SME
Physical Chemistry II: Quantum & Spectroscopy
Molecular orbital theory
Use this for “Two atomic orbitals have αA = −11 eV and αB = −15 eV, with β = −2.5 eV. Find the bonding and antibonding MO energies.”
Diagram
E±
E±
=
αA + αB
2
∓
√((
αA − αB
2
)²
+
β
2
)
E± = −13 ∓ 3.2 eV
E₊ = −16.2 eV (bonding, below both AOs); E₋ = −9.8 eV (antibonding); splitting 6.4 eV.
β = −2.5 eV; the overlap S is neglected.
This is an example. Type your own values to start a new problem.
αA
Coulomb integral of A
example · eV
αB
Coulomb integral of B
example · eV
β
Resonance integral
example · eV
E₊
Bonding energy
calculated · eV
E₋
Antibonding energy
calculated · eV
ΔE
Splitting
calculated · eV
Formulas
E₊
=
αA + αB
2
−
√((
αA − αB
2
)²
+
β
2
)
−16.2
=
−11 + (−15)
2
−
√((
−11 − (−15)
2
)²
+
(−2.5)
2
)
E₋
=
αA + αB
2
+
√((
αA − αB
2
)²
+
β
2
)
−9.798
=
−11 + (−15)
2
+
√((
−11 − (−15)
2
)²
+
(−2.5)
2
)
ΔE
=
2
×
√((
αA − αB
2
)²
+
β
2
)
6.403
=
2
×
√((
−11 − (−15)
2
)²
+
(−2.5)
2
)
ΔE = E₋ − E₊
6.403 = −9.798 − (−16.2)
Assumptions
•
Each MO is a mix of one orbital on A and one on B (LCAO), so E solves the secular determinant |αA − E, β; β, αB − E| = 0.
•
The overlap S is neglected. β is negative, so E₊, the lower root, is the bonding MO; equal α gives α ± β.
•
αA and αB are the orbitals’ own energies (Coulomb integrals), negative for a bound electron.
Step-by-step
WE KNOW
αA = −11 eV, αB = −15 eV, β = −2.5 eV
FIND
bonding energy (E₊), antibonding energy (E₋), splitting (ΔE)
Step 1 · Find bonding energy (E₊)
Use E₊ = (αA + αB) ÷ 2 − √(((αA − αB) ÷ 2)² + β²)
Expanding the determinant gives a quadratic in E; its lower root is the mean of the α values less the square root.
E₊
=
−11 + (−15)
2
−
√((
−11 − (−15)
2
)²
+
(−2.5)
2
)
E₊
=
−11 + (−15)
2
−
√(
(
4
2
)
2
+
(−2.5)
2
)
E₊
=
−11 + (−15)
2
−
√(
2
2
+
(−2.5)
2
)
E₊
=
−
26
2
−
√
4 + 6.25
E₊
=
−
26
2
−
√
10.25
E₊
=
−
26
2
−
3.202
E₊ = −13 − 3.202
E₊ = −16.2 eV → The bonding MO lies nearer the lower AO, so it is mostly B’s orbital (the more electronegative atom).
Step 2 · Find antibonding energy (E₋)
Use E₋ = (αA + αB) ÷ 2 + √(((αA − αB) ÷ 2)² + β²)
The upper root of the same quadratic: the antibonding MO, pushed above the higher AO.
E₋
=
−11 + (−15)
2
+
√((
−11 − (−15)
2
)²
+
(−2.5)
2
)
E₋
=
−11 + (−15)
2
+
√(
(
4
2
)
2
+
(−2.5)
2
)
E₋
=
−11 + (−15)
2
+
√(
2
2
+
(−2.5)
2
)
E₋
=
−
26
2
+
√
4 + 6.25
E₋
=
−
26
2
+
√
10.25
E₋
=
−
26
2
+
3.202
E₋ = −13 + 3.202
E₋ = −9.798 eV
Step 3 · Find splitting (ΔE)
Use ΔE = 2 × √(((αA − αB) ÷ 2)² + β²)
The two roots sit the same distance either side of the mean, so the gap is twice the square root.
ΔE
=
2
×
√((
−11 − (−15)
2
)²
+
(−2.5)
2
)
ΔE
=
2
×
√(
(
4
2
)
2
+
(−2.5)
2
)
ΔE
=
2
×
√(
2
2
+
(−2.5)
2
)
ΔE
=
2
×
√
4 + 6.25
ΔE
=
2
×
√
10.25
ΔE = 2 × 3.202
ΔE = 6.403 eV
Check
Put the numbers back in. Both sides should match.
−16.2
=
−11 + (−15)
2
−
√((
−11 − (−15)
2
)²
+
(−2.5)
2
)
✓
−9.798
=
−11 + (−15)
2
+
√((
−11 − (−15)
2
)²
+
(−2.5)
2
)
✓
6.403
=
2
×
√((
−11 − (−15)
2
)²
+
(−2.5)
2
)
✓
6.403 = −9.7984379 − (−16.201562) ✓
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