© copyright 2009 william a. goddard iii, all rights reservedeews-90.502-goddard-l04 1 nature of the...
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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
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EEWS-90.502-Goddard-L04 2© copyright 2009 William A. Goddard III, all rights reserved
Last time
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EEWS-90.502-Goddard-L04 3© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 4© copyright 2009 William A. Goddard III, all rights reserved
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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EEWS-90.502-Goddard-L04 5© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 6© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 7© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 8© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 9© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 10© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 11© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 12© copyright 2009 William A. Goddard III, all rights reserved
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)
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EEWS-90.502-Goddard-L04 13© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 14© copyright 2009 William A. Goddard III, all rights reserved
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)
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EEWS-90.502-Goddard-L04 15© copyright 2009 William A. Goddard III, all rights reserved
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)
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EEWS-90.502-Goddard-L04 16© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 17© copyright 2009 William A. Goddard III, all rights reserved
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 ~
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EEWS-90.502-Goddard-L04 18© copyright 2009 William A. Goddard III, all rights reserved
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)
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EEWS-90.502-Goddard-L04 19© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 20© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 21© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 22© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 23© copyright 2009 William A. Goddard III, all rights reserved
New material
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EEWS-90.502-Goddard-L04 24© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 25© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 26© copyright 2009 William A. Goddard III, all rights reserved
Plotting of orbitals: line cross-section vs. contour
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EEWS-90.502-Goddard-L04 27© copyright 2009 William A. Goddard III, all rights reserved
Contour plots of 1s, 2s, 2p hydrogenic orbitals
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EEWS-90.502-Goddard-L04 28© copyright 2009 William A. Goddard III, all rights reserved
Contour plots of 3s, 3p, 3d hydrogenic orbitals
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EEWS-90.502-Goddard-L04 29© copyright 2009 William A. Goddard III, all rights reserved
Contour plots of 4s, 4p, 4d hydrogenic orbtitals
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EEWS-90.502-Goddard-L04 30© copyright 2009 William A. Goddard III, all rights reserved
Contour plots of hydrogenic 4f orbitals
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EEWS-90.502-Goddard-L04 31© copyright 2009 William A. Goddard III, all rights reserved
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/
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EEWS-90.502-Goddard-L04 32© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 33© copyright 2009 William A. Goddard III, all rights reserved
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.
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EEWS-90.502-Goddard-L04 34© copyright 2009 William A. Goddard III, all rights reserved
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)
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EEWS-90.502-Goddard-L04 35© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 36© copyright 2009 William A. Goddard III, all rights reserved
Consider adding the 3rd electron to the 2p orbital
ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2pz(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
0.2A
1s
2.12A
2p
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EEWS-90.502-Goddard-L04 37© copyright 2009 William A. Goddard III, all rights reserved
Consider adding the 3rd electron to the 2s orbital
ΨLi(1,2,3) = A[(Φ1s)(Φ1s)(Φ2pz(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
0.2A
1s
2.12A2s
R~0.2A
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EEWS-90.502-Goddard-L04 38© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 39© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 40© copyright 2009 William A. Goddard III, all rights reserved
He, 2
Ne, 10
Ar, 18
Zn, 30
Kr, 36
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EEWS-90.502-Goddard-L04 41© copyright 2009 William A. Goddard III, all rights reserved
Many-electron configurations
General aufbau
ordering
Particularly stable
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EEWS-90.502-Goddard-L04 42© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 43© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 44© copyright 2009 William A. Goddard III, all rights reserved
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EEWS-90.502-Goddard-L04 45© copyright 2009 William A. Goddard III, all rights reserved
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EEWS-90.502-Goddard-L04 46© copyright 2009 William A. Goddard III, all rights reserved
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EEWS-90.502-Goddard-L04 47© copyright 2009 William A. Goddard III, all rights reserved
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
IP3d = (Zeff3d )2/2n2 = 10.17 eV Zeff
3d = 2.59;
IP4p = (Zeff4p )2/2n2 = 8.73 eV Zeff
4p = 3.20;
IP3d = (Zeff3d )2/2n2 = 24.75 eV Zeff
3d = 4.05; 3d<4s<4p
IP4s = (Zeff4s )2/2n2 = 21.58 eV Zeff
4s = 5.04;
IP4p = (Zeff4p )2/2n2 = 17.01 eV Zeff
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
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EEWS-90.502-Goddard-L04 48© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 49© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 50© copyright 2009 William A. Goddard III, all rights reserved
Transtion metal orbitals
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EEWS-90.502-Goddard-L04 51© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 52© copyright 2009 William A. Goddard III, all rights reserved
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
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EEWS-90.502-Goddard-L04 53© copyright 2009 William A. Goddard III, all rights reserved