Symmetry and Group Theory Notes, Point Groups, Character Tables & Short Tricks PDF for IIT JAM, BITSAT, CSIR NET & GATE

Symmetry & Group Theory

Complete Handwritten-Style Notes · Solved Examples · MCQs · 27 Practice Questions

📋 Table of Contents

1. Introduction — Why Symmetry Matters
2. Symmetry Elements & Operations
3. Point Groups — Assignment Algorithm
4. Low & High Symmetry Point Groups
5. Properties of Groups & Representations
6. Character Tables — Reading & Using Them
7. Chirality & Optical Activity
8. Molecular Vibrations & IR / Raman Spectroscopy
9. Solved Examples
10. MCQ Bank (Exam-Style)
11. Practice Question Set (27 Qs)
12. Exam Tips & Tricks

Why Does Symmetry Matter in Chemistry?

Think of a snowflake — it looks identical after every 60° rotation. That is symmetry in action. In chemistry, when a molecule possesses symmetry, vast amounts of information about its electronic structure, bonding, spectroscopic behaviour, and reactivity become predictable without any calculation.

🔬 What Symmetry Lets Us Do

• Predict the number of IR and Raman active vibrational modes
• Construct molecular orbital (MO) diagrams systematically
• Determine whether a molecule is optically active (chiral)
• Interpret UV-Visible spectra of coordination compounds
• Identify allowed/forbidden electronic transitions

"Before symmetry, you needed pages of mathematics.
After symmetry, a single glance at a character table is enough."

Core Vocabulary — Memorise These First

• Symmetry Element — A geometric entity (axis, plane, centre, or point) about which a symmetry operation is performed.
• Symmetry Operation — The actual physical action (rotation, reflection, inversion) that leaves the molecule looking identical.
• Point Group — The complete set of all symmetry operations for a molecule; they all pass through one common point.
• Character Table — A compact mathematical summary of all irreducible representations of a point group.

The Five Symmetry Elements & Their Operations

Every possible molecular symmetry can be described using just five types of symmetry elements. Understanding each one in detail is the foundation for everything that follows.

Symbol	Element	Operation	Key Example
E	Identity	Do nothing — all atoms stay in place	Every molecule has E
Cₙ	n-fold Rotation Axis	Rotate 360°/n (counterclockwise = positive)	CHCl₃ has C₃; H₂O has C₂
σ	Mirror Plane	Reflect every atom through the plane	H₂O
i	Inversion Centre	(x,y,z) → (−x,−y,−z) through the centre	Staggered ethane, SF₆
Sₙ	Improper Rotation Axis	Rotate 360°/n, then reflect ⊥ to that axis	CH₄ has S₄; staggered ethane has S₆

① The Identity Operation (E)

Completely trivial in appearance but mathematically essential. Every molecule must contain E, otherwise it cannot form a mathematical group. Think of E as "do nothing and come back to the start."

② Proper Rotation (Cₙ)

• Definition: Rotation by 360°/n around the rotation axis.
• Principal axis: The Cₙ with the largest value of n. Always placed along z by convention.
• Multiple rotations: Cₙᵏ = rotating k times by 360°/n. Note that Cₙⁿ ≡ E always.
• Snowflake magic: One snowflake possesses C₂, C₃, and C₆ axes simultaneously (60°, 120°, and 180° rotations).

⚡ Shortcut: Snowflake Rotation Table

For a regular hexagonal snowflake, rotations and their equivalences:

Angle	Operation	Equivalence
60°	C₆	—
120°	C₃	C₆²
180°	C₂	C₆³
240°	C₃²	C₆⁴
300°	C₆⁵	—
360°	C₆⁶	≡ E

③ Mirror Planes (σ) — Three Flavours

• σₕ (horizontal): Perpendicular to the principal axis Cₙ. The "flat" plane.
• σᵥ (vertical): Contains the principal axis; passes through atoms.
• σd (dihedral): Contains the principal axis but bisects angles between C₂ axes; passes between peripheral atoms.

✅ Quick Memory Aid for σ Labels

h = horizontal = ⊥ to Cₙ
v = vertical = contains Cₙ, through atoms
d = dihedral = contains Cₙ, between atoms
Trick: "h is the hat on top; v goes through the roof; d splits the difference."

④ Inversion Centre (i)

Every atom at position (x, y, z) maps to (−x, −y, −z). The molecule looks identical after this. Molecules like ethane (staggered), benzene, SF₆ have an inversion centre. Tetrahedral molecules like CH₄ do NOT — a common exam trap!

🚫 Common Exam Trap — Inversion

• Has i: Square, rectangle, octahedron, snowflake, staggered ethane, PtCl₄²⁻, SF₆
• No i: Triangle, pentagon, tetrahedron, CH₄, NH₃
• Note: S₂ ≡ i (improper rotation of order 2 is the same as inversion)
• Note: S₁ ≡ σ (improper rotation of order 1 is the same as reflection)

⑤ Improper Rotation (Sₙ) — The Compound Operation

Two-step operation: first rotate 360°/n about the axis, then reflect through the plane perpendicular to that axis. Both steps must be done — Sₙ is not a sum of Cₙ and σ, it is a single operation whose net effect is described by both.

• CH₄ has three S₄ axes — each bisects a pair of opposite edges of the tetrahedron.
• Two consecutive S₄ ≡ C₂: S₄² = C₂
• Staggered ethane has an S₆ coincident with its C₃ axis.

Assigning Point Groups — The 6-Step Algorithm

The point group of a molecule is determined by a systematic decision tree. Follow these steps in order, and you will never get it wrong.

STEP 1 — Is the molecule special low symmetry (C₁, Cₛ, Cᵢ) or special high symmetry (Td, Oh, Ih, C∞v, D∞h)?

↳ YES → Assign it directly and stop.

↳ NO → Continue below.

STEP 2 — Find the principal Cₙ axis (highest n).

STEP 3 — Are there n C₂ axes ⊥ to the principal axis?

YES → D groups  |  NO → C or S groups

STEP 4 — Is there a σₕ ⊥ to the principal axis?

YES: D → Dₙₕ ; C → Cₙₕ

STEP 5 — Are there σᵥ or σd planes containing the principal axis?

YES: D → Dₙd ; C → Cₙᵥ  |  NO: D → Dₙ

STEP 6 — Is there an S₂ₙ axis collinear with Cₙ?

YES → S₂ₙ group  |  NO → Cₙ group

The D vs C vs S Decision

Situation	Plane Present?	Point Group	Example
D-set: σₕ exists	σₕ	Dₙₕ	BF₃ (D₃ₕ), benzene (D₆ₕ)
D-set: σd exists	σd	Dₙd	Allene (D₂d), staggered ethane (D₃d)
D-set: no planes	None	Dₙ	[Co(en)₃]³⁺ (D₃)
C-set: σₕ exists	σₕ	Cₙₕ	Trans-N₂F₂ (C₂ₕ)
C-set: σᵥ exists	σᵥ	Cₙᵥ	NH₃ (C₃ᵥ), H₂O (C₂ᵥ)
C-set: S₂ₙ exists	None	S₂ₙ	1,3,5,7-F₄ COT (S₄)
C-set: nothing else	None	Cₙ	H₂O₂ gauche (C₂)

Low & High Symmetry Groups — Quick Reference

Low Symmetry Groups

Group	Symmetry Content	Molecular Example
C₁	Only E — no symmetry at all	CHFClBr (four different groups on carbon)
Cₛ	E and one mirror plane σ only	H₂C=CClBr (single plane of the molecule)
Cᵢ	E and inversion centre i only	HClBrC–CHClBr staggered

High Symmetry Groups

Group	Key Features	Example	Total Ops
C∞v	Linear, no inversion centre; ∞ C∞ and ∞ σᵥ	HCl, HF, CO, HCN	∞
D∞h	Linear, with inversion centre; adds C₂ ⊥ and σₕ	H₂, N₂, CO₂, C₂H₂	∞
Td	4 C₃, 3 C₂, 3 S₄, 6 σd — no C₄ axes	CH₄, CCl₄, NH₄⁺, SiF₄	24
Oh	All of Td plus 3 C₄, C₂ (=C₄²), i, S₄, S₆, σₕ, σd	SF₆, [PtCl₆]²⁻, cube	48
Ih	6 C₅, 10 C₃, 15 C₂, i, 6 S₁₀, 10 S₆, 15 σ	B₁₂H₁₂²⁻, C₆₀ (buckminsterfullerene)	120

💡 Distinguishing Td vs Oh — An Exam Favourite

• Td molecules have only C₃ and C₂ — they have S₄ but NO C₄.
• Oh molecules have C₃, C₂, AND C₄ — adding C₄ changes everything.
• Memory rule: "Four-fold symmetry → six-fold coordination → Oh. Three-fold symmetry → four-coordination → Td."

Properties of a Mathematical Group

For a collection of symmetry operations to constitute a group, it must satisfy four rules. Every point group in chemistry obeys all four.

• Closure: The product (combination) of any two operations in the group must also be a member of the group. E.g. in C₂v: σᵥ × C₂ = σᵥ′
• Associativity: A(BC) = (AB)C always holds for symmetry operations.
• Identity: E is present and EA = AE = A for any operation A.
• Inverse: Every operation A has an inverse A⁻¹ such that A × A⁻¹ = E. Mirrors are their own inverse (σ² = E). C₃ and C₃² are inverses of each other.

Matrix Representations of Operations

Each symmetry operation can be expressed as a square transformation matrix acting on coordinates (x, y, z). For the C₂v group, the four transformation matrices are:

C₂v Transformation Matrices

E	C₂	σᵥ(xz)	σᵥ′(yz)
[+1, 0, 0 / 0, +1, 0 / 0, 0, +1]	[−1, 0, 0 / 0, −1, 0 / 0, 0, +1]	[+1, 0, 0 / 0, −1, 0 / 0, 0, +1]	[−1, 0, 0 / 0, +1, 0 / 0, 0, +1]
χ = 3	χ = −1	χ = 1	χ = 1

The character (χ) = trace = sum of diagonal elements of the matrix.

Reducible vs Irreducible Representations

• Reducible representation (Γ): A combination of simpler (irreducible) ones. The 3×3 matrices for C₂v give characters 3, −1, 1, 1 — this Γ reduces to A₁ + B₁ + B₂.
• Irreducible representation: Cannot be broken down further; 1×1 matrices along the diagonal.
• Block diagonalisation is the process of converting reducible matrices to irreducible components.

The Reduction Formula — Most Important Equation

nᵢ = (1/h) Σ [N(R) × χ(R) × χᵢ(R)]

• nᵢ = number of times irreducible representation i appears
• h = order of the group (total number of symmetry operations)
• N(R) = number of operations in the class
• χ(R) = character of the reducible representation under operation R
• χᵢ(R) = character of the irreducible representation i under operation R
• The result must always be a non-negative integer — if you get a fraction, recheck your calculation.

Reading & Using Character Tables

A character table is the DNA of a point group — every symmetry property of the molecule is encoded in it.

Anatomy of the C₂v Character Table

C₂v	E	C₂	σᵥ(xz)	σᵥ′(yz)	Linear functions	Quadratic
A₁	1	1	1	1	z	x², y², z²
A₂	1	1	−1	−1	Rz	xy
B₁	1	−1	1	−1	x, Ry	xz
B₂	1	−1	−1	1	y, Rx	yz

Seven Properties of Irreducible Representations (Must Know!)

• Order (h): Total count of all symmetry operations = sum across the top row.
• Classes: Operations with identical characters are grouped; number of classes = number of columns = number of rows (character tables are always square).
• Number of irreps = number of classes.
• Sum of squares of dimensions = h: For C₂v: 1² + 1² + 1² + 1² = 4 = h. ✓
• Sum of squares of characters × N(class) = h for each irrep separately.
• Orthogonality: Σ χᵢ(R)·χⱼ(R) = 0 for i ≠ j (dot product of any two rows = 0).
• Totally symmetric representation (all characters = 1) is always present — it is A₁ (or Ag, A₁g etc.).

Label Decoding — A, B, E, T

Label	Dimension of irrep	Meaning
A	1	Symmetric (χ = +1) w.r.t. principal Cₙ rotation
B	1	Antisymmetric (χ = −1) w.r.t. principal Cₙ rotation
E	2	Doubly degenerate — two modes at same energy
T	3	Triply degenerate (only in cubic/icosahedral groups)
Subscript g	—	Symmetric to inversion (gerade)
Subscript u	—	Antisymmetric to inversion (ungerade)
Subscript 1	—	Symmetric to perpendicular C₂ (or to σᵥ if no C₂⊥)
Subscript 2	—	Antisymmetric to perpendicular C₂ (or to σᵥ)
Prime (′)	—	Symmetric to σₕ
Double prime (″)	—	Antisymmetric to σₕ

Chirality & Optical Activity Through Symmetry

A molecule is chiral (dissymmetric) if it is non-superimposable on its mirror image. The symmetry criterion is simple and powerful:

Chiral ↔ Molecule has ONLY E and/or proper rotation axes (Cₙ)
i.e., NO σ, NO i, NO Sₙ (n ≥ 2)

✅ Chiral Point Groups

A molecule is chiral if it belongs to: C₁, Cₙ (n ≥ 2), Dₙ

Non-chiral (achiral) point groups contain σ, i, or Sₙ: Cₛ, Cᵢ, Cₙₕ, Cₙᵥ, Dₙₕ, Dₙd, Td, Oh, Ih, S₂ₙ

• Optical activity: Chiral molecules rotate plane-polarised light. Dextrorotatory (+) = clockwise; levorotatory (−) = counterclockwise.
• [Ru(en)₃]²⁺ belongs to D₃ — chiral. Its mirror images look like left- and right-handed propellers.
• CBrClFI belongs to C₁ — chiral (four different substituents on carbon).
• A propeller has a C₃ axis but no mirror plane → chiral despite having rotational symmetry.
• Key insight: It is the absence of improper symmetry elements (σ, i, Sₙ) that makes a molecule chiral — not just the absence of all symmetry.

Molecular Vibrations, IR & Raman Activity

Degrees of Freedom

Molecule Type	Total DOF	Translational	Rotational	Vibrational
Linear (N atoms)	3N	3	2	3N − 5
Non-linear (N atoms)	3N	3	3	3N − 6
H₂O (non-linear, N=3)	9	3	3	3
CO₂ (linear, N=3)	9	3	2	4
NH₃ (non-linear, N=4)	12	3	3	6
XeF₄ (non-linear, N=5)	15	3	3	9

Steps to Find Vibrational Modes via Group Theory

• Step 1: Assign x, y, z coordinates to every atom; count 3N total vectors.
• Step 2: For each symmetry operation, find χ(R) for the reducible representation:
  — If atom moves position: contribute 0 per atom
  — If atom stays: contribute +1 for each unchanged vector direction, −1 for each reversed direction
• Step 3: Reduce Γtotal using the reduction formula to get irreducible representations.
• Step 4: Subtract Γtrans and Γrot (from the right column of the character table under x,y,z and Rx,Ry,Rz).
• Step 5: Remaining = Γvib — these are the vibrational modes.

H₂O — Worked Out

H₂O Complete Vibrational Analysis (C₂v)

Γtotal = 3A₁ + A₂ + 3B₁ + 2B₂ (from 9 motion vectors)

Γtrans = A₁ + B₁ + B₂ (matching z, x, y)

Γrot = A₂ + B₁ + B₂ (matching Rz, Ry, Rx)

Γvib = 3A₁ + A₂ + 3B₁ + 2B₂ − (A₁ + A₂ + 2B₁ + 2B₂) = 2A₁ + B₁

✅ 3 modes (= 3N−6 = 9−6 = 3). Both A₁ modes and the B₁ mode are IR active.

IR Activity Rule

A vibrational mode is IR active if it belongs to the same symmetry species as x, y, or z (i.e., as a translation)

Physically: the vibration must cause a change in dipole moment. Translations x, y, z represent motion of the centre of charge — if a vibrational mode transforms like them, it shifts charge.

Raman Activity Rule

A mode is Raman active if it transforms as one of: x², y², z², xy, xz, yz (or their combinations)

Physically: Raman activity requires a change in polarisability of the molecule during the vibration.

⚡ Mutual Exclusion Rule

For molecules with a centre of inversion (i), no mode can be both IR and Raman active. IR-active modes are u (ungerade); Raman-active modes are g (gerade). This is an extremely useful diagnostic: if a molecule shows mutual exclusion in its spectra, it must possess a centre of inversion.

Carbonyl Stretching — The Classic Diagnostic

• cis-ML₂(CO)₂ (C₂v): Γco = A₁ + B₁ → both are IR active → two CO bands in IR.
• trans-ML₂(CO)₂ (D₂h): Γco = Ag + B₃u → Ag is IR inactive, B₃u is active → one CO band in IR.
• fac-M(CO)₃L₃ (C₃v): Γco = A₁ + E → both IR active → two CO bands (E is degenerate).
• Diagnostic rule: Count CO bands in IR to determine geometry of metal carbonyl complexes.

Solved Examples

Assign the point group of H₂O

Determine the point group of the water molecule step by step.

STEP 1Is H₂O a special case? Not linear (C∞v or D∞h); not Td/Oh/Ih. → Proceed to step 2.

STEP 2Highest-order rotation axis: C₂ through the O atom bisecting the H–O–H angle. This is the principal axis (z).

STEP 3Any C₂ axes perpendicular to the principal C₂? No. → C or S group.

STEP 4Mirror plane perpendicular to C₂ (σₕ)? No.

STEP 5Mirror planes containing C₂? Yes — two: the plane of the molecule (xz) and one perpendicular to it (yz). → Cₙv group.

STEP 6n = 2, with two σᵥ planes.

Point Group = C₂v  |  Symmetry elements: E, C₂, σᵥ(xz), σᵥ′(yz)

Assign the point group of NH₃

Determine the point group of ammonia (pyramidal geometry).

STEP 1Not a special low or high symmetry case.

STEP 2Principal axis: C₃ through N and the centre of the H₃ triangle.

STEP 3Any C₂ ⊥ to C₃? No. → C or S group.

STEP 4σₕ? No (NH₃ is pyramidal, not planar).

STEP 5σᵥ planes? Yes — three vertical planes, each containing N, one H, and the C₃ axis.

Point Group = C₃v  |  Elements: E, 2C₃, 3σᵥ

Determine IR-active CO stretching modes for fac-Mo(CO)₃(RCN)₃

The molecule has C₃v symmetry. How many IR-active C–O stretches are expected?

BASISUse three C–O bond vectors as the basis set (one per CO group).

EAll 3 vectors unchanged → χ(E) = 3

C₃All 3 CO groups move to new positions → χ(C₃) = 0

σᵥEach σᵥ plane passes through one CO, leaving it unchanged; the other two interchange → χ(σᵥ) = 1

ReduceΓco: E=3, 2C₃=0, 3σᵥ=1 → nA₁ = (1/6)[3 + 0 + 3] = 1; nE = (1/6)[6 + 0 + 0] = 1. So Γco = A₁ + E.

IR?A₁ transforms as z (IR active). E transforms as (x,y) (IR active). Both modes are IR active.

Two IR-active C–O stretching bands expected (A₁ and E). E is degenerate → appears as one absorption. Observed at ~1920 and ~1790 cm⁻¹.

Reduce a Representation in C₂h

Reduce Γ = [4, 0, 2, 2] in C₂h (E, C₂, i, σₕ). Order h = 4.

AgnAg = (1/4)[(1)(4)(1) + (1)(0)(1) + (1)(2)(1) + (1)(2)(1)] = (1/4)[4+0+2+2] = 2

BgnBg = (1/4)[(1)(4)(1) + (1)(0)(−1) + (1)(2)(1) + (1)(2)(−1)] = (1/4)[4+0+2−2] = 1

AunAu = (1/4)[(1)(4)(1) + (1)(0)(1) + (1)(2)(−1) + (1)(2)(−1)] = (1/4)[4+0−2−2] = 0

BunBu = (1/4)[(1)(4)(1) + (1)(0)(−1) + (1)(2)(−1) + (1)(2)(1)] = (1/4)[4+0−2+2] = 1

Γ = 2Ag + Bg + Bu

Is XeO₄ (Td) consistent with 2 Raman bands for Xe–O stretch?

Raman spectroscopy shows two bands for Xe–O stretching at 776 and 878 cm⁻¹. Is this consistent with Td symmetry?

BASISUse 4 Xe–O bond vectors in Td. Γco: E=4, 8C₃=1, 3C₂=0, 6S₄=0, 6σd=2.

REDUCEnA₁ = (1/24)[4+8+0+0+12] = 1; nT₂ = (1/24)[12+0+0+0+12] = 1. Γ = A₁ + T₂.

RAMAN?A₁ transforms as x²+y²+z² — Raman active. T₂ transforms as (xy, xz, yz) — Raman active.

Both A₁ and T₂ are Raman active → two Raman bands expected. This is consistent with Td symmetry. ✓

Exam-Style Multiple Choice Questions

Q1How many symmetry operations does the C₂v point group contain?

• (A) 2
• (B) 3
• (C) 4
• (D) 6

Ans: C — E, C₂, σᵥ(xz), σᵥ′(yz) → h = 4

Q2Which of the following molecules has D∞h symmetry?

• (A) HCl
• (B) CO₂
• (C) HCN
• (D) SO₂

Ans: B — CO₂ is linear with inversion centre → D∞h. HCl and HCN are linear without i → C∞v.

Q3The symmetry operation S₂ is equivalent to which of the following?

• (A) C₂ followed by σᵥ
• (B) σₕ alone
• (C) Inversion (i)
• (D) C₂ alone

Ans: C — S₂ = C₂ then σₕ, which is equivalent to inversion through the centre.

Q4NH₃ belongs to which point group?

• (A) C₃h
• (B) C₃v
• (C) D₃h
• (D) D₃d

Ans: B — NH₃ has C₃ axis and 3σᵥ planes but no σₕ and no C₂ ⊥ axes → C₃v.

Q5A molecule with only a single mirror plane and no other symmetry element belongs to which point group?

• (A) C₁
• (B) Cₛ
• (C) Cᵢ
• (D) C₂

Ans: B — Cₛ contains only E and one σ plane. Example: H₂C=CClBr.

Q6How many vibrational modes does a non-linear molecule with 5 atoms have?

• (A) 10
• (B) 12
• (C) 9
• (D) 15

Ans: C — 3N−6 = 3(5)−6 = 9. (XeF₄ is a classic example with 9 vibrational modes.)

Q7Which irreducible representation label is used for a doubly degenerate vibrational mode?

• (A) A
• (B) B
• (C) E
• (D) T

Ans: C — E = 2-dimensional (doubly degenerate); T = triply degenerate.

Q8The "mutual exclusion rule" applies to molecules that possess:

• (A) A C₂ axis
• (B) A mirror plane
• (C) A centre of inversion
• (D) An S₄ axis

Ans: C — In centrosymmetric molecules, modes are either IR active (u) or Raman active (g), never both.

Q9For trans-ML₂(CO)₂ (D₂h symmetry), how many CO stretching bands appear in the IR spectrum?

• (A) 1
• (B) 2
• (C) 3
• (D) 4

Ans: A — Γco = Ag + B₃u. Ag is IR inactive (centrosymmetric); B₃u is IR active → only one band.

Q10CH₄ belongs to which point group?

• (A) Oh
• (B) Td
• (C) D₄h
• (D) Ih

Ans: B — CH₄ has 4 C₃ axes, 3 C₂ axes, 3 S₄ axes, 6 σd planes → Td. Note: Td ≠ Oh (no C₄ axis in Td).

Q11Which of the following molecules is chiral?

• (A) H₂O
• (B) NH₃
• (C) CHFClBr
• (D) CHCl₃

Ans: C — CHFClBr has C₁ symmetry (no σ, no i, no Sₙ) → chiral. NH₃ and H₂O have mirror planes.

Q12The character of the identity operation E in any irreducible representation equals:

• (A) Always 1
• (B) Always 0
• (C) The dimension of the representation
• (D) The order of the group

Ans: C — χ(E) = dimension. A = 1, E = 2, T = 3. This is why the sum of squares of χ(E) = h.

27-Question Practice Set

Based on IIT-JAM, GATE, CSIR-NET, BITSAT, TGT/PGT exam patterns. Answers to selected questions provided.

Determine the point group of staggered ethane (C₂H₆). How many σ planes does it possess?

How many C₂ axes does benzene (D₆h) possess in total? Classify them with prime/double-prime notation.

Verify that the irreducible representations B₁ and B₂ in C₂v are mutually orthogonal.

PF₅ has D₃h symmetry. List all its symmetry elements and operations.

[PtCl₄]²⁻ is square planar. Assign its point group. Does it have an inversion centre?

For CO₂ (D∞h), how many vibrational modes exist? How many are IR active?

Why does CH₄ (Td) have no centre of inversion while SF₆ (Oh) does? Explain using geometry.

Reduce the following representation in C₃v: Γ = [6, 3, 2] (E, 2C₃, 3σᵥ).

A molecule belongs to C₂h. Does it possess: (a) a centre of inversion? (b) a mirror plane? (c) improper rotation axis?

Determine the point group of allene (H₂C=C=CH₂). Justify by identifying each symmetry element.

cis-[Co(NH₃)₄Cl₂]⁺ vs trans-[Co(NH₃)₄Cl₂]⁺ — which has higher symmetry? Assign point groups.

List all the vibrational modes of XeF₄ (D₄h) and state which are IR active, Raman active, or both.

For the C₃v character table, verify property 5: that the sum of squares of characters multiplied by N(class) = h, for the E irreducible representation.

How does the IR spectrum of fac-Mo(CO)₃L₃ differ from mer-Mo(CO)₃L₃? Use group theory to determine the number of IR-active CO stretches for each.

The molecule trans-1,2-dichloroethylene has C₂h symmetry. List all symmetry operations and identify which vibrational modes are IR vs Raman active.

Cyclohexane exists in chair and boat conformations. Which has higher symmetry? Assign point groups for both.

Show by matrix multiplication that C₂ × σᵥ(xz) = σᵥ′(yz) for the C₂v group.

[Fe@Ge₁₀]³⁻ is known. If the cage has D₄d symmetry, how many C₂ axes perpendicular to the principal axis does it possess?

Define the terms "reducible representation" and "irreducible representation." Give one example of each for the C₂v point group.

The Raman spectrum of AsP₃ (Cs symmetry-broken tetrahedron) shows four bands. Confirm this is consistent with the point group C₃v of AsP₃.

What is the order of the Oh point group? How many classes of symmetry operations does it have?

Explain why linear molecules (C∞v and D∞h) are treated as special cases rather than following the standard 6-step algorithm.

Use the reduction formula to find how many times the T₂ representation appears in the reducible representation Γ = [4, 1, 0, 0, 2] in Td.

Ferrocene (staggered, D₅d) versus ferrocene (eclipsed, D₅h) — which has a centre of inversion? Justify your answer.

In the C₃v character table, why do C₃ and C₃² appear as "2C₃" in the same column? What does this mean about their characters?

A molecule of formula ML₄ has D₄h symmetry. Draw all its C₂ axes and classify them as C₂, C₂′, or C₂″.

For N₂H₄ in its gauche conformation (C₂ symmetry), is the molecule chiral? What is its point group and does it show optical activity?

📌 Selected Answers

• Q1: D₃d — three σd planes (each containing C–C bond and one pair of H atoms from opposite ends)
• Q5: D₄h — yes, inversion centre is present (opposite Cl atoms map onto each other)
• Q6: CO₂ has 4 modes (3N−5): symmetric stretch (IR inactive, Raman active), asymmetric stretch (IR active), 2× bending (IR active, degenerate)
• Q8: nA₁ = (1/6)[6+6+6] = 3; nA₂ = (1/6)[6+6−6] = 1; nE = (1/6)[12−6+0] = 1. So Γ = 3A₁ + A₂ + E
• Q21: h = 48; 10 classes of symmetry operations
• Q23: nT₂ = (1/24)[(3)(4)(1)+(0)(1)(0)+(0)(0)(−1)+(0)(0)(−1)+(6)(2)(1)] = (1/24)[12+0+0+0+12] = 1
• Q24: Staggered D₅d has an S₁₀ axis and inversion centre (i). Eclipsed D₅h has σₕ but no inversion centre.
• Q27: C₂ (gauche N₂H₄) — chiral, as it has only E and C₂ (no σ, no i). Shows optical activity.

Exam Tips, Tricks & High-Yield Points

🎯 High-Frequency Exam Points

• Point group of H₂O = C₂v; NH₃ = C₃v; CH₄ = Td; SF₆ = Oh; CO₂ = D∞h; HCl = C∞v — commit these to memory.
• S₂ ≡ i and S₁ ≡ σ — if you see S₂ in options, it means inversion!
• Mutual exclusion rule: if IR and Raman bands are at the same wavenumber, the molecule has no inversion centre.
• Number of IR-active CO bands: cis = 2; trans = 1. One of the most tested facts in GATE/CSIR-NET.
• Chiral point groups contain ONLY E and proper Cₙ rotations: C₁, Cₙ (n≥2), Dₙ.
• The character under E always equals the dimension of the irrep: A=1, E=2, T=3.
• Reduction formula must give non-negative integers — if you get a fraction, you've made an error.
• The order h of Oh = 48, Td = 24, D₄h = 16, C₂v = 4, C₃v = 6.
• For the principal axis: always use the z direction. The xy plane is perpendicular to z.

🚨 Classic Traps in Competitive Exams

• Trap 1: CHCl₃ looks like it could have Oh symmetry — it does not! It has C₃v (three Cl atoms equivalent, one H on C₃).
• Trap 2: Staggered ethane is D₃d, NOT D₃h. The hydrogen eclipsing in the eclipsed form gives D₃h.
• Trap 3: Allene (D₂d) is NOT the same as ethylene (D₂h). Allene has σd planes, not σₕ.
• Trap 4: "E" in the character table label (E irrep) is completely different from "E" the identity operation. Context matters!
• Trap 5: XeF₄ is square planar (D₄h), not tetrahedral. Xe has lone pairs pushing geometry flat.
• Trap 6: The totally symmetric representation (A₁, Ag, A₁g etc.) is always Raman active.

Quick Recall: How to Tell D vs C Groups

Memory Device: "D has Diagonals, C is Closed"

Look for C₂ axes that stick out perpendicular to the main axis like the spokes of a wheel. If you find them → D group. Count how many: there are always n of them for a Dₙ group. If no spokes → C or S group.

For a D group: σₕ → Dₙh. No σₕ but σd → Dₙd. No planes at all → Dₙ.

For a C group: σₕ → Cₙh. σᵥ → Cₙv. S₂ₙ only → S₂ₙ. Nothing → Cₙ.

Important Tables at a Glance

Molecule	Point Group	IR-active vibrations	Raman-active	Chiral?
H₂O	C₂v	All 3 (2A₁ + B₁)	All 3	No (σᵥ)
CO₂	D∞h	2 of 4 (πu, σu)	1 of 4 (σg+)	No (i)
NH₃	C₃v	All 6	All 6	No (σᵥ)
CH₄	Td	T₂ modes only	A₁ + E + T₂	No (σd)
SF₆	Oh	T₁u only	A₁g + Eg + T₂g	No (i)
CHFClBr	C₁	All modes	All modes	Yes!
Benzene	D₆h	Only u modes	Only g modes	No (i, σ)
Staggered ethane	D₃d	Only u modes	Only g modes	No (i)

Symmetry & Group Theory — Inorganic Chemistry Chapter 4 Study Notes

For academic use · IIT-JAM · GATE · CSIR-NET · BITSAT · TGT/PGT Preparation

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