Carbon sp hybrids: Acetylene and the Triple bond
C2H2 is H-CC-H
Form sp on each C leaving 2px, 2py “unused”
H
-
sp
C
sp
+
+ sp
C
sp
-
H
Predicts linear structure. Above are all σ bonds
H-C-C-H
Uses up 2 valence e- for each C in sp (and 2 from 1s H).
Leaves 2 valence e - for each C unused.
2py
2py
Put last valence
e- into π orbitals
formed from
2px, 2py
2px
H
2px
C
sp
C
H
2 e- in πx
2 e- in πy
Gives two π bonds connecting carbon atoms. π bonds are at right
angles to each other. Total C-C bonds are now 3, one σ, 2 π
Short Comparison of Bond Order, Bond Length, Bond Energy
Molecule
Ethane, C2H6
Ethylene, C2H4
Acetylene, C2H2
C-C
Bond Order
1 (1 σ)
2 (1σ, 1π)
3 (1σ, 2π)
C-C
Bond Length
1.54Å
1.35Å
1.21Å
C-C
Bond E,
kcal/mole
83
125
230
Conjugation and Delocalization of
Electrons and Bonds
Although energy of π* in ethylene < σ*, conjugated polylenes
have even lower energy π* levels. These absorb light at longer
wavelength- sometimes even in visible (human eye’s light
perception). Conjugated polyenes:
C=C-C-C=C-C
2 essentially independent double bonds.
C=C-C=C-C=C
polyene
Double bonds ALTERNATE!
Could draw π bonds between any 2 C’s
C
C
C
C
C
This gives delocalized structure
C
C
C
C
C
C
C
π* bonds
drawn are
not unique!
ψMO = (Const)[2py(1) + 2py(2)
+ 2py(3) + 2py(4) +2py(5)
+ 2py(6) +………]. Add 2py
Atomic Orbitals on each C
Delocalized Molecular
Orbital spans many atoms
Delocalizing electrons generally lowers their energy.
π
States in the 4 Carbon Polyene Butadiene
Energy
H2C=CH-HC=CH2
π* antibonding
levels
E=hν (photon)
2p orbitals
(one each on 4 carbons)
π bonding
levels
E=hν for butadiene << E=hν for ethylene Molecular
Orbital Energies
C
C
C
C
4 Delocalized π Bonding
States Constructed
From 4 2py
C atomic orbs
Benzene C6H6. This is a planar closed ring compound of
carbon which has sp2 carbon hybrids:
H
H
C
120˚
H
sp2
C
C
H
C
C
C
H
Electron
Delocalization
in Carbon
Ring
Compounds
H
Each C uses 3 of 4 valence e- in sp2 hybrid bond structure
to make 1 C-H bond 2 C-C bonds.
Are left with 1 valence e- and a 2p orbital for each C
π
or
Can make 3 π bonds in two different ways. Called two resonance
structures. System alternates between the two different forms.
This
Thisisislocalized
picture!
localizedpicture!
Benzene C6H6 planar
closed ring compound
with sp2 carbon hybrids
leaves 6 valence electrons
and 6 p orbitals unused
Delocalized Picture
H
H
Can form π bonds with
these p orbitals.
ψMO = (Const)[2py(1) + 2py(2)
+ 2py(3) + 2py(4) +2py(5)
+ 2py(6)]
Add 6 2py Atomic
Orbitals on each C
H
+
+
–
–
+
+
–
–
H
+
+
–
–
H
H
However,these π bonds
have the ability to delocalize
around the entire
ring of 6 carbons
There are 6 such combinations!
Can therefore think of
electrons as delocalized
ψMO = (Const)[2py(1) + 2py(2)
+ 2py(3) + 2py(4) +2py(5)
+ 2py(6)]
Add 6 2py Atomic
Orbitals on each C
And often see
symbol for C6H6 as
Circle indicates this delocalization. This delocalized orbital comes
from adding all 2p orbitals in phase ( + on top) and ( - on bottom).
Can actually form a total of 6 delocalized M.O.’s for benzene
(6 2p Atomic Orbital’s → 6 M.O.’s). →
Beginnings
of a “conduction”
↓
band
Energy
+
Antibonding orbitals
0 (isolated C 2p)
π 2b
_
π3b
Bonding orbitals
π1b
“Band
←
Gap”
↑ Beginnings
of a “valence” band
Energy of 3 π bonding orbitals lower than energy of 2p (isolated)
orbitals on C from which they come. π antibonding are higher
than isolated 2p.
When put 6 e- into delocalized orbital get ~ (1/2) π bond per C-C
pair (1e- in π). Each C-C pair has one σ and (1/2) π.
Find experimentally all C-C bonds are of equal length (1.390Å) and
between that of C-C bond and C=C σ, π bond lengths.
If now consider bond energy of benzene, might guess it equals
6 C-C sigma bonds, 6 C-H sigma bonds, and 3 (C-C) π bonds.
Actually find benzene is more stable than this!
Delocalization of e- over 6 orbitals gives set
of 3 M.O.’s which have lower energy than
3 π C-C bonds on ethylene. i.e.→
Energy of (π1b)2 + (π2b)2 + (π3b)2 < Energy of 3 π2ethylene
Difference is ~ 160 kjoules/mole. This turns out to be about the
energy of one whole ethylene π bond.
i.e. Energy of (π1b)2 + (π2b)2 + (π3b)2 ≅ Energy of 4 π2eth.
More accurately: Energy of (π2b)2 = (π3b)2 ≅ π2eth
And: Energy of (π1b)2 ≅ 2 π2eth
Beginnings
of a “conduction”
↓
band
Energy
+
Antibonding orbitals
0 (isolated C 2p)
π 2b
_
2∆EEth
π1b
“Band
←
Gap”
π3b
Bonding orbitals
↑ Beginnings
of a “valence” band
Bonding in Solids
Think of a solid as a single giant molecule with roughly 1023 atoms.
Electrons can travel over the whole solid via delocalized orbitals
that cover all 1023 atoms.
Consider first the situation where each individual atom of the
solid has just one orbital contributing to bonding.
In this case must get 1023 molecular orbitals because atomic
orbitals map into molecular orbitals, one for one.
Because there are so many “molecular” energy levels
(one for each molecular orbital), levels form a band in
energy space
→
E
Energy
1023
equivalent
Atomic
orbitals
Band of 1023 delocalized
molecular orbitals of
slightly diffent energies
Delocalized Bonding in Metals
Consider Lithium metal. The Lithium atom has the atomic
configuration 1s22s1 with the 2p level unfilled.
As in any molecule with a filled core shell like 1s2,
these electrons do not participate in bonding. Still, they
form a delocalized band with 1023 molecular orbitals that
are completely filled.
The valence electrons 2s1 and the unfilled 2p orbitals are more
interesting. The 1023 2s atomic orbitals form a band of
1023 “molecular” orbitals. This band is only half filled
because each 2s1 orbital has only one e- .
There are three 2p orbitals on each atom leading to
a band of 3×1023 molecular orbitals. This band is “empty”
but overlaps in energy the 2s band →
3x1023
equivalent
Energy, E
2p Atomic
orbitals
1023
equivalent
2s Atomic
orbitals
1023
equivalent
1s Atomic
orbitals
Lithium
1s22s1
3 x 1023 2p delocalized molecular
orbitals in band
Overlapping of 2s and 2p
orbital bands
1023 half filled 2s delocalized
molecular orbitals in band
1023 filled 1s delocalized
Molecular orbitals in band