Transcript Slide 1

Lecture 6, September 17, 2009
Nature of the Chemical Bond
with applications to catalysis, materials
science, nanotechnology, surface science,
bioinorganic chemistry, and energy
Course number: KAIST EEWS 80.502 Room E11-101
Hours: 0900-1030 Tuesday and Thursday
William A. Goddard, III, [email protected]
WCU Professor at EEWS-KAIST and
Charles and Mary Ferkel Professor of Chemistry,
Materials Science, and Applied Physics,
California Institute of Technology
Senior Assistant: Dr. Hyungjun Kim: [email protected]
Manager of Center for Materials Simulation and Design (CMSD)
Teaching Assistant: Ms. Ga In Lee: [email protected]
Special assistant:
Tod Pascal:[email protected]
EEWS-90.502-Goddard-L04
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2009 William A. Goddard III, all rights reserved
1
Last time
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Energy for 2 electron product of spinorbitals
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
ra(1)=| ψa(1)|2 and rb(2)=| ψb(2)|2
Thus Jab > 0
Jab 1/R if ra is centered on atom a while rb is centered on atom b
a large distance R far from a. For shorter distances at which the
densities overlap, the Jab decreases (shielding)
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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
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Both electrons have the same spin
ψa(1) = Φa(1)a(1)
ψb(2) = Φb(2)a(2)
<ψa|ψb>= 0 = < Φa| Φb><a|a> = < Φa| Φb>
the spatial orbitals for same spin must be orthogonal
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)]
= [Φa(1)Φb(2)- Φb(1)Φa(2)][a(1)a(2)]
The total energy is
Eaa = 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
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Assume electrons have the opposite spin
ψa(1) = Φa(1)a(1)
ψb(2) = Φb(2)b(2)
Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)b(2)]=
We obtain <ψa|ψb>= 0 = < Φa| Φb><a|b> = 0
Independent of the overlap of the spatial orbitals.
Thus spatial orbitals can overlap, <Φa|Φb> = S
The exchange term for spin orbitals with opposite spin is zero;
get exchange only between spinorbitals with the same spin
Thus the total energy is
Eab = haa + hbb + Jab + 1/R
Just as for the simple product wavefunction, Φa(1)Φb(2)
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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)][a(1)b(2)]=
=[Φa(1) a(1)][Φb(2)b(2)] – [Φb(1) b(1)][Φa(2)a(2)]
Interchanging the spins leads to
[Φa(1) b(1)][Φb(2)a(2)] – [Φb(1) a(1)][Φa(2)b(2)] =
= A[Φa(1)Φb(2)][b(1)a(2)]
Which is neither + or – times the starting wavefunction.
Thus must combine to obtain proper spatial and spin symmetry
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Correct space and spin symmetry for ab wavefunctions
[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)]=
A[Φa(1)Φb(2)][a(1)b(2)]-A[Φb(1)Φa(2)][a(1)b(2)]
Is symmetric in spin coordinates (the MS = 0 component of the
S=1 triplet)
[Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)]=
A[Φa(1)Φb(2)][a(1)b(2)]+A[Φb(1)Φa(2)][a(1)b(2)]
Is antisymmetric in spin coordinates (the MS = 0 component of the
S=0 triplet)
Thus for the ab case, two Slater determinants must be combined
to obtain the correct spin and space permutational symmetry
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The triplet state for 2 electrons
The wavefunction
[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)]
Leads directly to
3E
ab = haa + hbb + (Jab –Kab) + 1/R
Exactly the same as for
[Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)a(2)]
[Φa(1)Φb(2)-Φb(1)Φa(2)][b(1)b(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)][a(1)a(2)]
MS = 0 : [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)]
MS = -1: [Φa(1)Φb(2)-Φb(1)Φa(2)][b(1)b(2)]
where < Φa|Φb> = 0
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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)][a(1)b(2)-b(1)a(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
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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) is
1E = <ab|H|(ab+ba)>/<ab|(ab+ba)>
g
1E = {(h
2)/R}/(1 + S2)
+
h
+
(h
+
h
)
S
+
J
+
K
+
(1+S
g
aa
bb
ab
ba
ab
ab
Similary for the VB triplet we obtain
3E = <ab|H|(ab-ba)>/<ab|(ab-ba)>
u
3E = {(h
2)/R}/(1 - S2)
+
h
(h
+
h
)
S
+
J
K
+
(1-S
u
aa
bb
ab
ba
ab
ab
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
Then we can define the energy as
Egx = +Ex/(1 + S2)
1E = Ecl + E x
g
g
x = - Ex/(1 - S2)
where
E
u
3E = Ecl + E x
u
u
Ex = {(hab + hba) S + Kab –EclS2}
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The VB exchange energies for H2
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.
-Ex/(1 - S2)
+Ex/(1 + S2)
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1E
g
3E
u
= Ecl + Egx
= Ecl + Eux
Each energy is
referenced to
the value at
R=∞, which is
-1 for Ecl, Eu, Eg
0 for Exu and Exg
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12
Analysis of the VB exchange energy, Ex
where Ex = {(hab + hba) S + Kab –EclS2} = T1 + T2
Here T1 = {(hab + hba) S –(haa + hbb)S2} = 2St
Where t = (hab – Shaa) contains the 1e part
T2 = {Kab –S2Jab} contains the 2e part
Clearly the Ex is
dominated by T1
and clearly T1 is
dominated by the
kinetic part, TKE.
Thus we can
understand
bonding by
analyzing just the
KE part if Ex
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T2
T1
Ex
TKE
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13
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 + t/(1+S)
eu = ecl - t/(1-S)
ecl = (hLL + 1/R) is the energy for bringing a proton up to H atom
t = (hLR - ShLL) contains the terms that dominate the bonding
and antibonding character of these 2 states
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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) + t/(1+S) ≡ ecl + Egx
eu = (hLL + 1/R) - t/(1-S) ≡ ecl + Eux
ecl = (hLL + 1/R) is the classical energy
t = (hLR - ShLL) is the VB interference or
resonance energy that dominates the
bonding and antibonding of these states
t/(1+S)
t/(1+S)
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Contragradience
The above discussions show that the interference or exchange
part of KE dominates the bonding, tKE=KELR –S KELL
This decrease in the KE due to overlapping orbitals is
dominated by
tx = ½ [< (хL). ((хR)> - S [< (хL)2>
Dot product is
хL
large and negative
in the shaded
region between
atoms, where the L
and R orbitals have
opposite slope
(congragradience)
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хR
16
Comparison of exchange energies for 1e and 2e bonds
H2
Eg1x ~ +2St /(1 + S2)
Eu1x ~ -2St /(1 - S2)
H2+ case
egx ~ +t/(1 + S)
eux ~ -t/(1 - S)
For R=1.6bohr (near Re), S=0.7
Eg1x ~ 0.94t vs. egx ~ 0.67t
Eu1x ~ -2.75t vs. eux ~ -3.33t
For R=4 bohr, S=0.1
Eg1x ~ 0.20t vs. egx ~ 0.91t
Eu1x ~ -0.20t vs. eux ~ -1.11t
E(hartree)
Eu1x
Eg1x R(bohr)
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17
H atom, excited states
In atomic units, the Hamiltonian
h = - (Ћ2/2me)2 – Ze2/r
Becomes
h = - ½ 2 – 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)
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The excited angular states of H atom, 1 nodal plane
φnlm = Rnl(r) Zlm(θ,φ)
the excited angular functions, Zlm must have nodal
planes to be orthogonal to Zs
3 cases with one nodal plane
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)
z
+
pz
x
z
We find it useful to keep track of how often the
wavefunction changes sign as the φ coordinate is
increased by 2p = 360º
px
+
x
If m=0, denote it as s: pz = ps
If m=1, denote it as p: px, py = pp
If m=2 we call it a d function
If m=3 we call it a f function
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The excited angular states of H atom, 2 nodal planes
z
dz2
Z20 = dz2 = [3 – ]
m=0, ds
+
Z21 = dzx = zx =r2 (sinθ)(cosθ) cosφ
m = 1, dp - 57º
Z2-1 = dyz = yz =r2 (sinθ)(cosθ) sinφ
x
+
Z22 = dx2-y2 = x2 – y2 = r2 (sinθ)2 cos2φ
m = 2, dd
Z22 = dxy = xy =r2 (sinθ)2 sin2φ
Where we used [(cosφ)2 – (sinφ)2]=cos2φ and 2cosφ sinφ=sin2φ
Summarizing:
one s angular function (no angular nodes) called ℓ=0
three p angular functions (one angular node) called ℓ=1
five d angular functions (two angular nodes) called ℓ=2
seven f angular functions (three angular nodes) called ℓ=3
nine g angular functions (four angular nodes) called ℓ=4
ℓ is referred to as the angular momentum quantum number
There are (2ℓ+1) m values, Zℓm for each ℓ
20
z2
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r2
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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
Zr = 7.1
Zr = 2
Zr = 1.9
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
2/27 – 2(Zr)/3 + 1]exp(-Zr/3)
R
=
3s
=
[2(Zr)
30
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Size (a0)
name
total nodal planes
radial nodal planes
angular nodal planes
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
1s 0 0 0 1.0
2p are related, but it turns out that the total
2s 1 1 0 4.0
energy depends only on n.
2p 1 0 1 4.0
3s 2 2 0 9.0 Enlm = - Z2/2n2
3p 2 1 1 9.0
The potential energy is given by
3d 2 0 2 9.0
4s 3 3 0 16.0 PE = - Z2/2n2 ≡ -Z/ Rˉ , where Rˉ =n2/Z
4p 3 2 1 16.0
Thus Enlm = - Z/(2 Rˉ )
4d 3 1 2 16.0
22
4f
3 0 3 16.0
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New material
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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
1.7A
0.74A
H--H
H
H
C
4.8
H
H
H
H
H
H
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H
H
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Hydrogen atom excited states
Energy
zero
-0.033 h0 = -0.9 eV
-0.056 h0 = -1.5 eV
-0.125 h0 = -3.4 eV
-0.5 h0 = -13.6 eV
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4s
4p
4d
3s
3p
3d
2s
2p
1s
4f
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Plotting of orbitals:
line cross-section vs. contour
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Contour plots of 1s, 2s, 2p hydrogenic orbitals
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Contour plots of 3s, 3p, 3d hydrogenic orbitals
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Contour plots of 4s, 4p, 4d hydrogenic orbtitals
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Contour plots of hydrogenic 4f orbitals
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He atom
With 2 electrons, the ground state has both in the He+ 1s orbital
ΨHe(1,2) = A[(Φ1sa)(Φ1sb)]= Φ1s(1)Φ1s(2) (ab-ba)
EHe= 2<1s|h|1s> + J1s,1s
First lets review the energy for He+. Writing Φ1s = exp(-zr) we
determine the optimum z for He+ as follows
<1s|KE|1s> = + ½ z2 (goes as the square of 1/size)
<1s|PE|1s> = - Zz (linear in 1/size)
Applying the variational principle, the optimum z must satisfy
dE/dz = z - Z = 0 leading to z = Z,
KE = ½ Z2, PE = -Z2, E=-Z2/2 = -2 h0.
writing PE=-Z/R0, we see that the average radius is R0=1/z
Now consider He atom: EHe = 2(½ z2) – 2Zz + J1s,1s
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e-e energy of He atom
How can we estimate J1s,1s
Assume that each electron moves on a sphere
e1
R0
e2
withthe average radius R0 = 1/z
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 z
A rigorous calculation (notes chapter 3, appendix 3-C page 6)
Gives J1s,1s = (5/8) z
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The optimum atomic orbital for He atom
He atom: EHe = 2(½ z2) – 2Zz + (5/8)z
Applying the variational principle, the optimum z must satisfy
dE/dz = 0 leading to
2z - 2Z + (5/8) = 0
Thus z = (Z – 5/16) = 1.6875
KE = 2(½ z2) = z2
PE = - 2Zz + (5/8)z = -2 z2
E= - z2 = -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.
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Interpretation: The optimum atomic orbital for He atom
ΨHe(1,2) = Φ1s(1)Φ1s(2) with Φ1s = exp(-zr)
We find that the optimum z = (Z – 5/16) = 1.6875
With this value of z, the total energy is E= - z2 = -2.8477 h0
This wavefunction can be interpreted as two electrons moving
independently in the orbital Φ1s = exp(-zr) 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 = Z = 2
With E(He+) = -Z2/2 = - 2 h0. thus the ionization potential (IP)
is 0.8477 h0 = 23.1 ©eV
(exact
24.6III,eV)
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2009 value
William A. =
Goddard
all rights reserved
34
Now lets add a 3rd electron to form Li
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb)(Φ1sg)]
Problem with either g = a or g = b, 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[(Φ1sa)(Φ1sb) )(Φ2sa)]
or
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py)
First consider Li+ with ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb)]
Here Φ1s = exp(-zr) with z = Z-0.3125 = 2.69 and
E = -z2 = -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/z = 0.372 a0 = 0.2A
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Consider adding the 3rd electron to the 2p orbital
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py)
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
1s
0.2A
2p
2.12A
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Consider adding the 3rd electron to the 2s orbital
ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py)
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
1s
0.2A
2s
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2.12A
R~0.2A
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37
Li atom excited states
Energy
MO picture
State picture
zero
1st
excited
state
-0.125 h0 = -3.4 eV
-0.205 h0 = -5.6 eV
2p
2s
DE = 2.2 eV
17700 cm-1
564 nm
(1s)2(2p)
(1s)2(2s)
Ground
state
Exper
-2.723 h0 = 74.1 eV
671 nm
1s
DE = 1.9 eV38
EEWS-90.502-Goddard-L04
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Aufbau principle for atoms
Energy
Kr, 36
Zn, 30
Ar, 18
6
2
2
4s
2s
He, 2
EEWS-90.502-Goddard-L04
10
6
6
2
4p
14
4d
4f
3d
3p
3s
Ne, 10
2
10
1s
2p
Get generalized energy
spectrum for filling in the
electrons to explain the
periodic table.
Full shells at 2, 10, 18, 30,
36 electrons
© copyright 2009 William A. Goddard III, all rights reserved
39
Kr, 36
Zn, 30
Ar, 18
Ne, 10
He, 2
EEWS-90.502-Goddard-L04
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40
Many-electron configurations
General
aufbau
ordering
EEWS-90.502-Goddard-L04
Particularly stable
© copyright 2009 William A. Goddard III, all rights reserved
41
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
EEWS-90.502-Goddard-L04
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42
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
© copyright
William
A. Goddard
III, all
rights reserved
24.5 He, 21.6 Ne, 15.8
Ar, 2009
14.0
Kr,12.1
Xe,
10.7
Rn
EEWS-90.502-Goddard-L04
43
EEWS-90.502-Goddard-L04
© copyright 2009 William A. Goddard III, all rights reserved
44
EEWS-90.502-Goddard-L04
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45
EEWS-90.502-Goddard-L04
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46
Transition metals; consider [Ar] plus one electron
[IP4s = (Zeff4s )2/2n2 = 4.34 eV  Zeff4s = 2.26; 4s<4p<3d
K
IP4p = (Zeff4p )2/2n2 = 2.73 eV  Zeff4p = 1.79;
IP3d = (Zeff3d )2/2n2 = 1.67 eV  Zeff3d = 1.05;
IP4s = (Zeff4s )2/2n2 = 11.87 eV  Zeff4s = 3.74; 4s<3d<4p
Ca+
IP3d = (Zeff3d )2/2n2 = 10.17 eV  Zeff3d = 2.59;
IP4p = (Zeff4p )2/2n2 = 8.73 eV  Zeff4p = 3.20;
IP3d = (Zeff3d )2/2n2 = 24.75 eV  Zeff3d = 4.05; 3d<4s<4p
Sc++
IP4s = (Zeff4s )2/2n2 = 21.58 eV  Zeff4s = 5.04;
IP4p = (Zeff4p )2/2n2 = 17.01 eV  Zeff4p = 4.47;
As the net charge increases the differential shielding for 4s vs 3d
is less important than the difference in n quantum number 3 vs 4
47
EEWS-90.502-Goddard-L04
copyright 2009
A. Goddard III, all rights reserved
Thus
charged system©prefers
3dWilliam
vs 4s
Transition metals; consider Sc0, Sc+, Sc2+
3d: IP3d = (Zeff3d )2/2n2 = 24.75 eV  Zeff3d = 4.05;
Sc++
4s: IP4s = (Zeff4s )2/2n2 = 21.58 eV  Zeff4s = 5.04;
4p: IP4p = (Zeff4p )2/2n2 = 17.01 eV  Zeff4p = 4.47;
(3d)(4s): IP4s = (Zeff4s )2/2n2 = 12.89 eV  Zeff4s = 3.89;
Sc+
(3d)2:
IP3d = (Zeff3d )2/2n2 = 12.28 eV  Zeff3d = 2.85;
(3d)(4p): IP4p = (Zeff4p )2/2n2 = 9.66 eV  Zeff4p = 3.37;
(3d)(4s)2: IP4s = (Zeff4s )2/2n2 = 6.56 eV  Zeff4s = 2.78;
Sc
(4s)(3d)2: IP3d = (Zeff3d )2/2n2 = 5.12 eV  Zeff3d = 1.84;
(3d)(4s)(4p): IP4p = (Zeff4p )2/2n2 = 4.59 eV  Zeff4p = 2.32;
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.
EEWS-90.502-Goddard-L04
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48
Implications on transition metals
The simple Aufbau principle puts 4s below 3d
But 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
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49
Transtion metal orbitals
EEWS-90.502-Goddard-L04
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50
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
EEWS-90.502-Goddard-L04
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51
Consider the ground state of B: [Be](2p)x1
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
z
2pz
2py
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
52
and
we denote the overall
state
as 2PA. Goddard III, all rights reserved
EEWS-90.502-Goddard-L04
© copyright
2009 William
EEWS-90.502-Goddard-L04
© copyright 2009 William A. Goddard III, all rights reserved
53