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Lecture 25: Introduction to Molecular Orbital Theory The material in this lecture covers the following in Atkins.
14 Molecular structure Molecular Orbital Theory (a) Linear combinations of atomic orbitals (b) bonding orbitals (c) anti-bonding orbitals
Lecture on-line
Introduction to Molecular Orbital theory (PowerPoint) Introduction to Molecular Orbital Theory (PDF)
Handout for this lecture
Molecular Orbital Theory H2+
We shall now discuss ways toapproximately solve :He (re,RN)ψ(re,RN) =Ee(RN )ψ(re,RN)
and represent the many - electronwave - function ψ(re,RN)
Hereˆ H e = ˆ T e + ˆ VNe + ˆ Vee + ˜ VNN
We shall this time use :
Molecular Orbital Theory H2+
ξ1 =[A(1)B(2)+A(2)B(1)]×[α(1)β(2)−β(1)α(2)]BA
In valence bond theory we localized one electron to each atom
el 1. el 2.
or BA
In molecular orbital theory each electron moves over the whole molecule
BA
el 1.el 2.
BAand
el 2.el 1.
The two theories are limiting cases (see later) of a more exact theory
Both electrons can be on the same nuclei
Molecular Orbital Theory H2+
We used the orbitals of theone - electron hydrogen to build upwavefunctions for many - electronatoms
We shall use the orbitals of
the one-electron H2+ molecule
to describe many electrondiatomic molecules
Molecular Orbital Theory H2+
The hamiltonian for the H2+
molecule:is given by
H = −
h2
2me∇e
2
Kinetic energy of electron
−e2
4πεo
1
rA1
Attraction of el. 1 by A
−e2
4πεo
1
rB1
Attraction of el. 1 by B
+e2
4πεo
1
R
rA1
A B
1
rB1
R
Repulsion between A and B
Molecular Orbital Theory H2+
We could now solve : ˆ H (r1,R)ϕ(r1,R) =E(R)ϕ(r1,R)With
V =−e2
4πεo[1
rA1+
1rB1
−1R
Where
H =−h2
2me∇e2 +V(r1,R)
To this end we write ϕ(r1,R) as a linear combination ( )of atomic orbitals LCAO
This is possible but tedious
We shall instead find approximate soluions
The atomic orbitals are in general those centered onthe atoms of our molecule
Was first done by a Danish astronomer in 1926
1sA (r1A ) =A(r1A ) =e−r1A /ao
(πao3 )3 / 2
Molecular Orbital Theory H2+
For H2+ that is :
1sA (r1B ) =A(r1B ) =e−r1B /ao
(πao3 )3 / 2
Where r1A and r1B are related by:
r1B = r1A2+R2−2r1ARcosθ
rA1
A B
1
rB1
R
θ
We now assume the electron to be equallylikely to be on each nuclei and write the molecular orbitals as :
ϕ +(1) = A(1) +B(1) ϕ−(1) = A(1) −B(1)
Note : we have as many linear combinationsas we have atomic orbitals
Molecular Orbital Theory H2+
ϕ +(1) = A(1) +B(1) ϕ−(1) = A(1) −B(1)
We can normalize our orbitals for easy energycalculation and use in probability density
ϕ+(1)ϕ+(1)∫ dv=N2 (A(1)+B(1))(A(1)+B(1))dv=1∫
ϕ+(1)ϕ+(1)∫ dv=N2 (A(1)A(1)dv∫ +N2 (B(1)B(1)dv∫+2N2 (A(1)B(1)dv∫ =1
A normalized B normalized overlap S > 1
ϕ+(1)ϕ+(1)∫ dv = +1×N2 +1×N2
+2S×N2 =1
N =1
2(1+S)
Molecular Orbital Theory H2+
ϕ +(1) = A(1) +B(1) ϕ−(1) = A(1) −B(1)
After normalization
ϕ +(1) =1
2(1+ S)[A(1) +B(1)] ϕ−(1) =
1
2(1− S)[A(1) −B(1)]
The electron probability density:
ϕ +(1)ϕ+ (1) =1
2(1+ S)[A(1) +B(1)]2
=1
2(1+ S)A(1)2 +
1
2(1+ S)B(1)2 + 2
1
2(1+ S)AB(1)
Density at A(Reduced)
Density at B(Reduced)
Density between A and B(Increased)
(a)The amplitude of the bonding molecular orbital in a hydrogen molecule-ion in a plane containing the two nuclei and
(b) a contour representation of the amplitude.
Molecular Orbital Theory H2+
ϕ +(1) =1
2(1+ S)[A(1) +B(1)]
Molecular Orbital Theory H2+
A representation of the constructive interference that occurs when two H 1s orbitals overlap and form a bonding orbital. Compare this illustration with Fig.14.14.
ϕ +(1) =1
2(1+ S)[A(1) +B(1)]
Molecular Orbital Theory H2+
The boundary surface of a orbital encloses the region where the electrons that occupy the orbital are most likely to be found. Note that the orbital has cylindrical symmetry.
ρ(1) =1
2(1+ S)A(1)2 +
1
2(1+ S)B(1)2 + 2
1
2(1+ S)AB(1)
Molecular Orbital Theory H2+
The electron density calculated by forming the square of the wavefunction used to construct Fig.14.14. Note the accumulation of electron density in the internuclear region.
ρ(1) =1
2(1+ S)A(1)2 +
1
2(1+ S)B(1)2 + 2
1
2(1+ S)AB(1)
Molecular Orbital Theory H2+
H =−
h2
2me∇e
2−e2
4πεo[
1rA1
+1
rB1−
1R
] ϕ+(1)=1
2(1+S)[A(1)+B(1)]
We can now evaluate the energy : E+ = ϕ+(1) ˆ H(1)∫ ϕ+ (1)dv
E+=1
2(1+S)[A(1)+B(1)]ˆ H (1)∫ [A(1)+B(1)]dv
=
12(1+S)
] A(1)∫ [−h2
2me∇e
2−e2
4πεo
1rA1
]A(1)dv
+
12(1+S)
] B(1)∫ [−h2
2me∇e
2−e2
4πεo
1rB1
]B(1)dv
+1
2(1+S)] A(1)∫ [
e2
4πεo
1rB1
]A(1)dv
+1
2(1+S)] B(1)∫ [
e2
4πεo
1rA1
]B(1)dv
+
22(1+S)
] A(1)∫ [−h2
2me∇e
2−e2
4πεo[
1rA1
+1
rB1]]B(1)dv+
e2
4πεo
1R
rA1
A B
1
rB1
R
θ
Molecular Orbital Theory H2+
E+ =1
2(1+ S)[A(1) +B(1)] ˆ H(1)∫ [A(1) +B(1)]dv
=
1
2(1+S)] A(1)∫ [−
h2
2me∇e
2 −e2
4πεo
1rA1
]A(1)dv
+
1
2(1+S)] B(1)∫ [−
h2
2me∇e
2 −e2
4πεo
1rB1
]B(1)dv
+1
2(1+S)] A(1)∫ [−
e2
4πεo
1rB1
]A(1)dv
+1
2(1+S)] B(1)∫ [−
e2
4πεo
1rA1
]B(1)dv
+
2
2(1+S)] A(1)∫ [−
h2
2me∇e
2 −e2
4πεo[1
rA1+
1rB1
]]B(1)dv
rA1
A B
1
rB1
R
θ+e2
4πεo
1
R
E1sH
2(1+ S)
E1sH
2(1+ S)
J2(1+S)
J2(1+S)
K'(1+S)
Molecular Orbital Theory H2+
- J
2(1+S)=
12(1+S)
] A(1)∫ A(1)[−e2
4πεo
1rB1
]dv
-J
2(1+S)=
12(1+S)
] B(1)B(1∫ [−e2
4πεo
1rA1
])dv
A B
rB1
R
1
2(1+ S)] A (1)A(1)dv
Interaction (attraction) between nucleus B and the charge cloud A(1)A(1)/2(1+S)
Interaction (attraction) between nucleus A and the charge cloud B(1)B(1)/2(1+S)
Molecular Orbital Theory H2+
K'=
22(1+S)
] A(1)∫ [−h2
2me∇e
2−e2
4πεo[
1rA1
+1
rB1]]B(1)dv
rA1
A B
1
rB1
R
θ
=2
2(1+S)] A(1)∫ [−
h2
2me∇e
2−e2
4πεo
1rB1
]]B(1)dv
−2
2(1+S)] A(1)B(1)∫ [
e2
4πεo
1rA1
]dv
SE1sH/(1+S)
−K
Interaction (attraction) between nucleus A and the charge cloud A(1)B(1)/(1+S)
Molecular Orbital Theory H2+
+e2
4πεo
1
RE+ =
E1sH
2(1+ )S+
E1sH
2(1+ S)−
J2(1+S)
−J
2(1+S)−
K -SE1sH(1+S)
E+=E1sH+SE1sH
(1+S)−
J +K(1+S)
+e2
4πεo
1R
We get in a similar way that
E−=E1sH−J -K(1-S)
+e2
4πεo
1R
E+=E1sH−J +K(1+S)
+e2
4πεo
1R
Molecular Orbital Theory H2+
The calculated and experimental molecular potential energy curves for a hydrogen molecule-ion.
E+=E1sH+J +K(1+S)
+e2
4πεo
1R
E−=E1sH−J -K(1-S)
+e2
4πεo
1R
Molecular Orbital Theory H2+
A representation of the destructive interference that occurs when two H1s orbitals overlap and form an antibonding * orbital. Compare this illustration with Fig.14.20.
ϕ−(1) =1
2(1−S)[A(1)−B(1)]
Molecular Orbital Theory H2+
(a) The amplitude of the antibonding molecular orbital in a hydrogen molecule-ion in a plane containing the two nuclei and
(b) a contour representation of the amplitude. Note the internuclear node.
ϕ−(1) =1
2(1−S)[A(1)−B(1)]
Molecular Orbital Theory H2+
The electron density calculated by forming the square of the wavefunction used to construct Fig.14.20. Note the elimination of electron density from the internuclear region.
ρ−(1) =1
2(1− S)A(1)2 +
12(1− S)
B(1)2 − 21
2(1− S)AB(1)
Molecular Orbital Theory H2+
(b) In an antibonding orbital, the nuclei are attracted to an accumulation of electron density outside the internuclear region.
A partial explanation of the origin of bonding and antibonding effects.
(a) In a bonding orbital, the nuclei are attracted to the accumulation of electron density in the internuclear region.
ϕ +(1) =1
2(1+ S)[A(1) +B(1)] ϕ−(1) =
12(1−S)
[A(1)−B(1)]
A molecular orbital energy level diagram for orbitals constructed from the overlap of H1s orbitals; the separation of the levels corresponds to that found at the equilibrium bond length. The ground electronic configuration of H2 is obtained by accommodating the two electrons in the lowest available orbital (the bonding orbital).
Molecular Orbital Theory H2+
rA1
A B
1
rB1
R
What you should learn from this lecture
Be able to construct the Hamiltonian
H =−h2
2me∇e
2−e2
4πεo
1rA1
−e2
4πεo
1rB1
+e2
4πεo
1R
Understand the justification for constructingthe two molecular orbitals as linear combinationsof atomic orbitals ϕ+(1)=N+(A(1)+B(1)) and ϕ−(1)=N−(A(1)−B(1))and be able to evaluate the normalizationconstants from the overlap integral S= A(1)B(1)dv∫
You are not asked to derive the energy expressions
for ϕ+: E+=E1sH−J +K(1+S)
+e2
4πεo
1R
or ϕ- :E−=E1sH−J -K(1-S)
+e2
4πεo
1R
What you should learn from this lecture
However you should understand the meening of
J = A(1)∫ A(1)[−e2
4πεo
1rB1
]dv and
K= A(1)B(1)∫ [e2
4πεo
1rA1
]dv and undertand how K
stabilize ϕ+ and destabalizes ϕ− .
rA1
A B
1
rB1
R
It is expected that you can derive the expressionfor the density due to ϕ+ as:
ρ(1)=1
2(1+S)A(1)2+
12(1+S)
B(1)2+21
2(1+S)A(1)B(1)
You shoule further understand the meening of
A(1)2, B(1)2, and A(1)B(1). Also you should understand
how +21
2(1+S)A(1)B(1) influence the energy E+ as well
as the features of Fig 14.16
What you should learn from this lecture It is expected that you can derive the expressionfor the density due to ϕ- as:
ρ(1)=1
2(1−S)A(1)2+
12(1−S)
B(1)2−21
2(1−S)A(1)B(1)
You shoule further understand the meening of
A(1)2, B(1)2, and A(1)B(1). Also you should understand
how −21
2(1−S)A(1)B(1) influences the energy E - as well
as the features of Fig 14.21
Understand the features of the plots in Fig. 14.17 and 14.19for ϕ+(1)=N+(A(1)+B(1)) and ϕ−(1)=N−(A(1)−B(1)),respectively.
Make note of the potential energy curve in Fig 14.18 and observe that E+ is lowered less in energy (compared to E1s than E- is increased (see also Fig 14.24