Lecture6,April 17,2019 From CC bonds –Transition metals CC …...e= 95.0 + 2*7.3 = 109.6 kcal/mol...

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08

From CC bonds – Transition metals CC RedNature of the Chemical Bond

with applications to catalysis, materialsscience, nanotechnology, surface science,

bioinorganic chemistry, and energy

1

Lecture 6, April 17, 2019

© copyright 2017 William A. Goddard III, all rights reserved

William A. Goddard, III, [email protected] 316Beckman Institute, x3093

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

California Institute of Technology

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

Teaching Assistant: Yalu Chen <[email protected]>: (office hours: 3-4 monday)

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L07,08 2

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 for SiH2 and CF2 use 1A1 AH2: get pyramidal AH3

Use 3B1 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.

© copyright 2016 William A. Goddard III, all rights reservedCh120a-Goddard-L06 3

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

1.526A

1.095A1.086A120.0º


Bonding (GVB) orbitals of ethane (staggered)

Note nodal planes from

orthogonalization to CH bonds on

right C


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


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


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 = S(½Ћw) 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)


Snap Bond Energy: Break bond without relaxing the fragments

Snap

Adiabatic

DErelax = 2*7.3 kcal/mol

DsnapDesnap (109.6 kcal/mol) De (95.0kcal/mol)


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

© copyright 2016 William A. Goddard III, all rights reservedCh120a-Goddard-L06

new

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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


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


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.


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


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. Thus the adiabatic BE = 109.6-14.6=95

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)


CH2 +CH2 è 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)


Comparison of The GVB bonding orbitals of ethene

and methylene


Twisted etheneConsider 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 on same center)


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


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.


CC double bond energies

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

The bond energies for ethene are

De=180.0, D0 = 169.9, D298K = 172.3 kcal/mol


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


Bond energies double bondsAlthough 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.


C=C bond energies


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.


GVB orbitals of HCCH


GVB orbitals of CH 2P and 4S- state


CC triple bonds

Since the first CCs bond is De=95 kcal/mol and the first CCpbond 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

2P

4S-


Bond energies


Bonding at a transition metaal

Bonding to a transition metals can be quite covalent.

Examples: (Cl2)Ti(H2), (Cl2)Ti(C3H6), Cl2Ti=CH2

Here the two bonds to Cl remove ~ 1 to 2 electrons from the Ti, making is very unwilling to transfer more charge, certainly not to C or H (it would be the same for a Cp (cyclopentadienyl ligand)

Thus TiCl2 group has ~ same electronegativity as H or CH3

The covalent bond can be thought of as Ti(dz2-4s) hybrid spin paired with H1s

A{[(Tids)(H1s)+ (H1s)(Tids)](ab-ba)}


GVB orbitals for bonds to Ti

Covalent 2 electron TiH bond in Cl2TiH2

Covalent 2 electron CH bond in CH4

Ti ds character, 1 elect H 1s character, 1 elect

Csp3 character 1 elect H 1s character, 1 elect

Think of as bond from Tidz2 to H1s


But TM-H bond can also be s-like

Cl2TiH+

ClMnH

Ti (4s)2(3d)2

The 2 Cl pull off 2 e from Ti, leaving a d1

configuration

Mn (4s)2(3d)5

The Cl pulls off 1 e from Mn, leaving a d5s1

configurationH bonds to 4s because of exchange stabilization of d5

Ti-H bond character1.07 Tid+0.22Tisp+0.71H

Mn-H bond character0.07 Mnd+0.71Mnsp+1.20H


Bond angle at a transition metal

For two p orbitals expect 90°, HH nonbond repulsion increases it

H-Ti-H plane

76°

Metallacycle plane

What angle do two d orbitals want


Best bond angle for 2 pure Metal bonds using d orbitals

Assume that the first bond has pure dz2 or ds character to a ligand along the z axis

Can we make a 2nd bond, also of pure ds character (rotationally symmetric about the z axis) to a ligand along some other axis, call it z.

For pure p systems, this leads to q = 90°

For pure d systems, this leads to q = 54.7° (or 125.3°), this is ½ the tetrahedral angle of 109.7 (also the magic spinning angle for solid state NMR).


Best bond angle for 2 pure Metal bonds using d orbitals

Problem: two electrons in atomic d orbitals with same spin lead to 5*4/2 = 10 states, which partition into a 3F state (7) and a 3P state (3), with 3F lower. This is because the electron repulsion between say a dxy and dx2-y2 is higher than between sasy dz2 and dxy

.

Best is ds with dd because the electrons are farthest apart

This favors q = 90°, but the bond to the dd orbital is not as good

Thus expect something between 53.7 and 90°

Seems that ~76° is often best


How predict character of Transition metal bonds?Start with ground state atomic configuration

Ti (4s)2(3d)2 or Mn (4s)2(3d)5

Consider that bonds to electronegative ligands (eg Cl or Cp) take electrons from 4s

easiest to ionize, also better overlap with Cl or Cp, also leads to less reduction in dd exchange

(4s)(3d)5(3d)2

Now make bond to less electronegative ligands, H or CH3

Use 4s if available, otherwise use d orbitals


But TM-H bond can also be s-like

Cl2TiH+

ClMnH

Ti (4s)2(3d)2

The 2 Cl pull off 2 e from Ti, leaving a d1

configuration

Mn (4s)2(3d)5

The Cl pulls off 1 e from Mn, leaving a d5s1

configurationH bonds to 4s because of exchange stabilization of d5

Ti-H bond character1.07 Tid+0.22Tisp+0.71H

Mn-H bond character0.07 Mnd+0.71Mnsp+1.20H


Example (Cl)2VH3

+ resonance configuration


Example ClMo-metallacycle butadiene


Example [Mn≡CH]2+


Summary: start with Mn+ s1d5

dy2 s bond to H1sdx2-x2 non bondingdyz p bond to CHdxz p bond to CHdxy non bonding4sp hybrid s bond to CH


Compare chemistry of column 10


Ground state of group 10 column

Pt: (5d)9(6s)1 3D ground statePt: (5d)10(6s)0 1S excited state at 11.0 kcal/molPt: (5d)8(6s)2 3F excited state at 14.7 kcal/mol

Pd: (5d)10(6s)0 1S ground statePd: (5d)9(6s)1 3D excited state at 21.9 kcal/molPd: (5d)8(6s)2 3F excited state at 77.9 kcal/mol

Ni: (5d)8(6s)2 3F ground stateNi: (5d)9(6s)1 3D excited state at 0.7 kcal/molNi: (5d)10(6s)0 1S excited state at 40.0 kcal/mol


Salient differences between Ni, Pd, Pt

2nd row (Pd): 4d much more stable than 5s è Pd d10 ground state

3rd row (Pt): 5d and 6s comparable stability è Pt d9s1 ground state


Salient differences between Ni, Pd, Pt

Ni Pd Pt

4s more stable than 3d 5s much less stable than 4d 6s, 5d similar stability3d much smaller than 4s(No 3d Pauli orthogonality)Huge e-e repulsion for d10

4d similar size to 5s (orthog to 3d,4s

Differential shielding favors n=4 over n=5,

stabilize 4d over 5s èd10

2nd row (Pd): 4d much more stable than 5s è Pd d10 ground state

3rd row (Pt): 5d and 6s comparable stability è Pt d9s1 ground state

Relativistic effects of 1s huge èdecreased KE ècontraction è stabilize and contract all ns èdestabilize and expand nd


Next section

48

Theoretical Studies of Oxidative Addition and Reductive Elimination: J. J. Low and W. A. Goddard III J. Am. Chem. Soc. 106, 6928 (1984) wag 190

Reductive Coupling of H-H, H-C, and C-C Bonds from Pd Complexes J. J. Low and W. A. Goddard III J. Am. Chem. Soc. 106, 8321 (1984) wag 191

Theoretical Studies of Oxidative Addition and Reductive Elimination. II. Reductive Coupling of H-H, H-C, and C-C Bonds from Pd and Pt Complexes J. J. Low and W. A. Goddard III Organometallics 5, 609 (1986) wag 206


Mysteries from experiments on oxidative addition and reductive elimination of CH and CC bonds on Pd and Pt

Why are Pd and Pt so different


Mysteries from experiments on oxidative addition and reductive elimination of CH and CC bonds on Pd and Pt

Why is CC coupling so much harder than CH coupling?


Step 1: examine GVB orbitals for (PH3)2Pt(CH3)


Stopped Apr 17

52


Analysis of GVB wavefunction


Alternative models for Pt centers


energetics

Not agree with experiment


Possible explanation: kinetics


Consider reductive elimination of HH, CH and CC from Pd

Conclusion: HH no barrier

CH modest barrierCC large barrier


Consider oxidative addition of HH, CH, and CC to Pt

Conclusion: HH no barrier

CH modest barrierCC large barrier


Summary of barriers

But why?

This explains why CC coupling not occur for Pt while CH and HHcoupling is fast


How estimate the size of barriers (without calculations)


Examine HH coupling at transition state

Can simultaneously get good overlap of H with Pd sd hybrid and with the other H

Thus get resonance stabilization of TS è low barrier


Examine CC coupling at transition state

Can orient the CH3 to obtain good overlap with Pd sd hybrid OR can orient the CH3 to obtain get good overlap with the other CH3But CANNOT DO BOTH SIMULTANEOUSLY, thus do NOT get

resonance stabilization of TS è high barier


Examine CH coupling at transition state

H can overlap both CH3 and Pd

sd hybrid simultaneously but CH3 cannot

thus get ~ ½ resonance

stabilization of TS


Now we understand Pt chemistry

But what about Pd?

Why are Pt and Pd so dramatically different


Pt goes from s1d9 to d10 upon reductive eliminationthus product stability is DECREASED by 12 kcal/mol

Using numbers from QM


Ground state configurations for column 10

Ni Pd Pt


Pd goes from s1d9 to d10 upon reductive eliminationthus product stability is INCREASED by 20 kcal/mol

Using numbers from QM

Pd and Pt would be ~ same


Thus reductive elimination from Pd is stabilized by an extra 32 kcal/mol than for Pt due to the ATOMIC nature of the states

The dramatic stabilization of the product by 35 kcal/mol reduces the barrier from ~ 41 (Pt) to ~ 10 (Pd)

This converts a forbidden reaction to allowed


Summary energetics

Conclusion the atomic character of the metal can

control the chemistry

Lecture6,April 17,2019 From CC bonds –Transition metals CC …...e= 95.0 + 2*7.3 = 109.6 kcal/mol...

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Transcript of Lecture6,April 17,2019 From CC bonds –Transition metals CC …...e= 95.0 + 2*7.3 = 109.6 kcal/mol...