SME
Physical Chemistry II: Quantum & Spectroscopy · Problem type
Hückel π energy of a ring (Frost circle)
Use this for “Find the π energy and delocalization energy of the cyclopentadienyl anion, C₅H₅⁻, by Hückel theory. Is it aromatic?”
Diagram
A
ring
of
6
carbons
with
6
π
electrons:
levels
α
+
2β
cos(2π
k
6
)
at
the
vertices
of
a
polygon
in
a
circle
of
radius
2β,
one
vertex
down.
π energy beyond 6α: 8β; in isolated double bonds 6β; delocalization 2β.
Every bonding level full, no unpaired electrons: aromatic (4n + 2).
β is negative, so the lowest level is α + 2β.
This is an example. Type your own values to start a new problem.
N
Ring carbons
example
q
Charge on the ring
example
e
π electrons
calculated
Eπ
π energy (in β)
calculated
Eᵢₛₒ
Isolated double bonds (in β)
calculated
Edeloc
Delocalization energy (in β)
calculated
u
Unpaired electrons
calculated
Formulas
e
=
N
−
q
6 = 6 − 0
Eπ
=
Hückel
energy
of
e
electrons
in
a
ring
of
N
8 = Hückel energy of 6 electrons in a ring of 6
Eᵢₛₒ
=
2
×
min(
e
2,
floor(
N
2
))
6
=
2
×
min(
6
2,
floor(
6
2
))
Edeloc = Eπ − Eᵢₛₒ
2 = 8 − 6
u
=
unpaired
ring
electrons
of
e
in
a
ring
of
N
0 = unpaired ring electrons of 6 in a ring of 6
Assumptions
•
Hückel theory: each ring carbon gives one p orbital, and the levels are α + 2β cos(2πk ÷ N), k = 0 to N − 1; β is negative.
•
π electrons are one per carbon less the charge: C₅H₅⁻ has 6. They fill the lowest levels in pairs, one in each of a pair first.
•
Energies are counted in β beyond Nα. The reference puts the electrons in as many isolated C=C bonds as the ring can hold, the rest at α; the difference is the delocalization energy.
Step-by-step
WE KNOW
N = 6, q = 0
FIND
π electrons (e), π energy (in β) (Eπ), isolated double bonds (in β) (Eᵢₛₒ), delocalization energy (in β) (Edeloc), unpaired electrons (u)
Step 1 · Find π electrons (e)
Use e = N − q
Each ring carbon gives one π electron; a negative charge adds electrons, a positive one takes them away.
e
=
6
−
0
e
=
6
Step 2 · Find π energy (in β): Eπ
Use Eπ = Hückel energy of e electrons in a ring of N
Draw
the
ring
point
down
in
a
circle
of
radius
2β:
each
corner
is
a
level.
Fill
from
the
bottom
and
add
each
electron’s
2cos(2π
k
N
).
Eπ = Hückel energy of 6 electrons in a ring of 6
2 × 2 + 4 × 1
Eπ = 8
Step 3 · Find isolated double bonds (in β): Eᵢₛₒ
Use Eᵢₛₒ = 2 × min(e ÷ 2, floor(N ÷ 2))
Only
⌊
N
2
⌋
C=C
bonds
fit
in
the
ring,
each
holding
its
pair
at
α
+
β:
2β
a
bond;
any
pair
left
over
sits
at
α
and
adds
0.
Eᵢₛₒ
=
2
×
min(
6
2,
floor(
6
2
))
Eᵢₛₒ = 6
Step 4 · Find delocalization energy (in β): Edeloc
Use Edeloc = Eπ − Eᵢₛₒ
What the ring gains over the same electrons in isolated double bonds.
Edeloc = 8 − 6
Edeloc = 2 → Every level filled in pairs with energy to spare (4n + 2 π electrons): aromatic.
Step 5 · Find unpaired electrons (u)
Use u = unpaired ring electrons of e in a ring of N
Count the single electrons: a pair of equal levels takes one each before either takes a second.
u
=
unpaired
ring
electrons
of
6
in
a
ring
of
6
u
=
0
Check
Put the numbers back in. Both sides should match.
6 = 6 − 0 ✓
8 = Hückel energy of 6 electrons in a ring of 6 ✓
6
=
2
×
min(
6
2,
floor(
6
2
))
✓
2 = 8 − 6 ✓
0 = unpaired ring electrons of 6 in a ring of 6 ✓
Topic check
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