Lecture 5, October 5, 2016, 2nd CC bonds - wag.caltech.eduwag.caltech.edu/home/ch120/Lectures/Ch120a_FA2016/Ch120-L05-16-… · CC bonds. Ch120a-Goddard-L05 ... Notation: sz and sz

Nature of the Chemical Bondwith applications to catalysis, materials

science, nanotechnology, surface science,bioinorganic chemistry, and energy

Course number: Ch120aHours: 2-3pm Monday, Wednesday, Friday

WilliamA. Goddard, III, [email protected] Beckman Institute, x3093

Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics,

California Institute of Technology

TeachingAssistant: Shane Flynn [email protected]: Mon 3pm

Ch120a-Goddard-L05 © copyright 2016 William A. Goddard III, all rights reserved

Lecture 5, October 5, 2016, 2nd CC bonds

© copyright 2016 William A. Goddard III, all rights reservedCh120a-Goddard-L05

review of L4

2

© copyright 2016 William A. Goddard III, a

ll rights reserved 3

Ch120a-Goddard-L05

The GVB orbitals for the (3s)2 pair of Si atom

Long dashes indicate zero amplitude, solid lines indicate positive amplitude while short dashes indicate negative amplitude. The spacing between contours is 0.05 in atomic units

symmetric pairing.Notation: sz and sz bar or ℓ and ℓ bar. Cannot type bars. use zs to show the bar case

Ch120a-Goddard-L05 © c

opyright 2016 William A. Goddard III, all rights reserved

analyze the pooched or hybridized orbitals

4

z z

zx

zx

φ2s + λφpzφ2s - λφpzPooching of the 2s orbitals in

opposite directions leads

to a dramatic increase in the ee distance, reducing ee repulsion.

1-D

2-D

Schematic. The line shows

© copyright 2016 William A. Godda

rd III, all rights reserved

Ch120a-Goddard-L05

Role of pooched or hybridized atomic lobe orbitalsin bonding of BeH+

5

Consider the bonding of H to Be+

The simple VB combination of H1s with the 2s orbital of Be+ leads to a very small overlap and contragradience

In fact optimizing the wavefunction for BeH+ leads to pooching of the 2s toward the H1s with much improved overlap and contragradience.

© copyright 2016 WilliamA. Goddard

III, all rights reserve

dCh120a-Goddard-L05

The ground state for C atom

6

Based on our study of Be, we expect that the ground state of C isΨyz=A[(sx)(xs)+(xs)(sx)](αβ−βα)(yα)(zα)] which we visualize as

z

x

2s pair pooched+x and –x

yz open shellsx py

pzxs

Ψyx=A[(sz)(zs)+(zs)(sz)](αβ−βα)(yα)(xα)] which we visualize as

2s pair pooched+z and –z

xy open shell

zs sz

px py

Ψxz=A[(sy)(ys)+(ys)(sy)](αβ−βα)(xα)(zα)] which we visualize aspx

pzsy,ys

2s pair pooched+y and –y

xz open shell

Ch120a-Goddard-L05 © copyright 2016 William A. Goddard III, all

rights reserved 7

The GVB orbitals of Silicon atom


Ch120a-Goddard-L05 © copyright 2016 William A

. Goddard III, all rights reserved

Role of pooched or hybridized atomic lobe orbitalsin bonding of BeH neutral

8

At small R the H can overlap significantly more with sz thanwith zH, so that we can form abond pair just like in BeH+. This leads to the wavefunction

Thus the wave function isA{[(sz)(zs)+(zs)(sz)](αβ−βα)(Hα)}

where sz≡(s+λz) and zs ≡(s-λz)

Here the H overlaps slightly more with sz than with zs, but the spin on sz is half the time α

Thus at large R we obtain a slightly repulsive interaction.

At large R the the orbitals of Beare already hybridized

zs sz H

Hszzs

A{[(sz)(H)+(H)(sz)](αβ−βα)(zsα)}In which the zs hybrid must now get orthogonal to the sz and H bond pair. This weakens the bond from that of BeH+ by ~ 1 eV

Ch120a-Goddard-L05 © copyright 2016 William A. Goddard III,

all rights reserved 9

Compare bonding inBeH+ and BeH

BeH+

1eV Repulsive orthogonalization of

zs with sz H

Long range Repulsive interaction

with HH

TA’s check numbers,all from memory

BeH+ has long range attraction no short range repulsion

3 eV1 eV

Short range Attractive interaction sz with H

Be

2 eV


Now bond 2nd H to BeH

10

Expect linear bond in H-Be-H and much stronger than the 1st bond

Expect bond energy similar to BeH+, maybe stronger, because the zs orbital is already decoupled from the sz.

Cannot bind 3rd H because no singly occupied orbitals left.

1Σ+~3.1 eV

Ch120a-Goddard-L05 © copyright 2016 WilliamA. Goddard III, all rights reserved 11

Compare bonding in BeH and BeH2

TA’s check numbers, all from memory

BeH+ MgH+

3.1 eV R=1.31A

1.34 eV R=1.73A

Expect linear bond in H-Be-H and much stronger than the 1st bond

Expect bond energy similar to BeH+, maybe stronger, because the zs orbital is already decoupled from the sz.

1Σ+

2Σ+

1Σ+

linear

Cannot bind 3rd H because no singly occupied orbitals left.

2.1 eV R=1.65 A

2.03 eV R=1.34A

~3.1 eV ~2.1 e

Ch120a-Goddard-L05 © copyright 2016 William A. Goddard III, all

rights reserved 12

The GVB orbitals of Silicon atom


© copyright 2016 William A. Goddard III, all rights reserved

Ch120a-God

dard-L05

Bonding H atom to all 3 states of C

13

Now we can get a bond to the lobe orbital just as for BeH

Bring H1s along z axis to C and consider all 3 spatial states.

(2px)(2pz)

C 2pz singly occupied.

H1s can get bonding

Get S= ½ state,

Two degenerate states, denote as 2Π

(2px)(2py)

(2py)(2pz)


Is the 2Π state actually 2Π?

The presence of the lobe orbitals might seem to complicate the symmetryΨyz=A[(sx)(xs)+(xs)(sx)](αβ−βα)(yα) (zΗ bond)2]Ψxz=A[(sy)(ys)+(ys)(sy)](αβ−βα)(xα)(zΗbond)2)]

To see that there is no problem, rewrite in the CI form (and ignore the zH bond)Ψyz=A[(s2 – λ x2)](αβ−βα)(yα)]Now form a new wavefunction by adding - λ y2 to Ψyz

Φyz ≡ A[s2 – λ x2 – λ y2](αβ−βα)(yα)]But the 3rd term is A[y2](αβ−βα)(yα)]= – λ A[(yα)(yβ)(yα)]=0 Thus Φyz = Ψyz and similarlyΦxz = A[s2 – λ x2 – λ y2](αβ−βα)(xα)] = Ψxz

Thus the 2s term [s2 – λ x2 – λ y2] is clearly symmetric about the z axis, so that these wavefunctions have 2Π symmetry

skip

© copyright 2016 William A. Goddard I 15

II, all rights reserved

Ch120a-Goddard-L05

Bonding of H to lobe orbital of C, Long RAt large R the lobe orbitals of C are already hybridized

Thus the wave function isA{(pxα)(pyα)[(sz)(zs)+(zs)(sz)](αβ−βα)(Hα)}

2s pair pooched+z and –z

xy open shell

Unpaired H

Here the H overlaps slightly more with sz than with zs, but the spin on sz is half the time α

Thus at large R we obtain a slightly repulsive interaction.

At small R we obtain bonding


Bonding of H to lobe orbital of C, small R

16

At small R the H can overlap significantly more with sz than with zH, so that we can form a bond pair just like in BeH+. This leads to the wavefunctionA{[(sz)(H)+(H)(sz)](αβ−βα)(zsα)(pxα)(pyα)}

Sz-H bond pair nonbond orbitals

But now the zs hybrid must now get orthogonal to the sz and H bond pair. This destabilizes the bond by ~ 1 eVThe symmetries of the nonbond orbitals are: zs=σ, px=πx, py=πy

Since the nonbond orbitals, σ, πx, πy are orthogonal to each other the high spin is lowest (S=3/2 or quartet state)We saw for NH that (πxπy –πyπx)(αα) has 3Σ- symmetry. CH has one additional high spin nonbond σ orbital, leading to 4Σ-

Hszzs

pxpy

Ch120a-Goddard-L05 © copyright

2016 WilliamA. Goddard III, all rights reserved 17The bonding states of CH and SiH

The low-lying state of SiH are shown at the right.Similar results are obtained for CH.

The bond to the p orbital to form the 2Π state is best

CH SiHKcal/mol De(2Π)Δ(2Π−4Σ-)De(4Σ-)

The bond to the lobe orbital is weaker than the p, but it is certainly significant

p bond

sz bond

80.0 70.117.1 36.262.9 33.9

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yright 2016 WilliamA. Goddard III, all rights reserved 18

Ch120a-Goddard-L05

GVB orbitals of SiH4Σ- state

© cop

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yright 2016 WilliamA. Goddard III, all rights reserved 19Ch120a-Goddard-L05

GVB orbitals of SiHstate

H

sx

xs

pz

py


Analysis of bonding in CH and SiHBond to p orbital is still the best for C and Si but the lobe bond is also quite strong, especially for CHThus hydridization in the atom due to electron correlation leads naturally to the new 4Σ- bonding state.Note that although the (sx)(xs) lobe pair for the atom are at 180º in the atom, they bend to ~104º for CH and SiH

180º

The reason is that as the pH bond is formed, the incoming H orbital overlaps the 2s part of the lobe orbital. To remain orthogonal, the 2s orbital builds in some –z character along with the x character already there. This rotates the lobe orbital away from the incoming H. This destabilizes the lobe pair making it easier for the 2nd H to bond to the lobe pair.

104º


Bonding the 2nd H to CH and SiH

21

As usual, we start with the ground states of CH or SiH, 2Πxand 2Πy and bond bring an H along some axis, say x.

H

sx

xs

pz

py

2Πy

2Πx

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William A. Go

ddard III, all rights reserved 22

Ch

Bonding the 2nd H to CH and SiH

Now we get credible bound states from both components

H

sx

xs

pz

py

2Πy

2Πx

A bond to the sx lobe orbital of CH (2Πy)

A bond to the p orbital of CH (2Πx)

This leads to the 1A1state of CH2 and SiH2that has already been

discussed.

This leads to the 3B1state that is the ground

state of CH2

o

© copyright 2016 WilliamA. Goddard III, all rights reserved 23

Applying C2z or σ in the plane interchanges (sy) and (ys) but the (sy)(ys) pair is symmetric under this interchange. Thus the total symmetry is1A1.

Ch120a-Goddard-L05

The p bond leads t the 1A1 state

GVB orbitals for SiH(1A1)

Ψ=A{[(sy)(ys)+(ys)(sy)(αβ−βα)](SiHLbond)2(SiHRbond)2}The wavefunction is


Bonding the 2nd H to the lobe orbital

24

At large distances the bond to the lobe orbital will be slightly repulsive and at an angle of 128 to the already formed p bond.

However at short distances, we form a strong bond. After forming the bond, each bond pair readjusts to have equivalent character (but an average of lobe bond and p bond).

The wavefunction becomesΨ=A{(SiHLbond)2(SiHRbond)2 [(σℓα)(πxα)]}

Here the two bond pairs and the σℓ orbital have A1 while πx has b1symmetry so that the total spatial symmetry is B1.

This leads to both 3B1 and 1B1 states, but triplet is lower (since σℓ

σℓπx

and πx are orthogonal).


Analyze Bond angles of CH2 and SiH2θe Re

1A1 state. Optimum bonding,è pz orbital points at Hz whilepx orbital points at Hx, leading to a bond angle of 90º.We expect that HH repulsion increases this by slightly less than the 13.2º for NH2 and 14.5º for OH2 and indeed it increases by 12.4º. But for Si the increase from 90º is only 2º as for P and As.

θe Re3B1 state. Optimum bonding,è the two bonds at ~128º. Here HH orthogonalization should increase this a bit but Cmuch less than 12º since H’s are much farther apart. However now the σℓ orbital must get orthogonal to the two bond pairs è a bond angle decrease. The lone pair affect dominates for SiH2decreasing the bond angle by 10º to 118º while the HH affect dominates for CH2, increasing the bond angle by 5º to 133º

σℓ

or

© copyright 2016 WilliamA. Goddard III, all rights reserved 26Ch120a-Goddard-L05

The GVB orbitals fSiH2 (3B1)

Ch120a-Goddard-L05 © copyright 20

16 Willi

amA. Goddard III, all rights reserved 2

Analysis of bond energies of 1A1 state

7

Consider the first two p bonds. Ignoring the affect of the bonds on the lobe orbitals, the main difference arises from the exchange terms.For C or Si A[(2s)2(pzα)(pxα)] leads to a term in the energy of the form (Jxz –Kxz) since the x and z orbitals have the same spin.But upon bonding the first H to pz, the wavefunction becomes A{(2s)2[(pzH+Hpx)(αβ−βα)(pxα)}. Now the pz and px orbitals have the same spin only have the time, so that this exchange term is decreased to - ½ Kxz.However in forming the second bond, there is no additional correction.Since Kxz ~ 14 kcal/mol for C and ~9 kcal/mol for Si. This means that the 2nd bond should be stronger than the first by 7 kcal/molfor C and by 4.5 kcal/mol for Si. E(kcal/mol)1st bond 2nd bond

CSi

8070

9076.2

This is close to the observed differences.


Analysis of bond energies of CH2 and SiH2 state

28

CH

62.9ℓ70.1p

SiH56.9ℓ

3P

80.0p

4Σ-33.9ℓ

90.0p99.1ℓ

3B1

1A176.1p

CH2

SiH2

CH Lobe bonds: 63 and 9950% increase

SiH Lobe bonds: 35 and 5750% increase

Assume 50% increase in lobe bond is from the first p bond destabilizing the lone pair

4Σ-

2Π2Π

1A13B1


The Bending potential surface for CH2

29

3B1

1A1

1B1

-3Σg

1Δg

9.3 kcal/mol

The ground state of CH2 is the 3B1 state not 1A1.


Analysis of bond energies of 3B1 state

30

For CH the lobe bond is 17 kcal/mol weaker than the p bond while for SiH it is 37 kcal/mol weaker.Forming the lobe bond requires unpairing the lobe pair which is~1 eV for the C row and ~1.5 eV for the Si row. This accounts forthe main differences suggesting that p bonds and lobe bonds areotherwise similar in energy.Forming a lobe bond to CH or SiH should be easier than to C or Si, because the first p bond has already partially destabilized thelobe pair. Since the SiH2(3B1) state is 19 kcal/mol higher than SiH2(1A1) but SiH(4Σ-) is 35 kcal/mol higher than Si(2Π), we conclude that lobe bond has increased in strength by ~16 kcal/molIndeed for CH the 3B1 state is 9.3 kcal/mol lower than 1A1implying that the lobe bond has increased in strength relative to the p bond by 26 kcal/mol.


CH2 GVB orbitals

plane bonding leads to an extremely flat potential curve for CH3.Since the lobe orbital is already unpaired, get a very strong bond energy of 109kcal/mol.

Ch120a-Goddard-L05 © copyright 2016 William A. Godd

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ard III, all rights reserved 3

Add 3rd H to form CH3

2

For CH2 we start with the 3B1 state and add H. Clearly the best place is in the plane, leading to planar CH3, as observed.As this 3rd bond is formed, each bond pair readjusts to a mixture of p and lobe character to become equivalent ~sp2 orbitalsWe could also make a bond to the out-of-plane pπ orbital but this would lead to large HH repulsions.However the possibility of favorable out-of-

120º

133º

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ight 2016 WilliamA. Goddard III, all rights reserved 33

weak, 72 kcal/mol.Ch120a-Goddard-L05

Add 3rd H to 1A1 SiH2 to form SiH3.

For SiH2 we start with the 1A1 state and add H to the lobe pair.Clearly this leads to a pyramidalSiH3, with an the optimum bondangle of 111º.Since we must unpair the lobe orbital this 3rd bond is relatively


Add 4th H to form CH4 and SiH4.For CH3 we start with the planar molecule and bring the H up to the out of plane p orbital. As the new bond forms all four bonds readjust to become equivalent leading to a tetrahedral CH4molecule. This bond is 105 kcal/mol, slightly weaker than the 3rd

109 kcal/mol) since it is to a p orbital and must interact with the other H’s.For SiH2 we start with the pyramidal geometry (111º bond angle) and add to the remaining lobe orbital. As the new bond forms, all four bonds readjust tobecome equivalent, leading to a tetrahedral SiH4 molecule.No unpairing is required è a strong bond, 92 kcal/mol (the 3rd bond was 72)

f 4


GVB orbitalsCH3 and CH

o


GVB orbitals of SiH3

36

SiH bond pair

Dangling bond orbital


GVB orbitals of SiH4

. Goddard III, all rights reserved 38

Orthogonal and point to vertices of a tetrahedron.Rationalizes bonding in CH4.Assumes 75% p character GVB: CH4 is 70% pAtom: lobe is 13%p: total =226/4=57% p

Ch120a-Goddard-L05 © copyright 2016 William A

Hybridization ofGVB Orbitals

Idealized case.Tetrahedral: sp3

o

x Te t r

y

z

(s-x+y-z)/2

(s+x+y+z)/2(s-x-y+z)/2

(s+x-y-z)/2


Comparisons of successive bond energies SiHn and CHn

p lobe

p

lobep

p

lobe

lobe

© copyright 2016 William A. Goddard III, all r

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servedCh120a-Goddard-L05

The ground state for B atom

40

Based on our study of C, we expect that the ground state of B isΨyz=A[(sx)(xs)+(xs)(sx)](αβ−βα)(zα)] which we visualize as

sx

z

szx

2s pair pooched ±x (or ±y)

z open shell pz

xs

2s pair pooched ± z x open shell zs

Ψyx=A[(sz)(zs)+(zs)(sz)](αβ−βα)(xα)] which we visualize aspx

sz

Ψxz=A[(sz)(zs)+(zs)(sz)](αβ−βα)(yα)] which we visualize as

2s pair pooched ± z y open shellpy

zs


ights re

servedCh120a-Goddard-L05

form BH and AlH by bonding along the z axis

41

Bonding to the pz state of B we obtainΨyz=A[(sx)(xs)pair](zΗ+Ηz)(αβ−βα)]

2s pair pooched ±x (or ±y)

H-z covalent bond

pz H

sx

xs

H-sz covalent bond open shell

px szzs

Ψyx=A[(sz)(H)+(H)(sz)](αβ−βα)(xα)(zsα)]

Ψxz yx=A[(sz)(H)+(H)(sz)](αβ−βα)(yα)(zsα)]

H-sz covalent bond open shellpy sz H

zs

128º

H

1Σ+

3Πx

3Πy

Ch120a-Goddard-L05 © copyright 2016 Will

iam A. Goddard III, all rights reserved

The bonding states of BH and AlH

42

The ground state of BH and AlH is obtained by bonding to the p orbital (leading to the 1Σ+ state.

However the bond to the lobe orbital (leading to the 3Π state) is also quite strong.

The bond to the lobe orbital is weaker than the p, but it is certainly significant

1Σ+

3Π

1Π

3Σ+

2P+ H

© copyright 2016 William A. Goddard III, all rights reserved

Ch120a-Goddar

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BH2 and AlH2

43

128º

Starting from the ground state of BHor AlH, the second bond is to a lobeorbital, to form the 2A1 state.

Just as for the 3B1 state of CH2 and SiH2 the bond for BH2 opens by several degrees to 131º while the bond to theAlH2 closes down by ~9º.

θe Re


BH3 and AlH3

44

Bonding the 3rd H to the 2A1 state of BH2, leads to planar BH3 or AlH3.

But there is no 4th bond since there remain no additional unpaired orbitals to bond to.


Re-examine bonding in NH, OH, and FH

45

Why did we ignore hybridization of the (2s) pair for NH, OH, and FH?

The reason is that the ground state of N atom

A[(2sα)(2sβ)(xα)(yα)(zα)]

Already has occupied px,py,pz orbitals. Thus

Pauli annihilates any hybridization in the 2s orbital.


Re-examine bonding in NH, OH, and FH

46

However, the doublet excited state of N can have hybridization, egA(2s)2(y)2(zα)è A[(sx)(xs)+(xs)(sx)](αβ−βα)(y)2(zα)which leads to the 2A1 excited states of NH2 of the form

θe Re


Bonding to halides (AXn, for X=F,Cl,…

47

The ground state of F has just one singly-occuped orbital and hence bonding to C is in many ways similar to H, leading to CF, CF2, CF3, and CF4 species. However there are significant differences.

Thus CF leads to two type so bonds, p and lobe just like CH

2Π

p bond

4Σ−

lobe bond

Covalent bondexpect (CH) 80 kcal/mol

63 kcal/mol

actual bond (CF)

120 kcal/mol

63 kcal/mol

Ch120a-Goddard-L05 © copyright 2016 William A. Goddard III, all rights reserv

ed

How can CF lead to such a strong bond, 120 vs 80 kcal/mol?

48

Consider the possible role of ionic character in the bonding

In the extreme limit

+

C F C+ F-

E (R=∞) = 0 E (R=∞) =IP (C) – EA (F)

=11.3 – 3.4 =7.9 eVIP (C) = 11.3 eV = 260 kcal/mol EA (F) = 3.40 eV = 78.4 kcal/molCan Coulomb attraction make up this difference?


Estimate energy of pure ionic bond for CF

49

Covalent limit (C + F)

Ionic limit (C+ + F-)

7.9 eV14.4/R

Re=1.27AIonic estimate ignores shielding and pauli repulsion for small R. Thus too large

14.4/1.27 =11.3 eV

Net bond = 11.3-7.9 =3.4 eV= 78 kcal/mol

Units for electrostatic interactions E=Q1*Q2/(ε0*R) where ε0 converts units (called permittivity of free space) E(eV) = 14.4 Q1(e units)*Q2(e units)/R(angstrom)E(kcal/mol) = 332.06 Q1(e units)*Q2(e units)/R(angstrom)


CF has strong mixture of covalent and ionic character

50

+

Pure covalent bond ~ 80 kcal/mol (based on CH)Pure ionic bond ~ 78 kcal/mol (ignore Pauli and shielding)

Net bond = 120 kcal/mol is plausible for 2Π state

But why is the bond for the 4Σ- state only 63, same as for covalent bond?

To mix ionic character into the 4Σ− state the electron must be pulled from the sz lobe orbital.

This leads to the (2s)1(2p)2 state of C+ rather than the ground state (2s)2(2p)1 which is 123 kcal/mol = 5.3 eV higherThus ionic bond is NEGATIVE (78-123=-45 kcal/mol)


Consider ionic contribution to 4Σ− state

51

+

C F C+ F-

E (R=∞) = 0 E (R=∞) =IP (C) – EA (F)

=16.6 – 3.4 =13.2 eV


ights reserved

Ch120a-Goddard-L05

Bonding the 2nd F to CF

52

With the 4Σ- state at 57 kcal/mol higher than 2Π, we need only consider bonding to 2Π, leading to the 1A1 state.

Bad Pauli repulsion increases FCF angle to 105º

1A1

3B1

57 kcal/mol


Now bond 3rd F to form CF3

53

Get pyramidal CF3(FCF angle ~ 112º)

In sharp contrast to planar CH3

The 3rd CF bond should be much weaker than 1st two.

This strong preference for CF to use p character makes conjugated flourocarbons much less stabe than corresponding saturated compounds.

Thus C4H6 prefers butadiene but C4F6 prefers cylcobutene

Of course the 4th bond to formCF4 leads to a tetrahedral structure


Summary, bonding to form hydrides

54

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.


New Material lecture 5

55

Ch120a-Goddard-L05 © copyright 2016 William A. Goddard III, all rights reserved 5

6

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

107.7º

111.2º

1.095A

1.526A

120.0º 1.086A


Ch120a-Goddard-L05

Bonding (GVB) orbitals of ethane (staggered)

Note nodal planes from

orthogonalization to CH bonds on

right C

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Ch120a-Goddard

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


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 energieswithin the molecule, but rather the electrostatic energy forinteracting 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 rotationalbarrier of 0.03 kcal/mol (favoring eclipsed). Here the CH bondsare ~ 3 times that in CH3-CH3 so that electrostatic effects woulddecrease by only 1/3. However overlap decreases exponentially.

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Ch120a-Goddard-L05

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 areHCH = 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


Ch120a-Goddard-L05

Trends: geometries of alkanes

CH bond length = 1.095 ± 0.001A

CC bond length = 1.526 ± 0.001A

CCC bond angles

HCH bond angles

© copyright 2016 William A. Goddard III, all rights reservedCh120a-Goddard-L05 62

Bond energiesDe = EAB(R=∞) - EAB(Re)e for equilibrium)Get from QM calculations. Re is distance at minimum energy.


Bond energiesDe = EAB(R=∞) - EAB(Re)Get from QM calculations. Re is distance at minimum energyD0 = H0AB(R=∞) - H0AB(Re)H0=Ee + ZPE is enthalpy at T=0KZPE = Σ(½Ћω)This is spectroscopic bond energy from ground vibrational state (0K)Including ZPE changes bond distance slightly to R0


Bond energiesDe = EAB(R=∞) - EAB(Re)Get from QM calculations. Re is distance at minimum energyD0 = H0AB(R=∞) - H0AB(Re)H0=Ee + ZPE is enthalpy at T=0KZPE = Σ(½Ћω)This is spectroscopic bond energy from ground vibrational state (0K)Including ZPE changes bond distance slightly to R0Experimental 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)}. (H = E + pV assuming an ideal gas)


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 isD298(A-B) = H298(A) – H298(B) – H298(A-B)D298 – D0 = 0∫298 [Cp(A) +Cp(B) – Cp(A-B)] dT =2.4 kcal/molif 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)}.

Ch120a-Goddard-L05 © copyright 2016 William A. Goddar

d III, all rights reserved 66

Snap Bond Energy: Break bond without relaxing the fragments

Snap

ΔErelax = 2*7.3 kcal/mol

c

Dsnaapp De (95.0kcal/mol)

Adiabati

Desnap (109.6 kcal/mol)


Bond energies for ethane

D0 = 87.5 kcal/molZPE (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

68De,snap (CH3-CH3) = 95.0 + 2*7.3 = 109.6 kcal/molCh120a-Goddard-L05 © copyright 2016 William A. Goddard III, all rights reserved

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

© c

opyright 2016 WilliamA. Goddard III, all rights reserved 69

Ch120a-Goddard-L05

Substituent effects on Bond energiesThe 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


Ch120a-Goddard-L05

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.


71

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 PUT IN ACTUALRELAXATION ENERGIES


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 gains7.3 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)

Documents

Lecture 5, October 5, 2016, 2nd CC bonds - wag.caltech.eduwag.caltech.edu/home/ch120/Lectures/Ch120a_FA2016/Ch120-L05-16-… · CC bonds. Ch120a-Goddard-L05 ... Notation: sz and sz