SUPPLEMENTARY NOTES FOR NATURE OF THE CHEMICAL BOND

SUPPLEMENTARY NOTES FOR

NATURE OF THE CHEMICAL BOND

copyright 2009 William A. Goddard III, all rights reserved

Prepared by TAs for Caltech Course

Special KAIST 2009 version

Ch120 - Study Guide 01

Adam Griffith

September 26, 2005

In this guide: about study guides; comments on ch120; basic principles of quantum mechanics; applica-tions of chemical bonding to H+

2 ; suggested readingLast updated September 27, 2005.

1 About Study Guides

The purpose of this study guide is to attempt to elucidate and clarify topics that were covered in lecture.Unfortunately, there may be nuances or details in the lecture that cannot be adequately captured by thestudy guide. This guide should be used as a supplement to your own class notes, not as a substitutefor attending lectures.

2 Comments on Ch120

Professor Goddard’s handwriting can be difficult to decipher at times. If you are unsure what he haswritten and think that it might be important, then by all means ask him or the TAs. If you can’t read it,you can bet that there are others who can’t as well.

Concerns about the mathematical and physics prerequisites for the course are valid. However, while thelecutre material will sometiems seem particularly mathy, the math you need to know for the course ismostly algebra. Additionally, as we progress through the course material we will move away from mathand more toward chemistry. Familiarity with the notations and mathematical formalisms that ProfessorGoddard introduces is the most important thing for you to learn, and their importance supersedes anyquantum mechanical review you may think you need.

The purpose of this course is to give you the intuition to predict properties and structures of chemicals andmaterials. If you have any questions or comments please feel free to contact any of the TAs.

3 Basic Principles of Quantum Mechanics

• Why Do We Need Quantum Mechanics?

We began by discussing the importance of Quantum Mechanics (QM) and how it is necessary dueto the failure of Classical Mechanics (CM) to describe the existence of atoms and thus chemicalbonds. The basis for the argument against CM is dependent on using the definition of energy andthe hydrogenic atom as a model. The hydrogenic atom has one proton and one electron separatedby a distance R. We should be able to recall the energy components that comprise the total energy

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of this system: Potential Energy (PE) and Kinetic Energy. Solving for the total energy of thissystem is simply solving for its PE and KE and taking the sum of the two. The classical values forthese terms are:

PE =−e2R

(1)

KE =1

2mv2 (2)

We oftenfurther abbreviate PE and KE to the letters representing their operators in classical andquantum mechanics: V and T, respectively. If we were looking for the lowest energy of a system(i.e. its most probable state), we need to minimize the two components of the total energy E. Noticethat the classical values of V and T are independent of each other: V is a function of position whileT is a function of velocity. Thus in a classical system, energy is minimized by having the electronon top of the proton. According to Coulomb’s Law as R → 0, E → −∞, which is not physicallypossible. Therefore we need a different approach to understanding chemical bonding.

• Quantum Mechanics in a Very Small Nutshell

In order for us to develop a quantum mechanical intuition about chemical and material systems weneed to first recognize a few important details of QM. A rigorous QM course will discuss the fourpostulates of QM. If you are unfamiliar with these postulates, they should be in any standard QMtextbook. Instead of going through the postulates individually, we talked about the significance ofthe pstulates as a whole. These are the main points:

1. The wavefunction ψ(x, y, z, t) contains everything we want to know to solve for the energy ofa particular state.

2. The probability density of a wavefunction is∫ρ(x, y, z, t)dτ and must be normalized in order

to be relevant. I.e.,

P =

∫ρ(x, y, z, t)dτ =

∫ψ∗ψ = 〈ψ | ψ〉 = 1 (3)

3. Superposition allows us to make linear combinations of “good” wavefunctions to make other“good” wavefunctions.

4. The Hamiltonian (H) determines the time-evolution of ψ.

5. Phase factors do not change the state that a wavefunction represents.

6. We can show that if H is independent of time, then ψ is an eigenfunction for the Time-Independent Schrodinger Equation:

Hψ = Eψ (4)

7. By the variational principle, the energy of an approximate wavefunction describing a partic-ular state is greater than or equal to the energy of the exact wavefunction describing the state.This means that “better” wavefunctions have lower energy.

2

8. The expectation value for the classical operator V is − e2

R. Since R is the average value of R

over the entire wavefunction, it is dependent on the wavefunction. Similarly, the expectationvalue for T is dependent on 〈| ∂ψ

∂x|2〉, and it is greater when the gradient (or slope) of the

wavefunction is steep.

9. In quantum mechanics T and V are dependent on the same wavefunction ψ; whereas, in clas-sical mechanics T and V can vary independently.

By rigorously following these rules we are now ready to embark on using QM to make predictions.

• Properties of Characteristic Wavefunctions

We then learned how – based on the qualitative shape of a wavefunction – we can determine whatits energy components are. PE is most favorable when R is small, i.e. the average distance betweenthe electron and proton is small. Therefore, wavefunctions that are narrow and have a smaller R arefavored in terms of PE. Alternatively, KE is more favorable when the wavefunction is more smooth,as is the case with relatively flat wavefunctions (See Figure 1). Deeper discussion of this subject canbe found in the suggested reading (Goddard Book 1.2.1).

• The Atomic Radius of Hydrogen

With the previous information we can now determine the qualitative shape of an ideal wavefunction.The ideal wavefunction must minimize the sum of T and V. The T term has 1

R2 dependence, andthe V term has 1

Rdependence. Thus, at short R the T term dominates, but at long R the V term

dominates. The minimum energy of the system is obtained at an intermediate value of R. Theoptimal distance between an electron and a hydrogen nucleus can be found by taking the derivativeof the sum and setting it equal to zero:

E = T + V (5)

=�

2

2m

1

R2− e2

R(6)

∂E

∂R=

−2�2

2m

1

R3+e2

R2= 0 (7)

R =�

2

me2≡ a0 = 0.529A (8)

Using this value of R, the energy of the hydrogen atom is:

E =�

2

2m

1

a20

− e2

a0

= −13.6eV (9)

There are several different units we can use for energy. The energy unit in atomic units is theHartree, Eh, which is defined as:

1Eh = 27.2eV = 627.5kcal

mol≡ e2

a0

(10)

3

4 Applications of Chemical Bonding to H+2

Now that we have solved for the energy of the hydrogen atom (the energy of an electron paired with aproton), we can approach a more applical problem: the energy of an electron paired with two protons.The wavefunction for this system is not unlike that for the hydrogen atom. We considered two protonsnext to each other at an interatomic distance of R for R → ∞. One electron can be centered on either theleft proton or the right proton. The wavefunction with the electron centered on the left will henceforth belabeled ψl. Similarly, the wavefunction with the electron centered on the right will be labeled ψr.

To study the chemical bond of H+2 we want to consider how the electron behaves at varying R. As we

bring the two protons together (i.e. when R becomes finite), the electron is no longer confined to beon one particular proton, and it can begin to have some probability of being on both protons. Thus ourwavefunctions ψl and ψr are no longer realistic wavefunctions to describe the state of H+

2 at finite R. Wecan generate new (but approximate) wavefunctions from the superposition principle by taking a linearcombination of atomic orbitals (LCAO). We will be able to apply the variational principle to evaluatetheir relative “goodness” by qualitatively determining their relative energies. The LCAO method does notguarantee perfect wavefunctions, but it is a useful first-approximation that can be done without computerswith relative ease.

With the LCAO method, besides the two wavefunctions ψl and ψr, we can let one wavefunction be (ψl+ψr)and the other be (ψl − ψr). We then estimate how the energies of these wavefunctions vary with nuclearseparationR. Note: While what follows from this appears to be very similar to the hydrogen atom example,this example is being used to determine whether the diatomic H+

2 is bound, or if it will spontaneously fallapart into a H atom and a proton (H+).

A question appears in that we do not know whether (ψl + ψr) or (ψl − ψr) is better for describing thebonding of the system. To have good KE, we need a wavefunction that is as smooth as possible. Thewavefunction (ψl + ψr) fits that criteria, and since there is a relatively high probability of the electronbeing located between the two protons, its value of R is larger than in the ψl and ψr wavefunctions.

For clarity, we will define ψg = (ψl + ψr) and ψu = (ψl − ψr). The subscripts g and u will be explainedin more detail later. First, we can consider how ψg varies from the single proton-centered wavefunctionsψl and ψr. The KE is small and more favorable for the flatter wavefunction ψg. It can be said that ψgis stabilized when the wavefunction is flat over a large region of space (See Figure 2). This leads to theconcept of contragradience – the KE stabilization that arises when two atomic orbitals (ψl and ψr) haveopposing gradients over a large region of space. Consider the following implications at various distancesof R:

• R → ∞: small contragradience over a very large region → no effect

• R → 0: large contragradience over a very small region → no effect

• Intermediate R: large contragradience over a large region → favorable KE effect

In the case of ψu the effect is reversed. As R → ∞ the gradient of ψu at the center of the bond increasesgreatly causing the overall KE to increase substantially.

The effect of shifting charge density to and from the nuclei also has an effect on the PE of the wavefunc-tions ψg and ψu, however these effects are generally considered secondary to the effect in KE.

Additional discussion of this example can be found in the suggested reading (Chapter 2 of the GoddardBook). This is a very important concept that we will revisit many more times.

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5 Suggested Reading

• Quantum Mechanics Review – (Insert your favorite undergraduate physical chemistry text here,i.e. Atkins, Levine, McQuarrie, etc.)

• Mathematical Formalisms in Quantum Mechanics –

– Liboff, R; Introduction to Quantum Mechanics

– Shankar, R; Principles of Quantum Mechanics

– Szabo, A. and N. Ostlund; Modern Quantum Chemistry

• Class Notes Reading (Goddard Book) –

– Basic Principles of QM – Section 1.1.1, 1.1.4-1.1.7

– The Hydrogen Atom – Section 1.2.0-1.2.3

– The Nodal Theorem – Section 1.3.0-1.3.1

– The Chemical Bond of H+2 – Section 2.0, 2.1.0-2.1.1, 2.1.3

6 Figures

Figure 1: Atomic hydrogen orbitals with varying R.

5

Figure 2: An illustration of wavefunctions for the hydrogen atom (top) and the ψg (middle) and ψu (bot-tom) states of H+

2 .

6


Adam Griffith

September 28, 2005

In this guide: nodal theorem; molecular orbital theory; valence bond theoryLast updated September 30, 2005.

1 The Nodal Theorem

The nodal theorem is a qualitative rule we can apply to wavefunctions. For the wavefunctions ψg and ψu

in Figure 1 we can easily recognize that ψu has a node. (A node is a point where the probability density ofa wavefunction is zero.) A wavefunction with a node is less-smooth than its node-less counterpart. Thuswe can easily say that ψg is a better (lower energy) wavefunction than ψu.

2 Summary: MO vs. VB

When two electrons occupy the same bond at the same time, they repel each other, and their motionsare not perfectly independent. We consider two approaches: molecular orbital theory, which assumeselectrons do not repel each other and move independently, and valence bond theory, which assumeselectrons interact so strongly they cannot be on the same nuclei at the same time. The truth lies betweenthe two extremes. VB theory is better at describing dissociation, and MO is better at describing excitedstates and is useful for atomic spectrum calculations.

3 Molecular Orbital Wavefunctions

The hydrogen molecule Hamiltonian is given by:

H =

(−1

2∇2

1 −1

rL1

− 1

rR1

)+

(−1

2∇2

2 −1

rL2

− 1

rR2

)+

1

r12(1)

If we neglect 1/r12, the Hamiltonian becomes separable into two parts involving one coordinate only,both of which look like the H+

2 Hamiltonian. We create eigenstates of this approximate H by multiplyingeigenstates of H+

2 for each electron together. Recall that H+2 has two LCAO solutions (Figure 1). Then the

ground state MO wavefunction is Φgg = φgφg (Figure 2).

MO excited states can be formed similarly, but with a complication. In order for the wavefunctions tobe symmetric or antisymmetric under inversion, some of the product wavefunctions are combined intolinear combinations (Figure 3). We’ll see why this is necessary in upcoming lectures on symmetry andcommuting operators.

1

4 Deficiency of the MO Wavefunction

The MO wavefunction, though easy to understand conceptually, fails to describe H2 binding energetics,particularly at the dissociation limit (Figure 4).

The MO wavefunction fails because it does not consider electron correlation, which describes how twoor more electrons interact with each other. In leaving out the 1/r12 term, it assumes electrons moveindependently.

If we expand out the MO wavefunction, we see that electrons are just as likely to both be on the samenucleus (LL and RR, ionic terms) as they are to be on different nuclei (LR and RL, covalent terms):

ΦMO = (φL + φR)(φL + φR) = (2)

= φLφL + φLφR + φRφL + φRφR (3)

5 Valence Bond Wavefunction

The valence bond wavefunction eliminates the LL and RR terms, forcing the two electrons to be ondifferent nuclei (Figure 5).

ΦV Bg = χLχR + χRχL (4)

Our wavefunction now suffers from “correlation overkill”, but the resulting description is closer to theexact potential energy surface, suggesting that the electrons in the H2 bond are in fact strongly correlated.

6 Improving MO and VB Wavefunctions

The exact wavefunction lies between the extremes of the fully correlated VB wavefunction and the fullyuncorrelated MO wavefunction. We can improve the VB wavefunction by adding partial ionic character:

Φ′V B = φLφR + φRφL + λ(φLφL + φRφR) (5)

In the same way, we can improve the MO wavefunction by adding some of the uu excited state:

Φ′MO = (φL + φR)(φL + φR) + γ(φL − φR)(φL − φR) (6)

This procedure, called configuration interaction, produces a wavefunction identical to the improvedVB wavefunction for λ = (1 + γ)/(1 − γ). As an exercise, plot this “intermediate” wavefunction withcontours, and note how the electron-electron repulsion is reduced in comparison to the uncorrelated MOwavefunction.

2

7 Figures

Figure 1: The LCAO States of H+2 .

3

Figure 2: MO wavefunction contour plot.

Figure 3: Ground and excited MO wavefunctions.

4

Figure 4: Potential energy plot of MO, VB, and Exact wavefunctions.

Figure 5: VB wavefunction contour plot.

5

Study Guide for Lecture 3

Sanja Pudar

September 30, 2005

1 Study Guide Outline

1. Review of last lecture

2. Ionization potentials of atoms

3. ��

� VB and MO descriptions

4. �� VB and MO descriptions

5. Inversion and transposition

6. Figures

2 Guide

2.1 Review of Last Lecture

In our last lecture, Professor Goddard recalled that the time independent Schrodinger Equation is:

� ��

��

The Variational Principle lets us know that the energy for an exact wavefunction describing a par-ticular state will always be lower in energy than an approximate trial function that we can createthrough LCAO methods or other means. Additionally, we defined the wavefunction � to be

� � ��

where � is a parameter that is related to the “contractedness” of the wavefunction. In other words,when � is large, the hydrogenic wavefunction �will have a high probability density near the nucleus.Likewise, when � is small, the wavefunction’s probability density will be spread away from thenucleus.

We would like a general expression for the energy of one electron in any atom. With this equationwe could solve for the ionization potential for a one-electron atom. We may use the Schrodinger

1

Equation with the wavefunction �,but we also must recognize that the potential energy of the elec-tron will have an additional dependence on Z, the atomic number of the atom where the electron islocated:

� ��

��

We can then minimize this energy with respect to � and re-obtain an expression for the atomic radiusof the hydrogen atom (or any other one-electron atom):

�

��

��

��

�

��

��

��

which last time lead us to define �� as the Bohr radius

��

��

2.2 Ionization Potential of Atoms

For the sake of perspective, we then considered a general nucleus with charge �� with an electronand with an electron-nuclear distance of�. The potential energy of this system is simply Coulomb’sLaw: � � ��

��

. The Virial Theorem allows us to automatically state that ��

� � and thus

� � �

� � � ��

�

��

��

. By simply finding the atom’s optimal electron-nucleus separation, �, andinserting the atom’s value of �, we can solve for the ionization potential of the atom.

For the � atom: � � �� bohr, and so IP=13.6 eV. For the �� atom: �� bohr and so IP=54.4 eV.

2.3 ��

� VB and MO Descriptions

As a review from the previous lecture, we explained that with the LCAO wavefunctions of the �and atomic orbitals we can generate the �� and �� wavefunctions which in MO terms aredenoted as � and �, respectively. The � state is lower in energy than the � state due to the nodaltheorem. Since there are no other electrons for the one elecron in ��

� to interact with, the MO andVB pictures are completely identical.

2.4 �� VB and MO Descriptions

Now that we are ready to look at systems with two electrons, we start to see a distinction betweenVB and MO theories. VB and MO theories can be classified by their degrees of static correla-tion. VB theory incorporates static correlation in that the location of electron probability densitydepends on other electronic probability. As a result, electrons are usually kept far apart, and thusthe VB construct is preferred for situations where electrons would be far apart, e.g. situations wherebonds are forming or breaking. MO theory does not incorporate static correlation and so there is

2

no explicit restriction for electrons to be near or far from each other within the orbital picture. Thislack of restriction is beneficial for calculating excitation energies but is a major problem for bonddissociations (See Figure 1).

To analyze qualitative differences between VB and MO theories we represented thewavefunctionsfor the �� system as a 2-D grid showing the relative placement of each electron on each atom center.

(a) The VB description

Since VB formalism does not allow electrons to move freely with respect to other electrons,we need to specify a notation to express the wavefunction of��. The two VB wavefunctionswe used for �� are � �� and �� (See figure 2). Since � �� has no node, it isthe ground state.

(b) The MO description

The MO formalism has no static correlation and so it is possible to consider a state with bothelectrons being centralized on either atom of ��. Thisstate is denoted as the �� state, and itis a good wavefunction since inversion of the two electrons results in the same wavefunction.Alternatively, we find that the wavefunctions �� and �� are not good wavefunctions, howeverthe LCAO orbitals �� and �� are good wavefunctions. We find from the picture ofthe wavefunctions that both should have identical energy, but if we consider electron repulsionwe recognize that the �� state has both electrons on one of the nuclei (see Figure 3). Thisraises an important point about MO theory: even though MO theory allows electrons to be ontop of each other, this does not mean it is energetically favorable.

Considering the VB and MO binding curves we see that MO theory gives a much higherdissociation energy in part to the fact that some of the dissociated species will have ��

character (ionic state) in addition to � � � character (covalent state), while VB only allows� �� character in its dissociated species (See Figure 1).

Below are shown the binding curves for four MO wavefunctions.

2.5 Inversion and Transposition

Inversion is an operation that inverts each of the coordinates in a wave function from a positivevalue to a negative value, that is, �� become ��. We saw that applying the inversionoperation to our 2-electron wavefunction did not change the value of the Hamiltonian. Furthermore,if we apply the inversion operation to a wave function twice, we return to the initial wavefunction:

��

We are interested in determining the relationship between the wave functions before and after ap-plying the inversion operator, and for that we consider the cases where (i) the original state and theinverted state are the same state (the system is non-degenerate) and (ii) the two states are differentbut have the same energy (the system is degenerate). We know there are at most two states becauseapplying the inversion operator twice to a wave function gives us back the original function. If thesystem is non-degenerate, both states will differ at most by a phase factor, � � ��, where � is a realnumber. In this case we have:

��

3

from which we can deduce that � can only have the values of �� and ��. This means that whenwe apply the inversion operator to our wave function, it will either remain the same, or it willchange sign. We call functions that remain the same “gerade functions” (from the german wordgerade=even) and those that change sign upon inversion we call “ungerade” (uneven). Our analysisshows that all 2-electron wavefunctions must either be gerade or ungerade. This also holds for thecase when the system is degenerate and we have two different states,� and �, because even if �

and �� cannot be related by a phase factor, we can always write linear combinations of the form � � � �� or � � � ��, which are gerade and ungerade respectively as can be easilyverified.

Examples of gerade functions are ��, ��, ��, ��. An exampleof an ungerade function is ��, which upon inversion becomes ��),the negative of the original function.

We also considered another type of operation called transposition, where the coordinates of twoelectrons are exchanged, that is �� become �� and viceversa. Using the samereasoning as above, we can conclude that upon the application of transposition, our special wavefunctions will either remain the same or change sign. Those that remain the same we call symmetric,and those that change sign we call antisymmetric. In terms of notation, transposition does thefollowing:

��

that is, we xchange the electron numbers. Examples of symmetric functions are ��, ��, �� ,�� . Examples of antisymmetric functions are �� and �� , as can beeasily verified.

Suggested additional reading:

� Bonding basics in ��

� and �� - Goddard Book: 2.1.1-2.1.4

� VB and MO descriptions of ��

� and �� - Goddard Book: 2.3.3,2.3.4

� Two electron permutation symmetry - Appendix 3A - Goddard notes

4

2.6 Figures

Figure 1: The VB and MO binding curves of �� compared to the exact binding energy.

Figure 2: The creation of VB wavefunctions from �� and �� atomic functions.

5

Figure 3: The creation of MO wavefunctions from �� and �� states.

Figure 4: Energies for the states of ��.

6


John Keith

October 3, 2005

In this guide: Spin, Pauli Principle, Overlap and BondingLast updated October 5, 2005.

1 Special note

Although this course requires the student to only have a low level of math background, the lecturesfor this week are substatially math-focused. This will not be the norm for the course however,and in following weeks we will devote time most of our time to chemical reasoning rather thanmathematical reasoning. The object of this course is to lay a foundation that is rigorous (math-ematical) and build applicable (chemical) rationales for solving important problems in chemistryand materials science. If you are daunted by the math we show do not worry! These lectures wellnot compromise the entire course, and we will move into applications in the 3rd week, and yourTA’s will be more than happy to make the material presented as clear and accessible as possible.

2 Symmetry review

Last class we looked at the VB and MO formalisms for H2. Since the Hamiltonian is invariant un-der transposition, we needed a representation of electrons that would either be symmetric or anti-symmetric under transposition operations. Additionally, since they are describing the H2 molecule,the wavefunctions must also be symmetric or antisymmetric under the inversion operator.

Considering VB wavefunctions LL, RR, LR, RL, LR + RL and LR − RL and MO wave-functions φgg, φuu, φgu, φug, φgu+ug, φgu−ug. Which of these wavefunctions are valid, and what arethere symmetries under transposition and inversion?

3 Spin

So far we have only considered spatial wavefunctions. Now let us expand our scope and considerwhat if our electrons have a finite size (say a small sphere). Now, instead of simply expressingthe position of the sphere, we now need to express its orientation using variables σθ and σϕ. Nowwe have five different coordinates in our wavefunction: x,y,z, σθ, and σϕ. If the orientation of thesphere is totally independent of the position, the wavefunction will be factorable into two parts:

1

ψ (x, y, z, σθ, σϕ) = φ(x, y, z)χ(σθ, σϕ) = φ(r)χ(σ) (1)

It this electron is spinning about its axis with an angular momentum s, the fact it has a chargewould lead to a magnetic moment

μ = γs

which could be detected with an external magnetic field leading to

ΔE = −μzBz = −γBzsz

Classically, sz could have any value ±|s| where |s| is the total angular momentum. Experi-mentally, only two values of sz are found by the magnetic field: sz = ±1

2�. This is not to say

that an electron should be modeled as a sphere, however this internal angular momentum we justencountered does exist. The eigenvalues that come from the spin operator are orthogonal and inthis class we refer to them as spin functions. The spatial functions will be referred to as orbitals,and the products of orbitals and spin functions are called spin orbitals.

If we take the dot product of two states (spin orbitals), ψi and ψj , if both states are α, or one isα and one is β:

〈ψiα|ψjα〉 = 〈φi|φj〉〈α|α〉 = 〈φi|φj〉〈ψiα|ψjβ〉 = 〈φi|φj〉〈α|β〉 = 0

Therefore, states with different spin are always orthogonal, regardless of φi and φj . States withthe same spin are orthogonal when φi and φj are orthogonal.

Now consider our Hamiltonian and assume the it is not dependent on spin, but only on orbitals.Then:

Hψiα = (Hφi)α = Eiφiα = Eiψiα

Hψiβ = (Hφi)β = Eiφiβ = Eiψiβ

and we find there are two spin orbital eigenstates corresponding to each orbital eigenstate.Recall that this is based off of an assumption that the Hamiltonian is not dependent on spin, whenindeed it is, and these energies are called spin-orbit coupling energies. However, the effects ofspin-orbital coupling is minimal for the first 4 rows of the periodic table, and only important toconsider in rows 5 and below.

2

4 Pauli Principle

Assigning α and β to Looking at spin orbitals for a 2-electron system (and assigning every combi-nation of α and β to each) we have:

ψαα(1, 2) = (φ(1, 2))(α(1)α(2))

ψαβ(1, 2) = (φ(1, 2))(α(1)β(2))

ψβα(1, 2) = (φ(1, 2))(β(1)α(2))

ψββ(1, 2) = (φ(1, 2))(β(1)β(2))

Recall though that the orbitals were required to be either symmetric or antisymmetric undertransposition. The same is true for the spin orbitals. Therefore, our four spin orbitals with expandedspin orbits are not those above, but:

ψαα(1, 2) = (φ(1, 2))(αα)

ψαβ(1, 2) = (φ(1, 2))(αβ + βα)

ψβα(1, 2) = (φ(1, 2))(αβ − βα)

ψββ(1, 2) = (φ(1, 2))(ββ)

For a given set of orbitals, there are 4 uniquely possible states. However, only those that areantisymmetric under transposition are observed by experiment.

Spin Function Ms Number of states Symmetryα(1)α(2) 1 S=1 (triplet) symmetric

α(1)β(2) + β(1)α(2) 1 S=1 (triplet) symmetricβ(1)β(2) 1 S=1 (triplet) symmetric

α(1)β(2) − β(1)α(2) 0 S=0 (singlet) antisymmetric

A postulate is incorportated into QM to account for this called the Pauli Principle. Pauli’sprinciple states that overall wavefunctions must be antisymmetric – i.e. the wavefunction mustchange sign when any two electrons are exchanged.

That is to say:

ΨA(1, 2) ≡ ψa(1)ψb(2)

If we interchange electrons, we need a new wavefunction:

ΨB(1, 2) ≡ ΨA(2, 1) = ψb(1)ψa(2)

3

To have the property that

ΨB(1, 2) = −ΨA(2, 1)

so that

ψb(1)ψa(2) = −ψa(1)ψb(2)

This is not a valid wavefunction, but this linear combination is

ψa(1)ψb(2) − ψb(1)ψa(2) (2)

This spin orbital has some interesting properties:

• Consider when ψb = ψa. The wavefunction is zero, and we cannot place two electrons in thesame spin orbital.

• Consider a new function, ψc = ψb+γψa. Where γ is non-zero, and thus ψc is not orthogonalto ψa.

〈ψa|ψc〉 = 〈ψa|ψb〉 + γ 〈ψa|ψa〉 = γ �= 0

If we use ψc in place of ψb in equation X, we see

ψa(1)ψc(2) − ψc(1)ψa(2) = ψaψb + γψaψa − ψbψa − γψaψa = ψaψb − ψbψa

So we say that if a wavefunction as any overlap, the exact overlapping portion cancels itselfwhen it is antisymmetric.

We can define a special operator, called the antisymmetrizer, to make any spin orbital productan antisymmetric wavefunction. We know that the antisymmetrizer should be able to generate anantisymmetric wavefunction for any number of atoms. We can do this by employing a Slater (orStackel, as Bill enjoys mentioning) determinant.

∀Ψ(1, 2, . . . , N) ≡(

1

N !

)1/2

∣∣∣∣∣∣∣∣∣

φa(1) φb(1) · · · φz(1)φa(2) φb(2) · · · φz(2)

......

. . ....

φa(N) φb(N) · · · φz(N)

∣∣∣∣∣∣∣∣∣For example with the non-antisymmetric spin orbital made up of orthonormal components

Ψ = φaφbαα:

∀Ψ =

∣∣∣∣ φa(1) φb(1)φa(2) φb(2)

∣∣∣∣ = (φaα)(φbα) − (φbα)(φaα)

= (φaφb − φbφa)(αα)

4

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+ �� <� =�

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"


Sanja Pudar

October 9, 2005

Study Guide Outline

1. Hydrogen Atomic Orbitals

2. Shielding

3. The Aufbau Principle

1 Hydrogen Atomic Orbitals

Atomic orbitals of hydrogen atom can be obtained by solving the Schrdinger equation,which happens tobe separable into a radial and an angular components. Unfortunately, the Schrdinger equation can besolved exactly only for simple systems, such as hydrogenic atoms. Such orbitals are classified accordingto their energy levels (which depend on the total number of nodes -radial andangular) and also accordingto the number of angular nodes and symmetry. The suggestions for further reading at the end of this guidewill help you visualize these orbitals better (you can also find them in almost every quantum mechanicstextbook).The energy of hydrogen orbitals is calculated to be � � �

��

��. Knowing that �� and �� orbitals both

have the same total number of nodes, their energy is the same, where this is regarded to as ”accidentaldegeneracy.“ The same happens in case of ��, ��, and �� orbitals.In order to classify orbitals according to symmetry, we consider how they change sign when rotated aboutthe z axis. If there is no sign change (�� ) , then the symmetry of orbital is �. If there are two signchanges in order to get to the original function (�� ), then the symmetry is �. Finally, if thereare four sign changes in order to get to the original function (�� ), then the symmetry is Æ. It isvery important that you familiarize yourself with the different orbitals and their properties (geometry andsymmetry) since this course depends on understanding what orbitals do to form bonds.

2 Shielding

We looked at lithium as an example of a system with more than two electrons and said that the energyof each level doesn’t depend only on the total number ofnodes anymore, but that we must evaluate eachorbital in the presence of the others to make energy comparisons between orbitals that would be degeneratein a hydrogenic atom. It is particularly important to evaluate the degree of electron-electron repulsion (i.e.to which degree both electrons occupy the same space) and the degree of shielding of nuclear charge.In the case of lithium, the �� orbital is doubly occupied, so the third electron can go to either the �� or a ��orbital. The �� orbital has a node through the nucleus, while �� electron density is highest at the nucleus.

1

Because of this, the �� electrons completely shield the nucleus so that �� electron experiences effectivenuclear charge (�� ) of ��. In addition, the �� orbital has long lobes that would overlap less with thealready filled �� orbital, which means that there is less electron-electron repulsion.On the other hand, �� orbital is nonzero at the nucleus, and therefore �� electrons partially shield thenucleus, so that �� electron experiences larger effective nuclear charge. However, due to the significantoverlap of �� and �� electron density, the electron repulsion is greater than in a case of �� orbital.The calculated effective nuclear charges for �� and �� electrons are 1.28 and 1.00 respectively. Overall,for this system, the �� has a lower energy, which tells us that the larger effective charge dominates thehigher electron-electron repulsion effect. The ground state of Li atom is ��, followed by the firstexcited state ��.

3 The Aufbau Principle

Similar shielding effects cause splitting of states with the same n value resulting in the following orderingof orbital energies:

��

Higher number of angular nodes leads to less penetration of the core orbitals, which leads to greatershielding by core orbitals. Ground states of most atoms can be qualitatively predicted by the AufbauPrinciple which arises from shielding effects. Below is given the order in which atomic orbitals are filledaccording to the Aufbau Principle:

��

Notice that n-splitting is in some cases smaller than the l-splitting.

Suggested further reading:

� Excited states of H atoms 5.1

� Hydrogenlike wave functions 5.1

� Shielding and the aufbau principle 5.2.1a to 5.2.2

2


Adam Griffith

October 10, 2005

In this guide: Pauli Principle, atomic orbitals, Hund’s Rules, electronic configurations, suggested readingLast updated October 12, 2005.

1 Pauli Principle

A very quick review of the Pauli Principle. A spin-orbital has the form Ψ(r, σ) = Θ(r)χ(σ) with thespatial and spin functions being separable. The Pauli Principle states that the total wavefunction (spin-orbital) must be anitsymmetric with respect to exchange of electrons. That is, for the transposition operatorτij = τ spatial

ij τ spinij

τijΨ(1 . . . i . . . j . . . N) = Ψ(1 . . . j . . . i . . . N) = −Ψ(1 . . . i . . . j . . . N) (1)

We can ensure that our wavefunctions are antisymmetric by using the antisymmetrizer:

∀(ψa(1)ψb(2)...ψn(N)) =

∣∣∣∣∣∣∣∣

ψa(1) ψb(1) . . . ψn(1)ψa(2) ψb(2) . . . ψn(2). . . . . . . . . . . .

ψa(N) ψb(N) . . . ψn(N)

∣∣∣∣∣∣∣∣(2)

Because of the properties of determinants we can easily see that if ψa = ψb, then our total wavefunctionwill be 0. Additionally, transposition of electrons is equivalent to switching rows i and j in the determinant,which will give an additional factor of −1.

2 Atomic Orbitals

Dr. Goddard reviewed single-electron atomic orbitals in class. Rather than filling up several pageswith images of different orbitals and their wavefunctions, we would suggest checking out this website:http://www.shef.ac.uk/chemistry/orbitron/ (or any of the many other websites that you can find by usingGoogle).

3 Hund’s Rules and Term Symbols

Each atomic wavefunction can be described using term symbols of the form 2S+1LJ . S is the total spinquantum number, and 2S+1 is referred to as the spin multiplicity. L is the total orbital angular momentum.J is the total angular momentum (a combination of S and L). Hund’s Rules specify that:

1

1. The term with the maximum multiplicity (2S+1) lies lowest in energy. That is, the singlet state ishighest in energy, followed by the doublet state, the triplet state, the quartet state, etc.

2. For a given multiplicity, the term with the largest value of L lies lowest in energy. That is . . .< F <D < P < S.

3. For atoms with less than half-filled shells, the level with the lowest value of J lies lowest in energy.

For more detail go to: http://hyperphysics.phy-astr.gsu.edu/hbase/atomic/hund.html.

4 Ground State Electronic Configurations

The ground state electronic configurations of several orbitals are quickly described:

Be 1s22s2

B 1s22s22pC 1s22s22p2

N 1s22s22p3

O 1s22s22p4

F 1s22s22p5

Ne 1s22s22p6

Dr. Goddard’s unique way of drawing these orbitals was quickly described and will be shown in moredetail in coming lectures.

5 Suggested Reading

• Goddard book Chapter 4

• Goddard Book Chapter 5 (particularly 5.3)

• http://www.shef.ac.uk/chemistry/orbitron/

• http://hyperphysics.phy-astr.gsu.edu/hbase/atomic/hund.html

2


John Keith

October 12, 2005

In this guide: Term Symbols, hydride bonds, molecular structuresLast updated October 18, 2005.

1 Term Symbols

One of the main topics we discussed in lecture today is the means by which we identify states –term symbols. Term symbols are an abbreviated means to denote a precise electronic state of amolecule while stating its spin multiplicity, total spin angular momentum, and how its electronicspin is coupled with its angular momentum.Atomic term symbols have the form2S+1LJ . S is thetotal spin angular momentum (1/2 for a doublet state, 1 for a triplet). L is the total orbital angularmomentum and is related to the number of degenerate configurations we can arrange in a particularstate (values of L are assigned as0 = S, 1 = P, 2 = D, etc). J (typically called Russell-SaundersCoupling) corresponds to the vectoral sum of L and S. One can use GVB diagrams to identifydifferent terms.

Excercise: use the McQuarrie reading online to determine the term symbols for the followingatoms:

• N - electronic configuration =1s22s22p3. Hund’s rules tell us that threeα-spin electronsgive the best configuration for nitrogen atom. Threeα-spin electrons leads toS = 3

2, and

2S + 1 = 4, hence a quartet state. By following another of Hund’s rules we fill the threelone electrons into three differentp orbitals. This leads to anL = 0 or S state. Therefore, theground state term symbol (neglecting Russell-Saunders coupling) is4S. Excited states of Natom also can have a2P or a2D state. What electronic configurations lead to these terms?

• O - electronic configuration =1s22s22p4. Twoα-spins correspond toS = 1, so2S + 1 = 3,a triplet state. The occupations ofp orbitals by the four electrons leads to2L + 1 = 3, orL = 1, aP state. Overall, the ground state of O atom is3P .

• F - electronic configuration =1s22s22p5. Oneα-spins correspond toS = 12, so2S + 1 = 2,

a doublet state. The occupations ofp orbitals by the four electrons leads to2L + 1 = 3, orL = 1, aP state. Overall, the ground state of F atom is2P .

• Ne - electronic configuration =1s22s22p6. Noα-spins correspond toS = 0, so2S + 1 = 1,a singlet state. Fully occupiedp-orbitals leads to2L + 1 = 1, orL = 0, anS state. Overall,the ground state of Ne atom is1S.

1

It is in your best interest to become familiar with these term symbols since they are usedubiquitously in this course.

2 Hydride bonds

GVB diagrams are a useful tool for identifying electronic states of molecules and predict chemicalbonding. Now that we can identify lone atoms with atomic term symbols we can build fragmentsof molecules and piece these fragments together to form molecules. The easiest fragments to startwith are those where atoms are bound to hydrogen, i.e. hydrides.

Consider H2: Two 1s orbitals can overlap to form a bond. The wavefunction for the bondingstate is(lr+ rl)(αβ − βα). Since only two electrons are considered and they are of opposite spin,the antisymmetrizer will not orthogonalize any atomic orbitals since the overall wavefunction issatisfactorally antisymmetric.

Consider He-H: When He and H approach each other, the1s orbital of He is full with twoopposite-spinned electrons. Bringing a third electron from H into this space would force one ofthe electrons in the1s of He to become orthogonal to the other two electrons that would make abond. This orthogonalization is costly though, more than the benefit of forming a bond, and as aresult there is no bond in He-H. In general,σ bonds with 3 electrons will never form, and hencewe can rule out considering most bonds to the Nobel gases (i.e. Ne-H, Ar-H, etc.) We will seelater however that 3-electronπ bonds are not as bad as 3-electronσ bonds, and certain moleculesare best described with 3-electronπ bonds.

Consider F-H: The GVB picture of F has threep orbitals, two with a pair of electrons, andone with a single electron. Using the previous example as an analogy, only the F orbital withthe single electron will form a bond to H. This F-H state is a singlet state since all electronsare paired. Additionally, since all electrons are paired in a closed shell, rotating FH about thez-axis will not result in any sign changes of the atomic orbitals on F. Additionally, a plane ofreflection bisecting the molecule reflects the same doubly occupied orbital making it symmetricunder reflection. Therefore, FH is a1Σ+ state.

Consider FH2 : Note that the closed shell molecule of FH has no singly-occupied orbitals forwith to bond. Therefore, we predict that FH2 is not bound and that is correct.

Consider O-H: The GVB picture of O has threep orbitals, one with a pair of electrons, andtwo with a single electrons. Two unpaired electrons allow for two different orbitals the H atom canbond to. Once it is bond, one electron still remains in ap orbital on O. Since rotation about thez-axis results in a sign change of the atomic orbitals the state might be labeled aΠ state. This willalways be the case for all molecules that have only one electron in a non-pz p-orbital. Therefore,the state for OH is2Π. Note that we do not note its reflection symmetry since aΠ state will alwaysbe antisymmetric under reflection.

Consider OH2: OH from the previous excercise does have a singly-occupied orbital to make abond, so we expect that it will bond to H to form OH2. Identifying a term symbol for this moleculerequires us to recognize the symmetry point group of OH2 as beingC2v. TheC2v point group hasas operations a180◦ rotation, and two planes of reflection. Unique states in theC2v point groupwill either be symmetric or antisymmetric under operations of rotation or by the plane of reflectionthat contains fewer atoms. ’A’ or ’B’ corresponds to symmetry or antisymmetry under rotation

2

while the subscript ’1’ or ’2’ denote symmetry or antisymmetry under reflection.OH2 is a singlet since it has no unpaired electrons. It is also symmetric under rotation and

reflection since all the atomic orbitals on O are doubly occupied. Therefore we assign OH2 withthe1A1 state.

So far we have only encountered molecules with one or fewer electrons in a non-pz p-orbital.In cases were there are more we need to special care to assign the correct term.

Recall that the spatial part for the wavefunction for any orbital can be expressed in terms of itsradial and angular components, I.e. the expression:

ψn,1,ml(r, θ, φ) = Rn,1(r)Y

ml1 (θ, φ) (1)

corresponds to the wavefunction for ap orbital once we insert the quantum numbersn andml

to describe the state we desire. TheR term is the radial function (a function of distance), and theY term is a spherical harmonic (a function ofθ andφ). The spherical harmonics forp orbitals arecomplex since theml = ±1 values correspind to functions with the forme±iφ. Since we cannoteasily see the complex plane, we make linear combinations of the complex spherical harmonics toform new functions wecan visualize. As a result we come up with the following (see McQuarrieor Goddard text for details):

ψpx = Asinθcosφ (2)

ψpy = Asinθsinφ (3)

ψpz = Acosθ (4)

Where A is a constant. The wavefunction for a molecule with two electrons non-pz p-orbitalswould look something like this:

ψ = (xy − yx)(αα) (5)

Since we are interested in angular dependence, we can substitute the angular parts of thepxandpy orbitals above to have:

(xy − yx)(αα) = A((sinθ1cosφ1)(sinθ2sinφ2) − (sinθ1sinφ1)(sinθ2cosφ2))(αα) (6)

Indeed, rotation about the z-axis is a rotation aboutφ, and any functions ofθ would remainconstant also so we have:

(xy − yx)(αα) = A((cosφ1)(sinφ2) − (sinφ1)(cosφ2))(αα) (7)

= A(sin(φ1 − φ2))(αα) (8)

which is unvariant if bothφ1 andφ2 are both changed by the same amount at the same time.Therefore, two same-spin electrons in differentp-orbitals will lead to aΣ state.

3

Consider N-H: Recognizing the NH has two unpaired electrons and one antisymmetric reflect-ing p-orbital, and noting the conclusions of the statements above, the ground state for NH is3Σ−.

Consider H2: Adding another H bond is possible since NH has two singly-occupied atomicorbitals on N. This new state will create aC2v molecule that has the same symmetry operations asthe case for OH2. The ground state of NH2 is 2B1. Prove this to yourself with the same analysiswe did for OH2.

Consider H3: Adding another H bond is still possible to form NH3. This molecule however ispart of a new symmetry point group,C3v. C3v has120◦ rotation about one axis, and three planesof reflection. Symmetry and antisymmetry of rotation is denoted with an ’A’ or an ’E’ in this pointgroup, where symmetry and antisymmetry of reflection is denoted with a ’1’ or a ’2’ as a subscript(just as withC2v). The NH3 ground state is1A1.

3 Molecular structures

We finished class by introducing some molecular structures of hydrides in differnt groups of theperiodic table. The GVB orbital paradigm implies that most molecular bonds would occur at90◦

angles from one another, and indeed for row 2 and below of the periodic table this is the case!However, bond angles for first row molecules deviate substantially, and we will address how toexplain this phenomenon in following classes.

4 Suggested reading

• Atomic p Orbitals: McQuarrie: Hydrogen Atom Chapter 6.10

• Term Symbols and Symmetry operation for diatomic molecules: Goddard Book - 10.1.1-10.1.3

• General Symmetry operations and point groups: Miessler and Tarr: Chapter 4

4


Sanja Pudar

October 14, 2005

Study Guide Outline

1. Bond Angles in Hydrides

2. Lobe Orbitals

3. Introduction to CH2

1 Bond Angles in Hydrides

We looked at the bond angles of the compounds within a vertical group and saw that for smaller atomsthe bond angles are greater than 90Æ, but as the atoms get larger, the bond angle tends to 90 degrees.VB predicts 90Æ angles for all these bonds (consider for example H2O�H2S�H2Se�H2Te) but when theatoms are small, the � orbitals of the two hydrogens on the molecule are close enough that they runinto one another. The reason why H-O-H angle in H2O is 104.5Æ is that while the � orbitals on oxygenare orthogonal to each other, when they form bonds to the � orbitals on hydrogen, the � orbitals in thehydrogens are not naturally orthogonal to each other. As a result, the bond angle for H2O opens to a valuelarger than 90Æ due to the Pauli principle. In the cases of 2nd row atoms and those lower,the hydridebond involves p orbitals from higher orbital shells (e.g. ��, ��, etc.), and since the higher level orbitalsextend further from the nucleus the resulting bonds keep the hydrogens far enough apart that only minimaloverlap occurs.However, a reasonable question to ask is why the angle for NH2 smaller than that for H2O. Recall that aswe move right across the periodic table, �� increases, and the size of atoms decrease. So atomistically,N is larger than O, and so the bond angles need to be opened slightly less than for the case with oxygen.

2 Lobe Orbitals

Consider BeH: From everything we have learned thus far, there should not be a bound state since wewould assume bringing a hydrogen towards a doubly occupied �� orbital is analagous to the He� Hexample which is unbound. Alternatively, BeH� should form a bond. Actually, both BeH and BeH� formbonds (with energies 46 kcal/mol and 71 kcal/mol, respectively). The reason for this unexpected bond isthat since the �� and �� on Be are relatively close in energy, those orbitals can hybridize or “pooch”.Pooching only occurs between one � orbital and one empty � orbital of the same n quantum number. Ifone of the electrons in the �� orbital mixes �� orbital character, it forms what will be referred to as a lobeorbital. Lobe orbital can allow for more overlap as a bonding orbital since its mixture of � and � characterallows it to bond to species that would otherwise never bond due to orthogonality conditions. Additionally,

1

the lobe orbital has more � character which has better penetration to the nucleus, and extends further fromthe nucleus than a pure � orbital.In order to describe the mathematical basis for pooching consider two electrons in the �� orbital with theoverall wavefunction of

��

This wavefunction is simply the antisymmetric product of two �� orbitals, labeled �� and ��, where�� and �� . Upon mixing some �� character into both of these orbitals we canexpress �� and �� now as linear combinations of the quantities of � character we are adding (omitting spinand electron number labels):

��

��

These new �� and �� wavefunctions are shown in Figure 1.

Figure 1: �� and �� lobe orbital wavefunctions.

We can see from the amplitude of the wavefunctions that each lobe orbital will form bonds on oppositesides of the nucleus. The value for is calculated by optimizing for the best energy of the lobe orbitals,and its best value is � ��.The dissociation curve of BeH shows that Be-H bond is not as good as the bond in BeH�, which is aconsequence of the need to unpair the electrons in the pooched orbitals. When the first bond forms, therewill be two spin-paired electrons forming the bond plus one additional electron on the remaining side ofthe hybrid orbital. This additional electron will experience repulsion from the bond electrons and has toget orthogonal to them. This costs about 1.5 eV in energy, which is deducted from the total energy thatwe gain by forming the bond. Hence, the well depth of this bond is shallower than that of a normal bond.When we form the second bond to the remaining lobe orbital we don’t have to pay this energy penaltyagain because the orbitals have already been unpaired.

2

3 Introduction to CH2

Before we introduced lobe orbitals, we found out that the ground state of CH2 is singlet. However, it iswell known that the g.s. is a triplet. This can be explained using the lobe orbitals.It is found that the first hydrogen bonds to a � orbital, resulting in � state. The second hydrogen bondsto a lobe orbital, leaving the system in a triplet state. These two hydrogen atoms have no preference overeither � or lobe orbital, thus the two bonds become equal.During the next class, we’ll talk about where third hydrogen prefers to bind, and we’ll also compare theformation of CH4 to SiH4.

Suggested reading:

� Chapter 6: GVB Model of Bonding

3


Adam Griffith

October 17, 2005

In this guide: Symmetry; Diatomic Term Symbols; Molecular Term SymbolsLast updated October 27, 2005.

1 The Origin of ml States and Symmetry

We are familiar that there are three different ways we can arrange four electrons into three different porbitals, and this denotes the 3P states of O. (See Figure 1.) From quantum mechanics we know thereare associated values of ml that describe the relative angular momentum of the three degenerate states.Assigning theseml values to the familiar labels px, py, and pz is not trivial, however. Recall that the spatialpart for the wavefunction for any orbital can be expressed in terms of its radial and angluar components.I.e. the expression,

ψn,l,ml(r, θ, φ) = Rn,l(r)Y

mll (θ, φ) (1)

corresponds to the wavefunction for a p orbital once we insert the quantum numbers associated with thestate we wish to describe. The R term is the radial function (a function of distance), and the Y term isa spherical harmonic (a function of θ and φ). The spherical harmonics for p orbitals are complex sincethe ml = ±1 values correspond to functions with the form e±iφ. (See McQuarrie or the Goddard book inthe suggested reading section). Since we cannot see the complex plane, we cannot easily visualize whatthese orbitals look like, however it is perfectly acceptable to create linear combinations of these complexspherical harmonics to form real functions that we can visualize. As a result, we can make new expressionsthat describe the 3 degenerate states with the following real wavefunctions:

ψpx = A sin θ cosφ (2)

ψpy = A sin θ sinφ (3)

ψpz = A cos θ (4)

Where A is a coefficient that arises with particular values of Z and quantum number n.

2 Diatomic Term Symbols

Now that we have three real functions to describe the p orbitals, we can think of how each of the three statesof O behaves under rotation about the z-axis. We start with the xyz2 state and assign a wavefunction to thisstate. Since there is one electron in each of the px and py orbitals, and assuming that Hund’s rules apply,

1

we can write its wavefunction as (xy − yx)(αα). However, we know that both px and py have angulardependencies, and in order to see how these orbitals behave under rotation we substitute: x = cosφ andy = sinφ since these are the only φ-dependent parts of the px and py wavefunctions. Therefore:

(xy − yx)(αα) = (cosφ1 sinφ2 − sinφ1 cosφ2)(αα) = sin(φ2 − φ1)(αα) (5)

which is invariant under rotation of the z-axis by φ. Since it is invariant, we do not expect it to have anyangular momentum. We assign this xyz2 state as ml = 0. We then arbitrarily assing the x2yz and xy2zstates to have ml values of -1 and +1, respectively. We can see that the x2yz state transforms in the samemanner as a pure py orbital. Even though the px orbital does change with rotation, the doubly-occupied pxorbital does not change since the two electrons contribute to two sign-changes overall. The same argumentcan be used to say that the xyz state behaves like a pure px orbital under rotation about the z-axis. Thesame technique we just used to assign the spatial functions of a p-orbital to values of ml can be used for dorbitals as well.

Now that we know how to identify states by their characteristic orbitals, we can assign terms (Σ, Π, and∆) for diatomic molecules that resemble the different types of atomic orbitals we encounter (σ, π, and δ).We use these capital Greek characters as terms for only diatomic molecules because the nuclei of any twoatoms side by side will always have the same basic symmetry element: a C∞ axis of rotation (i.e. it isinvariant of rotation about its z-axis).

A few additional symmetries must be specified when dealing with diatomic molecules. First, the symmetryof the wavefunction will be either symmetric or anti-symmetric with respect to reflection through a mirrorplane parallel to the axis of rotation. We denote symmetry of the wavefunction by a superscript “+” andantisymmetric with a superscript “-”. Since Π states are never symmetric with respect to reflection we donot bother writing “+” or “-” for the Π states.

We also need to consider the effect of inversion symmetry in diatomics. Recall that the effect of inversionis to take (x, y, z) → (-x, -y, -z). If the wavefunction remains the same sign after application of theinversion operator, then it is said to be symmetric under inversion, and is denoted with a “g”. Likewise, ifthe sign of the wavefunction changes after application of the inversion operator, then it is antisymmetricunder inversion and is denoted with a “u”.

3 Molecular Term Symbols

When we think about molecules that consist of more than two atoms we need to devise different termsto identify states since these molecules do not have the same C∞ element that all diatomics have. Thisnew paradigm for molecular terms needs to account for the structural geometry of any molecule, and soagain we invoke symmetry operations on the Hamiltonian to do so. Our new system for determiningmolecular terms depend on the fundamental symmetry elements that exist in the symmetry point groupfor the molecule. To illustrate, we use the water molecule. (See Figure 2.) Water is of the C2v point groupbecause it has certain characteristic symmetry operators. These operators operate on the wavefunctionof water and are the following:

1. A C2 axis of rotation – A Cn axis shows a completely identical representation of the moleculeoccurs n times during a rotation of 360◦ about a particular axis. In the case of water, rotationabout the z-axis results in an identical representation after rotations of 180◦ and 360◦. Since two

2

representations of the molecule occur under this rotation, water has a C2 axis of symmetry. (SeeFigure 3.)

A Quick Excercise for the Reader: What are all of the Cn’s for:

(a) a book with no distinguishable markings on its cover?

(b) an OREOTM cookie with no markings on the cookie?

(c) a four-sided die in the shape of a pyramid without any numbers?

In the case of water, a C2 operation rotates the water molecule 3602

= 180◦. If we find that thewavefunction is symmetric upon application of this C2 operator then the molecule is assigned theterm “A”. If it is antisymmetric then it is assigned the term “B”. Application of the C2 operator twiceresults in the exact same orientation we started with. This operation, the identity operator (“E” foreinheit) is also a symmetry element.

2. An identity operator, E – All molecules have an identity operator. If the only symmetry operationthat exists for a molecule is E, then it is said to be of the C1 point group. Application of theE operator on a wavefunction will always result in the same wavefunction, so wavefunctions arealways symmetric under the E operator.

3. One mirror plane that is parallel to the axis of rotation (σv = σxz) – The mirror plane alongthe xz-plane will take (x, y, z) → (x, -y, z). If application of the σxz operator results in the samewavefunction, then it is symmetric under σxz. If σxz results in the negative of the wavefunction, thenit is antisymmetric under σxz. (See Figure 4.)

A Quick Excercise for the Reader:

(a) How many planes of symmetry are in an octagonal STOP sign, neglecting the word “STOP”?

(b) How may are parallel to its highest order axis of rotation (C8 in this case)?

(c) How may are perpendicular to this axis of rotation? (Planes of symmetry parallel to the axisof rotation are σv’s. Planes perpendicular to the axis of rotation are σh’s.)

4. Another mirror plane that is parallel to the axis of rotation (σv = σyz) – The mirror plane alongthe yz-plane will take (x, y, z) → (-x, y, z). If application of the σyz operator results in the samewavefunction, then it is symmetric under σyz. If σyz results in the negative of the wavefunction, thenit is antisymmetric under σyz. (See Figure 5.)

These symmetry elements are not independent! One can test to see if any other symmetry elements exist inour point group by taking products of the operators to see if any new operations have not been accountedfor. Taking the product of E with any other operator will always result in the operator itself. Additionally,in this point group, the product of any operator with the same operator also results in E. The product ofσxz ∗ C2 = σyz.

Once we have all of the operators (also called generators), we can create a character table that containsall of the symmetry operations and their corresponding effect on the wavefunction (Figure 6). With thischaracter table, by matching how the C2 and σxz operators affect H2O, we can identify its state as 1A1.

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Depending on the term symbol of a molecule, we can identify its internal composition (i.e. whether themolecule has a pooched orbital or not, and thus how it prefers to bond to other atoms).

NOTE: The C2v point group has a unique property compared to other point groups. Generally, symmetryor antisymmetry of the wavefunction under reflection of a mirror plane is denoted by the subscripts of the“A” or “B” terms. The C2v point group has two unique mirror planes, so therefore we need to decide whichplane we choose to determine the unique terms. Standard convention is to define the primary mirror planeas the plane that intersects the most atoms, and so convention would be to place the H2O molecule in thexz plane. Professor Goddard prefers to define the primary mirror plane as the plane that reflects the mostatoms onto other atoms, but in doing so he places the atoms in the yz plane. This is a subte detail that onlyarises in C2v point groups, so be aware of the different conventions that are used.

4 Suggested Reading

• Atomic p Orbitals: McQuarrie: Hydrogen Atom Chapter 6.10

• Term Symbols and Symmetry Operations for Diatomic Molecules: Goddard Book – 10.1.1-10.1.3

• General Symmetry Operations and Point Groups: Miessler and Tarr: Chapter 4

5 Figures

Figure 1: The x2yz, xyz2, xy2z, states of 3Π oxygen

Figure 2: The water molecule and our selected coordinate axes

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Figure 3: The C2 symmetry operator applied to water

Figure 4: The σxz symmetry operator applied to water

Figure 5: The σyz symmetry operator applied to water

Figure 6: The symmetry character table for the C2v point group

5


Sanja Pudar

October 19, 2005

Study Guide Outline

1. Homonuclear Diatomics

2. The Unified Atom Limit

1 Homonuclear Diatomics

We have already seen how symmetry plays an extremely important role in understanding how bonds areformed. Due to their molecular symmetry, diatomic molecules of one element (homonuclear diatomics)are a simple, but general test case to analyze bonding between two atoms.

When scientists tried to predict the ground state configuration for O2, VB theory falsely predicted it tobe a �� state instead of a ��

�state. Although MO theory correctly predicted O2, it failed (along with

VB theory as well) to predict the correct state for N�

2 (both predicted �� instead of ��

). Scientistswere curious what could be failing in explaining the correct ground states for some diatomics. Mullikanproposed that though atomic orbitals for individual atoms were well characterized by quantum mechanics,one could not assume that molecular orbital energy levels correlated directly the understood ordering ofatomic orbitals. Instead, he proposed a different scheme where MOs were formed by atomic orbitalsmatched by symmetry. In order for us to visualize this and understand the consequences of this, this guidewill outline the process for creating the unified atom limit diagram for homonuclear molecules.

2 The Unified Atom Limit

The unified atom limit diagram depicts how atomic orbital symmetries match into molecular orbitals ofhomonuclear diatomic molecules. According to our lecture, the right side of the diagram correspondsto two separated atoms, each with nuclear charge �, while the left side corresponds to the two atomscompletely super-imposed on top of each other, a species with nuclear charge ��. Somewhere betweenthese two limits lie the actual molecular orbital energy spacing for a particular homonuclear diatomic.

Starting with the energy levels of the unified atom limit, we consider only the bonding orbital combinationsof similar atomic orbitals (�� and ��. and NOT �� and �� . In doing this, the �-orbitals would bondwith each other to form �� orbitals. Likewise, the �-bonding p-orbital, �� orbitals would bond other ��orbitals to form �� orbitals. The �-bonding p-orbitals, �� and �� would bind with like p-orbitals to formpu molecular orbitals. �-orbitals also need to be considered, and due to the symmetries of the differentd-orbitals we get �� , �� and �� , and �� and �� Æ� orbitals.

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As we start to pull the unified atoms apart, the energies of each unified orbital state changes. From a first-order approximation (i.e., neglecting orbital relaxation), energy of the �-orbitals from the unified atom willincrease since we are moving nuclei away from the point of maximum electron density, and thus makingfor less favorable potential energy. However, the energy of the �� orbitals (which have �� symmetry) willdecrease since we are moving nuclei from the region of node into the region rich with electronic density.The �� and �� orbitals will have their energies slightly increase since we are pulling the nuclei from thep-orbital node across a constant node region.

The fact that there was splitting of the p-orbital energies means the degeneracy of the �-orbital was broken.Now that the unified atom has split, the resulting bonding orbital is the molecular orbital for the homonu-clear diatomic. These molecular orbitals are labeled by the atomic symmetries their component atomicorbitals have, (i.e. �� , �� , �� , �� and �� a doubly degenerate ��, etc).

Now consider how these orbitals are when the atoms are at infinite separation. Each atom that comprisedthe homonuclear diatomic is considered a lone atom with relative atomic orbital energies that follow theAufbau Principle. When these atoms start to approach, their atomic orbitals split to form bonding andantibonding molecular orbitals. Taking the symmetric combination of two �� wavefunctions results in a�� bonding orbital, just like in the case for H2. Likewise, the antisymmetric combination of the �� orbitalslead to the �� antibonding orbital. The same can be done for �� orbitals to obtain the �� bonding and�� antibonding orbitals.

Again, the same process of taking linear combinations of wavefunctions causes the �� orbitals to form�� and �� molecular orbital states. The antisymmetric and symmetric combinations �� and �� orbitalsresult in the bonding and antibonding states, �� and �� respectively,both of which are doubly degenerate.Since the �� orbitals form � bonds, they inherently have better overlap, and as a result the effects oftheir bonding and antibonding in the �� and �� states are greater than the corresponding bonding andantibonding of the �� and �� orbitals.

Now that the two sides of the diagram have been expressed, we use the non-crossingrule to connect thelowest energy state on the left side of the diagram with the lowest energy state on the right that has the samesymmetry. The non-crossing rule simply states that energy levels of one symmetry will not cross a statewith the same symmetry. Matching the lowest like-symmetry states will ensure that the non-crossing ruleholds. The resulting map of energy levels is called a correlation diagram. It is possible to use a correlationdiagram built between reactants and products to analyze the intermediate steps of a chemical reaction,and to see whether individual steps of a reaction are energetically allowed or forbidden. Experimentalresults allow us to determine where different symmetry states cross and thus which molecular orbitals areoccupied before others, which in turn allows us to determine electronic configurations of the diatomics.With this correlation diagram, we can see how the ordering of molecular orbitals varied for N�

2 , and wecan see that it correctly predicts the ��

�as its ground state.

Suggested reading:

� Molecular Orbitals and Correlation Diagrams - ��

2


Adam Griffith

October 21, 2005

In this guide: Bonding in HydrocarbonsLast updated October 29, 2005.

1 Things to Consider When Bonding

In this guide we will summarize the most important points to keep in mind when studying more complexmolecules.

Consider the fragments that need to be put together to form the molecules. Pay special attention to what theground state of the system is and how that changes for different substituents. You need to be very familiarwith the order in which the different carbon orbitals form bonds with atoms like hydrogens and fluorines(hydrogens bond to both lobe and p-orbitals, while fluorines have a strong preference for p-orbitals). Thishas a drastic influence bond energies and molecular reactivity.

Consider the types of orbitals that are necessary to form the bonds that the new molecule contains. Whenthere are double or triple bonds, only the first bond is a sigma bond. The second and third bonds are pi-bonds. Sigma bonds can be made with lobe orbitals or p-orbitals while pi-bonds require p-orbitals. This isespecially important in the cases where the fragments’ ground states do not have the orbital configurationnecessary to form the desired molecule. In these cases we need to consider the energy costs in promotingthe system to the desired configuration. This may include, for example, forcing a fluorine atom to bondto a lobe orbital instead of a p-orbital. This will cost energy, since fluorines are the most electronegativeatoms in the periodic table and prefer to bond to orbitals that allow them to obtain good overlap with theirown p-orbitals and attract the electrons towards their nuclei (they form bonds with high ionic character).

Consider the type of orbitals that are involved in a bond to assess the strength and length of each bond.The carbon atom can be in the sp1, sp2, and sp3 hybridization. In the first case, there are only two hybridorbitals and two p-orbitals; in the second case there are three hybrid orbitals and one p-orbital and inthe third case all orbitals are hybrid. In general, the less hybrid orbitals there are per atom, the mores-character each one of them will have, and the shorter the bonds will be (because s-orbitals are smallerthan p-orbitals). This is one of the primary reasons why the bond length decreases from ethane to ethyne.

2 Strongly Suggested Reading

• Goddard Book - Chapter 6, 7, 10

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SUPPLEMENTARY NOTES FOR NATURE OF THE CHEMICAL BOND

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