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experimental helium E = -79.0 eV = 1st + 2nd IP = 24.6 eV + 54.4 eV = -79.0 eV
Note that this value is greater than the actual E by 5.3%.
Approximation Methods ― Perturbation
Ĥ0y = E0y E0 = -108.8 eV
<E′> = yĤ‘y dt = y e2/(4peor12) y dt = 34.0 eV
Ĥ = Ĥ0 + Ĥ'
‘Effective’ Nuclear Charge, Z′ replaces Z accounts for ‘shielding’ of nuclear attraction due to other e-
1s(1) = (p)-1/2 (Z′/a)3/2 exp(-Z′r/a)
Variation: = N • 1s(1) 1s(2)
1s(1) = (p)-1/2 (Z/a)3/2 exp(- Zr/a)
see example 12.11 on page 396 Adjust Z′ (1.6875) to minimize <E> to -77.5 eV
the error is now down to 1.9%
Approximation Methods ― VariationVariation theory often introduces a term in the wavefunction, f, that can be continuously adjusted in numerous iterations until the minimum energy (best) is obtained.
= N • 1s(1) 1s(2)
Use H-like wave functions to represent each electron in the He atom.
This variational wave function only accounts for 3 q#s for each electron: n, ℓ, mℓ. It does not account for spin which is not evident in basic QM, but added as the 6th postulate based on Dirac’s Relativistic QM.
spin angular momentum quantum # (s)
fermions ― s = ½ & ms = ± ½
1s(1) = (p)-1/2 (Z′/a)3/2 exp(-Z′r/a)
Pauli Exclusion PrincipleWe must now introduce the spin function to our He wave function
2-1/2[(1) (2) + (1) (2)]
2-1/2[(1) (2) - (1) (2)]
(1) (2) (1) (2)
(1) (2) (2) (1)
Can’t distinguish electrons – use linear combination
Multi-electron wavefunctions can be represented as n x n matrices where n is the number of electrons in the atom (which is also = to the number of columns and rows in the matrix.The renormalization constant is n!-1/2 .
Cross-multiplying the matrix will give you the determinant of the matrix (called the Slater determinant) which is also the ground state wavefunction for the atom.
+
-
e- #1
e- #2
1s 1s
He,gs ≠1s(1) (1) 1s(1) a(1)
1s(2) (2) 1s(2) a(2)2-1/2 = 0
If any column or row of the matrix is identical, the determinant is 0. Therefore the wavefunction would also have to be 0, which is not valid.
Lithium atom matrix
Pauli Exclusion Principle ― antisymmetric regarding exchange of electrons no more than two electrons in an orbital no two electrons can have the same set of 4 quantum #s
Hartree–Fock Wave Functions (linear variation theory p398)
Variational method using non-hydrogen-likewave functions that retain ‘orbital’ analogy
Satisfy F fi = ei fi ^
Hartree-Fock Operator – equivalent to Ĥ i = quantum number related to orbital ei = energy of ith state fi = Hartree-Fock variational function for ith state
Hartree-Fock treatment of He ― <E> = -77.9 eV
Real He atom = -79.0 eV Hydrogen-like f = -77.5 eV
HF neglects instantaneous correlations in motions of electrons – improved on byconfiguration interactions (CI)
Hartree–Fock Wave Functions (linear variation theory p398)
Variational method using non-hydrogen-likewave functions that retain ‘orbital’ analogy
Satisfy F fi = ei fi ^
Hartree-Fock Operator – equivalent to Ĥ i = quantum number related to orbital ei = energy of ith state fi = Hartree-Fock variational function for ith state
Hartree-Fock treatment of He ― <E> = -77.9 eV
Real He atom = -79.0 eV Hydrogen-like f = -77.5 eV
HF neglects instantaneous correlations in motions of electrons – improved on byconfiguration interactions (CI)
Li,gs
1s(1) (1) 1s(1) (1) 2s(1) (1)
1s(2) (2) 1s(2) (2) 2s(2) (2)
1s(3) (3) 1s(3) (3) 2s(3) (3)
6-1/2
The Slater determinant for Li will have 6 terms
Note that putting 1s again in the third column forces two identical columns
For atoms with increased numbers of electrons it is the total energy that determines the filling order not the individual energy of the H-like orbital. This is why the 4s orbital is lower in energy than 3d and that there are many ‘exceptions’ to the filling rules.
Once again the paired concepts of penetration and shielding provides a rationale for these energy changes (see fig. 12.5 on p385)
Total Orbital Angular Momentum L = S ℓ (for all e-)
gs: ℓ(1) = 0 & ℓ(2) = 0; L = 0 same for 1s 2s es
1s 2p excited states: L = 0 + 1 = 1
Total Orbital Angular Momentum = [L(L + 1)]1/2ħ
Spectroscopic Term Symbols for es
L = symbol e.g.
0123
SPDF
1s 2s1s 2p1s 3d
Total Spin Angular Momentum S = |S ms | (for all e-)
gs: ms = ½ & ms = - ½ ; S = 0
multiplicity = 2S + 1 if S = 0 then 2S + 1 = 1 (singlet) paired
if S = 1 then 2S + 1 = 3 (triplet) unpaired
1S0 = singlet ground state1S1 = 1st singlet excited state electrons still have opposite spins3S = triplet excited state electrons still have like spins
Variation functions are often obtained by taking a basis set of functions, yi, each of which meets the requirements for any acceptable wavefunction, y, and may be related to multiple states of some related or model system. Each of the basis set of functions is multiplied by an adjustable constant such that the overall variational wavefunction f is ……
f = Si ci • yi
The constants (ci) fulfill three separate requirements: Combined, they serve to normalize the variational function. Individually they act as weighting factors that indicate the magnitude of the contribution for each function in the basis set. They become the adjustable factors used to minimize the system energy to give the best variational function such that …..
dE/dc1 = dE/dc2 = …. dE/dci = 0
This minimization process leads to the best value for the energy of the system, as well as the values for the coefficient for each basis set function.
Linear variation theory f = Si ci • yiUsing a general example that has only two basis set functions ……
f = c1y1 + c2y2
Set up the expression for <E> …… ∫ *f Ĥ f dt ≥ E ∫ *f f dt
The solution involves (see 12.28 on p398) energy integral …. H11, H22, H12/H21
overlap integrals ... S11, S22, S12/S21
Only needed if f is un-normalized function
∫ (c1y1 + c2y2 )* Ĥ (c1y1 + c2y2 ) dt ≥ E ∫ (c1y1 + c2y2 )* (c1y1 + c2y2 ) d t
e.g. H11= ∫ y1*Ĥy1 dt etc.
e.g. S12 = S21 = ∫ y1*y2 dt etc.
Linear variation theory f = Si ci • yi
Using a general example that has only two basis set functions ……
f = c1y1 + c2y2
Matrix solution for <E> ……
H11 – ES11 H12 – ES12
H21 – ES21 H22 – ES22
= 0
e.g. H11= ∫ y1*Ĥy1 dt etc.
e.g. S12 = S21 = ∫ y1*y2 dt etc.
Set up the expression for <E> …… = ∫ *f Ĥ f dt ≥ E ∫ *f f dt
Using a general example that has only two basis set functions ……
f = c1y1 + c2y2
The determinant for this matrix gives a binomial function in E,
which can be solved to give 2 possible values … E1 and E2.
Example 12.12 2 orthonormal wavefunctions H11 = -15, H22 = -4, H12 = H21 = -1Find E and E* and the coefficients of each variational wavefunction
-15 – E -1
-1 -4 – E = 0
E2 +19E + 59 = 0 & E = -15.09 & -3.91
f = 0.996y1 + 0.0896y2 and
f* = -.0896y1 + 0.996y2
Molecular Properties
What data would you want to know to understand the properties of a diatomic molecule?
What additional information would you want to extend this to a polyatomic molecule?
Bond distance Bond energy Dipole moment
Molecular geometry or bond angles
How did General Chemistry guide you in predicting this data?
Molecular Properties
Geometry ― bond lengths
bond angles
Stability and reactivity ― bond energiesdipole moments (e- distribution)
VSEPR sp, sp2, sp3 hybrids
“A hybrid AO is a LCAO from same atom”“The purpose of hybrid AOs is to rationalize the observed geometry on a molecule”
Empirical treatmentTables of ‘averaged’ bond energiesThe Pauling electronegativity scale
Theoretical Molecular Methods
Empirical ― uses ‘averaged’ experimental data
Ab initio ― uses QM theory only ‘from scratch’
Semi-empirical ― uses ‘averaged’ empirical data as a starting point and makes adjustments
using theoretical QM methods.
Using Empirical Values (e.g bond length/energy)
accumulate experimental data from many compounds.find average values for the same type of bond.establish a table with these ‘averaged’ values.assume value is the same regardless of the larger molecular context.
AVG Bond Energy in kJ mol-1
X is C-X C=X CX X-X X-H
C 344 615 812 344 415
H 415 436 436
O 350 725 143 463
N 292 615 890 159 391
F 441 158
Cl 328 243
Br 276
I 240
S 259 266 368
BOND N=N O=O NºN O-N
Energy 418 498 946 175
Average Bond Dipole Moments in Debyes (1 D = 3.335641 Cm)
H - O 1.5 C - Cl 1.5 C = O 2.5
H - N 1.3 C - Br 1.4 C - N 0.5
H - C 0.4 C - O 0.8 C º N 3.5
e = 1.6022 x 10-19
Dipole Moments & Electronegativity
In MO theory the charge on each atom is related to the probability of finding the electron near that nucleus, which is related to the coefficient of the AO in the MO
CH2O Geometry
Ĥ = KN + Ke + VNN + VNe + Vee
QM treatment of molecules ― Hamiltonian
Born-Oppenheimer Approximation ― = e N
nucleus is slow & electron fast1. electron “sees” nucleus as stationary2. nucleus “sees” electrons as time- averaged cloud
Electrical Energy at fixed R ― <E> = ∫eĤe
Ĥe = Ke + VNN + VNe + Vee
ĤN = KN + VN
ĤN yN = (KN yN + Ee yN) = EN yN
VN = Ee
H2+ Molecular Ion
Basis set = 1s1 and 1s2 LCAO = f1 = c1(1s1 + 1s2) f f2 = c2(1s1 - 1s2)