Lecture 1, September 26, 2016 Elements QM, stability H, H2+€¦ · Transition metal systems: organometallic reaction mechanisms. (oxidative-addition, reductive elimination, metathesis)

© copyright 2016 William A. Goddard III, all rights reservedCh120a-Goddard-L01 1

Nature of the Chemical Bond with applications to catalysis, materials

science, nanotechnology, surface science, bioinorganic chemistry, and energy

William A. Goddard, III, [email protected] Beckman Institute, x3093

Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics,

California Institute of Technology

Teaching Assistant: Shane Flynn <[email protected]>

Lecture 1, September 26, 2016Elements QM, stability H, H2+

Course number: Ch120aHours: 2-3pm Monday, Wednesday, Friday


OverviewThis course aims to provide a conceptual understanding of the chemical bond sufficient to predict semi-quantitatively the structures, properties, and reactivities of materials, without computations

The philosophy is similar to that of Linus Pauling, who in the 1930’s revolutionized the teaching of chemistry by including the concepts from quantum mechanics (QM), but not its equations.

We now include the new understanding of chemistry and materials science that has resulted from QM studies over the last 50 years.

We develop an atomistic QM-based understanding of the structures and properties of chemical, biological, and materials systems.


Intended audienceThis course is aimed at experimentalists and theorists in chemistry, materials science, chemical engineering, applied physics, biochemistry, physics, electrical engineering, and mechanical engineering with an interest in characterizing and designing catalysts, materials, molecules, drugs, and systems for energy and nanoscale applications. Courses in QM too often focus more on applied mathematics rather than physical concepts. Instead, we start with the essential differences between quantum and classical mechanics (the description of kinetic energy) which is used to understand why atoms are stable and why chemical bonds exist. We then introduce the role of the Pauli Principle and spin and proceed to use these basic concepts to predict the structures and properties of various materials, including molecules and solids spanning the periodic table.


Applications:Organics: Resonance, strain, and group additivity. Woodward-Hoffman rules, and reactions with dioxygen, and ozone.Carbon Based systems: bucky balls, carbon nanotubes, graphene; mechanical, electronic properties, nanotech appl.Semiconductors, Surface Science: Si and GaAs, donor and acceptor impurities, surface reconstruction, and surface reactions.Ceramics: Oxides, ionic materials, covalent vs. ionic bonding, concepts ionic radii, packing in determining structures and properties. Examples: silicates, perovskites, and cuprates.Hydrogen bonding, Hypervalent systems: XeFn, ClFn, IBX.Transition metal systems: organometallic reaction mechanisms. (oxidative-addition, reductive elimination, metathesis)Bioinorganics: Electronic states, reactions in heme molecules.Organometallic catalysts: CH4 CH3OH, ROMP, MetallocenesMetal oxide catalysts: selective oxidation, ammoxidationMetals and metal alloys: chemisorption, Fuel cell catalysts Superconductors: mechanisms: organic and cuprate systems.


Course detailsHomework every week, hand out on Wednesday 3pm, due Wednesday 2pm, graded and back 1pm Friday.

OK to collaborate on homework, but indicate who your partners were and write your own homework (no xerox from partners)

Exams: no collaboration, open book for everything distributed in course, no internet or computers except for course materials

Grade: Final 48%, Midterm 24%, Homework 28% (best 7 of 8)

No late homework or exams

Will have TA office hour in this room after the Friday lecture

lectures will often start with a review of the important stuff from previous lecture.

Ask questions during lectures


Course details

On line: old course notes, 16 chapters, from 1974-75; can download and print

Lectures this year on powerpoint, will be on line after the lecture

ppt from last year also on line

Schedule: MWF 2-3pm

Occasionally I will add an extra lecture from 3-4pm to make up for missing a lecture while on a trip

© copyright 2016 William A. Goddard III, all rights reserved

Scheduling

Ch120a-Goddard-L01 7

Friday September 30, wag at conference. TA recitationMake up lecture Monday Oct. 3 at 3pm


Need for quantum mechanicsConsider the classical description of the simplest atom, hydrogen-like atom with

1 nucleus of charge qn = +Ze

1electron with charge qe = –e

separated by a distance R between them

PE = potential energy = ?

Nucleusqn =+Ze






Nucleusqn =+Ze

PE = potential energy = Zqeqp/R = -Ze2/RKE = kinetic energy = ?

assume that the proton is sitting stillIf e is distance R from proton the PE is






Nucleusqn =+Ze

PE = potential energy = qeqp/R = -Ze2/RKE = kinetic energy = ½ mev2 = p2/2me where p = me vWhat is the lowest energy (ground state) of this system?


Need for quantum mechanics

Consider the classical description of the simplest atom, hydrogen with 1 proton of charge qp = +e and one electron with charge qe = –e separated by a distance R between them

PE = potential energy = qeqp/R = -e2/RKE = kinetic energy = mev2/2 = p2/2me where p = me vWhat is the lowest energy (ground state) of this system?

PE: R = 0 PE = - ∞KE: p = 0 KE = 0

Total Energy = E = KE + PE = - ∞ (cannot get any lower)

assume electron has velocity v(t) and that the proton is sitting still


Problem with classical mechanics

Ground state for H atom has the electron sitting on the proton (R=0) with velocity = 0.

Thus electron and proton move together

Since their charges cancel there is no interaction of the H atom with anything else in the universe (except gravity)



Ground state for H atom has the electron sitting on the proton (R=0) with velocity = 0.



Thus there is no H2 molecule



Ground state for H atom has the electron sitting on the proton (R=0) with v=0.




Similarly the carbon atom would have all electrons at the nucleus; thus no hydrocarbons, no amino acids, no DNA,



Ground state for H atom has the electron sitting on the proton (R=0) with v=0.





Thus no people.


Problem with classical mechanicsGround state for H atom has the electron sitting on the proton (R=0) with v=0.





Thus no people.

This classical world would be a very dull universe with no room for us


Quantum Mechanics to the rescue

The essential element of QM is that all properties that can be known about the system are contained in the wavefunction, Φ(x,y,z,t) (for one electron), where the probability of finding the electron at position x,y,z at time t is given by

P(x,y,z,t) = | Φ(x,y,z,t) |2 = Φ(x,y,z,t)* Φ(x,y,z,t)

Note that ∫Φ(x,y,z,t)* Φ(x,y,z,t) dxdydz = 1

since the total probability of finding the electron somewhere is 1.


PE of H atom, QMIn QM the total energy can be written as E = KE + PE where for the H atom PE = the average value of (-e2/r) over all positions of the electron. Since the probability of the electron at xyz is P(x,y,z,t) = | Φ(x,y,z,t) |2 = Φ(x,y,z,t)* Φ(x,y,z,t) We can write

PE = ∫Φ(x,y,z,t)* Φ(x,y,z,t) (-e2/r) dxdydz or

PE = ∫Φ(x,y,z,t)* (-e2/r) Φ(x,y,z,t) dxdydz which we write asPE = < Φ| (-e2/r) |Φ> = -e2/Where is the average value of 1/rR

_ R_





_ R_

Thus PE in QM is very similar to CM, just use average distance





_ R_

Thus PE in QM is very similar to CM, just use average distanceNow what is the best value of KE in QM?


Best value for PE in QM of H atom

Consider possible shapes of wavefunction Φ(x,y,z,t)* of H atom

We plot the wavefunction along the z axis with the proton at z=0Which has the lowest PE?




We plot the wavefunction along the z axis with the proton at z=0Which has the lowest PE?Since PE = -Ze2/ , it is case c.Indeed the lowest PE is for the wavefunction to be a delta function with = 0 Leading to a ground state with PE = - ∞ just as for Classical Mechanics

R_

R_




We plot the wavefunction along the z axis with the proton at z=0Which has the lowest PE?Since PE = -e2/ , it is case c.Indeed the lowest PE is for a delta function with = 0 Leading to a ground state with PE = - ∞ just as for Classical Mecha

R_

R_

For PE. QM is the same as CM, just average over P= |Φ(x,y,z,t)|2

PE scales as 1/ R_


What about KE?In Classical Mechanics the position and momentum of the electron can be specified independently, Thus the lowest energy had R=0 and v=0, But in QM both the KE and PE are derived from the SAME wavefunction.

In CM, KE = p2/2me

In QM the KE for a one dimensional system isKEQM = (Ћ2/2me)<(dΦ/dx)*| (dΦ/dx)> = <p2>/2me

where p =(Ћ/i) (dΦ/dx) and * indicates complex conjugate

Thus KE increases as the average slope increasesSince it depends on the slope squared KE will not allow the electron to localize at a point. For a periodic function, KE increases as the period decreases. Thus fast wiggles much worse than slow wiggles


Comment about KE

KE = (Ћ2/2me)<(dΦ/dx)| (dΦ/dx)>

This is not the usual form for KE. Most books writeKE = -(Ћ2/2me) ∫ Φ* (d2Φ/dx2) dx = ∫ Φ* [-(Ћ2/2me)(d2/dx2)] Φ

= ∫Φ*[px]2 Φ where px = (Ћ/i) (d/dx)This is the form that Schrodinger came up with and that is in essentially all QM books.

I consider that the fundamental form for the KE is

To compare to standard QM books, write u= (dΦ/dx) and dv= (dΦ/dx)* dx and integrate by parts ∫u dv = u(∞)v(∞) - u(-∞)v(-∞) - ∫v duThus since u and v must be 0 on the boundaries (otherwise get infinite total probability <Φ|Φ> rather than 1) the standard form converts to my form


Interpretation of QM form of KE

KE = (Ћ2/2me)<(dΦ/dx)| (dΦ/dx)>

KE proportional to the average square of the gradient or slope of the wavefunction

Thus the KE in QM prefers smooooth wavefunctions

In 3-dimensions

KE = (Ћ2/2me)<(∇Φ ∇Φ> =

=(Ћ2/2me) ∫ [(dΦ/dx)2 + (dΦ/dx)2 + (dΦ/dx)2] dxdydz

Still same interpretation:

.

the KE is proportional to the average square of the gradient or slope of the wavefunction


Best value for KE in QM of H atom


Which has the lowest KE?




Which has the lowest KE?clearly it is case a.Indeed the lowest KE is for a wavefunction with ∞ Leading to a ground state with KE = 0 just as for Classical Mechani

R_


But in QM the same wavefunction must be used for KE and PEKE wants ∞ whereas PE wants = 0. Who wins?



Which has the lowest KE?clearly it is case a.Indeed the lowest KE is for a wavefunction with ∞ Leading to a ground state with KE = 0 just as for Classical Mechani

R_

R_

R_


The compromise between PE and KE

How do PE and KE scale with , the average size of the orbital?

PE ~ -C1/

KE ~ +C2/ 2

Now lets find the optimum

R_

R_

R_

R_


R_

Consider very large Here PE is small and negative, while KE is (small)2 but positive, thus PE wins and the total energy is negative

Analysis for optimum R_


Now consider very small Here PE is large and negative, while KE is (large)2 but positive, thus KE wins and the total energy is positive

R_


R_

R_




R_


R_

Thus there must be some intermediate for which the total energy is most negativeThis is the for the optimum wavefunction

R_

R_




R_


R_

Thus there must be some intermediate for which the total energy is most negativeThis is the for the optimum wavefunction

R_

R_


Conclusion in QM the H atom has a finite size,


Discussion of KE

In QM KE wants to have a smooth wavefunction but electrostatics wants the electron concentrated at the nucleus.

Since KE ~ 1/R2 , KE always keeps the wavefunction finite, leading to the finite size of H and other atoms. This allows the formation of molecules and hence to existence of life

In QM it is not possible to form a wavefunction in which the position is exactly specified simultaneous with the momentum being exactly specified. The minimum value is

<(δx)(δp)> ≥ Ћ/2 (The Heisenberg uncertainty principle)

Sometimes it is claimed that this has something to do with the finite size of the atom.

It does but I consider this too hand-wavy.


Implications

QM leads to a finite size for the H atom and for C and other atoms

This allows formation of bonds to form H2, benzene, amino acids, DNA, etc.

Allowing life to form

Thus we owe our lives to QM

The essence of QM is that wavefunctions want to be smooth, wiggles are bad, because they increase KE

© copyright 2016 William A. Goddard III, all rights reserved 37

The wavefunction for H atom

In this course we are not interested in solving for wavefunctions, rather we want to deduce the important properties of the wavefunctions without actually solving any equations

However it is useful to know the analytic form. The ground state of H atom has the form (here Z=1 for H atom, Z=6 for C 5+

For the atom, we use spherical coordinates r,Ө,Φ not x,y,z

N0 is the normalization constant, <Φ|Φ>=1

As Z increases the orbital gets smaller (more attraction to nucleus)

© copyright 2016 William A. Goddard III, all rights reserved 38

The wavefunction for H atom

In this course we are not interested in solving for wavefunctions, rather we want to deduce the important properties of the wavefunctions without actually solving any equations

However it is useful to know the analytic form. The ground state of H atom has the form (here Z=1 for H atom, Z=6 for C 5+

For the atom, we use spherical coordinates r,Ө,Φ not x,y,z

a0 = Ћ2/me2 = 0.529 A =0.053 nm is the Bohr radius (the average size of the H atom)

N0 is the normalization constant, <Φ|Φ>=1


Atomic Units

a0 = Ћ2/me2 = = 0.529 A =0.053 nm is the Bohr radius

E = -Ze2/2 R_

Z=1 for Hydrogen atom

R_

For Z ≠ 1, = a0/Z

Charge of nucleus = ZeCharge of electron = -eAverage distance = R

_PE = - e (Ze)/ R

_

Total E = PE/2



The PE = -Ze2/ while KE= Ze2/ 2

Thus the total energy E = -Ze2/(2 ) = PE/2 = -KE

This is called the Virial Theorem and is general for all molecules

The energy for H atom

R_

R_

R_


Atomic Units

Atomic units:me = 1, e = 1, Ћ = 1 leads to unit of length = a0 and unit of energy = h0In atomic units: KE= <∇Φ.∇Φ>/2 (leave off Ћ2/me)PE = <Φ|-1/r|Φ>/2 (leave off e2)

a0 = Ћ2/me2 = = 0.529 A =0.053 nm is the Bohr radius

h0 = e2/a0 = me e4/ Ћ2 = Hartree = 27.2116 eV = 627.51 kcal/mol = 2625.5 kJ/mol


E = -0.5 h0Classical turning point r = a0/√2

Local PE and KE of H atom

Local KE positive Local KE negativeLocal KE negative

Local PE, negativeLocal KE, positive

PE = -Ze2/

KE= Ze2/ 2 Φ(z)

R_

R_


2 ways to plot orbitals

1-dimensional Iine plot of orbital along z axis

2-dimensional contour plot of orbital in xz plane, adjacent contours differ by 0.05 au


Bring a proton up to an H atom to form H2+

Is the molecule bound?That is does it have a lower energy at finite R than at R = ∞Several possibilitiesElectron is on the left proton, L

Electron is on the right proton, R

Or we could combine them

Now consider H2+ molecule

L R

rLrR

e

At R = ∞ these are have the same energy, but not for finite RIn QM we always want the wavefunction with the lowest energy. Question: which combination is lowest?


Combine Atomic Orbitals for H2+ molecule

Two extreme possibilities

Antisymmetric combination

Symmetric combination

the Dg = Sqrt[2(1+S)] and Du = Sqrt[2(1-S)] factors above are the constants needed to ensure that <Φg|Φg> =∫ Φg|Φg dxdydz = 1 (normalized)<Φu|Φu> =∫ Φu|Φu dxdydz = 1 (normalized)I will usually eschew writing such factors, leaving them to be understood

Which is best (lowest energy)?


Energies of of H2+ Molecule

LCAO = Linear Combination of Atomic Orbitals

Good state: g

Ungood state: u

g state is bound since starting the atoms at any distance between arrows, the molecule will stay bonded, with atoms vibrating forth and back


But WHY is the g state bound?



Common rational :

Superimposing two orbitals and squaring to get the probability leads to moving charge into the bond region.

This negative charge in the bond region attracts the two positive nuclei

Sounds reasonable,

Why is it wrong

+ +-



+ +-

Moving charge to the bond midpoint decreases density near atoms, thus electrons move from attractive region near nuclei to less attractive region near bond midpoint, this INCREASES the PE

Common rational :

Superimposing two orbitals and squaring to get the probability leads to moving charge into the bond region.

This negative charge in the bond region attracts the two positive nuclei


The change in electron density for molecular orbitals

The densities ρg and ρu for the g and u LCAO wavefunctions of H2

+ compared to superposition of ρL + ρRatomic densities (all densities add up to one electron)

Adding the two atomic orbitals to form the g molecular orbital increases the electron density in the bonding region, as expected. This is because in QM, the amplitudes are added and then squared to get probability density


Compare change in density with local PE function

The local PE for the electron is lowered at the bond midpoint from the value of a single atom

But the best local PE is still near the nucleusThus the Φg = χL + χR wavefunction moves charge to the bond region AT THE EXPENSE of the charge near the nuclei, causing an increase in the PE, and opposing bonding

PE(r) = -1/ra – 1/rb


The PE of H2+ for g and u states

The total PE of H2+ for the Φg = χL + χR and Φu = χL - χR

wavefunctions (relative to the values of Vg = Vu = -1 h0 at R = ∞)


If the bonding is not due to the PE, then it must be KE


If the bonding is not due to the PE, then it must be KE

The shape of the Φg = χL + χR and Φu = χL - χR wavefunctions compared to the pure atomic orbital (all normalized to a total probability of one).

We see a dramatic decrease in the slope of the g orbital along the bond axis compared to the atomic orbital.

This leads to a dramatic decrease in KE compared to the atomic orbital

This decrease arises only in the bond region.

It is this decrease in KE that is responsible for the bonding in H2

+


The KE of g and u wavefunctions for H2+

The change in the KE as a function of distance for the g and u wavefunctions of H2

+

(relative to the value at R=∞ of KEg=KEu=+0.5 h0)

Use top part of 2-7

Comparison of the g and u wavefunctions of H2

+ (near the optimum bond distance for the g state), showing why g is so bonding and u is so antibonding


R too short leads to a big decrease in slope but over a very short region, little bondingR is too large leads to a decrease in slope over a long region, but the change in slope is very small little bondingOptimum bonding occurs when there is a large region where both atomic orbitals have large slopes in the opposite directions (contragradient). This leads to optimum bonding

Why does KEg has an optimum?


KE dominates PE

Changes in the total KE and PE for the g and u wavefunctions of H2

+ (relative to values at R=∞ of KE :+0.5 h0PE: -1.0 h0E: -0.5 h0

The g state is bound between R~1.5 a0 and ∞(starting the atoms at any distance in this range leads to atoms vibrating forth and back. Exciting to the u state leads to dissociation

Good state: g

Ungood state: u


KE dominates PE, leading to g as ground state

Calculations show this, but how could we have predicted that g is better than u without calculations?

Answer: the nodal theorem: The ground state of a QM systems has no nodes. Thus g state lower E than u state


The nodal Theorem

The ground state of a system has no nodes (more properly, the ground state never changes sign).

This is often quite useful in reasoning about wavefunctions.

For example the nodal theorem immediately implies that the g wavefunction for H2

+ is the ground state (not the u state)


The nodal Theorem 1DSchrodinger equation, H Φk = Ek ΦkOne dimensional: H =- ½ d2/dx2 + V(x)Consider the best possible eigenstate of Hwith a node, Φ1 and construct a nonnegative function Ө0 =|Φ1| as in bFor every value of x, V(x)[Φ1]2 = V(x)[Ө0]2 so that V0 = ∫ [Ө0]*V(x)[Ө0] = ∫ [Φ1]*V(x)[Φ1]2 = V1

a

b

Φ1

Ө0

c Φ0

Also |dӨ0/dx|2 = |dΦ1/dx|2 for every value of x except the single point at which the node occurs (does not count since no length)

Thus T0 = ½ ∫ |dӨ0/dx|2 = ½ ∫ |dΦ1/dx|2 = T1. Hence E0 = T0 + V0 = T1 + V1 = E1.


We just showed that for the best possible eigenfunction of H with a node, H Φ1 = E1 Φ1

Ө0 =|Φ1| has the same energy as Φ1E0 = T0 + V0 = T1 + V1 = E1. However Ө0 is just a special case of a nodeless wavefunction that happens to go to 0 at one point. Thus we could smooth out Ө0 in the region of the node as in c, decreasing the KE and lowering the energy. Thus the optimum nodeless wavefunction Φ0leads to E0 < E1. Only for a potential so repulsive at some point, that all wavefunctions are 0, do we get E0 = E1

The nodal Theorem 1D a

b

Φ1

Ө0

c Φ0


The nodal Theorem for excited states in 1DFor one-dimensional finite systems, we can order all eigenstates by the number of nodes

E0 < E1 < E2 .... En < En+1

(where only a sufficiently singular potential can lead to an = sign )

The argument is the same as for the ground state.

Consider best wavefunction Φn with n nodes and flip the sign at one node to get a wavefunction Өn-1 that changes sign only n-1 times.

Show that En-1 = En

But Өn-1 is not the best with n-1 sign changes.

Thus we can smooth out Өn-1 in the region of the extra node to decrease the KE and lower the energy for the Φn-1,. Thus the optimum n-1 node wavefunction leads to En-1 < En.


The nodal Theorem 3DIn 2D a wavefunction that changes size once will have a line of points with Φ1=0 (a nodal line) For 3D there will be a 2D nodal surface with Φ1=0. In 3D the same argument as for 1D shows that the ground state is nodeless. We start with Φ1 the best possible eigenstate with a nodal surface and construct a nonnegative function Ө0 =|Φ1|For every value of x,y,z, V(x,y,z)[Φ1]2 = V(x,y.z)[Ө0]2 so that V0 = ∫ [Ө0]*V(xyz)[Ө0] = ∫ [Φ1]*V(xyz)[Φ1]2 = V1

Also |∇Ө0|2 = |∇Φ1|2 everywhere except along a 2D plane Thus T0 = ½ ∫ |∇Ө0|2 = ½ ∫ |∇Φ1|2 = T1. Hence E0 = T0 + V0 = T1 + V1 = E1. As before E1 is the best possible energy for an eigenstate with a nodal plane. However Ө0 can be improved by smoothing Thus the optimum nodeless wavefunction Φ0 leads to E0 < E1.


The nodal Theorem for excited states in 3DFor 2D and 3D, one cannot order all eigenstates by the number of nodes. Thus consider the 2D wavefunctions

+ Φ00 Φ10

Φ01 Φ20+-

+-

+-+

It is easy to show as in the earlier analysis that

E00 < E10 < E20< E21

E00 < E01 < E11 < E21

But the nodal argument does not indicate the relative energies of E10 and E20 versus E01

- Φ11+-

++-+

+- -Φ21


Back to H2+

Nodal theorem The ground state must be the g wavefunction

g state u state


H2 molecule, independent atomsStart with non interacting H atoms, electron 1 on H on earth, χΕ(1)the other electron 2 on the moon, χΜ(2)What is the total wavefunction, Ψ(1,2)?




Maybe Ψ(1,2) = χΕ(1) + χΜ(2) ?




Maybe Ψ(1,2) = χΕ(1) + χΜ(2) ?

Since the motions of the two electrons are completely independent, we expect that the probability of finding electron 1 at some location on earth to be independent of the probability of finding electron 2 at some location on the moon.

Thus




Maybe Ψ(1,2) = χΕ(1) + χΜ(2) ?

Since the motions of the two electrons are completely independent, we expect that the probability of finding electron 1 at some location on earth to be independent of the probability of finding electron 2 at some location on the moon.

Thus P(1,2) = PE(1)*PM(2)

This is analogous to the joint probability, say of rolling snake eyes (two ones) in dice

P(snake eyes)=P(1 for die 1)*P(1 for die 2)=(1/6)*(1/6) = 1/36

Question what wavefunction Ψ(1,2) leads to P(1,2) = PE(1)*PM(2)?© copyright 2009 William A. Goddard III, all rights reserved


Ψ(1,2) = χΕ(1)χΜ(2) leads to P(1,2) = |Ψ(1,2)|2 = Ψ(1,2)* Ψ(1,2) =

= [χΕ(1)χΜ(2)]* [χΕ(1)χΜ(2)] =

= [χΕ(1)* χΕ(1)] [χΜ(2)* χΜ(2)] =

= PE(1) PM(2)

Answer: product of amplitudes

Conclusion: the wavefunction for independent electrons is the product of the independent orbitals for each electron



= [χΕ(1)χΜ(2)]* [χΕ(1)χΜ(2)] =

= [χΕ(1)* χΕ(1)] [χΜ(2)* χΜ(2)] =

= PE(1) PM(2)


Conclusion the wavefunction for independent electrons is the product of the independent orbitals for each electron

Back to H2, ΨEM(1,2) = χΕ(1)χΜ(2)

But ΨME(1,2) = χΜ(1)χΕ(2) is equally good since the electrons are identical



= [χΕ(1)χΜ(2)]* [χΕ(1)χΜ(2)] =

= [χΕ(1)* χΕ(1)] [χΜ(2)* χΜ(2)] =

= PE(1) PM(2)


Conclusion the wavefunction for independent electrons is the product of the independent orbitals for each electron

Back to H2, ΨEM(1,2) = χΕ(1)χΜ(2)

But ΨME(1,2) = χΜ(1)χΕ(2) is equally good since the electrons are identicalAlso we could combine these wavefunctions χΕ(1)χΜ(2) ± χΕ(1)χΜ(2)

Which is best?


Consider H2 at R=∞

ΦRL

Two equivalent wavefunctions

At R=∞ these are all the same, what is best for finite R

The wavefunction for H2 at long R

ΦLR

ΦLR

ΦLR

ΦRL

ΦRL


Plot of two-electron

wavefunctions along molecular

axis

z

x

LR wavefunction

L

LR

L

R

R


Plot of two-electron wavefunctions along molecular axis

z

x

LR wavefunction RL wavefunction

L

L

R

R

L

L

R

R


Combine the LR and RL

wavefunctions

z

x

Lower KE (good)

Higher PE (bad)

EE poor

higher KE (bad)

Lower PE (good)

EE good

L

L

R

R

L

L

R

R

LR+RL LR-RL

Who wins?


Combine the LR and RL

wavefunctions

z

x

Lower KE (good)

Higher PE (bad)

EE poor

higher KE (bad)

Lower PE (good)

EE good

L

L

R

R

L

L

R

R

LR+RL LR-RL

Who wins? LR+RL is best because no nodes


Valence Bond wavefuntion of H2

Valence bond: start with ground state at R=∞ and build molecule by combining best state of atoms

Nodal theorem says ground state is nodeless, thus Φg is better than Φu

LR+RL

LR-RL


Details

E(H2+)1 = KE1 + PE1a + PE1b + 1/Rab for electron 1

E(H2+)2 = KE2 + PE2a + PE2b + 1/Rab for electron 2

E(H2) = (KE1 + PE1a + PE1b) + (KE2 + PE2a+PE2b) + 1/Rab+EE

EE = <Φ(1,2)|1/r12| Φ(1,2)> assuming <Φ(1,2)|Φ(1,2)>=1

Note that EE = ∫dx1dy1dz1 ∫dx2dy2dz2[| Φ(1,2)|2/r12] > 0 since all terms in integrand > 0


Summary: bonding in H2 is due to decrease in KE as add the L(1)R(2) and R(1)L(2) two electron

wavefunctions


Stability of H atom, bonding in H2+ and bonding in H2 all due to KESmooth wavefunctions better than wiggly wavefunctions

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Lecture 1, September 26, 2016 Elements QM, stability H, H2+€¦ · Transition metal systems: organometallic reaction mechanisms. (oxidative-addition, reductive elimination, metathesis)