Nature of the Chemical Bond with applications to ... - KOCw

Nature of the Chemical Bond

with applications to catalysis, materials science, nanotechnology, surface science,

bioinorganic chemistry, and energy

William A. Goddard, III, wag3@kaist.ac.kr WCU Professor at EEWS-KAIST and

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

California Institute of Technology

Course number: KIAST EEWS 80.502 Room E11-101 Hours: 0900-1030 Tuesday and Thursday

Senior Assistant: Dr. Hyungjun Kim: linus16@kaist.ac.kr Postdoctoral fellow EEWS

Overview

This course is patterned after the Course on Nature of the

Chemical Bond that Goddard has taught at Caltech since 1972.

It 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

Intended audience

This 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

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.

Hypervalent systems: XeF_n, ClF_n, IBX chemistry.

Transition metal systems: organometallic reaction mechanisms.

(oxidative-addition, reductive elimination, metathesis)

Bioinorganics: Electronic states, reactions in heme molecules.

Organometallic catalysts: CH₄ → CH3OH, ROMP, Metallocenes Metal oxide catalysts: selective oxidation, ammoxidation

Metals and metal alloys: chemisorption, Fuel cell catalysts

Need for quantum mechanics

Consider the classical description of the simplest atom,

hydrogen with 1 proton of charge q_p = +e and one electron with charge q_e = –e separated by a distance R between them

PE = potential energy = q_eq_p/R = -e²/R

KE = kinetic energy = ½ m_ev² (assume proton is sitting still) What is the lowest energy (ground state) of this system?

PE:

Need for quantum mechanics

Consider the classical description of the simplest atom,

hydrogen with 1 proton of charge q_p = +e and one electron with charge q_e = –e separated by a distance R between them

PE = potential energy = q_eq_p/R = -e²/R

KE = kinetic energy = ½ m_ev² (assume proton is sitting still) What is the lowest energy (ground state) of this system?

PE: R = 0 ➔ PE = -

∞

KE:

Need for quantum mechanics

Consider the classical description of the simplest atom,

hydrogen with 1 proton of charge q_p = +e and one electron with charge q_e = –e separated by a distance R between them

PE = potential energy = q_eq_p/R = -e²/R

KE = kinetic energy = m_ev²/2 = p²/2m_e where p = m_e v What is the lowest energy (ground state) of this system?

PE: R = 0 ➔ PE = -

∞

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 v=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)

Problem with classical mechanics

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

Thus there is no H₂ molecule

Problem with classical mechanics

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

Thus there is no H₂ molecule

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

Problem with classical mechanics

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

Thus there is no H₂ molecule

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

Thus no people.

Problem with classical mechanics

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

Thus there is no H₂ molecule

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

Thus no people.

This 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 is 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.

In 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 as PE = < Φ| (-e²/r) |Φ> = -e²/

Where is the average value of 1/r R

_ R

_

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=0 Which 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 Mechanics R

_

R _

What about KE?

In CM the position and momentum of the electron can be

specified independently, R=0 and v=0, but in QM both the KE and PE are derived from the SAME wavefunction.

In QM the KE for a one dimensional system is

KE = [

∫

^(pΦ)†(pΦ) dx]/2m_e = [

∫

^{(Ћ dΦ/idx)* (Ћ} dΦ/idx) dx]/2m_e

= (Ћ²/2m_e)

∫

^(dΦ/dx)* (dΦ/dx) dx Thus

KE = (Ћ²/2m_e)<(dΦ/dx)| (dΦ/dx)>

Interpretation of QM form of KE

KE = (Ћ²/2m_e)<(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 = (Ћ²/2m_e)<(Φ Φ> =

=(Ћ²/2m_e)

∫

^{[(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

My form for KE differs from usual one in QM

One dimensional, usual form

KE = - (Ћ²/2m_e)

∫

^{(Φ*)(d2Φ/dx2}) dx = - (Ћ²/2m_e) <Φ|(d²/dx²)| Φ>

= <Φ| - (Ћ²/2m_e) (d²/dx²)| Φ>

Where KE operator is - (Ћ²/2m_e) (d²/dx²) Now integrate by parts:

-∫

u(dv/dx)dx =

∫

(du/dx)(v)dx if u,v ➔ 0 at boundaries Let u = Φ^* and dv/dx = d²Φ/dx² then get Goddard form KE = (Ћ²/2m_e)<(dΦ/dx)| (dΦ/dx)>

Both forms of the KE are correct, I consider the second form more fundamental and more useful

It can be used to derive the 1^st form by integrating by parts

Clearly KE is always positive and decreasing slopes decreases KE

Best value for KE in QM of H atom

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

Which has the lowest KE?

But in QM the same wavefunction must be used for KE and PE

Best value for KE in QM of H atom

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

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 Mechanics R

_

_ _

The compromise between PE and KE

How do PE and KE scale with ? PE ~ -C₁/

KE ~ +C₂/ ²

Now lets find the optimum

R _ R

_ R

_

R _

Now consider very small

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

R _

Consider very large

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

R _

Thus there must be some intermediate for which the total energy is most negative

This is the for the optimum wavefunction R

_

R _

Analysis for optimum 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 to 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, with the minimum value being

<(dx)(dp> > Ћ/2 (The Heisenberg uncertainty principle)

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

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

Where = 0.529 A =0.053 nm is the Bohr radius (the average size of the H atom), =a₀/Z

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

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 form, which for the ground state of H atom has the form (here Z=1 for H atom, Z=6 for C ⁵⁺

Here we use r,Ө,Φspherical coordinates rather than x,y,z

R _

The PE = -Ze²/ while KE= ZeR ²/ 2 _

R _ _

Atomic Units

= 0.529 A =0.053 nm is the Bohr radius

E = -Ze²/2 R _

Z=1 for Hydrogen atom

= - Z² e²/2a₀ = - m_eZ² e⁴/2 Ћ² = Z² h₀/2 R

For Z ≠ 1, _ = a₀/Z

where h₀ = e²/a₀ = m_e e⁴/ Ћ² = Hartree = 27.2116 eV

= 627.51 kcal/mol = 2625,5 kJ/mol Atomic units:

m_e = 1, e = 1, Ћ = 1 leads to

unit of length = a₀ and unit of energy = h₀

In atomic units: KE= <Φ^.Φ>/2 (leave off Ћ²/m_e)

E = -0.5 h₀ Classical turning point r = a₀/√2

Local PE and KE of H atom

Local KE positive Local KE negative Local KE negative

Local PE, negative Local KE, positive

2 views of orbitals

Iine plot of orbital along z axis

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 a lower energy at finite R than at R = ∞ Two possibilities

Electron is on the left proton Electron is on the right proton Or we could combine them At R = ∞

These are all the same, but not for finite R

Now consider H

₂⁺

molecule

Combine Atomic Orbitals for H

₂⁺

molecule

Two extreme possibilities

Antisymmetric combination Symmetric combination

the D_g = Sqrt[2(1+S)] and D_u = Sqrt[2(1-S)] factors above are the constants needed to ensure that

<Φ_g|Φ_g> =

∫

^Φ_g^|Φ_g dxdydz = 1 (normalized)

<Φ_u|Φ_u> =

∫

^Φ_u^|Φ_u dxdydz = 1 (normalized)

I will usually writing such factors, but they are understood Which is best (lowest energy)?

Energies of of H

₂⁺

Molecule

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, but increasing the density in bond region ➔ decrease density near atoms, thus moves electrons from very attractive region near nuclei to less

attractive region near bond midpoint, this INCREASES the

+ - +

The change in electron density for molecular orbitals

The densities r_g and r_u for the g and u LCAO

wavefunctions of H₂⁺ compared to superposition of r_L + r_R atomic densities (all densities add up to one electron)

Adding the two atomic orbitals to form the g molecular orbital

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 nucleus

Thus the Φ_g = _L + _R wavefunction moves charge to the bond region AT THE EXPENSE of the charge near the nuclei, causing PE(r) = -1/r_a – 1/r_b

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

The shape of the Φ_g = _L + _R and Φ_u = _L - _R wavefunctions

compared to the pure atomic

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

Show 2-5

The total PE of H₂⁺ for the Φ_g = _L + _R and Φ_u = _L - _R

wavefunctions (relative to the values of V_g = V_u = -1 h₀ at R = ∞)

The KE of g and u wavefunctions

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

KE_g=KE_u=+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

R too short leads to a big decrease in slope but over a very short region, ➔ little bonding

R is too large leads to a decrease in

slope over a long region, but the change in slope is very small ➔ little bonding

Optimum bonding occurs when there is a large region where both atomic orbitals have large slopes in the opposite

directions (contragradient).

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 h₀

PE: -1.0 h₀ E: -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

H

₂

molecule, independent atoms

Start with non interacting H atoms, electron 1 on H on earth, _E(1) the other electron 2 on the moon, _M(2)

What is the total wavefunction, Ψ(1,2)?

Maybe Ψ(1,2) = _E(1) + _M(2) ?

Since the motions of the two electrons are completely

independent, we expect that the probability of finding electron 1 somewhere to be independent of the probability of finding electron 2 somewhere.

Thus P(1,2) = P_E(1)*P_M(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

Ψ(1,2) = _E(1)_M(2) leads to

P(1,2) = |Ψ(1,2)|² = Ψ(1,2)^* Ψ(1,2) =

= [_E(1)_M(2)]^* [_E(1)_M(2)] =

= [_E(1)^* _E(1)] [_M(2)^* _M(2)] =

= P_E(1) P_M(2)

Answer: product of amplitudes

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

Back to H₂, Ψ_EM(1,2) = _E(1)_M(2)

But Ψ_ME(1,2) = _M(1)_E(2) is equally good since the electrons are identical

Also we could combine these wavefunctions

End of Lecture 1