Transcript Slide 1

Lecture 6 January 18, 2012
CC Bonds diamond, ΔHf, Group additivity
Nature of the Chemical Bond
with applications to catalysis, materials
science, nanotechnology, surface science,
bioinorganic chemistry, and energy
Course number: Ch120a
Hours: 2-3pm Monday, Wednesday, Friday
William A. Goddard, III, [email protected]
316 Beckman Institute, x3093
Charles and Mary Ferkel Professor of Chemistry, Materials
Science, and Applied Physics,
California Institute of Technology
Teaching Assistants: Caitlin Scott <[email protected]>
Hai Xiao [email protected]; Fan Liu <[email protected]>
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Ch120a1
Last time
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Summary, bonding to form hydrides
General principle: start with ground state of AHn and add H to
form the ground state of AHn+1
Thus use 1A1 AH2 for SiH2 and CF2 get pyramidal AH3
Use 3B1 for CH2 get planar AH3.
For less than half filled p shell, the presence of empty p
orbitals allows the atom to reduce electron correlation of the
(ns) pair by hybridizing into this empty orbital.
This has remarkable consequences on the states of the Be,
B, and C columns.
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Now combine Carbon fragments to form larger molecules
(old chapter 7)
Starting with the ground state of CH3 (planar),
we bring two together to form ethane, H3C-CH3.
As they come together to bond, the CH bonds
bend back from the CC bond to reduce overlap
(Pauli repulsion or steric interactions between
the CH bonds on opposite C).
At the same time the 2pp radical orbital on each
C mixes with 2s character, pooching it toward
the corresponding hybrid orbital on the other C
120.0º
1.086A
107.7º 1.095A
1.526A
111.2º
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Bonding (GVB) orbitals of ethane (staggered)
Note nodal planes
from
orthogonalization
to CH bonds on
right C
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Staggered vs. Eclipsed
There are two
extreme cases for the
orientation about the
CC axis of the two
methyl groups
The salient difference between these is the overlap of the CH
bonding orbitals on opposite carbons.
To whatever extent they overlap, SCH-CH Pauli requires that they
be orthogonalized, which leads to a repulsion that increases
exponentially with decreasing distance RCH-CH.
The result is that the staggered conformation is favored over
eclipsed by 3.0 kcal/mol
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Alternative interpretation
The bonding electrons are
distributed over the molecule, but it
is useful to decompose the
wavefunction to obtain the net
charge on each atom.
This leads to qH ~ +0.15 and qC ~ -0.45.
qH ~ +0.15
qC ~ -0.45
These charges do NOT indicate the electrostatic energies
within the molecule, but rather the electrostatic energy for
interacting with an external field.
Even so, one could expect that electrostatics would favor
staggered.
The counter example is CH3-C=C-CH3, which has a rotational
barrier of 0.03 kcal/mol (favoring eclipsed). Here the CH bonds
are ~ 3 times that in CH3-CH3 so that electrostatic effects would
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decrease
by only 1/3.©However
decreases
exponentially.
7
Propane
Replacing an H of ethane with CH3,
leads to propane
Keeping both CH3 groups staggered
leads to the unique structure
Details are as shown. Thus the bond
angles are
HCH = 108.1 and 107.3 on the CH3
HCH =106.1 on the secondary C
CCH=110.6 and 111.8
CCC=112.4,
Reflecting the steric effects
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Trends: geometries of alkanes
CH bond length = 1.095 ± 0.001A
CC bond length = 1.526 ± 0.001A
CCC bond angles
HCH bond angles
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Bond energies
De = EAB(R=∞) - EAB(Re)
e for equilibrium)
Get from QM calculations. Re is
distance at minimum energy.
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Bond energies
De = EAB(R=∞) - EAB(Re)
Get from QM calculations. Re is
distance at minimum energy
D0 = H0AB(R=∞) - H0AB(Re)
H0=Ee + ZPE is enthalpy at T=0K
ZPE = S(½Ћw)
This is spectroscopic bond energy from
ground vibrational state (0K)
Including ZPE changes bond distance
slightly to R0
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Bond energies
De = EAB(R=∞) - EAB(Re)
Get from QM calculations. Re is
distance at minimum energy
D0 = H0AB(R=∞) - H0AB(Re)
H0=Ee + ZPE is enthalpy at T=0K
ZPE = S(½Ћw)
This is spectroscopic bond energy from
ground vibrational state (0K)
Including ZPE changes bond distance
slightly to R0
Experimental bond enthalpies at 298K and atmospheric pressure
D298(A-B) = H298(A) – H298(B) – H298(A-B)
D298 – D0 = 0∫298 [Cp(A) +Cp(B) – Cp(A-B)] dT =2.4 kcal/mol if A and
B are nonlinear molecules (Cp(A) = 4R). {If A and B are atoms D298
– D0 = 0.9 kcal/mol (Cp(A) = 5R/2)}.
(HCh120a-Goddard-L07,08
= E + pV assuming© an
ideal gas)
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Bond energies, temperature corrections
Experimental measurements of bond energies, say at 298K,
require an additional correction from QM or from spectroscopy.
The experiments measure the energy changes at constant
pressure and hence they measure the enthalpy,
H = E + pV (assuming an ideal gas)
Thus at 298K, the bond energy is
D298(A-B) = H298(A) – H298(B) – H298(A-B)
D298 – D0 = 0∫298 [Cp(A) +Cp(B) – Cp(A-B)] dT =2.4 kcal/mol
if A and B are nonlinear molecules (Cp(A) = 4R).
{If A and B are atoms D298 – D0 = 0.9 kcal/mol (Cp(A) = 5R/2)}.
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Snap Bond Energy: Break bond without relaxing the fragments
Snap
DErelax = 2*7.3 kcal/mol
Adiabatic
D
Desnap (109.6snap
kcal/mol) De (95.0kcal/mol)
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Bond energies for ethane
D0 = 87.5 kcal/mol
ZPE (CH3) = 18.2 kcal/mol,
ZPE (C2H6) = 43.9 kcal/mol,
De = D0 + 7.5 = 95.0 kcal/mol (this is calculated from QM)
D298 = 87.5 + 2.4 = 89.9 kcal/mol
This is the quantity we will quote in discussing bond breaking
processes
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The snap Bond energy
In breaking the CC bond of ethane the geometry changes from
CC=1.526A, HCH=107.7º, CH=1.095A
To CC=∞, HCH=120º, CH=1.079A
Thus the net bond energy involves both breaking the CC bond
and relaxing the CH3 fragments.
We find it useful to separate the bond energy into
The snap bond energy (only the CC bond changes, all other
bonds and angles of the fragments are kept fixed)
The fragment relaxation energy.
This is useful in considering systems with differing substituents.
For CH3 this relation energy is 7.3 kcal/mol so that
De,snap (CH3-CH3) = 95.0 + 2*7.3 = 109.6 kcal/mol
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Substituent effects on Bond energies
The strength of a CC bond changes from 89.9 to 70 kcal/mol as
the various H are replace with methyls.Explanations given include:
•Ligand CC pair-pair
repulsions
•Fragment relaxation
•Inductive effects
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Ligand CC pair-pair repulsions:
Each H to Me substitution leads to 2 new CH bonds gauche to
the original CC bond, which would weaken the CC bond.
Thus C2H6 has 6 CH-CH interactions lost upon breaking the
bond,
But breaking a CC bond of
propane loses also two addition
CC-CH interactions.
This would lead to linear changes
in the bond energies in the table,
which is approximately true.
However it would suggest that the
snap bond energies would
decrease, which is not correct.
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Fragment relaxation
Because of the larger size of Me compared to H, there will be
larger ligand-ligand interaction energies and hence a bigger
relaxation energy in the fragment upon relaxing form
tetrahedral to planar geometries.
In this model the snap bond enegies are all the same.
All the differences lie in the relaxation of the fragments.
This is observed to be approximately correct
Inductive effect
A change in the character of the CC bond orbital
due to replacement of an H by the Me.
Goddard believes that fragment relaxation is the correct
explanation
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Bond energies: Compare to CF3-CF3
For CH3-CH3 we found a snap bond energy of
De = 95.0 + 2*7.3 = 109.6 kcal/mol
Because the relaxation of tetrahedral CH3 to planar gains
7.3 kcal/mol
For CF3-CF3, there is no such relaxation since CF3 wants
to be pyramidal, FCF~111º
Thus we might estimate that for CF3-CF3 the bond energy
would be De = 109.6 kcal/mol, hence D298 ~ 110-5=105
Indeed the experimental value is D298=98.7±2.5 kcal/mol
suggesting that the main effect in substituent effects is
relaxation (the remaining effects might be induction and
steric)
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New material lecture 6, January 18, 2012
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CH2 +CH2  ethene
Starting with two methylene radicals (CH2) in the
ground state (3B1) we can form ethene
(H2C=CH2) with both a s bond and a p bond.
The HCH angle in CH2 was 132.3º, but Pauli Repulsion with
the new s bond, decreases this angle to 117.6º (cf with 120º
for CH3)
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Comparison of The GVB
bonding orbitals of ethene
and methylene
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Twisted ethene
Consider now the case where the plane of one CH2 is rotated by
90º with respect to the other (about the CC axis)
This leads only to a s bond. The
nonbonding pl and pr orbitals can be
combined into singlet and triplet states
Here the singlet state is referred to as N (for Normal) and the
triplet state as T.
Since these orbitals are orthogonal, Hund’s rule suggests that T is
lower than N (for 90º). The Klr ~ 0.7 kcal/mol so that the splitting
should be ~1.4 kcal/mol.
Voter, Goodgame, and Goddard [Chem. Phys. 98, 7 (1985)] showed that N is
below
T by 1.2 kcal/mol, due
to Intraatomic
Exchange
(s,p
same center) 24
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Goddard III, all
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Twisting potential surface for ethene
The twisting potential surface for ethene is shown below. The
N state prefers θ=0º to obtain the highest overlap while the T
state prefers θ=90º to obtain the lowest overlap
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geometries
For the N state (planar) the CC bond
distance is 1.339A, but this increases to
1.47A for the twisted form with just a
single s bond.
This compares with 1.526 for the CC bond of ethane.
Probably the main effect is that twisted ethene has very little CH
Pauli Repulsion between CH bonds on opposite C, whereas
ethane has substantial interactions.
This suggests that the intrinsic CC single bond may be closer to
1.47A
For the T state the CC bond for twisted is also 1.47A, but
increases to 1.57 for planar due to Orthogonalization of the
triple coupled pp orbitals.
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CC double bond energies
The bond energies for ethene are
De=180.0, D0 = 169.9, D298K = 172.3 kcal/mol
Breaking the double bond of ethene, the HCH bond angle
changes from 117.6º to 132.xº, leading to an increase of 2.35
kcal/mol in the energy of each CH2 so that
Desnap = 180.0 + 4.7 = 184.7 kcal/mol
Since the Desnap = 109.6 kcal/mol, for H3C-CH3,
The p bond adds 75.1 kcal/mol to the bonding.
Indeed this is close to the 65kcal/mol rotational barrier.
For the twisted ethylene, the CC bond is De = 180-65=115
Desnap = 115 + 5 =120. This increase of 10 kcal/mol compared to
ethane might indicate the effect of CH repulsions
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bond energy of F2C=CF2
The snap bond energy for the double bond of ethene od
Desnap = 180.0 + 4.7 = 184.7 kcal/mol
As an example of how to use this consider the bond energy
of F2C=CF2,
Here the 3B1 state is 57 kcal/higher than 1A1 so that the
fragment relaxation is 2*57 = 114 kcal/mol, suggesting that
the F2C=CF2 bond energy is Dsnap~184-114 = 70 kcal/mol.
The experimental value is D298 ~ 75 kcal/mol, close to the
prediction
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Bond energies double bonds
Although the ground state of CH2 is 3B1 by 9.3 kcal/mol,
substitution of one or both H with CH3 leads to singlet ground
states. Thus the CC bonds of these systems are weakened
because of this promotion energy.
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C=C bond energies
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CC triple bonds
Starting with two CH radicals in the 4S- state we can form
ethyne (acetylene) with two p bonds and a s bond.
This leads to a CC bond length of 1.208A compared to 1.339
for ethene and 1.526 for ethane.
The bond energy is
De = 235.7, D0 = 227.7, D298K = 229.8 kcal/mol
Which can be compared to De of 180.0 for H2C=CH2 and
95.0 for H3C-CH3.
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GVB orbitals of HCCH
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GVB orbitals of CH 2P and 4S- state
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CC triple bonds
Since the first CCs bond is De=95 kcal/mol and the first CCp
bond adds 85 to get a total of 180, one might wonder why the
CC triple bond is only 236, just 55 stronger.
The reason is that forming the triple bond
requires promoting the CH from 2P to 4S-,
which costs 17 kcal each, weakening the
bond by 34 kcal/mol. Adding this to the 55
would lead to a total 2nd p bond of 89
kcal/mol comparable to the first
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2P
4S-
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Bond energies
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Diamond
Replacing all H atoms of ethane and with methyls, leads to with a
staggered conformation
Continuing to replace H with methyl groups
forever, leads to the diamond crystal
structure, where all C are bonded
tetrahedrally to four C and all bonds on
adjacent C are staggered
A side view is
This leads to the diamond crystal structure. An expanded view
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37
Infinite structure from tetrahedral bonding plus staggered bonds
on adjacent centers
2nd layer
3
1
1
1st layer
1
2
02
2
2nd layer
0
0
1c
1st layer
1
1
1
1
2nd layer
1st layer
Chair
configuration
of cylcohexane
Not shown: zero layer just like 2nd layer but above layer 1
rd layer just like the 1st layer but below layer 2
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c
The unit cell of diamond crystal
An alternative view of the
diamond structure is in terms of
cubes of side a, that can be
translated in the x, y, and z
directions to fill all space.
c
f
c
i
c
i f
f
f
f
i c
i
Note the zig-zag chains c-i-f-i-c
f
and cyclohexane rings (f-i-f)-(i-f-i) c
c
There are atoms at
•all 8 corners (but only 1/8 inside the cube): (0,0,0)
•all 6 faces (each with ½ in the cube): (a/2,a/2,0), (a/2,0,a/2),
(0,a/2,a/2)
•plus 4 internal to the cube: (a/4,a/4,a/4), (3a/4,3a/4,a/4),
(a/4,3a/4,3a/4), (3a/4,a/4,3a/4),
Thus each cube represents 8 atoms.
All other atoms of the infinite crystal are obtained by translating
this
cube by multiples©of
a in 2011
theWilliam
x,y,zA. Goddard
directions
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c
39
Diamond Structure
Now bond one of these
atoms, C2, to 3 new C
so that the bond are
staggered with respect
to those of C1.
5a
3a
1a
4b
2b
5
6
3
4
2
1b
4a
2a
1
Start with C1 and
make 4 bonds to form
a tetrahedron.
5b
3b
1c
7
Continue this process.
Get unique structure:
diamond
Note: Zig-zag chain
1b-1-2-3-4-5-6
Chair cyclohexane
ring: 1-2-3-3b-7-1c
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Properties of diamond crystals
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Properties of group IV molecules (IUPAC group 14)
1.526
There are 4 bonds to each atom, but each bond connects two
atoms.
Thus to obtain the energy per bond we take the total heat of
vaporization and divide by two.
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William
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Note
for Si, that the average
bond
isA.much
different
than for Si H42
Comparisons of successive bond energies SiHn and CHn
p
lobe
lobe
lobe
p
lobe
p
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p
43
Redo the next sections
Talk about heats formation first
Then group additivity
Then resonance etc
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Benzene and Resonance
referred to as Kekule or VB structures
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Resonance
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Benzene wavefunction
is a superposition of the VB structures in (2)
benzene as
≡
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+
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47
More on resonance
That benzene would have a regular 6-fold symmetry is not
obvious. Each VB spin coupling would prefer to have the
double bonds at ~1.34A and the single bond at ~1.47 A (as
the central bond in butadiene)
Thus there is a cost to distorting the structure to have equal
bond distances of 1.40A.
However for the equal bond distances, there is a
resonance stabilization that exceeds the cost of distorting
the structure, leading to D6h symmetry.
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Cyclobutadiene
For cyclobutadiene, we have the same situation, but here the
rectangular structure is more stable than the square.
That is, the resonance energy does not balance the cost of
making the bond distances equal.
1.34 A
1.5x A
The reason is that the pi bonds must be orthogonalized,
forcing a nodal plane through the adjacent C atoms, causing
the energy to increase dramatically as the 1.54 distance is
reduced to 1.40A.
For benzene only one nodal plane makes the pi bond
orthogonal
to both other
bonds,
leading
to lower
cost
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A. Goddard
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49
graphene
Graphene: CC=1.4210A
Bond order = 4/3
Benzene: CC=1.40 BO=3/2
Ethylene: CC=1.34 BO = 2
CCC=120°
Unit cell has 2 carbon atoms
1x1 Unit cell
This is referred to as graphene
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Graphene band structure
1x1 Unit cell
Unit cell has 2 carbon atoms
Bands: 2pp orbitals per cell
2 bands of states each with N states
where N is the number of unit cells
2 p electrons per cell  2N electrons for
N unit cells
The lowest N MOs are doubly occupied,
leaving N empty orbitals.
The filled 1st band touches
the empty 2nd band at the
Fermi energy
Get semi metal
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2nd band
1st band
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51
Graphite
Stack graphene layers as ABABAB
Can also get ABCABC Rhombohedral
AAAA stacking much higher in energy
Distance between layers = 3.3545A
CC bond = 1.421
Only weak London dispersion
attraction between layers
De = 1.0 kcal/mol C
Easy to slide layers, good lubricant
Graphite: D0K=169.6 kcal/mol, in plane bond = 168.6
Thus average in-plane bond = (2/3)168.6 = 112.4 kcal/mol
112.4 = sp2 s + 1/3 p
Diamond: average CCs = 85 kcal/mol  p = 3*27=81 kcal/mol
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energetics
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Allyl Radical
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Allyl wavefunctions
It is about 12 kcal/mol
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Cn
What is the structure of C3?
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Cn
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Energetics Cn
Note extra stability of odd Cn by 33 kcal/mol, this is because odd
Cn has an empty px orbital at one terminus and an empty py on the
other, allowing stabilization of both p systems
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Stability of odd Cn
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Bond energies and thermochemical calculations
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Bond energies and thermochemical calculations
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Heats of Formation
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Heats of Formation
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Heats of Formation
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Heats of Formation
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Bond energies
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Bond energies
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Bond energies
Both secondary
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Average bond energies
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Average bond energies
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Real bond energies
Average bond
energies of little
use in predicting
mechanism
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Group values
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Group functions of propane
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Examples of using
group values
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Group values
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Strain
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Strain energy cyclopropane from Group values
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Strain
energy
c-C3H6
using real
bond
energies
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Stained GVB orbitals of cyclopropane
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Benson Strain energies
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Resonance in thermochemical Calculations
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Resonance in thermochemical Calculations
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Resonance energy butadiene
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Allyl radical
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Benzene resonance
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Benzene resonance
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Benzene resonance
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Benzene resonance
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Benzene resonance
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