6

Molecular Symmetry

Understanding the systematic treatment of molecular symmetry through group theory, character tables, and their applications to spectroscopy and bonding

An Introduction to Symmetry Analysis

Symmetry and bonding of molecules are intimately linked. Symmetry considerations are essential for constructing molecular orbitals and analysing molecular vibrations, particularly where these are not immediately obvious. They also enable us to extract information about molecular and electronic structure from spectroscopic data.

Key Concepts The systematic treatment of symmetry uses group theory, a branch of mathematics that allows us to classify molecules by their symmetry properties, construct molecular orbitals, and analyse molecular vibrations and selection rules.

Group theory enables us to draw general conclusions about molecular properties such as polarity and chirality without performing any calculations at all.

6.1 Symmetry Operations, Elements, and Point Groups

Key Points Symmetry operations are actions that leave the molecule apparently unchanged; each symmetry operation is associated with a symmetry element. The point group of a molecule is identified by noting its symmetry elements and comparing these elements with the elements that define each group.

Symmetry Operations and Elements

A symmetry operation is an action, such as rotation through a certain angle, that leaves the molecule apparently unchanged. Associated with each symmetry operation is a symmetry element—a point, line, or plane with respect to which the symmetry operation is performed.

Symmetry Operation Symmetry Element Symbol
Identity 'whole of space' E
Rotation by 360°/n n-fold symmetry axis Cn
Reflection mirror plane σ
Inversion centre of inversion i
Rotation + Reflection n-fold improper rotation axis Sn

Note: S1 = σ and S2 = i

Interactive Symmetry Operations

🔄 Symmetry Operation Demonstrations
Identity (E)
Rotation (Cn)
Reflection (σ)
Inversion (i)
Improper (Sn)

The identity operation E consists of doing nothing to the molecule. Every molecule has at least this operation.

O H H

H₂O - unchanged under identity

An n-fold rotation Cn is a symmetry operation if the molecule appears unchanged after rotation by 360°/n.

C₂ O H H

H₂O has a C₂ axis bisecting the HOH angle

A reflection σ in a mirror plane leaves the molecule unchanged. Mirror planes are classified as:

  • σv (vertical) - contains the principal axis
  • σh (horizontal) - perpendicular to the principal axis
  • σd (dihedral) - bisects angle between C₂ axes
σᵥ O H H

H₂O has two vertical mirror planes (σᵥ and σᵥ')

The inversion operation i projects each atom through a central point to an equal distance on the other side. Under inversion, an atom at (x, y, z) moves to (−x, −y, −z).

S F₁ F₆ F₂ F₅ i

SF₆ has a centre of inversion at the S atom

An improper rotation Sn consists of a rotation by 360°/n followed by a reflection in the plane perpendicular to the rotation axis.

C S₄

CH₄ has S₄ axes (90° rotation + reflection)

Example 6.1: Identifying Symmetry Elements

Problem: Identify the symmetry elements in the eclipsed conformation of an ethane molecule.

Answer: The eclipsed conformation of CH₃CH₃ has the elements: E, C₃, 3C₂, σh, 3σv, S₃

The staggered conformation additionally has the elements i and S₆.

Sketch the S₄ axis of an NH₄⁺ ion. How many of these axes does the ion possess?

Point Groups

By identifying the symmetry elements of a molecule, we can assign it to its point group. The name of the point group is its Schoenflies symbol, such as C3v for ammonia.

Common Point Groups

Point Group Symmetry Elements Examples
C₁ E SiHClBrF
C₂ E, C₂ H₂O₂
Cs E, σ NHF₂
C2v E, C₂, σᵥ, σᵥ' H₂O, SO₂Cl₂
C3v E, 2C₃, 3σᵥ NH₃, PCl₃, POCl₃
C∞v E, 2Cφ, ∞σᵥ OCS, CO, HCl
D2h E, 3C₂, i, 3σ N₂O₄, B₂H₆
D3h E, 2C₃, 3C₂, σh, 2S₃, 3σᵥ BF₃, PCl₅
D4h E, 2C₄, C₂, 2C₂', 2C₂'', i, 2S₄, σh, 2σᵥ, 2σd XeF₄, trans-[MA₄B₂]
D∞h E, ∞C₂', 2Cφ, i, ∞σᵥ, 2Sφ CO₂, H₂, C₂H₂
Td E, 8C₃, 3C₂, 6S₄, 6σd CH₄, SiCl₄
Oh E, 8C₃, 6C₂, 6C₄, 3C₂, i, 6S₄, 8S₆, 3σh, 6σd SF₆

Interactive Point Group Identifier

🔍 Point Group Decision Tree
Is the molecule linear?
Example 6.2: Identifying the Point Group of a Molecule

Problem: To what point groups do H₂O and XeF₄ belong?

Answer:

H₂O: Possesses E, C₂, σᵥ, σᵥ' → Point group C2v

XeF₄: Possesses E, C₄, 2C₂, 2C₂', 2C₂'', σh, 2σᵥ, 2σd, i, S₄ → Point group D4h

Identify the point groups of (a) BF₃, a trigonal-planar molecule, and (b) the tetrahedral SO₄²⁻ ion.

Molecular Examples

O

H₂O

C2v
N

NH₃

C3v
Xe

XeF₄

D4h
C

CH₄

Td
S

SF₆

Oh

6.2 Character Tables

Key Point The systematic analysis of the symmetry properties of molecules is carried out using character tables. A character table displays all the symmetry elements of the point group together with a description of how various objects or mathematical functions transform under the corresponding symmetry operations.

Structure of a Character Table

Component Description
First column Symmetry species (irreducible representations)
Top row Symmetry operations arranged by class
Body Characters (χ) showing how orbitals transform
Right columns Linear functions (x, y, z) and quadratic functions (xy, z², etc.)

Character Meanings

Character Significance
+1 The orbital is unchanged
−1 The orbital changes sign
0 The orbital undergoes a more complicated change

Example Character Tables

C2v
C3v
Td

C2v Character Table (h = 4)

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

C3v Character Table (h = 6)

C3v E 2C₃ 3σᵥ Linear Quadratic
A₁ 1 1 1 z x²+y², z²
A₂ 1 1 −1 Rz
E 2 −1 0 (x, y), (Rx, Ry) (x²−y², xy), (zx, yz)

Td Character Table (h = 24)

Td E 8C₃ 3C₂ 6S₄ 6σd Linear Quadratic
A₁ 1 1 1 1 1 x²+y²+z²
A₂ 1 1 1 −1 −1
E 2 −1 2 0 0 (2z²−x²−y², x²−y²)
T₁ 3 0 −1 1 −1 (Rx, Ry, Rz)
T₂ 3 0 −1 −1 1 (x, y, z) (xy, yz, zx)

Symmetry Labels and Degeneracy

Symmetry Label Degeneracy
A, B 1 (non-degenerate)
E 2 (doubly degenerate)
T 3 (triply degenerate)
Example 6.3: Identifying the Symmetry Species of Orbitals

Problem: Identify the symmetry species of each oxygen valence-shell orbital in H₂O (C2v).

Answer:

  • O 2s and O 2pz: characters (1,1,1,1) → A₁
  • O 2px: characters (1,−1,1,−1) → B₁
  • O 2py: characters (1,−1,−1,1) → B₂
Example 6.4: Determining Degeneracy

Problem: Can there be triply degenerate orbitals in BF₃?

Answer: BF₃ belongs to D3h. The maximum character in the E column is 2, so the maximum degeneracy is 2. No orbitals can be triply degenerate.

The SF₆ molecule is octahedral. What is the maximum possible degree of degeneracy of its orbitals?

6.3 Polar Molecules

Key Point A molecule cannot be polar if it belongs to any group that includes a centre of inversion, any of the groups D and their derivatives, the cubic groups (T, O), the icosahedral group (I), or their modifications.

A polar molecule has a permanent electric dipole moment. The conditions that prevent polarity are:

Rules for Polarity
  • A molecule cannot be polar if it has a centre of inversion
  • A molecule cannot have a dipole moment perpendicular to any mirror plane
  • A molecule cannot have a dipole moment perpendicular to any axis of rotation
Example 6.5: Judging Polarity

Problem: The ruthenocene molecule is a pentagonal prism with Ru sandwiched between two C₅H₅ rings. Can it be polar?

Answer: A pentagonal prism belongs to D5h. D groups have perpendicular C₂ axes that rule out dipoles in all directions. Therefore, ruthenocene must be nonpolar.

Examples of Polar vs Nonpolar

Polar Molecules

  • H₂O (C2v)
  • NH₃ (C3v)
  • HCl (C∞v)
  • CHCl₃ (C3v)

Nonpolar Molecules

  • CO₂ (D∞h)
  • BF₃ (D3h)
  • SF₆ (Oh)
  • CCl₄ (Td)

6.4 Chiral Molecules

Key Point A molecule cannot be chiral if it possesses an improper rotation axis (Sn).

A chiral molecule cannot be superimposed on its own mirror image. Chiral molecules and their mirror images are called enantiomers. Chiral molecules that do not interconvert rapidly are optically active—they can rotate the plane of polarized light.

The Sn Test for Chirality

A molecule with any improper rotation axis Sn cannot be chiral. Remember:

  • S₁ = σ (mirror plane)
  • S₂ = i (inversion centre)
Example 6.6: Judging Chirality

Problem: The complex [Mn(acac)₃] has the D₃ point group. Is it chiral?

Answer: D₃ consists of elements (E, C₃, 3C₂) and does not contain an Sn axis. The complex is chiral and therefore optically active.

Important: CHClFBr belongs to C₁, not Td. It has tetrahedral geometry but not tetrahedral symmetry, so it is chiral.

Is the conformation of H₂O₂ with a dihedral angle of ~90° chiral?

6.5 Molecular Vibrations

Key Points
  • If a molecule has a centre of inversion, none of its vibrations can be both IR and Raman active (exclusion rule)
  • A vibrational mode is IR active if it has the same symmetry as x, y, or z (electric dipole)
  • A vibrational mode is Raman active if it has the same symmetry as a quadratic function (xy, x², etc.)

Counting Vibrational Modes

Nonlinear molecule: 3N − 6 vibrations
Linear molecule: 3N − 5 vibrations

The Exclusion Rule

Exclusion Rule

If a molecule has a centre of inversion, none of its modes can be both IR and Raman active. A mode may be inactive in both.

CO₂ Vibrations (D∞h)

ν₁ Symmetric Stretch
Raman
ν₃ Asymmetric Stretch
IR
ν₂ Bend (2× degenerate)
IR
Example 6.7: Using the Exclusion Rule

Problem: Which vibrations of CO₂ are IR or Raman active?

Answer:

  • ν₁ (symmetric stretch): Dipole unchanged → IR inactive, Raman active
  • ν₃ (asymmetric stretch): Dipole changes → IR active, Raman inactive (exclusion rule)
  • ν₂ (bend): Dipole changes → IR active, Raman inactive (exclusion rule)

cis vs trans Isomer Distinction

Consider cis-[PdCl₂(NH₃)₂] (C2v) vs trans-[PdCl₂(NH₃)₂] (D2h):

Isomer Point Group IR Bands Raman Bands
cis C2v 2 (A₁ + B₂) 2 (A₁ + B₂)
trans D2h 1 (B2u) 1 (Ag)
Example 6.9: Reducing a Representation ([Ni(CO)₄])

Problem: The tetrahedral [Ni(CO)₄] molecule has Td symmetry. Determine which CO stretching modes are IR or Raman active.

Answer: The CO displacements transform as A₁ + T₂

  • A₁: transforms like x²+y²+z² → Raman active only
  • T₂: transforms like (x,y,z) and (xy,yz,zx) → Both IR and Raman active

Result: 1 IR band, 2 Raman bands

6.6 Symmetry-Adapted Linear Combinations (SALCs)

Key Point Symmetry-adapted linear combinations of orbitals are combinations of atomic orbitals that conform to the symmetry of a molecule and are used to construct molecular orbitals of a given symmetry species.

Molecular orbitals are constructed from atomic orbitals of the same symmetry. This fundamental principle means that σ, π, or δ bonds can only be formed from atomic orbitals of the same symmetry species.

SALCs in NH₃ (C3v)

The three H1s orbitals of NH₃ span A₁ + E:

A₁
φ₁ = ψA1s + ψB1s + ψC1s
+ + +

All same phase

E
φ₂ = 2ψA1s − ψB1s − ψC1s
+ − −

Larger coefficient on A

E
φ₃ = ψB1s − ψC1s
0 + −

B and C opposite

Example 6.10: Identifying SALCs

Problem: Identify the symmetry species of SALCs from H1s orbitals in NH₃.

Answer: Under E, all 3 orbitals unchanged (χ=3). Under C₃, none unchanged (χ=0). Under σᵥ, one unchanged (χ=1).

Characters (3,0,1) reduce to A₁ + E

Example 6.11: SALC Symmetry in NO₂

Problem: Identify the symmetry species of φ = ψO' − ψO'' (O 2px orbitals in C2v NO₂).

Answer:

  • Under C₂: φ → φ (character +1)
  • Under σᵥ: both change sign, φ → −φ (character −1)
  • Under σᵥ': φ → −φ (character −1)

Characters (1, 1, −1, −1) → A₂

6.7 The Construction of Molecular Orbitals

Key Point Molecular orbitals are constructed from SALCs and atomic orbitals of the same symmetry species.

Procedure for Constructing MO Schemes

  1. Assign a point group to the molecule
  2. Look up the shapes of SALCs
  3. Arrange SALCs in increasing order of energy (s < p < d)
  4. Combine SALCs of the same symmetry type
  5. Estimate relative energies from overlap considerations
  6. Verify with computational methods

NH₃ Molecular Orbital Diagram

NH₃ MO Energy Levels

N orbitals
MOs
H₃ SALCs
3a₁
2e
2p
2a₁ (LP)
E
1e
2s
1a₁
A₁

● Bonding   ● Nonbonding (Lone Pair)   ● Antibonding

Example 6.12: Constructing MOs from SALCs

Problem: Which oxygen orbitals can form MOs with the H₂O SALCs φ₁ = ψA + ψB and φ₂ = ψA − ψB?

Answer:

  • φ₁ has A₁ symmetry → combines with O 2s and O 2pz
  • φ₂ has B₂ symmetry → combines with O 2py

Linear combinations:

a₁: ψ = c₁ψO2s + c₂ψO2pz + c₃φ₁
b₂: ψ = c₄ψO2py + c₅φ₂

6.8 The Vibrational Analogy

Key Point The shapes of SALCs are analogous to stretching displacements. SALCs in orbital diagrams correspond to normal vibrational modes of molecules.
Example 6.13: IR and Raman Bands of Octahedral Molecules

Problem: For an AB₆ molecule (Oh), sketch the normal modes of A−B stretches and identify IR/Raman activity.

Answer: The stretching SALCs have symmetry A1g + Eg + T1u

  • A1g: Raman active (transforms like x²+y²+z²)
  • Eg: Raman active
  • T1u: IR active (transforms like x, y, z)

Octahedral AB₆ Stretching Modes

A1g
Raman
Eg
Raman
T1u
IR

6.9 The Reduction of a Representation

Key Point A reducible representation can be resolved into its constituent irreducible representations using the reduction formula.

The Reduction Formula

ci = (1/h) ΣC g(C) χi(R) χ(R)

Where:

Reduction Formula Calculator

📊 Reduction Formula Calculator (C2v)

Result: Γ = ?

Enter characters and click Calculate

Example 6.14: Using the Reduction Formula

Problem: For cis-[PdCl₂(NH₃)₂] (C2v), what symmetry species are spanned by atomic displacements?

Answer: Characters: χ(E)=15, χ(C₂)=−1, χ(σᵥ)=1, χ(σᵥ')=5

Using the reduction formula: Γ = 5A₁ + 2A₂ + 3B₁ + 5B₂

Subtracting translations (A₁+B₁+B₂) and rotations (A₂+B₁+B₂):

Vibrations = 4A₁ + A₂ + B₁ + 3B₂

6.10 Projection Operators

Key Point A projection operator is used to generate SALCs from a basis of orbitals.

The Projection Operator Formula

φ = ΣR χi(R) Rψ

Where χi(R) is the character of operation R for the desired symmetry species, and Rψ is the orbital generated by applying operation R to basis orbital ψ.

Example 6.15: Generating SALCs for [PtCl₄]²⁻

Problem: Generate the SALC of Cl σ orbitals for [PtCl₄]²⁻ (D4h).

Answer: Starting with ψ₁ and applying all 16 operations of D4h:

Show transformation table

R: E C₄ C₄³ C₂ C₂' C₂' C₂'' C₂'' i S₄ S₄³ σh σᵥ σᵥ σd σd
Rψ₁: ψ₁ ψ₂ ψ₄ ψ₃ ψ₁ ψ₃ ψ₂ ψ₄ ψ₃ ψ₂ ψ₄ ψ₁ ψ₁ ψ₃ ψ₂ ψ₄

Results:

φ(A1g) = ¼(ψ₁ + ψ₂ + ψ₃ + ψ₄)
φ(B1g) = ¼(ψ₁ − ψ₂ + ψ₃ − ψ₄)
φ(Eu) = ½(ψ₁ − ψ₃) and ½(ψ₂ − ψ₄)

The Cl σ SALCs span: A1g + B1g + Eu

[PtCl₄]²⁻ SALCs Visualization

A1g + + + +

All same phase

B1g + − + −

Alternating phases

Eu + − 0 0

Doubly degenerate

Chapter Summary

Symmetry Operations
  • Identity (E)
  • Rotation (Cn)
  • Reflection (σ)
  • Inversion (i)
  • Improper rotation (Sn)
Applications
  • Determine polarity
  • Determine chirality
  • Predict IR/Raman activity
  • Construct MO diagrams
  • Generate SALCs

Key Equations

Vibrational modes: 3N − 6 (nonlinear) or 3N − 5 (linear)

Reduction formula: ci = (1/h) Σ g(C) χi(R) χ(R)

Projection operator: φ = Σ χi(R) Rψ