4 h}$
Note
Square planar is the archetype D₄ₕ molecule. The C₄ principal axis is perpendicular to the plane; four C₂ axes lie in the plane; a σₕ is the plane itself; σᵥ and σd fill out the mirror population; inversion is through the central Xe. Consequences: no dipole moment (inversion centre — this is universal for centrosymmetric molecules), not chiral, and vibrational modes split cleanly between IR-active and Raman-active by the mutual exclusion rule. XeF₄'s symmetric stretch shows in Raman but not IR; the antisymmetric stretch shows in IR but not Raman. This is the mutual exclusion rule: molecules with an inversion centre have no vibrational mode that is simultaneously IR- and Raman-allowed.
Result
$X e F_{4} \in D_{4 h}$

Symmetry to physics — three predictions

Once you have the point group, three properties follow directly:

Dipole moment. A polar molecule has a nonzero dipole moment. Requirement: the point group must not have (a) an inversion centre, (b) a horizontal mirror plane perpendicular to the principal axis, or (c) more than one Cₙ axis or a perpendicular C₂ axis. The polar point groups are: C₁, Cs, Cₙ, Cnv, and C∞v. Every other point group has zero dipole moment — including all Dnh, Dnd, Oh, Td, Ih. Water (C₂ᵥ) is polar; benzene (D₆ₕ) is not.

Chirality. A chiral molecule has no improper rotation axis (no Sₙ, no σ, no i). Equivalently: the point group is Cₙ or Dₙ only. Every other point group contains at least one σ or Sₙ, and the molecule has an internal mirror that makes it superimposable on its mirror image. This is why “no mirror plane” from introductory organic chemistry is a special case of the more general symmetry rule.

Spectroscopic activity. A vibrational mode is IR-active if it transforms as x, y, or z (i.e., changes the dipole moment). It is Raman-active if it transforms as one of the quadratic functions (x², y², z², xy, xz, yz). The point-group character table tells you exactly which modes are which. For centrosymmetric molecules (Dnh with n even, Ih, Oh), the mutual exclusion rule says a mode cannot be both IR- and Raman-active — the two techniques give complementary information.

The mutual exclusion rule, quickly

If a molecule has an inversion centre (i in its point group), every vibrational mode is either “g” (gerade, symmetric under i) or “u” (ungerade, antisymmetric under i). IR selection rules require u modes (because dipole moment is odd under inversion); Raman selection rules require g modes (because polarisability is even under inversion). No mode can be both g and u, so no mode is both IR- and Raman-active.

Practically: for centrosymmetric molecules, count IR and Raman fundamentals separately, and their sum should equal 3N − 6 (total vibrational modes for a nonlinear molecule) or 3N − 5 (linear). Any coincidence between IR and Raman frequencies is either an artefact of overtones or evidence that the assumed structure is not centrosymmetric.

Common pitfalls

Confusing bent AX₂ with linear. Water is bent (C₂ᵥ, polar); CO₂ is linear (D∞h, non-polar). Both are AX₂E_n systems but the electron-pair count changes everything.

Assigning D∞h to slightly bent molecules. SO₂ is bent (C₂ᵥ), even though “sulfur is expanding octet.” Its S-O bonds are 120°-ish, not 180°.

Missing the principal axis. The principal axis is the highest-order Cₙ. Benzene has three C₂ axes and one C₆ — the C₆ is principal, not C₂. Getting this wrong flips the entire assignment.

Overlooking σₕ vs σᵥ vs σd. σₕ is perpendicular to the principal axis; σᵥ contains it (aligned with C₂ subaxes); σd bisects two σᵥ. Confusing them mis-classifies D-groups (Dnh vs Dnd).

Assuming symmetry means “looks symmetric”. BF₃ has D₃ₕ symmetry; NF₃ has C₃ᵥ. Both look “trigonal” but the pyramidal shape of NF₃ (nitrogen sticks out of the F-plane) breaks σₕ.

Practice

Problem 1 Easy What is the point group of methane (CH₄)?

Answer: Td. Methane has four C₃ axes (each along a C-H bond), three C₂ axes (bisecting H-H pairs), six σd planes, and three S₄ axes. Multiple high-order rotation axes (four C₃), so step 2 of the decision tree lands you at Td immediately. Consequences: no dipole moment, not chiral, and Raman/IR polarisation studies of methane's fundamentals were among the first experimental confirmations of the tetrahedral geometry.

Problem 2 Intermediate Predict the point group of a chair-conformation cyclohexane (all-carbon skeleton, ignoring the H atoms' individual positions).

Answer: D₃d. Chair cyclohexane has a C₃ axis (perpendicular to the average plane of the six carbons), three C₂ axes (each through a pair of opposite C-C midpoints), three σd planes (bisecting opposite C-C bonds), an S₆ axis, and an inversion centre. No σₕ (a horizontal mirror would map axial to equatorial, which are non-equivalent). The presence of inversion means chair cyclohexane has no dipole moment and obeys the mutual exclusion rule. This is why cyclohexane is symmetric enough to be a common NMR reference solvent — its axial and equatorial protons appear at different chemical shifts, but only because they are separately in different chemical environments; the ring itself is highly symmetric.

Problem 3 Hard Distinguish between trans-1,2-dichlorocyclohexane and cis-1,2-dichlorocyclohexane by point group. Which is chiral?

Symmetry-based chirality of 1,2-dichlorocyclohexanes

Formula
$trans has C_{2} through midpoints of C-C bonds, but no σ$
Substitute
$trans: point group C_{2}; cis: point group C_{s} (has σ plane)$
Note
For a single chair conformation: trans-1,2-dichlorocyclohexane has only E and a C₂ (through the two C-C midpoints between the ClCH-CHCl and the opposite side). No σ, no i, no S — so it belongs to the pure rotational group C₂ and is *chiral*. The cis isomer has a σ plane bisecting the C1-C2 bond and containing the two Cl atoms, so it is C_s — achiral. This is the standard 'trans dibromides are chiral, cis is meso' result for 1,2-disubstituted cyclohexanes, derived purely from point-group analysis. In practice, rapid ring-flipping in cis-1,2-dichlorocyclohexane averages the two half-chair conformations into an achiral time-averaged structure.
Result
$trans is C_{2} (chiral); cis is C_{s} (achiral, meso-like)$

Where this leads

Symmetry is the entrance ramp to real inorganic and physical chemistry:

Character tables — every point group has a table specifying how each symmetry element affects each type of function (x, y, z, x², xy, etc.), and reading these predicts every spectral selection rule.
SALC construction — symmetry-adapted linear combinations of atomic orbitals build the MO diagrams of polyatomics that we hand-waved through in MO theory.
Selection rules in spectroscopy — IR, Raman, UV-vis, EPR, and NMR all have symmetry-based rules for which transitions are allowed.
Crystallography — space groups extend point groups to include translation, giving the 230 possible 3D crystal symmetries.

Every one of them starts by listing the symmetry elements of the molecule.

Sources

Cotton, F. A., Chemical Applications of Group Theory, 3rd ed. (1990) — the canonical inorganic reference
Miessler, Fischer & Tarr, Inorganic Chemistry, 5th ed. (2014), Ch. 4
Housecroft & Sharpe, Inorganic Chemistry, 5th ed. (2018), Ch. 3
Vincent, Molecular Symmetry and Group Theory, 2nd ed. (2001) — accessible introduction
IUPAC Recommendations on Notation for Point Groups (Schoenflies system)