N2, O2, and F2

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N2, O2, and F2
- No 1s orbitals, because these are core orbitals.
- The dimension of the problem has to be reduced
by symmetry.
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N2, O2, and F2 (continued)
Intermezzo The 2p orbitals.
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N2, O2, and F2 (continued)
Intermezzo The 2p orbitals.
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N2, O2, and F2 (continued)
Symmetry elements
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N2, O2, and F2 (continued)
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N2, O2, and F2 (continued)
Make plus and minus combination from
orbitals that transform into each other.
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N2, O2, and F2 (continued)
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N2, O2, and F2 (continued)
group I
group II
group III
group IV
group V
group VI
- There is no interaction between orbitals of
different groups.
- The eight-dimensional problem is reduced to six
smaller problems.
- The one-dimensional problems directly give
solutions of the Fock-equation.
- Groups III and V, and groups IV and VI give
degenerate solutions.
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N2, O2, and F2 (continued)
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N2, O2, and F2 (continued)
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N2, O2, and F2 (continued)
Bond-order ( bonding electrons - antibonding
electrons)/2 (8 - 2) / 2 3
N2
N
N
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N2, O2, and F2 (continued)
MO of N2
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N2, O2, and F2 (continued)
MO of N2
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N2, O2, and F2 (continued)
MO of N2
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N2, O2, and F2 (continued)
MO of N2
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N2, O2, and F2 (continued)
MO of N2
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N2, O2, and F2 (continued)
MO of N2
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N2, O2, and F2 (continued)
Bond-order ( bonding electrons - antibonding
electrons)/2 (8 - 4) / 2 2
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N2, O2, and F2 (continued)
Hunds second rule The Slater-determinant has
the lowest energy if electrons are distributed as
much as possible over degenerated orbitals.
Hunds third rule The Slater-determinant has the
lowest energy if electrons are placed
in degenerated orbitals with the same spin.
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N2, O2, and F2 (continued)
Bond-order ( bonding electrons - antibonding
electrons)/2 (8 - 6) / 2 1