Concept of a basis. Based on this treatment we can assign the basis to one of the irreducible representations of the point group.

Similar documents
Construction of the C 2v character table

Chemistry 431. Lecture 14. Wave functions as a basis Diatomic molecules Polyatomic molecules Huckel theory. NC State University

Molecular orbitals, potential energy surfaces and symmetry

Also interested only in internal energies Uel (R) only internal forces, has symmetry of molecule--that is source of potential.

PAPER No. : 8 (PHYSICAL SPECTROSCOPY) MODULE NO. : 23 (NORMAL MODES AND IRREDUCIBLE REPRESENTATIONS FOR POLYATOMIC MOLECULES)

B F N O. Chemistry 6330 Problem Set 4 Answers. (1) (a) BF 4. tetrahedral (T d )

Degrees of Freedom and Vibrational Modes

The heart of group theory

Symmetry. Chemistry 481(01) Spring 2017 Instructor: Dr. Upali Siriwardane Office: CTH 311 Phone Office Hours:

Quiz 5 R = lit-atm/mol-k 1 (25) R = J/mol-K 2 (25) 3 (25) c = X 10 8 m/s 4 (25)

Lecture 12. Symmetry Operations. NC State University

LECTURE 3 DIRECT PRODUCTS AND SPECTROSCOPIC SELECTION RULES

Symmetrical: implies the species possesses a number of indistinguishable configurations.

Chemistry 431. NC State University. Lecture 17. Vibrational Spectroscopy

Chapter 6 Answers to Problems

Types of Molecular Vibrations

Symmetry: Translation and Rotation

Ch120 - Study Guide 10

5.4. Electronic structure of water

26 Group Theory Basics

Using Symmetry to Generate Molecular Orbital Diagrams

Molecular Orbital Theory

ChemWiki BioWiki GeoWiki StatWiki PhysWiki MathWiki SolarWiki

Hybridization of Atomic Orbitals. (Chapter 1 in the Klein text)

5 Irreducible representations

Symmetry and Group Theory

Molecular Symmetry. Symmetry is relevant to: spectroscopy, chirality, polarity, Group Theory, Molecular Orbitals

Problem Set 2 Due Thursday, October 1, & & & & # % (b) Construct a representation using five d orbitals that sit on the origin as a basis:

PAPER No.13 :Applications of molecular symmetry and group theory Module No.35: Symmetry and chemical bonding part-6 MO

Problem Set 2 Due Tuesday, September 27, ; p : 0. (b) Construct a representation using five d orbitals that sit on the origin as a basis: 1

Molecular Symmetry 10/25/2018

The symmetry properties & relative energies of atomic orbitals determine how they react to form molecular orbitals. These molecular orbitals are then

13 Applications of molecular symmetry and group theory

7. Arrange the molecular orbitals in order of increasing energy and add the electrons.

SYMMETRY IN CHEMISTRY

CH 612 Advanced Inorganic Chemistry Structure and Reactivity. Exam # 2 10/25/2010. Print name:

130 points on 6 pages + a useful page Circle the element/compound most likely to have the desired property. Briefly explain your choice

Molecular orbitals for σbonding in T d complexes

Physical Chemistry I Fall 2016 Second Hour Exam (100 points) Name:

Inorganic Chemistry with Doc M. Fall Semester, 2012 Day 9. Molecular Orbitals, Part 4. Beyond Diatomics, continued

where, c is the speed of light, ν is the frequency in wave numbers (cm -1 ) and µ is the reduced mass (in amu) of A and B given by the equation: ma

Group Theory: Matrix Representation & Consequences of Symmetry

Determining the Normal Modes of Vibration

ADVANCED INORGANIC CHEMISTRY QUIZ 4 November 29, 2012 INSTRUCTIONS: PRINT YOUR NAME > NAME.

Tables for Group Theory

13, Applications of molecular symmetry and group theory

Tables for Group Theory

A Quantum Mechanical Model for the Vibration and Rotation of Molecules. Rigid Rotor

Chapter 5 Equations for Wave Function

Chemistry 5325/5326. Angelo R. Rossi Department of Chemistry The University of Connecticut. January 17-24, 2012

Notation. Irrep labels follow Mulliken s convention: A and B label nondegenerate E irreps, doubly T degenerate, triply degenerate

In the fourth problem set, you derived the MO diagrams for two complexes containing Cr-Cr bonds:

Figure 1: Transition State, Saddle Point, Reaction Pathway

Molecular Orbitals in Inorganic Chemistry. Dr. P. Hunt Rm 167 (Chemistry)

Complex Numbers; Digression. Handouts - Download and Read! Cyclic Groups. Character Tables for Cyclic Groups

Name CHM 4610/5620 Fall 2016 November 15 EXAMINATION TWO SOLUTIONS

130 points on 6 pages + a useful page 7

The Unshifted Atom-A Simpler Method of Deriving Vibrational Modes of Molecular Symmetries

5.80 Small-Molecule Spectroscopy and Dynamics

5.80 Small-Molecule Spectroscopy and Dynamics

Advanced Spectroscopy. Dr. P. Hunt Rm 167 (Chemistry) web-site:

Chapter 12. Linear Molecules

Chem Spring, 2017 Assignment 5 - Solutions

b) For this ground state, obtain all possible J values and order them from lowest to highest in energy.

b) For this ground state, obtain all possible J values and order them from lowest to highest in energy.

Chem Symmetry and Introduction to Group Theory. Symmetry is all around us and is a fundamental property of nature.

Review of Matrices. L A matrix is a rectangular array of numbers that combines with other such arrays according to specific rules.

C 2 '' σ v ' C 2 ' "side on" "in-plane" 2S determine how the MO transforms under each symmetry operation, Figure 3.

LECTURE 2 DEGENERACY AND DESCENT IN SYMMETRY: LIGAND FIELD SPLITTINGS AND RELATED MATTERS

Chapter 6 Vibrational Spectroscopy

Homework Problem Set 1 Solutions

6.2 Polyatomic Molecules

Mo 2+, Mo 2+, Cr electrons. Mo-Mo quadruple bond.

Chem 673, Problem Set 5 Due Thursday, November 29, 2007

B7 Symmetry : Questions

A. General (10 points) 2 Points Each

Little Orthogonality Theorem (LOT)

Chem 673, Problem Set 5 Due Thursday, December 1, 2005

Symmetry and states of self stress in triangulated toroidal frames

Problem Set 5 Solutions

Determining the Normal Modes of Vibration

Midterm Exam I. CHEM 181: Introduction to Chemical Principles September 24, 2015 Key

5.111 Principles of Chemical Science

Lecture 4 Model Answers to Problems

Theoretical Concepts of Spin-Orbit Splitting

Degrees of Freedom and Vibrational Modes

Symmetry Adapted Orbitals

Chem 673, Problem Set 5 Due Tuesday, December 2, 2008

Renner-Teller Effect in Tetra-Atomic Molecules

5.04 Principles of Inorganic Chemistry II

CHEMISTRY 4021/8021 MIDTERM EXAM 1 SPRING 2014

1) For molecules A and B, list the symmetry operations and point groups.

Inorganic Chemistry with Doc M. Day 18. Transition Metals Complexes IV: Ligand Field Theory continued

Spectroscopic Selection Rules

f lmn x m y n z l m n (1) 3/πz/r and Y 1 ±1 = 3/2π(x±iy)/r, and also gives a formula for Y

Translation Groups; Introduction to Bands and Reciprocal Space. Timothy Hughbanks, Texas A& M University

Quantum Numbers. principal quantum number: n. angular momentum quantum number: l (azimuthal) magnetic quantum number: m l

EXPLAINING THE GEOMETRY OF SIMPLE MOLECULES USING MOLECULAR ORBITAL ENERGY-LEVEL DIAGRAMS BUILT BY USING SYMMETRY PRINCIPLES

Rotational Motion. Chapter 4. P. J. Grandinetti. Sep. 1, Chem P. J. Grandinetti (Chem. 4300) Rotational Motion Sep.

Wave Equations of Polyatomic Molecules

Transcription:

Concept of a basis A basis refers to a type of function that is transformed by the symmetry operations of a point group. Examples include the spherical harmonics, vectors, internal coordinates (e..g bonds, angles, torsions), translations, rotations and any other function needed to describe the electronic or nuclear properties of a molecule. The spherical harmonics include the orbitals, s, p, d etc. and can have more than one dimension. Thus, we need to examine how those functions are changed by the operations. Based on this treatment we can assign the basis to one of the irreducible representations of the point group.

Orbital basis

Oxygen s-orbitals in water, I C 2v E C 2 σ v σ v ' Γ O(s) +1 +1 +1 +1 Γ O(s) = a 1 s-orbitals on central elements will always transform as the totally symmetric representation but are not included in character tables

p z E C 2 σ v(xz) σ v(yz)

p z E C 2 σ v(xz) σ v(yz) 1 1 1 1

p x E C 2 σ v(xz) σ v(yz)

p x E C 2 σ v(xz) σ v(yz) 1-1 1-1

Oxygen p-orbitals in water, p x C 2v E C 2 σ v (xz) σ v '(yz) p z +1 +1 +1 +1 a 1 p x +1-1 +1-1 b 1 p y +1-1 -1 +1 b 2 Γ p 3-1 1 1 Thus, Γ p = a 1 + b 1 + b 2. The px orbital is said to form the basis for the b 1 representation, have b 1 symmetry, or transform as b 1 p z p y

S-orbitals as a basis I II III E each atom has 1 and not 3 labels so each operation is a 3x3 matrix as opposed to the 9x9 matrices of the Cartesian basis. In addition, there can be no sign change for an s- orbital. The resulting representation is, E = 1 0 0 1 0 0 1 0 0 0 1 0 ; C 2 = 0 0 1 ; σ = v 0 1 0 ; 0 0 1 0 1 0 0 0 1 σ ' v = 1 0 0 0 0 1 0 1 0

s-orbitals as a basis I II III E each atom has 1 and not 3 labels so each operation is a 3x3 matrix as opposed to the 9x9 matrices of the Cartesian basis. In addition, there can be no sign change for an s- orbital. The resulting representation is, E = 1 0 0 1 0 0 1 0 0 0 1 0 ; C 2 = 0 0 1 ; σ v = 0 1 0 ; 0 0 1 0 1 0 0 0 1 σ ' v = 1 0 0 0 0 1 0 1 0 Γ 3 1 3 1

Intuitive approach to finding the basis vectors in the s-orbital space Consider the symmetry adapted linear combinations (SALC's) A C for water s-orbitals I II III II III A = I B = II + III C = II - III A B = C 1 0 0 0 1 1 0 1 1 I II = III I II + III II III

In this basis, no basis vector is changed into another by a symmetry operation, i.e., this basis is symmetry adapted. Old 3x3 is now three 1x1 matrices and reducible representation Γ S is now three irreducible representations, Γ A, Γ B and Γ C. I II III Γ S = 3 1 3 1 i.e., Γ S = Γ A + Γ B + Γ C E C 2 σ σ' Γ A 1 1 1 1 Γ B 1 1 1 1 Γ C 1-1 1-1 Γ S 3 1 3 1 I A Γ A = 1 1 1 1 II III Γ B = 1 1 1 1 II III Γ C = 1-1 1-1 B C

Rotation basis

R z E C 2 σ v(xz) σ v(yz)

R z E C 2 σ v(xz) σ v(yz) 1 1-1 -1

R y E C 2 σ v(xz) σ v(yz)

R y E C 2 σ v(xz) σ v(yz) 1-1 1-1

Rotation of the water molecule, H z O x y O H H H ΓR(z) = 1 1-1 -1 C 2v E C 2 σ v (xz) σ v '(yz) R z +1 +1-1 -1 a 2 R x +1-1 -1 +1 b 2 R y +1-1 +1-1 b 1 Γ rot 3-1 1 1 Thus, Γ rot = a 2 + b 1 + b 2

Translation basis Cartesian basis

Translations along the x, y and z directions (x, y, z) transform in the same way as px, py and pz. z O x yx y O ΓT(x) = 1-1 1-1 H H H H C 2v E C 2 σ v (xz) σ v '(yz) T z +1 +1 +1 +1 a 1 T y +1-1 -1 +1 b 2 T x +1-1 +1-1 b 1 Γ trans 3-1 1 1 Thus, Γ trans = a 1 + b 1 + b 2

Motions of H 2 O The basis vectors are shown for the three atoms of water. Also shown is the result of the C 2 rotation: Using analysis of the 3N Cartesian vectors we can determine a reducible representation. We have already seen that the method involves counting the characters of the unmoved atoms. The important realization is that all atoms that are unmoved by a particular symmetry operation must have the same character.

We showed previously that C 2 has a character of -1 and σ v (yz) has a character of 3. For a reflection through the plane bisecting the H-O-H bond angle, χ(σv') = +1 since only the O is unshifted and a plane contributes +1 for each unshifted atom. The character for the identity element will always be the dimension of the basis since all labels are unchanged. For water then, χ(e) = 9. The representation (Γ) for water in this Cartesian basis is: E C 2 σ v (yz) σ v '(xz) Γ 9-1 3 1

Reducible and irreducible representations We call the representation Γ a reducible representation. Here we write the reducible representation as: E C 2 σ v (yz) σ v '(xz) Γ 9-1 3 1 The irreducible representations form the basis of the point group in the same way that the vectors along x, y and z form the basis for three dimensional space. H 2 O belongs the point group C 2v. In this point group There are 4 irreducible representations, A 1, B 1, A 2, B 2. The decomposition of the reducible representation is a Unique determination of the irreducible reps (or irreps)z spanned by Γ.

Internal coordinates as a basis The internal coordinates are Stretch r Bend θ Torsion τ Wag ω The advantages of this coordinate system are: Translation and rotation are eliminated. Force constant are defined in terms of bond stretches, valence angle bends, torsions, and wags. These quantities can be related to bond strengths and barriers for internal rotation.

Example of H 2 O For example, for H 2 O we have the following internal coordinates. r 1 r 2 θ 1 The bond coordinates r 1 and r 2 transform as: C 2v E C 2 σ v(xz) σ v(yz) Γ 2 0 2 0 This is a reducible representation.