CSIR-NET Group Theory and Molecular Symmetry — Reduction and Selection Rules

By Aniket Bhardwaj · 13 September 2026 · CSIR-NET Chemistry

Assigning a point group is the easy half of molecular symmetry, and most students stop there. CSIR-NET asks for the other half: taking a reducible representation, breaking it into irreducible components, and turning that into a testable statement about a spectrum. This guide assumes you can already reach the point group and concentrates on what comes next — characters, the reduction formula, direct products, and the selection rules that decide whether a vibration appears in the infrared, in the Raman spectrum, in both, or in neither. Every reduction below is carried out in full.

A one-paragraph recap of point groups

The five symmetry elements are the identity E, proper axes C_n, mirror planes σ (σ_v contains the principal axis, σ_h is perpendicular to it, σ_d bisects two C₂ axes), the inversion centre i, and improper axes S_n. Collect the elements a molecule possesses and the point group follows: H₂O is C_2v, NH₃ is C_3v, BF₃ is D_3h, CH₄ is T_d, SF₆ is O_h, CO₂ is D_∞h, HCl is C_∞v, and a molecule with no element other than E is C₁. Two consequences are worth carrying in your head: a molecule is chiral only if it has no improper axis (no σ, no i, no S_n), and it can be polar only if it belongs to C₁, C_s, C_n or C_nv.

The four rules a character table encodes

A character table is a compressed statement of the Great Orthogonality Theorem. You do not need to prove the theorem for the exam, but you must be able to use its consequences to check your own work.

The number of irreducible representations equals the number of classes of operations.
The sum of the squares of the dimensions equals the order of the group: Σ l_i² = h.
Rows are orthogonal, and each row's characters squared, weighted by the number of operations in the class, sum to h.
Mulliken labels: A means symmetric under the principal rotation, B antisymmetric, E is doubly degenerate, T triply degenerate; subscripts g and u mean symmetric or antisymmetric to inversion; 1 and 2 refer to a perpendicular C₂ or a σ_v.

C_2v	E	C₂	σ_v(xz)	σ_v′(yz)	Linear	Quadratic
A₁	1	1	1	1	z	x², y², z²
A₂	1	1	−1	−1	R_z	xy
B₁	1	−1	1	−1	x, R_y	xz
B₂	1	−1	−1	1	y, R_x	yz

Check it: four classes, four irreducible representations; 1² + 1² + 1² + 1² = 4 = h. Both rules hold.

Characters are traces — and unmoved atoms do all the work

Each symmetry operation can be written as a matrix acting on the 3N Cartesian displacement vectors of the molecule. The character χ(R) is the trace of that matrix. Any atom that moves to a different position contributes zero to the trace, so only atoms left in place count — which reduces a 9 × 9 matrix problem to counting.

χ(R) = (number of unmoved atoms) × (contribution per unmoved atom)
proper rotation C_n: 1 + 2cos(360°/n) · improper operation S_n: −1 + 2cos(360°/n)

Operation	E	C₂	C₃	C₄	C₆	σ (= S₁)	i (= S₂)	S₃	S₄	S₆
Contribution per unmoved atom	3	−1	0	1	2	1	−3	−2	−1	0

The reduction formula

n_i = (1/h) Σ_classes g_c · χ(R) · χ_i(R)

h is the order of the group, g_c the number of operations in that class, χ(R) the character of the reducible representation and χ_i(R) that of the irreducible one. Every n_i must come out a non-negative integer — if it does not, you have made an arithmetic error, and that is the most valuable self-check in this whole chapter.

Worked example 1 — the vibrations of H₂O (C_2v, h = 4). Convention: the molecule lies in the yz plane, so σ_v′(yz) is the molecular plane. (Books that place the molecule in xz get B₁ where we get B₂; the physics is identical, so always state your choice.)

Step 1 — count unmoved atoms and build Γ_total (3N = 9).
E: all 3 atoms unmoved → 3 × 3 = 9
C₂: only O unmoved → 1 × (−1) = −1
σ_v(xz), perpendicular to the molecule: only O lies in it → 1 × 1 = 1
σ_v′(yz), the molecular plane: all 3 atoms lie in it → 3 × 1 = 3
So Γ_total = (9, −1, 1, 3).

Step 2 — reduce.
n(A₁) = ¼[9(1) + (−1)(1) + 1(1) + 3(1)] = ¼[9 − 1 + 1 + 3] = 12/4 = 3
n(A₂) = ¼[9(1) + (−1)(1) + 1(−1) + 3(−1)] = ¼[9 − 1 − 1 − 3] = 4/4 = 1
n(B₁) = ¼[9(1) + (−1)(−1) + 1(1) + 3(−1)] = ¼[9 + 1 + 1 − 3] = 8/4 = 2
n(B₂) = ¼[9(1) + (−1)(−1) + 1(−1) + 3(1)] = ¼[9 + 1 − 1 + 3] = 12/4 = 3

Check: 3 + 1 + 2 + 3 = 9 = 3N. Good. So Γ_total = 3A₁ + A₂ + 2B₁ + 3B₂.

Step 3 — subtract translations and rotations, read off the character table:
Γ_trans = z + x + y = A₁ + B₁ + B₂
Γ_rot = R_z + R_y + R_x = A₂ + B₁ + B₂
Γ_vib = (3A₁ + A₂ + 2B₁ + 3B₂) − (A₁ + B₁ + B₂) − (A₂ + B₁ + B₂) = 2A₁ + B₂

That is three modes, exactly 3N − 6 = 9 − 6 = 3 for a non-linear molecule: the symmetric stretch and the bend (both A₁) and the antisymmetric stretch (B₂).

Worked example 2 — the vibrations of NH₃ (C_3v, h = 6). Classes: E, 2C₃, 3σ_v. Characters: A₁ = (1, 1, 1), A₂ = (1, 1, −1), E = (2, −1, 0).

Γ_total, 3N = 12:
E: 4 unmoved × 3 = 12
C₃: only N unmoved × 0 = 0
σ_v: each plane holds N and one H, so 2 unmoved × 1 = 2
Γ_total = (12, 0, 2).

Reduce, remembering g_c = 1, 2, 3:
n(A₁) = ⅙[1(12)(1) + 2(0)(1) + 3(2)(1)] = ⅙[12 + 0 + 6] = 18/6 = 3
n(A₂) = ⅙[1(12)(1) + 2(0)(1) + 3(2)(−1)] = ⅙[12 + 0 − 6] = 6/6 = 1
n(E) = ⅙[1(12)(2) + 2(0)(−1) + 3(2)(0)] = ⅙[24 + 0 + 0] = 24/6 = 4

Dimension check: 3(1) + 1(1) + 4(2) = 3 + 1 + 8 = 12 ✓ — note that each E counts twice because it is two-dimensional. Missing this is the commonest slip in degenerate groups.

Γ_trans = z(A₁) + (x,y)(E) = A₁ + E; Γ_rot = R_z(A₂) + (R_x,R_y)(E) = A₂ + E.
Γ_vib = 3A₁ + A₂ + 4E − A₁ − E − A₂ − E = 2A₁ + 2E, i.e. 2 + 2(2) = 6 modes = 3N − 6 = 12 − 6 = 6 ✓.

Selection rules — turning symmetry into a spectrum

The intensity of a transition depends on an integral, and an integral over all space vanishes unless its integrand contains the totally symmetric representation. For a fundamental vibration starting from the (totally symmetric) ground state, that condition simplifies to two rules you can apply straight off the character table.

IR active if the mode transforms as x, y or z (the dipole components)
Raman active if the mode transforms as a quadratic function — x², y², z², xy, xz, yz (the polarisability components)

Molecule	Point group	Γ_vib	IR active	Raman active
H₂O	C_2v	2A₁ + B₂	All 3 (A₁ has z, B₂ has y)	All 3 (A₁ has x²,y²,z²; B₂ has yz)
NH₃	C_3v	2A₁ + 2E	All 4 (A₁ has z, E has x,y)	All 4 (A₁ and E both carry quadratics)
CO₂	D_∞h	σ_g⁺ + σ_u⁺ + π_u	σ_u⁺ and π_u only	σ_g⁺ only

CO₂ demonstrates the rule of mutual exclusion: in a molecule with a centre of symmetry, no vibration can be both IR and Raman active, because g modes carry the quadratics and u modes carry x, y, z. This is a genuinely useful structural tool. If a spectrum shows bands at the same wavenumber in both the IR and the Raman, the molecule has no inversion centre — which is how one distinguishes, for example, a centrosymmetric trans isomer from a cis one.

Note the linear-molecule count: CO₂ has 3N − 5 = 9 − 5 = 4 vibrations, and the two bends are a degenerate π_u pair, which is why only three distinct wavenumbers are observed.

Reducing a smaller basis — just the O–H stretches

You are not obliged to start from 3N. Choosing a basis of bonds, or of ligand orbitals, gives the same machinery with far less arithmetic. An unmoved bond contributes 1; a bond that moves contributes 0.

Worked example 3 — the two O–H stretching modes of water. Unmoved bonds: E → 2; C₂ → 0 (the two bonds swap); σ_v(xz) → 0 (they swap again); σ_v′(yz) → 2 (both lie in the plane). So Γ_stretch = (2, 0, 0, 2).

n(A₁) = ¼[2 + 0 + 0 + 2] = 1; n(A₂) = ¼[2 + 0 − 0 − 2] = 0; n(B₁) = ¼[2 − 0 + 0 − 2] = 0; n(B₂) = ¼[2 − 0 − 0 + 2] = 1.
Γ_stretch = A₁ + B₂ — the symmetric and antisymmetric O–H stretches.

Consistency check: Γ_vib was 2A₁ + B₂, so subtracting the stretches leaves A₁, which must be the bend. The two calculations agree, and that agreement is worth writing out in an answer script.

Direct products and the vanishing-integral rule

Multiply characters operation by operation to get the direct product. In C_2v:

B₁ ⊗ B₂ = (1,−1,1,−1) × (1,−1,−1,1) = (1, 1, −1, −1) = A₂
A₂ ⊗ B₁ = (1,1,−1,−1) × (1,−1,1,−1) = (1, −1, −1, 1) = B₂

Two rules follow. First, the product of any representation with itself contains the totally symmetric representation — which is why an integral of the form ∫ψ²dτ never vanishes by symmetry. Second, the transition ∫ψ_f μ̂ ψ_i dτ is allowed only if Γ_f ⊗ Γ(μ̂) ⊗ Γ_i contains the totally symmetric representation. Applied to electronic spectra this is the origin of the Laporte rule: in a centrosymmetric complex, d→d transitions are g→g and therefore forbidden, which is why octahedral complexes are pale while tetrahedral ones — having no inversion centre — are intensely coloured.

Common mistakes

Forgetting g_c, the number of operations in a class. In C_3v the reduction must weight C₃ by 2 and σ_v by 3.
Counting a degenerate irrep once in the dimension check. An E counts 2, a T counts 3.
Subtracting only translations. Rotations must go too — three of them for a non-linear molecule, two for a linear one, which is where 3N − 5 comes from.
Not stating the axis convention. H₂O gives 2A₁ + B₂ in the yz convention and 2A₁ + B₁ in the xz one; both are right, an unlabelled answer is not.
Reading "IR active" off the quadratic column. Linear functions give IR, quadratic functions give Raman.
Applying mutual exclusion to a molecule with no centre of symmetry. It only applies when i is present.
Accepting a fractional n_i. It is always an arithmetic error — go back rather than round it.

Exam relevance

Exam	Typical demand from symmetry and group theory
IIT-JAM	Identifying symmetry elements; chirality and polarity from symmetry
GATE Chemistry	Point group assignment, order of the group, number of IR-active modes
CSIR-NET	Full reduction of Γ, IR/Raman activity, mutual exclusion, direct products, Laporte and selection rules
CUET-PG	Symmetry elements and point group recognition

The CSIR-NET paper comprises a general aptitude Part A and subject Parts B and C. Question counts, marks and negative marking are stated in the current official notification — check it there.

Practise the reductions by hand, then verify the arithmetic. Each n_i is a short weighted sum, and the fastest way to catch a slip is to recompute it independently — the suite's scientific calculator is enough for that, and the matrix tool is useful when you want to work explicitly with the transformation matrices whose traces the characters are. Group theory has no dedicated tool in the suite, so this button honestly opens the suite itself rather than pretending otherwise.

Open the ABC Chemistry Calculator Suite →

Preparing for CSIR-NET, GATE, IIT-JAM or CUET-PG chemistry? ABC Chemistry runs dedicated competitive-exam batches at its coaching centre and fully online for students across India — details at abcchemistry.in.