William A. Goddard, III, [email protected]

EEWS-90.502-Goddard-L04 1© copyright 2009 William A. Goddard III, all rights reserved

Nature of the Chemical Bond with applications to catalysis, materials

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

William A. Goddard, III, [email protected] Professor at EEWS-KAIST and

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

California Institute of Technology

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

Senior Assistant: Dr. Hyungjun Kim: [email protected] of Center for Materials Simulation and Design (CMSD)

Teaching Assistant: Ms. Ga In Lee: [email protected] assistant: Tod Pascal:[email protected]

Lecture 7, September 24, 2009


Schedule changes

There was no lecture on Sept. 22 because of the EEWS conference

Goddard will be traveling Oct 2-11 and will not give the lectures scheduled for Oct. 6 and 8

Consequently an extra lecture will be added at 2pm on Wednesday Sept. 30 and another at 2pm Wednesday Oct 14

L8: Sept. 29, as scheduled

L9: Sept. 30, new replaces Oct 6

L10: Oct. 1, as scheduled


L12: Oct. 14, new replaces Oct 8



Last time


Contour plots of 3s, 3p, 3d hydrogenic orbitals


The ground state of He atomPut both electrons in 1s orbitals

ΨHe(1,2) = A[(Φ1s)(Φ1sΦ1s(1)Φ1s(2) (

Φ1s = exp(- r)

EHe = 2(½ 2) – 2Z(5/8)whereJ1s,1s = (5/8)

Applying the variational principle, dE/d = 0

get = (Z – 5/16) = 1.6875

E= 2(-½2) = - 2 = -2.8477 h0

Interpretation: two electrons move independently in the orbital Φ1sexp(-r) which has been adjusted to account for the average shielding due to the other electron in this orbital.

On the average this other electron is closer to the nucleus about 31% of the time so that the effective charge seen by each electron is 2-0.31=1.69

The total energy is just the sum of the individual energies.


Now 3rd electron to form Li

ΨLi(1,2,3) = A[(Φ1s)(Φ1s(Φ1sby Pauli principle

Thus the 3rd electron must go into 2s or 2p atomic orbital

ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2s

ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2pz(or 2px or 2py)

Li+: Φ1sexp(-r) with = Z-0.3125 = 2.69

R1s = 1/ = 0.372 a0 = 0.2A


Add 3rd electron to the 2p orbital


The 2p orbital sees effective charge of Zeff = 3 – 2 = 1,

since it goes to zero at z=0, so that there is no shielding of 1s

Get size: R2p = n2/Zeff = 4 a0 = 2.12A

Energy: e2p = -(Zeff)2/2n2 = -1/8 h0 = -3.40 eV

0.2A

1s

2.12A

2p


Add the 3rd electron to the 2s orbital


The 2s orbital must be orthogonal to the 1s, which means that it must have a spherical nodal surface at ~ 0.2A, the size of the 1s orbital. Thus the 2s has a nonzero amplitude at z=0 so that it is not completely shielded by the 1s orbitals.

The result is Zeff2s = 3 – 1.72 = 1.28

Size: R2s = n2/Zeff = 3.1 a0 = 1.65A

Energy: e2s = -(Zeff)2/2n2 = -0.205 h0 = 5.57 eV

0.2A

1s

2.12A2s

R~0.2A


Li atom excited states

1s

2s

-0.125 h0 = -3.4 eV2p

Energy

zero

-0.205 h0 = -5.6 eV

-2.723 h0 = 74.1 eV

MO picture State picture

(1s)2(2s)

(1s)2(2p)

E = 2.2 eV17700 cm-1

564 nm

Ground state

1st excited state

Exper671 nm

E = 1.9 eV


He, 2

Ne, 10

Ar, 18

Zn, 30Kr, 36

Aufbau principle for atoms

Particularly stable atoms, closed shells

Xe, 54

Rn, 86

Uuo, 118


Many-electron configurations

General aufbau

ordering

Particularly stable


General trends along a row of the periodic table

As we fill a shell, thus

B(2s)2(2p)1 to Ne (2s)2(2p)6

Zeff increases leading to a decrease in the radius ~ n2/Zeff

And an increase in the IP ~ (Zeff)2/2n2

Example Zeff2s=

1.28 Li, 1.92 Be, 2.28 B, 2.64 C, 3.00 N, 3.36 O, 4.00 F, 4.64 Ne

Thus R(2s Li)/R(2s Ne) ~ 4.64/1.28 = 3.6


General trends along a column of the periodic table

As we go down a colum

Li [He}(2s) to Na [Ne]3s to K [Ar]4s to Rb [Kr]5s to Cs[Xe]6s

Things get more complicated

The radius ~ n2/Zeff

And the IP ~ (Zeff)2/2n2

But the Zeff tends to increase, partially compensating for the change in n so that the atomic sizes increase only slowly as we go down the periodic table and

The IP decrease only slowly (in eV):

5.39 Li, 5.14 Na, 4.34 K, 4.18 Rb, 3.89 Cs

(13.6 H), 17.4 F, 13.0 Cl, 11.8 Br, 10.5 I, 9.5 At

24.5 He, 21.6 Ne, 15.8 Ar, 14.0 Kr,12.1 Xe, 10.7 Rn


Transition metals; consider [Ar] plus one electron

[IP4s = (Zeff4s )2/2n2 = 4.34 eV Zeff

4s = 2.26; 4s<4p<3d

IP4p = (Zeff4p )2/2n2 = 2.73 eV Zeff

4p = 1.79;

IP3d = (Zeff3d )2/2n2 = 1.67 eV Zeff

3d = 1.05;

IP4s = (Zeff4s )2/2n2 = 11.87 eV Zeff

4s = 3.74; 4s<3d<4p


3d = 2.59;


4p = 3.20;


3d = 4.05; 3d<4s<4p

IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff

4s = 5.04;


4p = 4.47;

K

Ca+

Sc++

As the net charge increases the differential shielding for 4s vs 3d is less important than the difference in n quantum number 3 vs 4Thus charged system prefers 3d vs 4s


Transition metals; consider Sc0, Sc+, Sc2+

3d: IP3d = (Zeff3d )2/2n2 = 24.75 eV Zeff

3d = 4.05;

4s: IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff

4s = 5.04;

4p: IP4p = (Zeff4p )2/2n2 = 17.01 eV Zeff

4p = 4.47;

Sc++

For high charge (Sc++) 3d more stable than 4s (3d)1

For neutral system (Sc) fill 4s: (3d)(4s)2 This is because increased charge makes the differential shielding for 4s vs 3d less important than the in n quantum number 3 vs 4.

(3d)(4s): IP4s = (Zeff4s )2/2n2 = 12.89 eV Zeff

4s = 3.89;

(3d)2: IP3d = (Zeff3d )2/2n2 = 12.28 eV Zeff

3d = 2.85;

(3d)(4p): IP4p = (Zeff4p )2/2n2 = 9.66 eV Zeff

4p = 3.37;

Sc+

(3d)(4s)2: IP4s = (Zeff4s )2/2n2 = 6.56 eV Zeff

4s = 2.78;

(4s)(3d)2: IP3d = (Zeff3d )2/2n2 = 5.12 eV Zeff

3d = 1.84;

(3d)(4s)(4p): IP4p = (Zeff4p )2/2n2 = 4.59 eV Zeff

4p = 2.32;

Sc


Implications on transition metals

The simple Aufbau principle puts 4s below 3dBut increasing the charge tends to prefers 3d vs 4s. Thus Ground state of Sc 2+ , Ti 2+ …..Zn 2+ are all (3d)n

For all neutral elements K through Zn the 4s orbital is easiest to ionize.

This is because of increase in relative stability of 3d for higher ions

The character of transition metals Sc-Cu columns depends on circumstances


Transtion metal orbitals


Next More detailed description of first row atoms

Li: (2s)

Be: (2s)2

B: [Be](2p)1

C: [Be](2p)2

N: [Be](2p)3

O: [Be](2p)4

F: [Be](2p)5

Ne: [Be](2p)6


Consider the ground state of B: [Be](2p)1

Ψ(1,2,3,4,5) = A[Φ

Ignore the [Be] core then

Can put 1 electron in 2px, 2py, or 2pz each with either up or down spin. Thus get 6 states.

We will depict these states by simplified contour diagrams in the xz plane, as at the right.

Of course 2py is zero on this plane. Instead we show it as a circle as if you can see just the front part of the lobe sticking out of the paper.

2px

2pz

2py

z

x

Because there are 3 degenerate states we denote this as a P state. Because the spin can be +½ or –½, we call it a spin doublet and we denote the overall state as 2P


New material


Consider the ground state of C: [Be](2p)2


Can put 2 electrons in 2px, 2py, or 2pz each with both up and down spin.

Or can put one electron in each of two orbitals: (2px)(2py), (2px)(2px), (2py)(2pz),

We will depict these states by simplified contour diagrams in the xz plane, as at the right.

Which state is better? The difference is in the electron-electron repulsion: 1/r12

z

x

z

x

z

x

(2px)2

(2pz)2

(2px)(2pz)

Clearly two electrons in the same orbital have a much smaller average r12 and hence a much higher e-e repusion. Thus the ground state has each electron in a different 2p orbital


Consider the states of C: formed from (x)(y), (x)(z), (y)(z)

Consider first (x)(y): can form two spatial products: Φx(1)Φy(2) and Φy(1)Φx(2)

These are not symmetric, thus must combine

Φ(1,2)s=φx(1) φy(2) + φy(1) φx(2)

Φ(1,2)a= φx(1) φy(2) - φy(1) φx(2)

y

x

(2px)(2py)

Which state is better? The difference is in the electron-electron repulsion: 1/r12

To analyze this, expand the orbitals in terms of the angular coordinates, r,θ,φ

Φ(1,2)s= f(r1)f(r2)(sinθ1)(sinθ2)[(cosφ1)(sinφ2)+(sinφ1) (cosφ2)]

=f(r1)f(r2)(sinθ1)(sinθ2)[sin(φ1+φ2)]

Φ(1,2)a= f(r1)f(r2)(sinθ1)(sinθ2)[(cosφ1)(sinφ2)-(sinφ1) (cosφ2)]

=f(r1)f(r2)(sinθ1)(sinθ2)[sin(φ2 -φ1)]


Consider the symmetric and antisymmetric combinations of (x)(y)

y

x

(2px)(2py)Φ(1,2)s= f(r1)f(r2)(sinθ1)(sinθ2)[sin(φ1+φ2)]

Φ(1,2)a= f(r1)f(r2)(sinθ1)(sinθ2)[sin(φ2 -φ1)]

The big difference is that Φ(1,2)a = 0 when φ2 = φ1 and is a maximum for φ2 and φ1 out of phase by /2.

But for Φ(1,2)s the probability of φ2 = φ1 is comparable to that of being out of phase by /2.

Thus the best combination is Φ(1,2)a

Thus for 2 electrons in orthogonal orbitals, high spin is best because the electrons can never be at same spot at the same time

Combining with the spin parts we get

[φx(1) φy(2) + φy(1) φx(2or spin = 0

[φx(1) φy(2) - φy(1) φx(2also andor spin = 1


Summarizing the states for C atom

Ground state: three triplet states=2L=1. Thus L=1, denote as 3P (xy-yx) ≡ [x(1)y(2)-y(1)x(2)] (xz-zx)(yz-zy)Next state: five singlet states=2L+1. Thus L=2, denote as 1D (xy+yx)(xz+zx)(yz+zy)(xx-yy)(2zz-xx-yy)Highest state: one singlet=2L+1. thus L=0. Denote as 1S(zz+xx+yy)

y

x

(2px)(2py)

Hund’s rule. Given n electrons distributed among m equivalent orgthogonal orbitals, the ground state is the one with the highest possible spin. Given more than one state with the highest spin, the highest orbital angular momentum is the GS


Calculating energies for C atom

The energy of xy is

Exy = hxx + hyy + Jxy = 2hpp +Jxy

Thus the energy of the 3P state is

E(3P) = Exy – Kxy = 2hpp +Jxy - Kxy

For the (xy+yx) component of the 1D state, we get

E(1D) = Exy + Kxy = 2hpp +Jxy + Kxy

Whereas for the (xx-yy) component of the 1D state, we get

E(1D) = Exx - Kxy = 2hpp +Jxx - Kxy

This means that Jxx - Kxy = Jxy + Kxy so that

Jxx = Jxy + 2Kxy

Also for (2zz-xx-yy) we obtain E = 2hpp +Jxx - Kxy

For (zz+xx+yy) we obtain

E(1S) = 2hpp + Jxx + 2 Kxy

y

x

(2px)(2py)


Summarizing the energies for C atom

E(1S) = 2hpp + Jxx + 2Kxy

E(1D) = 2hpp +Jxx - Kxy = 2hpp +Jxy + Kxy

E(3P) = 2hpp + Exy – Kxy = 2hpp +Jxy - Kxy

2Kxy

3Kxy


Comparison with experiment

E(1S)

E(1D)

E(3P)

2Kxy

3Kxy

C Si Ge Sn Pb

TA’s look up data and list excitation energies in eV and Kxy in eV.

Get data from Moore’s tables or paper by Harding and Goddard Ann Rev Phys Chem ~1985)


Summary ground state for C atom

z

x

(2px)(2pz)

Ψ(1,2,3,4,5,6)xz= A[(1s)(1s)(2s)(2s)(2px)(2pz)] =

= A[(1s(2s)2(2px)(2pz)] =A[(Be)(2px)(2pz)] =

= A[(x)(z)] = which we visualize as

Ψ(1,2,3,4,5,6)xy = A[(x)(y)]

which we visualize as

Ψ(1,2,3,4,5,6)yz = A[(y)(z)]


z

x

Note that we choose to use the xz plane for all 3 wavefunctions, so that the py orbitals look like circles (seeing only the + lobe out of the plane

z

x


Consider the ground state of N: [Be](2p)3


Can put one electron in each of three orbitals: (2px)(2py)(2pz)

Or can put 2 in 1 and 1 in another:

(x)2(y), (x)2(z), (y)2(x), (y)2(z), (z)2(x), (z)2(y)

As we saw for C, the best state is (x)(y)(z) because of the lowest ee repulsion.

xyz can be combined with various spin functions, but from Hund’s rule we expect A[(x)(y)(z)] = [Axyz] to be the ground state. (here Axyz is the antisymmetric combination of x(1)y(2)x(3)]The four symmetric spin functions areWith Mswhich refer to as S=3/2 or quartetSince there is only one xyz state = 2L+1 with L=0, we denote it as L=0, leading to the 4S state.


Energy of the ground state of N: A[(x)(y)(z)] = [Axyz]

Simple product xyz leads to

Exyz = 3hpp + Jxy + Jxz + Jyz

E(4S) = <xyz|H|A[xyz] <xyz|A[xyz]

Denominator = <xyz|A[xyz] = 1

Numerator = Exyz - Kxy - Kxz – Kyz =

E(4S) = 3hpp + (Jxy - Kxy) + (Jxz - Kxz) + (Jyz – Kyz)

E(2P)

E(2D)

E(4S)

4Kxy

2Kxy

=3hpp + 3Jxy - 3Kxy

=3hpp + 3Jxy + 1Kxy = 3hpp + 2Jxy + Jxx - 1Kxy

=3hpp + 2Jxy + Jxx + 1Kxy

TA’s check this

z

x

Pictorial representation of the N ground

state

Since Jxy = Jxz = Jyz

and Kxy= Kxz = Kyz


Comparison with experiment

E(1S)

E(1D)

E(3P)

4Kxy

2Kxy

N P As Sb Bi




Consider the ground state of O: [Be](2p)4

Only have 3 orbitals, x, y, and z. thus must have a least one doubly occupied

Choices: (z)2(x)(y), (y)2(x)(z), (x)2(z)(y)

and (z)2(x)2, (y)2(x)2, (y)2(z)2

Clearly it is better to have two singly occupied orbitals.

Just as for C atom, two singly occupied orbitals lead to both a triplet state and a singlet state, but the high spin triplet with the same spin for the two singly occupied orbitals is best

z

x

z

x

(2px)2(2py)(2pz)

(2pz)2(2px)2


Ψ(1,2,3,4,5,6)xz= A[(1s)(1s)(2s)(2s)(2py)(2py)(2px)(2pz)]

= A[(1s(2s)2(2py)(2py)(2px)(2pz)] =

= A[(y)(y)(x)(z)] = which we visualize as

Summary ground state for O atom

z

x

Ψ(1,2,3,4,5,6)xy = A[(z)(z)(((x)(y)]


Ψ(1,2,3,4,5,6)yz = A[(x)(x)((y)(z)]

which we visualize as (2px)2(2py)(2pz)

z

x

z

x

We have 3 = 2L+1 equivalent spin triplet (S=1) states that we denote as L=1 orbital angular momentum, leading to the 3P state

(2py)2(2px)(2pz)

(2pz)2(2px)(2py)


Calculating energies for O atom

is

Exz = E(Be) + 2hyy + hxx + hzz + Jyy+ 2Jxy+ 2Jyx+ Jxz – Kxy – Kyz – Kxz

Check: 4 electrons, therefore 4x3/2 = 6 coulomb interactions

3 up-spin electrons, therefor 3x2/2 = 3 exchange interactions

Other ways to group energy terms

Exz = 4hpp + Jyy + (2Jxy – Kxy) + (2Jyx – Kyz) + (Jxz– Kxz)

Same energy for other two components of 3P state

(2py)2(2px)(2pz)The energy of Ψ(1,2,3,4,5,6)xz= A[(1s)(1s)(2s)(2s)(2py)(2py)(2px)(2pz)]

= A[[Be](y)(y)(x)(z)]


Comparison of O states with C statesO (2p)4 3P C (2p)2 3PNe (2p)6 1S

Compared to Ne, we have

Hole in x and z

Hole in y and z

Hole in x and y

Compared to Be, we have

Electron in x and z

Electron in y and z

Electron in x and y

z

x

Thus holes in O map to electrons in C

z

x


Summarizing the energies for O atom

E(1S)

E(1D)

E(3P)

2Kxy

3Kxy

O S Se Te Po




Consider the ground state of F: [Be](2p)5

Only have 3 orbitals, x, y, and z. thus must have two doubly occupied

Choices: (x)2(y)2 (z), (x)2(y)(z)2, (x)(y)2(z)2

Clearly all three equivalent give rise to spin doublet. Since 3 = 2L+1 denote as L=1 or 2P

Ψ(1,2,3,4,5,6,7)z= A[(1s)(1s)(2s)(2s)(2py)(2py)(2px)(2pz)]

= A[(1s(2s)2(2py)(2py)(2px)(2pz)] =

= A[[Be](x)(x)(y)(y)(z)]

which we visualize asz

x


Comparison of F states with B statesF (2p)5 2P B (2p)1 2PNe (2p)6 1S


Hole in z

Hole in x

Hole in y


Electron in z

Electron in x

Electron in y

Thus holes in F map to electrons in B

z

x

z

x


Calculating energies for F atom

is

Exz = 5hpp + Jxx+ Jyy+ (4Jxy – 2Kxy) + (2Jxz – Kxz) + (2Jyx– Kyz)

Check: 5 electrons, therefore 5x4/2 = 10 coulomb interactions

3 up-spin electrons, therefore 3x2/2 = 3 exchange interactions

2 down-spin electrons, therefore 2x1/2 = 1 exchange interaction

Other ways to group energy terms

Same energy for other two components of 2P state

(2px)2(2py)2(2pz)The energy of

Ψ(1-9)z= A{[Be](2px)(2px)(2py)(2py)(2pz)]


Comparison of F states with B statesF (2p)5 2P B (2p)1 2PNe (2p)6 1S


Hole in z

Hole in x

Hole in y


Electron in z

Electron in x

Electron in y

Thus holes in F map to electrons in B

z

x

z

x


Consider the ground state of Ne: [Be](2p)6

Only have 3 orbitals, x, y, and z. thus must have all three doubly occupied

Choices: (x)2(y)2(z)2

Thus get spin singlet, S=0

Since just one spatial state, 1=2L+1 L=0.

denote as 1S

Ψ(1-10)z = A[[Be](x)(x)(y)(y)(z)(z)]

which we visualize asNe (2p)6 1S

z

x


Calculating energy for Ne atom

is

Exz = 6hpp +Jxx+Jyy+ Jzz + (4Jxy – 2Kxy) + (4Jxz – 2Kxz) + (4Jyx– 2Kyz)

Check: 6 p electrons, therefore 6x5/2 = 15 coulomb interactions

3 up-spin electrons, therefore 3x2/2 = 3 exchange interactions

3 down-spin electrons, therefore 3x2/2 = 3 exchange interaction

Since Jxx = Kxx we can rewrite this as

Exz = 6hpp +(2Jxx-Kxx ) +(2Jyy-Kyy ) +(2Jzz-Kzz ) + 2(2Jxy – Kxy) +

2(2Jxz – Kxz) + 2(2Jyx– Kyz)

Which we will find later to be more convenient for calculating the wavefunctions using the variational principle

(2px)2(2py)2(2pz)2

The energy of

Ψ(1-9)z= A{[Be](2px)(2px)(2py)(2py)(2pz)(2pz)}


Summary of ground states of Li-Ne

N 4S (2p)3

O 3P (2p)4

F 2P (2p)5

Ne 1S (2p)6

Li 2S (2s)1

Be 1S (2s)2

B 2P (2p)1

C 3P (2p)2

Ignore (2s)2


Bonding H atom to He

Starting with the ground state of He, (1s)2 = A(He1s)(He1s) and bringing up an H atom (H1s), leads to

R

HeH: A[(He1s)(He1s)(H1s)]

But properties of A or alternatively the Pauli Principle tells us that the H1s must get orthogonal to the He 1s since both have an spin.

Consequently H1s must get a nodal plane, increasing its KE.

The smaller R the larger increase in KE. Get a repulsive interaction, no bond


Bonding H atom to Ne

Starting with the ground state of Ne, (1s)2(2s)2(2p)6 Ψ(Ne)= A{[Be](2px)(2px)(2py)(2py)(2pz)(2pz)}

and bringing up an H atom (H1s) along the z axis, leads to

A{[Be](2px)2(2py)2(Ne2pz)(Ne2pz)(H1s}

Where we focus on the Ne2pz orbital that overlaps the H atom

R

Consequently H1s must get a nodal plane, increasing its KE.

The smaller R the larger increase in KE.

Get a repulsive interaction, no bond

The properties of A or alternatively the Pauli Principle tells us that the H1s must get orthogonal to the Ne 2pz since both have an spin.


Now consider Bonding H atom to all 3 states of F

R

Bring H1s along z axis to F and consider all 3 spatial states.

z

x

William A. Goddard, III, [email protected]

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