© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L04 1 Nature of the Chemical Bond with applications to catalysis, materials.

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 6, September 17, 2009


Last time


For H2, H(1,2) = h(1) + h(2) +1/r12 + 1/R

Product wavefunction Ψ(1,2) = ψa(1)ψb(2)

E = haa + hbb + Jab + 1/R

haa ≡ <ψa(1)|h(1)|ψa(1)> ≡ <a|h|a>

hbb ≡ <ψb(2)|h(2)|ψb(2)> ≡ <b|h|b>

Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)> = ∫| ψa(1)|2 |ψb(2)|2 /r12 is the total Coulomb interaction between the electron density

a(1)=| ψa(1)|2 and b(2)=| ψb(2)|2

Thus Jab > 0

Jab 1/R if a is centered on atom a while b is centered on atom b a large distance R far from a. For shorter distances at which the densities overlap, the Jab decreases (shielding)

Energy for 2 electron product of spinorbitals


Energy for antisymmetrized product of 2 spinorbitals

A ψa(1)ψb(2)

E = haa + hbb + (Jab –Kab) + 1/R

Kab = < ψaψb|1/r12|ψb ψa > = ∫[ψa*(1)ψb(1)][ψb

*(2)ψa(2)] /r12 is the exchange integral for ψa and ψb Jab > Kab > 0


Both electrons have the same spin

ψa(1) = Φa(1)(1)

ψb(2) = Φb(2)(2)

<ψa|ψb>= 0 = < Φa| Φb><|> = < Φa| Φb>

the spatial orbitals for same spin must be orthogonal

Aψa(1)ψb(2)= A[Φa(1)Φb(2)][(1)(2)]

= [Φa(1)Φb(2)- Φb(1)Φa(2)][(1)(2)]

The total energy is

E = haa + hbb + (Jab –Kab) + 1/R

haa ≡ <Φa(1)|h(1)|Φa(1)> and similarly for hbb

Jab= <Φa(1)Φb(2) |1/r12 |Φa(1)Φb(2)>

Kab ≡ <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)>

Jab > Kab > 0


ψa(1) = Φa(1)(1)

ψb(2) = Φb(2)(2)

Aψa(1)ψb(2)= A[Φa(1)Φb(2)][(1)(2)]=

We obtain <ψa|ψb>= 0 = < Φa| Φb><|> = 0Independent of the overlap of the spatial orbitals. Thus spatial orbitals can overlap, <Φa|Φb> = SThe exchange term for spin orbitals with opposite spin is zero; get exchange only between spinorbitals with the same spinThus the total energy is

E = haa + hbb + Jab + 1/R Just as for the simple product wavefunction, Φa(1)Φb(2)

Assume electrons have the opposite spin


For spinorbitals with opposite spin, must combine Slater Determinants to obtain full permutational symmetry

The antisymmetrized wavefunction leads to

Aψa(1)ψb(2)= A[Φa(1)Φb(2)][(1)(2)]=

=[Φa(1) (1)][Φb(2)(2)] – [Φb(1) (1)][Φa(2)(2)]

Interchanging the spins leads to

[Φa(1) (1)][Φb(2)(2)] – [Φb(1) (1)][Φa(2)(2)] =

= A[Φa(1)Φb(2)][(1)(2)]

Which is neither + or – times the starting wavefunction.

Thus must combine to obtain proper spatial and spin symmetry


Correct space and spin symmetry for wavefunctions

[Φa(1)Φb(2)-Φb(1)Φa(2)][(1)(2)+(1)(2)]= A[Φa(1)Φb(2)][(1)(2)]-A[Φb(1)Φa(2)][(1)(2)]

Is symmetric in spin coordinates (the MS = 0 component of the S=1 triplet)

[Φa(1)Φb(2)+Φb(1)Φa(2)][(1)(2)-(1)(2)]= A[Φa(1)Φb(2)][(1)(2)]+A[Φb(1)Φa(2)][(1)(2)]

Is antisymmetric in spin coordinates (the MS = 0 component of the S=0 triplet)Thus for the case, two Slater determinants must be combined to obtain the correct spin and space permutational symmetry


The triplet state for 2 electrons

The wavefunction

[Φa(1)Φb(2)-Φb(1)Φa(2)][(1)(2)+(1)(2)]

Leads directly to 3E = haa + hbb + (Jab –Kab) + 1/R Exactly the same as for

[Φa(1)Φb(2)-Φb(1)Φa(2)][(1)(2)]

[Φa(1)Φb(2)-Φb(1)Φa(2)][(1)(2)]

These three states are collectively referred to as the triplet state and denoted as having spin S=1. It has 3 components

MS = +1: [Φa(1)Φb(2)-Φb(1)Φa(2)][(1)(2)]

MS = 0 : [Φa(1)Φb(2)-Φb(1)Φa(2)][(1)(2)+(1)(2)]

MS = -1: [Φa(1)Φb(2)-Φb(1)Φa(2)][(1)(2)]

where < Φa|Φb> = 0


The singlet state for two electrons

The other combination of MS=0 determinants leads to the singlet state and is denoted as having spin S=0

[Φa(1)Φb(2)+Φb(1)Φa(2)][(1)(2)-(1)(2)]

We will analyze the energy for this wavefunction next.

It is more complicated since < Φa|Φb> ≠ 0

1E = <ab|H|(ab+ba)>/<ab|(ab+ba)>

1E = {(haa + hbb + (hab + hba) S2 + Jab + Kab + (1+S2)/R}/(1 + S2)

This is the general energy for the ab+ba singlet, but most relevant to us is for H2, where Φa=XL and Φb=XR

In this case it is convenient to write the triplet state in terms of the same overlapping orbitals, even though they could be orthogonalized for the triplet state


Analysis of singlet and triplet energies for H2

Taking Φa=XL and Φb=XR for H2, the VB energy for the bonding

state (g, singlet) is1Eg = <ab|H|(ab+ba)>/<ab|(ab+ba)>1Eg = {(haa + hbb + (hab + hba) S + Jab + Kab + (1+S2)/R}/(1 + S2)

Similary for the VB triplet we obtain3Eu = <ab|H|(ab-ba)>/<ab|(ab-ba)>3Eu = {(haa + hbb - (hab + hba) S + Jab - Kab + (1-S2)/R}/(1 - S2)

We find it useful to define a classical energy, with no exchange or interference or resonance

Ecl = <ab|H|ab>/<ab|ab> = haa + hbb + Jab + 1/R

Egx = +Ex/(1 + S2)

Eux = - Ex/(1 - S2)

Ex = {(hab + hba) S + Kab –EclS2}

where

Then we can define the energy as

1Eg = Ecl + Egx

3Eu = Ecl + Eux


The VB exchange energies for H2

1Eg = Ecl + Egx

3Eu = Ecl + Eux

For H2, the classical energy is slightly attractive, but again the difference between bonding (g) and anti bonding (u) is essentially all due to the exchange term.

Each energy is referenced to the value at

R=∞, which is

-1 for Ecl, Eu, Eg 0 for Ex

u and Exg

+Ex/(1 + S2)

-Ex/(1 - S2)


Analysis of the VB exchange energy, Ex

where Ex = {(hab + hba) S + Kab –EclS2} = TT

Here T{(hab + hba) S –(haa + hbb)S2} = 2S

Where = (hab – Shaa) contains the 1e part

T{Kab –S2Jab} contains the 2e part

Clearly the Ex is dominated by T and clearly T is dominated by the kinetic part, T. E

x

T2

T1

TThus we can understand bonding by analyzing just the KE part if Ex


ecl = (hLL + 1/R) is the energy for bringing a proton up to H atom

= (hLR - ShLL) contains the terms that dominate the bonding and antibonding character of these 2 states

Re-examine the energy for H2+

The same kinetic term important for H2 is also the critical part of the energy for H2

+ the VB wavefunctions are

Φg = (хL + хR) and

Φu = (хL - хR) (ignoring normalization)

where H = h + 1/R. This leads to

eg = <L+R|H|L+R>/ <L+R|L+R> = 2 <L|H|L+R>/ 2<L|L+R>

= (hLL + hLR)/(1+S) + 1/R = (hLL+ShLL+hLR - ShLL)/(1+S) + 1/R

= (hLL + 1/R) + (hLR-ShLL)/(1+S)

eg = ecl + /(1+S)

eu = ecl - /(1-S)


The VB interference or resonance energy for H2+

The VB wavefunctions for H2+ Φg = (хL + хR)

and Φu = (хL - хR) lead to

eg = (hLL + 1/R) + /(1+S) ≡ ecl + Egx

eu = (hLL + 1/R) - /(1-S) ≡ ecl + Eux

ecl = (hLL + 1/R) is the classical energy

= (hLR - ShLL) is the VB interference or resonance energy that dominates the bonding and antibonding of these states

/(1+S)

/(1+S)


Contragradience

The above discussions show that the interference or exchange part of KE dominates the bonding, KE=KELR –S KELL

This decrease in the KE due to overlapping orbitals is dominated by

Dot product is large and negative

in the shaded region between

atoms, where the L and R orbitals have

opposite slope (congragradience)

x = ½ [< (хL). ((хR)> - S [< (хL)2>

хL хR


Comparison of exchange energies for 1e and 2e bonds

H2

Eg1x ~ +2S /(1 + S2)

Eu1x ~ -2S /(1 - S2)

H2+ case

egx ~ +/(1 + S)

eux ~ -/(1 - S)

Eu1x

Eg1x

R(bohr)

E(hartree)

For R=1.6bohr (near Re), S=0.7 Eg

1x ~ 0.94vs. egx ~

Eu1x ~ -2.75vs. eu

x ~ For R=4 bohr, S=0.1

Eg1x ~ 0.20vs. eg

x ~

Eu1x ~ -0.20vs. eu

x ~


H atom, excited states

In atomic units, the Hamiltonian

h = - (Ћ2/2me)– Ze2/r

Becomes

h = - ½ – Z/r

Thus we want to solve

hφk = ekφk for all excited states k

r

+Ze

φnlm = Rnl(r) Zlm(θ,φ) product of radial and angular functions

Ground state: R1s = exp(-Zr), Zs = 1 (constant)


The excited angular states of H atom, 1 nodal planeφnlm = Rnl(r) Zlm(θ,φ) the excited angular functions, Zlm must have nodal planes to be orthogonal to Zs

3 cases with one nodal planex

z

+

-

pz

Z10=pz = r cosθ (zero in the xy plane)Z11=px = rsinθcosφ (zero in the yz plane)Z1-1=py = rsinθsinφ (zero in the xz plane)

x

z

+-

px

We find it useful to keep track of how often the wavefunction changes sign as the φ coordinate is increased by 2 = 360º

If m=0, denote it as : pz = p

If m=1, denote it as : px, py = p

If m=2 we call it a function

If m=3 we call it a function


The excited angular states of H atom, 2 nodal planes

x

z+

-

dz2

-

+

57º

Z20 = dz2 = [3 z2 – r2 ] m=0, d

Z21 = dzx = zx =r2 (sinθ)(cosθ) cosφ

Z2-1 = dyz = yz =r2 (sinθ)(cosθ) sinφ

Z22 = dx2-y2 = x2 – y2 = r2 (sinθ)2 cos2φ

Z22 = dxy = xy =r2 (sinθ)2 sin2φ

m = 1, d

m = 2, d

Summarizing:one s angular function (no angular nodes) called ℓ=0three p angular functions (one angular node) called ℓ=1 five d angular functions (two angular nodes) called ℓ=2seven f angular functions (three angular nodes) called ℓ=3nine g angular functions (four angular nodes) called ℓ=4ℓ is referred to as the angular momentum quantum numberThere are (2ℓ+1) m values, Zℓm for each ℓ

Where we used [(cosφ)2 – (sinφ)2]=cos2φ and 2cosφ sinφ=sin2φ


Excited radial functions

Clearly the KE increases with the number of angular nodes so that s < p < d < f < g

Now we must consider radial functions, Rnl(r)The lowest is R10 = 1s = exp(-Zr)All other radial functions must be orthogonal and hence must have one or more radial nodes, as shown here

Note that we are plotting the cross section along the z axis, but it would look exactly the same along any other axis. Here

R20 = 2s = [Zr/2 – 1]exp(-Zr/2) and R30 = 3s = [2(Zr)2/27 – 2(Zr)/3 + 1]exp(-Zr/3)

Zr = 2 Zr = 1.9

Zr = 7.1


Combination of radial and angular nodal planes

Combining radial and angular functions gives the various excited states of the H atom. They are named as shown where the n quantum number is the total number of nodal planes plus 1

The nodal theorem does not specify how 2s and 2p are related, but it turns out that the total energy depends only on n.

Enlm = - Z2/2n2

The potential energy is given by

PE = - Z2/2n2 ≡ -Z/ , where =n2/Z

Thus Enlm = - Z/(2 )

1s 0 0 0 1.02s 1 1 0 4.02p 1 0 1 4.03s 2 2 0 9.03p 2 1 1 9.03d 2 0 2 9.04s 3 3 0 16.04p 3 2 1 16.04d 3 1 2 16.04f 3 0 3 16.0

nam

e

tota

l nod

al p

lane

sra

dial

nod

al p

lane

san

gula

r no

dal p

lane

s

R̄ R̄

R̄

Siz

e (a

0)


New material


Sizes hydrogen orbitals

Hydrogen orbitals 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f

Angstrom (0.1nm) 0.53, 2.12, 4.77, 8.48

H--H C

0.74A

H

H

H

H

1.7A

H H

H H

H H

4.8


Hydrogen atom excited states

1s-0.5 h0 = -13.6 eV

2s-0.125 h0 = -3.4 eV

2p

3s-0.056 h0 = -1.5 eV

3p 3d

4s-0.033 h0 = -0.9 eV

4p 4d 4f

Energy zero


Plotting of orbitals: line cross-section vs. contour


Contour plots of 1s, 2s, 2p hydrogenic orbitals


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


Contour plots of 4s, 4p, 4d hydrogenic orbtitals


Contour plots of hydrogenic 4f orbitals


He atom

With 2 electrons, the ground state has both in the He+ 1s orbital

ΨHe(1,2) = A[(Φ1s)(Φ1sΦ1s(1)Φ1s(2) (

He<1s|h|1s> + J1s,1s

Now consider He atom: EHe = 2(½ 2) – 2ZJ1s,1s

First lets review the energy for He+. Writing Φ1sexp(-r) we determine the optimum for He+ as follows

<1s|KE|1s> = + ½ 2 (goes as the square of 1/size)

<1s|PE|1s> = - Z(linear in 1/size)

Applying the variational principle, the optimum must satisfy dE/d = -Z = 0 leading to =Z,

KE = ½ Z2, PE = -Z2, E=-Z2/2 = -2 h0.

writing PE=-Z/R0, we see that the average radius is R0=1/


e-e energy of He atom

How can we estimate J1s,1s

Assume that each electron moves on a sphere

withthe average radius R0 = 1/

And assume that e1 at along the z axis (θ=0)

Neglecting correlation in the electron motions, e2 will on the average have θ=90º so that the average e1-e2 distance is ~sqrt(2)*R0

Thus J1s,1s ~ 1/[sqrt(2)*R0] = 0.71

A rigorous calculation (notes chapter 3, appendix 3-C page 6)

Gives J1s,1s = (5/8)

R0

e1

e2


The optimum atomic orbital for He atom

He atom: EHe = 2(½ 2) – 2Z(5/8)

Applying the variational principle, the optimum must satisfy dE/d = 0 leading to

2- 2Z + (5/8) = 0

Thus = (Z – 5/16) = 1.6875

KE = 2(½ 2) = 2

PE = - 2Z(5/8) = -2 2

E= - 2 = -2.8477 h0

Ignoring e-e interactions the energy would have been E = -4 h0

The exact energy is E = -2.9037 h0 (from memory, TA please check). Thus this simple approximation accounts for 98.1% of the exact result.


Interpretation: The optimum atomic orbital for He atom

ΨHe(1,2) = Φ1s(1)Φ1s(2) with Φ1sexp(-r)

We find that the optimum = (Z – 5/16) = 1.6875

With this value of , the total energy is E= - 2 = -2.8477 h0

This wavefunction can be interpreted as two electrons moving 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.

Ionizing the 2nd electron, the 1st electron readjusts to = Z = 2

With E(He+) = -Z2/2 = - 2 h0. thus the ionization potential (IP) is 0.8477 h0 = 23.1 eV (exact value = 24.6 eV)


Now lets add a 3rd electron to form Li

ΨLi(1,2,3) = A[(Φ1s)(Φ1s(Φ1s

Problem with either or , we get ΨLi(1,2,3) = 0

This is an essential result of the Pauli principle

Thus the 3rd electron must go into an excited orbital

ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2s

or

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

First consider Li+ with ΨLi(1,2,3) = A[(Φ1s)(Φ1s

Here Φ1sexp(-r) with = Z-0.3125 = 2.69 and

E = -2 = -7.2226 h0. Since the E (Li2+)=-9/2=-4.5 h0 the IP = 2.7226 h0 = 74.1 eV

The size of the 1s orbital is R0 = 1/ = 0.372 a0 = 0.2A


Consider adding the 3rd electron to the 2p orbital


Since the 2p orbital goes to zero at z=0, there is very little shielding so that it sees an effective charge of

Zeff = 3 – 2 = 1, leading to

a size of R2p = n2/Zeff = 4 a0 = 2.12A

And an energy of e = -(Zeff)2/2n2 = -1/8 h0 = 3.40 eV

0.2A

1s

2.12A

2p


Consider adding 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 orbtials.

The result is Zeff2s = 3 – 1.72 = 1.28

This leads to a size of R2s = n2/Zeff = 3.1 a0 = 1.65A

And an energy of e = -(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


Aufbau principle for atoms

1s

2s2p

3s3p

3d4s

4p 4d 4f

Energy

2

2

6

2

62

6

10

10 14

He, 2

Ne, 10

Ar, 18Zn, 30

Kr, 36

Get generalized energy spectrum for filling in the electrons to explain the periodic table.

Full shells at 2, 10, 18, 30, 36 electrons


He, 2

Ne, 10

Ar, 18

Zn, 30

Kr, 36


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

For each atom we add one more proton to the nucleus and one more electron to the valence shell

But the valence electrons only partially shield each other.

Thus 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 (2s Li)/(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++

As increase net charge the increases in the differential shielding for 4s vs 3d is less important than the difference 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


Transtion metal orbitals


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

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


© copyright 2009 William A. Goddard III, all rights reservedEEWS-90.502-Goddard-L04 1 Nature of the Chemical Bond with applications to catalysis, materials.

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