Physical Chemistry II (CHEM F213):
Molecules
Dr. Jayadevan. K. P.
BITS Pilani
K K Birla Goa Campus
Review:Atomic units and SI
units
1.602
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Review:Atomic units and SI
units
Example: The unit of energy in atomic units is Hartree.
Express 1 Hartree in the unit of J, kJ/mol, wavenumber
(cm-1) and in eV
𝑚𝑒 4
𝐸=
(4𝜋𝜀𝑜 )2 ℏ2
Ans: [1 Hartree = 27.2 eV ≡ 4.3595×10-18 J ≡ 2625
kJ/mol ≡ 2.195 × 105 cm-1]
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Review: Linear combination
For a linear combination of f1 and f2:
𝜙 = 𝑐1 𝑓1 + 𝑐2 𝑓2
𝜏𝑑𝜙𝐻𝜙 (𝑐1 𝑓1 + 𝑐2 𝑓2 )𝑑𝜏
= 𝑐(1 𝑓1 + 𝑐2 𝑓2 )𝐻
1 𝑑𝜏 + 𝑐1 𝑐2 𝑓 1 𝐻𝑓
𝑐12 𝑓 1 𝐻𝑓 2 𝑑𝜏 + 𝑐1 𝑐2 𝑓 2 𝐻𝑓
1 𝑑𝜏 + 𝑐22 𝑓 2 𝐻𝑓
2 𝑑𝜏
= 𝑐12 𝐻11 + 𝑐1 𝑐2 𝐻12 + 𝑐1 𝑐2 𝐻21 + 𝑐22 𝐻22
1 𝑑𝜏 ; 𝐻12 = 𝑓 1 𝐻𝑓
𝐻11 = 𝑓 1 𝐻𝑓 2 𝑑𝜏 ; 𝐻22 = 𝑓 2 𝐻𝑓
2 𝑑𝜏
𝑗 𝑑𝜏 Because 𝐻
𝐻𝑖𝑗 = 𝑓𝐻 𝑖𝑓 is Hermitian Hij = Hji
𝑆𝑖𝑗 = 𝑆𝑗𝑖 = | 𝜏𝑑 𝑗𝜙 𝑖𝜙 Hij = Hji, 𝑆𝑖𝑗 = 𝑆𝑗𝑖 are the matrix
elements
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Hydrogen molecule
Hamiltonian
r12
Full Hamiltonian operator for a hydrogen
Molecule will have the following terms: r1B r2B
r1A
Nuclear kinetic energy
r2A
Electron kinetic energy
HA HB
Attractive potential of electron 1 with
R
Nuclei A and B
Attractive potential of electron 2 with Nuclei A and B
Electron-electron repulsion energy
Nuclear potential energy
As per Born-Oppenheimer approximation Nuclei are
considered stationary and therefore no kinetic energy
terms
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Hydrogen molecule Valence
bond wavefunction
1 1 1 1 1 1 1
𝐻 = − 𝛻12 + 𝛻22 − − − − + +
2 𝑟1𝐴 𝑟1𝐵 𝑟2𝐴 𝑟2𝐵 𝑟12 𝑅
(Hamiltonian simplified in atomic units)
Each electron in H-atom in H2 molecule has been
labelled 1 and 2 though they are indistinguishable.
𝜓1 = 1𝑠𝐴 1 1𝑠𝐵 2 -----------(A)
𝜓2 = 1𝑠𝐴 2 1𝑠𝐵 1 ------------(B)
Both (A) and (B) can be expressed in a linear
combination as:
𝜓 = 𝑐1 𝜓1 + 𝑐2 𝜓2 𝑏𝑒 𝑎 𝑡𝑟𝑖𝑎𝑙 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 for solving
𝜏𝑑𝜓𝐻𝜓 𝑐12 𝐻11 + 2𝑐1 𝑐2 𝐻12 + 𝑐22 𝐻22
𝐸 𝑐1 , 𝑐2 = = 2
2
𝜏𝑑 𝜓 𝑐1 𝑆11 + 2𝑐1 𝑐2 𝑆12 + 𝑐22 𝑆22
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Expansion of H11,H12, H22, S11,
S12, S22
For a linear combination of f1 and f2:
𝜙 = 𝑐1 𝑓1 + 𝑐2 𝑓2
1 𝑑𝜏 ; 𝐻12 = 𝑓 1 𝐻𝑓
𝐻11 = 𝑓 1 𝐻𝑓 2 𝑑𝜏 ; 𝐻22 = 𝑓 2 𝐻𝑓
2 𝑑𝜏
𝑗 𝑑𝜏 Because 𝐻
𝐻𝑖𝑗 = 𝑓𝐻 𝑖𝑓 is Hermitian Hij = Hji
𝑆𝑖𝑗 = 𝑆𝑗𝑖 = | 𝜏𝑑 𝑗𝜙 𝑖𝜙 Hij = Hji, 𝑆𝑖𝑗 = 𝑆𝑗𝑖 are the matrix
elements.
𝑆11 = 𝑆22 = 1, 𝑆12 = 𝑆 2
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Hydrogen molecule Valence
bond wavefunction
𝜏𝑑𝜓𝐻𝜓 𝑐12 𝐻11 +2𝑐1 𝑐2 𝐻12 +𝑐22 𝐻22
𝐸 𝑐1 , 𝑐2 = = ----------(C)
𝜓 2 𝑑𝜏 𝑐12 𝑆11 +2𝑐1 𝑐2 𝑆12 +𝑐22 𝑆22
𝐸 𝑐1 , 𝑐2 is a function of variational parameters c1 and c2
Rewriting (C):
𝐸 𝑐1 , 𝑐2 𝑐12 𝑆11 + 2𝑐1 𝑐2 𝑆12 + 𝑐22 𝑆22 = 𝑐12 𝐻11 + 2𝑐1 𝑐2 𝐻12 + 𝑐22 𝐻22 ----(D)
Differentiate (D) w.r.t c1:
𝜕𝐸
𝑐12 𝑆11 + 2𝑐1 𝑐2 𝑆12 + 𝑐22 𝑆22 + 𝐸 2𝑐1 𝑆11 + 2𝑐2 𝑆12 = 2𝑐1 𝐻11 +
𝜕𝑐1
2𝑐2 𝐻12 )----------------(E)
𝜕𝐸 𝜕𝐸
At minimum: = 0, rearranging (E) Similarly = 0 => (H)
𝜕𝑐1 𝜕𝑐2
𝑐1 𝐻11 − 𝐸𝑆11 + 𝑐2 𝐻12 − 𝐸𝑆12 = 0 −−−−−−−−−− −(F)
𝑐1 𝐻12 − 𝐸𝑆12 + 𝑐2 𝐻22 − 𝐸𝑆22 = 0 ---------------------------(H)
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Hydrogen molecule Valence
bond wavefunction
From (F) and (H): Note: An overlap integral is a
Secular determinant: direct measure of extent of
overlap between 1sA and 1sB
𝐻11 − 𝐸𝑆11 𝐻12 − 𝐸𝑆12
=0
𝐻12 − 𝐸𝑆12 𝐻22 − 𝐸𝑆22
(Heitler-London secular determinant)
In polar coordinates and r as variable:
𝑺𝟏𝟏 = )𝟐( 𝑩𝒔𝟏 𝟏 𝑨𝒔𝟏 )𝟐( 𝑩𝒔𝟏 𝟏 𝑨𝒔𝟏 𝟐𝒓𝒅 𝟏𝒓𝒅
= 𝟏 = 𝟐 𝑩𝒔𝟏 𝟐 𝑩𝒔𝟏 𝟐𝒓𝒅 )𝟏( 𝑨𝒔𝟏 𝟏 𝑨𝒔𝟏 𝟏𝒓𝒅
𝑺𝟏𝟐 = )𝟏( 𝑩𝒔𝟏 𝟐 𝑨𝒔𝟏 )𝟐( 𝑩𝒔𝟏 𝟏 𝑨𝒔𝟏 𝟐𝒓𝒅 𝟏𝒓𝒅
= 𝟐𝑺 = 𝟐 𝑨𝒔𝟏 𝟐 𝑩𝒔𝟏 𝟐𝒓𝒅 𝟏 𝑩𝒔𝟏 𝟏 𝑨𝒔𝟏 𝟏𝒓𝒅
𝑺 = ( 𝟏 𝑩𝒔𝟏 𝟏 𝑨𝒔𝟏 𝟏𝒓𝒅 Nonzero for nonzero overlap: OVERLAP
INTEGRAL)
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Overlap Integral (S)
Overlap integral (S) depends on internuclear distance R
𝑅2
0≤𝑆 𝑅 ≤1; 𝑆 𝑅 = 𝑒 −𝑅 1+𝑅+
3
Q.1: Find the overlap integral at an internuclear separation R = 𝟐𝒂𝒐 for
two hydrogen atoms.
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Overlap integral: Example
Note: R = 2 in atomic units Ξ
2ao/ao
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Valence Bond Energy of H2
1𝑠𝐴 1 1𝑠𝐵 2
𝐻11 = 𝑟𝑑 1 𝑑𝑟2 1𝑠𝐴 1 1𝑠𝐵 (2)𝐻
Hamiltonian:
𝟏 𝟐 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏
=−
𝑯 𝟐
𝜵 + 𝜵𝟐 − − − − + +
𝟐 𝟏 𝒓𝟏𝑨 𝒓𝟏𝑩 𝒓𝟐𝑨 𝒓𝟐𝑩 𝒓𝟏𝟐 𝑹
1 1 1
− 𝛻12 − 1𝑠𝐴 1 = − 1𝑠𝐴 1 −−− −(I)
2 𝑟1𝐴 2
1 2 1 1
− 𝛻2 − 1𝑠𝐵 2 = − 1𝑠𝐵 2 ----------(II)
2 𝑟2𝐵 2
𝟏 𝟏𝟐
RHS of (I) and (II) are energy eigen values of 1s electron: 𝑬𝟏 = − × a.u. Z = 1 and n =1
𝟐 𝟏𝟐
1 1 11
𝐴 1 1𝑠𝐵 2 = −1 −
𝐻1𝑠 − + + 1𝑠𝐴 1 1𝑠𝐵 2
𝑟1𝐵 𝑟2𝐴 𝑟12 𝑅
1 1 1 1
𝐻11 = −1 + 𝐽 = −1 + − − + +
𝑟1𝐵 𝑟2𝐴 𝑟12 𝑅
𝟏 𝟏 𝟏 𝟏
𝑱 = 𝟐 𝑩𝒔𝟏 𝟏 𝑨𝒔𝟏 𝟐𝒓𝒅 𝟏𝒓𝒅 − − + + 𝟏𝒔𝑨 𝟏 𝟏𝒔𝑩 𝟐
𝒓𝟏𝑩 𝒓𝟐𝑨 𝒓𝟏𝟐 𝑹
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Valence Bond Energy of H2
𝟏 𝟏 𝟏 𝟏
𝑱 = 𝟐) 𝟏 𝑨𝒔𝟏( 𝟐𝒓𝒅 𝟏𝒓𝒅 − − + + (𝟏𝒔𝑩 𝟐 )𝟐
𝒓𝟏𝑩 𝒓𝟐𝑨 𝒓𝟏𝟐 𝑹
𝑑𝑟1 [1𝑠𝐴 (1)]2 𝑑𝑟2 [1𝑠𝐵 (2)]2 1𝑠𝐴 (1) 2 1𝑠𝐵 (2) 2 1
𝐽= − − + 𝑟𝑑 1 𝑑𝑟2 + -----(III)
𝑟1𝐵 𝑟2𝐴 𝑟12 𝑅
1 5 3 𝑅2
𝐽= 𝑒 −2𝑅 + − 𝑅 − (From (III): Expression will be given
𝑅 8 4 6
for reference for solving problems)
(Coulomb Integral as a function of R)
Terms in (III) in atomic units:
𝒅𝒓𝟏 [𝟏𝒔𝑨 (𝟏)]𝟐
−Charge density of electron (1) around A interacting with B
𝒓𝟏𝑩
𝑑𝑟2 [1𝑠𝐵 (2)]2
𝑟2𝐴
− Charge density of electron (2) around B interacting with A
1𝑠𝐴 (1) 2 1𝑠𝐵 (2) 2
𝑟𝑑 1 𝑑𝑟2 −Coulombic interaction of two electron densities
𝑟12
1
− Nuclear-Nuclear coulombic interaction
𝑅
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Coulomb integral J: Example
Note: R = 2
in atomic units
Ξ 2ao/ao
− −
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Valence Bond Energy of H2
Q.2. Prove that 𝐻11 = 𝐻22
1𝑠𝐴 1 1𝑠𝐵 2 = −1 + 𝐽
𝐻11 = 𝑟𝑑 1 𝑑𝑟2 1𝑠𝐴 1 1𝑠𝐵 (2)𝐻
1𝑠𝐴 2 1𝑠𝐵 1 = −1 + 𝐽
𝐻22 = 𝑟𝑑 1 𝑑𝑟2 1𝑠𝐴 2 1𝑠𝐵 (1)𝐻
Since electrons are indistinguishable when exchanged
between two atoms both 𝐻11 𝑎𝑛𝑑 𝐻22 𝑎𝑟𝑒 𝑠𝑎𝑚𝑒 𝑎𝑛𝑑 𝑡ℎ𝑒 𝐻
gives same energy terms on RHS.
1𝑠𝐴 2 1𝑠𝐵 1
𝐻12 = ඵ 𝑑𝑟1 𝑑𝑟2 1𝑠𝐴 1 1𝑠𝐵 (2)𝐻
1 1 1
− 𝛻12 − 1𝑠𝐵 1 = − 1𝑠𝐵 1 −−− −(IV)
2 𝑟1𝐵 2
1 1 1
− 𝛻22 − 1𝑠𝐴 2 = − 1𝑠𝐴 2 ----------(V)
2 𝑟2𝐴 2
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Valence Bond Energy of H2
1 1 1 1
𝐴 2 1𝑠𝐵 1 = −1 −
𝐻1𝑠 − + + 1𝑠𝐴 2 1𝑠𝐵 1 … … (VI)
𝑟1𝐴 𝑟2𝐵 𝑟12 𝑅
Multiply (VI) with 1𝑠𝐴 1 1𝑠𝐵 2 and integrate over 𝑑𝑟1 𝑎𝑛𝑑 𝑑𝑟2 gives 𝐻12 .
𝐻12 = 𝑟𝑑 1 𝑑𝑟2 1𝑠𝐴 1 1𝑠𝐵 (2)𝐻 1𝑠𝐴 2 1𝑠𝐵 1 = −𝑆 2 + 𝐾……(VII)
Since: 𝑆12 = 𝑟𝑑 1 𝑑𝑟2 1𝑠𝐴 1 1𝑠𝐵 (2) 1𝑠𝐴 2 1𝑠𝐵 (1) = 𝑆 2 and the
K is the coulombic interaction when electrons are exchanged and
negative sign from energy operator.
𝟏 𝟏 𝟏 𝟏
K= 𝟐 𝑩𝒔𝟏 𝟏 𝑨𝒔𝟏 𝟐𝒓𝒅 𝟏𝒓𝒅 − − + + 𝟏𝒔𝑨 𝟐 𝟏𝒔𝑩 𝟏
𝒓𝟏𝑨 𝒓𝟐𝑩 𝒓𝟏𝟐 𝑹
𝟏 𝟏 𝟏 𝟏
K= 𝟏 𝑩𝒔𝟏 𝟏 𝑨𝒔𝟏 𝟐𝒓𝒅 𝟏𝒓𝒅 − − + + 𝟏𝒔𝑨 𝟐 𝟏𝒔𝑩 𝟐
𝒓𝟏𝑨 𝒓𝟐𝑩 𝒓𝟏𝟐 𝑹
(Exchange integral: Also K is a function of R)
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Coulomb integral J and K vs R
Sketch of variation of J and K versus R:
0.05
J
2 4
0.0 R/ao
-0.2
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Elements of the secular
determinant
Heitler-London Secular determinant:
𝐻11 − 𝐸𝑆11 𝐻12 − 𝐸𝑆12
=0
𝐻12 − 𝐸𝑆12 𝐻22 − 𝐸𝑆22
𝐻11 = −1 + 𝐽 = 𝐻22 ; 𝐻12 = −𝑆 2 + 𝐾 = 𝐻21
Heitler-London Secular determinant with all energy terms included:
−1 + 𝐽 − 𝐸 −𝑆 2 +𝐾 − 𝐸𝑆 2
=0
−𝑆 2 + 𝐾 − 𝐸𝑆 2 −1+𝐽−𝐸
𝐸 + 1 − 𝐽 2 = −𝑆 2 + 𝐾 − 𝐸𝑆 2 2
𝐸 + 1 − 𝐽 = ± −𝑆 2 + 𝐾 − 𝐸𝑆 2 or
𝐸± 1 ± 𝑆 2 = −1 + 𝐽 ± 𝐾 − 𝑆 2 = − 1 ± 𝑆 2 + 𝐽 ± 𝐾
| divide by 1 ± 𝑆 2
𝑱±𝑲
𝑬± = −𝟏 +
𝟏±𝑺𝟐
𝑱±𝑲
𝐰𝐡𝐞𝐫𝐞 − 𝟏 𝐢𝐬 𝐭𝐡𝐞 𝐞𝐧𝐞𝐫𝐠𝐲 𝐨𝐟 𝐭𝐡𝐞 𝐆. 𝐒. 𝐨𝐟 𝐢𝐬𝐨𝐥𝐚𝐭𝐞𝐝 𝐇𝐚𝐭𝐨𝐦𝐬 . ∆𝑬± = 𝟏±𝑺𝟐
|Energy change when H2 molecule forms
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Heitler-London Energy:
Derivation notes
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Heitler-London Energy:
Derivation notes
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Normalized symmetric
Wavefunction for H2
𝑐1 𝐻11 − 𝐸𝑆11 + 𝑐2 𝐻12 − 𝐸𝑆12 = 0 −−−−−−−− −(F)
𝑐1 𝐻12 − 𝐸𝑆12 + 𝑐2 𝐻22 − 𝐸𝑆22 = 0 ---------------------(H)
𝑐1 −1 + 𝐽 − 𝐸+ + 𝑐2 −𝑆 2 + 𝐾 − 𝐸+ 𝑆 2 = 0
𝑐1 = 𝑐2
𝜓+ = 𝑐1 (𝜓1 +𝜓2 )
Normalization process:
𝑐1 2 𝑟𝑑 1 𝑑𝑟2 𝜓12 + 2𝜓1 𝜓2 + 𝜓22 = 1
𝑐12 2 + 2𝑆 2 = 1 where 𝑆 2 ≡ 𝑆12
1 1
𝑐1 = =>𝜓+ = (𝜓1 +𝜓2 )
2(1+𝑆 2 ) 2(1+𝑆 2 )
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Normalization: Notes on
symmetric part
Similar steps
for
antisymmetric
part
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Normalized Antisymmetric
wavefunction for H2
𝑐1 𝐸− + 1 − 𝐽 + 𝑐2 −𝑆 2 + 𝐾 − 𝐸− 𝑆 2 = 0
𝐽−𝐾
Substitute for 𝐸− = −1 + and simplify
1−𝑆 2
𝐾−𝐽𝑆 2 𝐾−𝐽𝑆 2
𝑐1 + 𝑐2 =0
1−𝑆 2 1−𝑆 2
𝑐2 = −𝑐1
𝜓− = 𝑐1 𝜓1 − 𝜓2
1
Repeat Normalization: 𝑐1 =
2(1−𝑆 2 )
1
𝜓− = 𝜓1 − 𝜓2
2(1−𝑆 2 )
| Normalized Antisymmetric wavefunction
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Internuclear Potential energy
Sketch of internuclear potential energy curves for H2
calculated by Heitler-London Valence bond method
1
E Antisymmetric 𝜓− = 𝜓1 − 𝜓2
2(1 − 𝑆2)
+4
1
𝚫𝑬− 𝑆𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐: 𝜓+ = (𝜓1 +𝜓2 )
Symmetric 2(1+𝑆 2 )
0
𝚫𝑬+
-4
Experimental
R/ao
0 2 4
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Example Problems
1. Find the overlap integral at an internuclear separation R = 3𝑎𝑜 for
two hydrogen atoms.
As per atomic units: R = 3
𝑅 2
0 ≤ 𝑆 𝑅 ≤ 1 ; 𝑆 𝑅 = 𝑒 −𝑅 1 + 𝑅 +
3
Substitute in the above equation R = 3
Ans: S = 0.348
2.Given the coulombic interaction J of two hydrogen atoms in H2 as:
−2𝑅 1 5 3 𝑅2
𝐽=𝑒 + − 𝑅 − Find J at an internuclear separation R = 3𝑎𝑜 .
𝑅 8 4 6
Substitute in the above equation R = 3
Ans: ~ -0.007
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Example Problems
3. Given the coulombic integral J as -0.008 and K as -0.16
in atomic units for two hydrogen atoms forming H2
molecule with overlap integral S = 0.55, using Heitler-
London method, find (a) E+ (b) E- (c) ∆E+ (d) ΔE- (e) H11
(f) H12 in atomic units
𝐽+𝐾
𝐸+ = −1 +
1+𝑆 2
𝐽−𝐾
𝐸− = −1 +
1−𝑆 2
Antisymmetric means when electron coordinates are interchanged and
energy is higher than symmetric part)
𝑱±𝑲 𝐽+𝐾
∆𝑬± = => ∆𝐸+ =
𝟏±𝑺𝟐 1+𝑆 2
𝐽−𝐾
∆𝐸− = ; 𝐻11 = −1 + 𝐽 ; 𝐻12 = −𝑆 2 + 𝐾
1−𝑆 2
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Example Problems
𝐽+𝐾
(a)𝐸+ = −1 + = -1.128 (energy of symmetric part)
1+𝑆 2
𝐽−𝐾
(b)𝐸− = −1 + = −0.782 (𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑎𝑛𝑡𝑖𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑝𝑎𝑟𝑡)
1−𝑆 2
Antisymmetric means when electron coordinates are interchanged and
energy is higher than symmetric part)
𝑱±𝑲 𝐽+𝐾 −0.008−0.16
(c) ∆𝑬± = => ∆𝐸+ = = = −0.128
𝟏±𝑺𝟐 1+𝑆 2 1+(0.55×0.55)
𝐽−𝐾 −0.008+0.16
(d) ∆𝐸− = = = 0.218 (Energy is more positive and
1−𝑆 2 1−(0.55×0.55)
destabilizes when spatial exchange is antisymmetric)
(e) 𝐻11 = −1 + 𝐽 = −1 − 0.008 = −1.008
(f) 𝐻12 = −𝑆 2 + 𝐾 = − 0.55 × 0.55 − 0.16 = −0.4625
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Molecular orbital theory of H2+
Hamiltonian for single electron H2+ :
= − 1 𝛻2 −
𝐻
1
−
1
+
1
Single electron Hamiltonian
2 𝑟𝐴 𝑟𝐵 𝑅
𝜓 = 𝑐1 1𝑠𝐴 + 𝑐2 1𝑠𝐵 Linear combination
𝐻 −𝐸 𝐻𝐴𝐵 − 𝐸𝑆
𝐴𝐴 =0 Secular determinant
𝐻𝐴𝐵 − 𝐸𝑆 𝐻𝐵𝐵 − 𝐸
𝐴 = 𝐻𝐵𝐵
𝐻𝐴𝐴 = 𝑟𝑑 1𝑠𝐴 𝐻1𝑠
𝐵 = 𝑟𝑑 1𝑠𝐵 𝐻1𝑠
𝐻𝐴𝐵 = 𝑟𝑑 1𝑠𝐴 𝐻1𝑠 𝐴
𝑆 = 𝑟𝑑 1𝑠𝐴 1𝑠𝐵 Overlap integral
BITS Pilani, K K Birla Goa Campus
Molecular orbital theory of H2+
1 2 1 1
− 𝛻1 − 1𝑠𝐴 = − 1𝑠𝐴
2 𝑟𝐴 2 e- rB
1 2 1 1 rA
− 𝛻1 − 1𝑠𝐵 = − 1𝑠𝐵 HA HB
2 𝑟𝐵 2
𝐻𝐴𝐴 = −1/2 + 𝐽/ Total energy (1 e-) R
1 1
𝐽/ = 𝑟𝑑 1𝑠𝐴 − + 1𝑠𝐴 Coulomb integral
𝑟 𝑅 𝐵
𝑆
𝐻𝐴𝐵 = − + 𝐾/ Total exchange energy (1 e-)
2
1 1
𝐾 / = 𝑟𝑑 1𝑠𝐴 − + 1𝑠𝐵
𝑟 𝑅 𝐵
Secular determinant expansion:
𝐻𝐴𝐴 − 𝐸 2 − 𝐻𝐴𝐵 − 𝐸𝑆 2 =0
BITS Pilani, K K Birla Goa Campus
Molecular orbital theory of H2+
𝐻𝐴𝐴 ±𝐻𝐴𝐵 1 𝐽 / ±𝐾/ 𝐽/ ±𝐾/
𝐸± = ; 𝐸± = − + ; ∆𝐸± =
1±𝑆 2 1±𝑆 1±𝑆
𝑅2
𝑆= 𝑒 −𝑅 1+𝑅+ Overlap integral
3
1
𝐽/ = 𝑒 −2𝑅 1+ Coulomb integral
𝑅
𝑆
𝐾 / = − 𝑒 −𝑅 1 + 𝑅 Coulomb exchange
𝑅
integral
Example Problem: Given the coulombic integral J/ as 0.03 and K/ as -0.13 in
atomic units for H2+ molecule with overlap integral S = 0.55, using MO
method, find (a) E+ (b) E- (c) ∆E+ (d) ΔE- (e) HAA (f) HAB in atomic units
[Home work practice] [Hint: Substitute in the appropriate expressions given
under MOT of H2+ ]
BITS Pilani, K K Birla Goa Campus
Molecular orbital theory of H2+
Sketch of Energy as a function of internuclear separation
R:
E
+0.10 𝟏
𝝍− = (𝟏𝒔𝑨 −𝟏𝒔𝑩 )
𝟐(𝟏 − 𝑺)
+0.05 𝚫𝑬−
𝟏
𝚫𝑬+ 𝝍+ = (𝟏𝒔𝑨 +𝟏𝒔𝑩 )
-0.05 𝟐(𝟏 + 𝑺)
-0.10 R/ao
2 4 6
𝚫𝑬+ : 𝐒table molecule with equilibrium bond length calculated as 2.49 a.u. = 0.132 nm bond length
from experiment = 0.106 nm
BITS Pilani, K K Birla Goa Campus
Molecular orbital theory of H2+
Determine the Molecular orbitals that correspond to the
energies 𝐸+ 𝑎𝑛𝑑 𝐸−
𝑐1 𝐻𝐴𝐴 − 𝐸 + 𝑐2 𝐻𝐴𝐵 − 𝐸𝑆 = 0 ---------(1)
𝑐1 𝐻𝐴𝐵 − 𝐸𝑆 + 𝑐2 𝐻𝐵𝐵 − 𝐸 = 0----------(2)
1 𝐽/ +𝐾 /
𝐸+ = − + into eqn. (1)
2 1+𝑆
and then solve for c1 and c2 simultaneously.
𝐽/ 𝑆−𝐾 / 𝐾/ −𝐽/ 𝑆
𝑐1 + 𝑐2 = 0 satisfies RHS when 𝑐1 = 𝑐2
1+𝑆 1+𝑆
Molecular orbital corresponding to 𝐸+ (Bonding)
1
𝜓+ = 𝑐1 1𝑠𝐴 + 1𝑠𝐵 𝑤𝑖𝑡ℎ 𝑛𝑜𝑟𝑚𝑎𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑁 = 𝑐1 = ;
2(1+𝑆)
Antibonding:𝜓− = 𝑐1 1𝑠𝐴 − 1𝑠𝐵 and 𝑐1 = −𝑐2 for antibonding.
Q: Find 𝑐1 for antibonding 𝜓−
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Sketches of bonding and antibonding
probability density maps
Bonding Antibonding
BITS Pilani, K K Birla Goa Campus
Molecular orbital wavefunction
of H2
Spatial part of Molecular orbital wavefunction (𝜓𝑀𝑂 ):
1
𝜓𝑀𝑂 = 1𝑠𝐴 1 + 1𝑠𝐵 1 [1𝑠𝐴 2 + 1𝑠𝐵 2 ]
[2 1+𝑆 ]
MO as LCAO:
𝜓𝑀𝑂 = 1𝑠𝐴 1 + 1𝑠𝐵 1 1𝑠𝐴 2 + 1𝑠𝐵 2
𝜓𝑀𝑂 = [1𝑠𝐴 1 1𝑠𝐵 2 +1𝑠𝐴 2 1𝑠𝐵 1 ]
+[1𝑠𝐴 1 1𝑠𝐴 2 + 1𝑠𝐵 1 1𝑠𝐵 2 ]
𝜓 = 𝑐1 𝜓𝑉𝐵 + 𝑐2 𝜓𝑖𝑜𝑛𝑖𝑐 ; Linear combination of 𝜓𝑉𝐵 and 𝜓𝑖𝑜𝑛𝑖𝑐
Molecular wavefunction can be obtained by a linear combination of valence
bond and ionic part and then by variational method energy minimum can be
obtained and compared with experimentally measured values of energy and
internuclear distance.
BITS Pilani, K K Birla Goa Campus
Summary Points
Heitler-London VB Method:
= − 1 𝛻12 + 𝛻22 −
𝐻
1
−
1
−
1
−
1
+
1
+
1
2 𝑟1𝐴 𝑟1𝐵 𝑟2𝐴 𝑟2𝐵 𝑟12 𝑅
(H2 molecule Hamiltonian simplified in atomic units)
Secular determinant:
𝐻11 − 𝐸𝑆11 𝐻12 − 𝐸𝑆12
=0
𝐻12 − 𝐸𝑆12 𝐻22 − 𝐸𝑆22
Heitler-London Secular determinant with all energy terms included:
−1 + 𝐽 − 𝐸 −𝑆 2 +𝐾 − 𝐸𝑆 2
2 2 =0
−𝑆 + 𝐾 − 𝐸𝑆 −1+𝐽−𝐸
𝑱±𝑲 𝑱±𝑲
𝑬± = −𝟏 + ; ∆𝑬± =
𝟏±𝑺𝟐 𝟏±𝑺𝟐
𝑐1 𝐻11 − 𝐸𝑆11 + 𝑐2 𝐻12 − 𝐸𝑆12 = 0
𝑐1 𝐻12 − 𝐸𝑆12 + 𝑐2 𝐻22 − 𝐸𝑆22 = 0
𝑐1 = 𝑐2 ; 𝜓+ = 𝑐1 (𝜓1 +𝜓2 ) ; 𝑐2 = −𝑐1 ; 𝜓− = 𝑐1 𝜓1 − 𝜓2
BITS Pilani, K K Birla Goa Campus
Summary Points
1
Symmetric: 𝜓+ = (𝜓1 +𝜓2 )
2(1+𝑆 2 )
1
Antisymmetric: 𝜓− = 𝜓1 − 𝜓2
2(1−𝑆 2 )
Molecular orbital theory of H2+ :
𝐻𝐴𝐴 ±𝐻𝐴𝐵 1 𝐽 / ±𝐾/ 𝐽/ ±𝐾/
𝐸± = ; 𝐸± = − + ; ∆𝐸± =
1±𝑆 2 1±𝑆 1±𝑆
𝟏 𝟏
𝝍+ = (𝟏𝒔𝑨 +𝟏𝒔𝑩 ) ; 𝝍− = (𝟏𝒔𝑨 −𝟏𝒔𝑩 )
𝟐(𝟏+𝑺) 𝟐(𝟏−𝑺)
Molecular orbital theory of H2 :
𝜓 = 𝑐1 𝜓𝑉𝐵 + 𝑐2 𝜓𝑖𝑜𝑛𝑖𝑐 | Linear combination of 𝜓𝑉𝐵 and
𝜓𝑖𝑜𝑛𝑖𝑐
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