Relief for the VB Theory:
Promotion and Hybridization
Prediction: CH2
H
C—H
|
H
Þ Trick 1: Promote a valence electron
O
C: [He]2s22px12py1 → [He]2s12px12py12pz1
↓
H Predicted: 90o
4 σ bonds can now form: CH4
Þ Trick 2: Describe the
Þ Overall huge gain in energy
same electron distribution
Þ Tetravalent carbon leads to
as composed of four singly
organic chemistry
occupied hybrid orbitals
BUT: One bond is from a 2s orbital,
h1 = s + p x + p y + p z
the other three from 2p orbitals
h2 = s − p x − p y + p z
3 hybrid orbitals
Þ
four
sp
h3 = s − p x + p y − p z
h4 = s + p x − p y − p z
Nils Walter: Chem 260
Now VB Theory Gets the Molecular
Geometries Right
Methane CH4:
sp3, tetrahedral,
109.5o
C: [He]2s22px12py1
→
[He]2s12px12py12pz1
Ethene C2H4:
sp2, trigonal
pyramidal, 120o
C: [He]2s22px12py1
→
[He]2s12px12py12pz1
Ethyne C2H2:
sp, linear, 180o
C: [He]2s22px12py1
→
[He]2s12px12py12pz1
h1 = s + p z
h2 = s − p z
Þ hybridization of N atomic orbitals
results in N hybrid orbitals
h1 = s + 2 p x
h2 = s +
3
1
3
1
px −
p y h3 = sNils
− Walter:
p x −Chemp260
y
2
2
2
2
But What About Polar Molecules?
Resonance!
E.g., HCl: Ψcovalent = ΨH (1) × ΨCl (2) + ΨH (2) × ΨCl (1)
for a purely covalent bond: electrons only can exchange
Þ Allow for both electrons to be on Cl, i.e., H+Cl-:
Ψion = ΨCl (1) × ΨCl (2)
Þ In reality: superposition Ψ = Ψcovalent + λΨion
Ionic-covalent
resonance
λ2 = relative proportion (probability) of the ionic contribution
To find λ the Variation Theorem is used:
Similar: benzene
Ψ = ΨKek1 + ΨKek2
1
The energy of a trial
wavefunction is never
less than the true energy
2
Resonance hybrid
Þ Resonance stabilizationNils Walter: Chem 260
Even Better for Polar Molecules:
Molecular Orbital Theory
Basic ideas:
• Every electron contributes to every bond
• Electrons spread through the entire molecule
• Each electron may be found in any of the
atomic orbitals involved
..
Oh no!
Þ Approximation: Linear Combination of Atomic Orbitals (LCAO)
Example: H2+
+
Ψ ( H 2 ) = c A ΨA + cB ΨB Þ cA2, cB2 = relative contributions
Homonuclear diatomic
molecule Þ cA2 = cB2
Þ Ψ ( H 2 + ) = ΨA ± ΨB
of the atomic orbitals to the
molecular orbital (MO)
Þ 1 σ MO
Nils Walter: Chem 260
Bonding and Antibonding Orbitals
Electron density
between nuclei
+
Ψ ( H 2 ) = ΨA − ΨB
1 σ MO:
bonding
+
Ψ ( H 2 ) = ΨA + ΨB
2 σ* MO:
antibonding
Nuclei repel each other
Þ antibonding MOs are typically
slightly more antibonding than
bonding MOs are bonding
No electron density between
nuclei (nodal plane) Þ
electrons
nuclei
Nils Walter:pull
Chem
260 apart