experimental helium E = -79.0 eV = 1st + 2nd IP = 24.6 eV + 54.4 eV = -79.0 eV

E = -54.4 + -54.4 = -108.8 eV

note that E < ‘real’ can’t be variation

Ĥ = (-ħ2/2 12 – Ze'2/r1) + (-ħ2/2 2

2 – Ze'2/r1) + e′2/r12

y = 1s1 • 1s2

Ĥy = Ey = E1 • 1s11s2 + E2 • 1s11s2 + (e′2/r12 • 1s11s2)

drop e- repulsion term — crude “perturbation”

Helium: experimental E = -79.0 eV

y = 1s(1) 1s(2) = (1/p)(1/a)3 exp(-2r/a)

<E> = <E>˚ + <E′> = -108.8 + 34.0 = -74.8 eV

<E'> = 0 y Ĥ y dt

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

total spin =

= 1s(1)a 1s(2)a +1 = 1s(1) 1s(2) • 2-1/2[(1) (2) + (1) (2)] 0

= 1s(1) 1s(2) • 2-1/2[(1) (2) - (1) (2)] 0

= 1s(1)b 1s(2)b -1

Angular momenta of charged particles can be distinguished using magnetic fields. Experiments show He to have spin = 0.

Pauli Exclusion Principle

If 2 electron wave function....

must be antisymmetric with respect to interchange of electrons. If the electrons are exchanged the wavefunction must change signs.

= spatial • spin functions = sym • anti-sym

= 1s(1) 1s(2) • 2-1/2[(1) (2) - (1) (2)]

= 1s(1) 1s(2) • 2-1/2[(1) (2) + (1) (2)]

= 1s(1) 1s(2) • 2-1/2[(1) (2) - (1) (2)]

12.10 Show that this is antisymmetric with respect to electron exchange

Pauli Exclusion Principle

= spatial • spin functions = sym • anti-sym

= 1s(1) 1s(2) • 2-1/2[(1) (2) - (1) (2)]

12.10 Show that this is antisymmetric with respect to electron exchange

= 1s(2) 1s(1) • 2-1/2[(2) (1) - (2) (1)] = …..

= 1s(1) 1s(2) • - 2-1/2[-(2) (1) + (2) (1)] = …..

= 1s(1) 1s(2) • - 2-1/2[(1) (2) - (1) (2)] = …..

= 1s(1) 1s(2) • - 2-1/2[-b(1) a(2) + a(1) b(2)] = …..

= - {1s(1) 1s(2) • 2-1/2[(1) (2) - (1) (2)]} = …..

He,gs = 1s(1) 1s(2) 2-1/2[(1) (2) - (1) (2)]

He,gs =1s(1) (1) 1s(1) (1)

1s(2) (2) 1s(2) (2)2-1/2

represented in matrix form .......

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)

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

12.8

Li2+ 1s•1s•1s 2s 1s•1s•2s actualE (eV) -122.4 -367.2 -30.6 -275.4 203E (J) -1.96E-17 -5.88E-17 -4.90E-18 -4.41E-17 3.26E-17

E = -Z2e4m/(8eo2h2n2)

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 number of terms in the wavefunction is n!where n is the # of electrons in the atom. 12.14

He Li C Na Si P#e = 2 3 6 11 14 15

# terms 2 6 720 3.99E+07 8.718E+10 1.308E+12

Helium excited states 1s 2s and the Pauli exclusion principle …

2-1/2 [1s(1) 2s(2)+1s(2) 2s(1)] 2-1/2[(1) (2) - (1) (2)]

2-1/2 [1s(1) 2s(2) - 1s(2) 2s(1)] 2-1/2[(1) (2) + (1) (2)] (1) (2)

b(1) b(2)

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

2-1/2 [1s(1) 2s(2) - 1s(2) 2s(1)] 2-1/2[(1) (2) + (1) (2)] (1) (2)

b(1) b(2)

Linear variation theory (p-398)

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.

H11 – ES11 H12 – ES12

H21 – ES21 H22 – ES22

= 0

Then solve for c1 and c2 by ……

c1/c2 = - (H12 – ES12)/(H11 – ES11) and … c1/c2 = - (H22 – ES22)/(H21 – ES21)

These apply to both E1 and E2.If y1 and y2 are orthonormal, then …. c1

2 + c22 = 1

Example 12.12

See figures 12.9 and 12.10 on page 402

f = c1y1 + c2y2 and f* = c1y1 - c2y2

H11 – ES11 H12 – ES12

H21 – ES21 H22 – ES22

= 0

c1/c2 = - (H12 – ES12)/(H11 – ES11)c1/c2 = - (H22 – ES22)/(H21 – ES21)

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)

f*

∫ f12 dt = c1

2 ∫ (1s1 + 1s2) 2 = 1 = c1

2 (∫ 1s12 + ∫2 1s11s2 + ∫1s2

2) = c1

2 (1 + ∫2 1s11s2 + 1) = c1

2 (2 + 2∫1s11s2) = c12 (2 + 2S) = 1

c1 = 1/(2 + 2S)½

H2+ Molecular Ion

true Re 1.06 1.32 1.07 ÅDe 2.8 1.8 2.35 eV

s bond ― f is symmetrical about bonding axis

f = C (1s1 + 1s2)

ò f*Ĥf dt ò f*f dt

< > = E C2 • ò (1s1 + 1s2) Ĥ (1s1 + 1s2) dt = ò1s1Ĥ1s1 + ò1s2Ĥ1s2 + 2ò1s1Ĥ1s2

ò (1s1 + 1s2)2 dt ò 1s1

2 + ò 1s22 + 2ò1s11s2

= H11 + H22 + H12 + H21 = 2(H11 + H12) = H11 + H12

2 + 2S12 1 + S12

<E>′ = H22 - H12

1 - S12

H1H2

s

*s

↑-13.6 eV

-15.4 eV

12.36 R S12 H12 C1 C2 R E1 E21.00 0.616 -20.26 0.556 1.142 1.00 -20.95 17.3491.15 0.540 -17.16 0.570 1.043 1.15 -19.98 7.7511.25 0.492 -15.30 0.579 0.992 1.25 -19.37 3.3451.32 0.459 -14.09 0.585 0.962 1.32 -18.97 0.9051.40 0.424 -12.80 0.593 0.932 1.40 -18.54 -1.3901.45 0.403 -12.04 0.597 0.915 1.45 -18.28 -2.6081.60 0.344 -9.99 0.610 0.873 1.60 -17.56 -5.4981.8 0.275 -7.73 0.626 0.830 1.8 -16.73 -8.102

2.00 0.218 -5.93 0.641 0.799 2.00 -16.04 -9.809

0.8 1.0 1.2 1.4 1.6 1.8 2.0-25.0

-20.0

-15.0

-10.0

-5.0

0.0

5.0

10.0

15.0

20.0

E1 E2

E1 should reach minimum at 1.32?

2AO’s give 2 MO’s - one bonding & one anti-bonding

Ee

r

'

H1H2

s

*s

<E> = H11 + H12

1 + S12

<E>′ = H22 - H12

1 - S12

-13.6 eV

-19.0 eV

0.9 eV

Bond Types

arise from m quantum # & orbital angular momentum along z axis (= mħ) (z axis = bonding axis)

bond types , , , etc.

s - s; s - pz; pz - pz

px - px or py - py

requires d orbitals

2nd period diatomic molecules

2pz 2py 2px

2s

2px 2py 2pz

2s

*s

*p

s

*s

s

p

AO #1 AO #2MO

Diatomic MOs

More complete treatments make MOs from larger AO sets.

AOs used must be same type s vs. p. mix s and pz as s px and py remain as p

AOs used must have same symmetry. g vs. u

34

Homonuclear Diatomic molecules

H2 (g1s)2

He2 (g1s)2 (u*1s)2

C2 (g1s)2 (u*1s)2 (g2s)2 (u

*2s)2 (u2p)4(g2p)0(g*2p)0

N2 (g1s)2 (u*1s)2 (g2s)2 (u

*2s)2 (u2p)4(g2p)2(g*2p)0

O2 (g1s)2 (u*1s)2 (g2s)2 (u

*2s)2 (g2p)2(u2p)4 (g*2p)2

F2 (g1s)2 (u*1s)2 (g2s)2 (u

*2s)2 (g2p)2(u2p)4 (g*2p)4

Diatomic MOs

More complete treatments make MOs from larger AO sets.

AOs used must be same type s vs. p. mix s and pz as s px and py remain as p

AOs used must have same symmetry. g vs. u

Semi-empirical treatment of N2 from Spartan (AM1) -14.5 eV MO: 1 2 3 4 5Eigenvalues:-1.52116 -0.78755 -0.59504 -0.59504 -0.52636 (ev): -41.39299 -21.43028 -16.19199 -16.19199 -14.32295 Sg+ Su+ ??? ??? Sg+ 1 N2 S 0.62096 0.64963 0.00000 0.00000 0.33825 2 N2 PX 0.00000 0.00000 0.63756 -0.30581 0.00000 3 N2 PY 0.00000 0.00000 -0.30581 -0.63756 0.00000 4 N2 PZ 0.33825 -0.27924 0.00000 0.00000 -0.62096 5 N1 S 0.62096 -0.64963 0.00000 0.00000 0.33825 6 N1 PX 0.00000 0.00000 0.63756 -0.30581 0.00000 7 N1 PY 0.00000 0.00000 -0.30581 -0.63756 0.00000 8 N1 PZ-0.33825 -0.27924 0.00000 0.00000 0.62096 MO: 6 7 8Eigenvalues: 0.03684 0.03684 0.22165 (ev): 1.00254 1.00254 6.03130 ??? ??? Su+ 1 N2 S 0.00000 0.00000 0.27924 2 N2 PX -0.66629 0.23675 0.00000 3 N2 PY 0.23675 0.66629 0.00000 4 N2 PZ 0.00000 0.00000 0.64963 5 N1 S 0.00000 0.00000 -0.27924 6 N1 PX 0.66629 -0.23675 0.00000 7 N1 PY -0.23675 -0.66629 0.00000 8 N1 PZ 0.00000 0.00000 0.64963

Semi-empirical treatment of N2 from Spartan (AM1) -14.5 eV

1) -41.4 eV sg 0.621 N22s + 0.338 N22pz + 0.621 N12s – 0.338 N12pz :

2) -21.4 eV su 0.650 N22s - 0.279 N22pz - 0.650 N12s – 0.279 N12pz :

3) -16.2 eV p 0.638 N22px - 0.306 N22py + 0.638 N12px - 0.306 N12py :

4) -16.2 eV p - 0.306 N22px - 0.638 N22py - 0.306 N12px - 0.638 N12py :

5) -14.3 eV sg 0.338 N22s - 0.621 N22pz + 0.338 N12s + 0.621 N12pz :

6) +1.00 eV p - 0.666 N22px + 0.237 N22py + 0.666 N12px - 0.237 N12py :

7) +1.00 eV p 0.237 N22px + 0.666 N22py - 0.237 N12px - 0.666 N12py :

8) +6.03eV su 0.279 N22s + 0.650 N22pz - 0.279 N12s + 0.650 N12pz :

Heat of Formation: 46.7 kJ/mol

Semi-empirical treatment of O2 from Spartan (AM1) -13.6 eV

MO: 1 2 3 4 5 Eigenval : -1.69510 -1.12412 -0.71524 -0.70967 -0.69421 (ev): -46.12594 -30.58885 -19.46255 -19.31098 -18.89045 Sg+ Su+ ??? Sg+ ???1 O2 S -0.61708 -0.67817 0.00000 -0.34528 0.000002 O2 PX 0.00000 0.00000 0.21578 0.00000 0.673383 O2 PY 0.00000 0.00000 0.67338 0.00000 -0.215784 O2 PZ -0.34528 0.20020 0.00000 0.61708 0.000005 O1 S -0.61708 0.67817 0.00000 -0.34528 0.000006 O1 PX 0.00000 0.00000 0.21578 0.00000 0.673387 O1 PY 0.00000 0.00000 0.67338 0.00000 -0.215788 O1 PZ 0.34528 0.20020 0.00000 -0.61708 0.00000 MO: 6 7 8 Eigenvalues: -0.38542 -0.01916 0.23853 (ev): -10.48773 -0.52126 6.49061 ??? ??? Su+ 1 O2 S 0.00000 0.00000 0.20020 2 O2 PX -0.21578 -0.67338 0.00000 3 O2 PY -0.67338 0.21578 0.00000 4 O2 PZ 0.00000 0.00000 0.67817 5 O1 S 0.00000 0.00000 -0.20020 6 O1 PX 0.21578 0.67338 0.00000 7 O1 PY 0.67338 -0.21578 0.00000 8 O1 PZ 0.00000 0.00000 0.67817

Semi-empirical treatment of O2 from Spartan (AM1) -13.6 eV

1) -46.1 eV sg - 0.617 O22s - 0.345 O22pz - 617 O12s + 0.345 O12pz :

2) -30.6 eV su - 0.678 O22s - 0.200 O22pz + 0.678 O12s + 0.200 O12pz :

3) -19.5 eV p 0.216 O22px + 0.673 O22py + 0.216 O12px - 0.673 O12py :

4) -19.3 eV sg - 0.345 O22s + 0.617 O22pz - 0.617 O12s - 0.345 O12pz :

5) -18.9 eV p 0.673 O22px - 0.216 O22py + 0.673 O12px - 0.216 O12py :

6) -10.5 eV p - 0.216 O22px - 0.673 O22py + 0.216 O12px + 0.673 O12py :

7) -0.52 eV p - 0.673 O22px + 0.216 O22py + 0.673 O12px - 0.216 O12py :

8) +6.49 eV su 0.200 O22s + 0.678 O22pz - 0.200 O12s + 0.678 O12pz :

Heteronuclear Diatomic molecules

HF 1s2 2s2 3s2 (1p22p2) 4s0

Basis set s = H1s (-13.6 eV), F1s (??), F2s (-40.2 eV?), F2pz (-18.6 eV)

p = F2px and F2py (-18.6 eV)One simpler treatment of HF is given in Atkins on page 428 gives the following results....

4 = s 0.98 (H1s) - 0.19(F2pz) -13.4 eV

px = py = F2px and F2py -18.6 eV

3s = 0.19(H1s) + 0.98(F2pz) -18.8 eV

2s = F2s ~ -40.2 eV1s = F1s << -40.2 eV

A more complicated system might mix all s type AOs in the following fashion....4 = s ???px = py = F2px and F2py

3s = -0.023(F1s) -0.411(F2s) +0.711(F2pz) +0.516(H1s)

2s = -0.018(F1s) +0.914(F2s) +0.090(F2pz) +0.154(H1s)

1s = 1.000(F1s) +0.012(F2s) +0.002(F2pz) -0.003(H1s)

MO: 1 2 3 4 5Eigenvalues:-1.82822 -0.63290 -0.51768 -0.51768 0.24632 (ev): -49.74849 -17.22198 -14.08688 -14.08688 6.70261 A1 A1 ??? ??? A1 1 H2 S 0.37583 -0.46288 0.00000 0.00000 0.80281 2 F1 S 0.91940 0.29466 0.00000 0.00000 -0.260523 F1 PX 0.00000 0.00000 -0.78600 0.61823 0.000004 F1 PY 0.00000 0.00000 0.61823 0.78600 0.000005 F1 PZ -0.11597 0.83601 0.00000 0.00000 0.53631

HF 1s2 2s2 3s2 (1p22p2) 4s0

Semi-empirical treatment of HF from Spartan (AM1)

One simpler treatment of HF is given in Atkins on page 428 gives the following results....4 = s 0.98 (H1s) - 0.19(F2pz) -13.4 eV

px = py = F2px and F2py -18.6 eV

3s = 0.19(H1s) + 0.98(F2pz) -18.8 eV

2s = F2s ~ -40.2 eV1s = F1s << -40.2 eV

H1s

F2s

F2p