Harvard Chem 106 Lecture 1

Chem 106E. Kwan Lecture 1: Bonding and Hybridization

multielectronatoms the variational

principle

August 31, 2011.

Bonding and Hybridization

Scope of Lecture

Eugene E. Kwan

*

Helpful References1. Coulson's Valence, 3rd ed. McWeeny, R. Oxford: Oxford University Press, 1979.

2. Quantum Chemistry, 5th ed. Levine, I.N. Upper Saddle River, New Jersey: Prentice-Hall, Inc., 2000.

3. Valency and Bonding: A Natural Bond Orbital Donor-Acceptor Perspective, Cambridge, Cambridge University Press, 2005.

LCAOmethod

basis functionsvs. orbitals

I thank Professor Frank Weinhold (Wisconsin) for manyhelpful discussions in the preparation of this lecture.

bonding andhybridization

equivalent vs.inequivalenthybridization

shape oflone pairs

the virialtheorem

Course Details:

Instructor: Dr. Eugene Kwan ([email protected])TF: Dr. Christian Gampe ([email protected])

Textbook/Readings: Ref. 3 and online notes

Website: http://isites.harvard.edu/icb/icb.do?keyword=k82275 (lecture videos available)

Problem Sets: five questions every friday

Grading Scheme:

problem sets (10%) 2x midterms (20%) writing assignment (20%) final exam (30%)

Prerequisite: Chem 30 or equivalent (undergrads welcome)

Overview of Topics:

(1) Covalent vs. Non-covalent Bonding(2) Pericyclic Reactions(3) Conformational Analysis(4) Reaction Mechanisms (kinetics, computations, isotopes...)(5) Reactive Intermediates (cations, anions, radicals, carbenes)(6) Catalysis (metal, secondary amine, hydrogen-bond, NHC)

Goals:

By the end of the course, I would like you to be able to think andwrite intelligently about any organic reaction.

Chem 106E. Kwan Lecture 1: Bonding and HybridizationHydrogenlike AtomsLet us start at the beginning, with the hydrogen atom that you(hopefully) have come to know and love. The hydrogen atomhas one proton and one electron. In hydrogenlike atoms, thereare Z protons and one electron:

+Z protonsone electron

This includes H,He+, Li2+, etc.

r

Here, we imagine the nucleus is fixed at the origin. Inspherical coordinates, the position of the electron is (r, , ).Here are the key facts about the hydrogenlike atoms:

(1) Radial Distribution Functions

Q: Where is the electron?A: Don't ask.

The electron isn't anywhere exactly, any more than the soundwave from a violin is anywhere specific. More precisely, ifwe prepare a bunch of hydrogen atoms in exactly the sameway, and then we measure the position of the electron, thenwe will not get the same answer every time. Rather, we willget some distribution of answers. This is a probabilitydistribution.

To make this more concrete, here is an example. One of thestates you can prepare the hydrogen atom in is the "1s state,"which is its lowest energy, or ground, state. The "radialprobability distribution" plots the probability of finding theelectron in a spherical shells extending out from the nucleus.

1s radial distribution

In this state, we will find the electron at a radius of <r> onaverage (but not every time).

In quantum mechanics, we postulate that there is an abstractmathematical object called the wavefunction, which, whensquared, gives the probability distribution (from which one can calculate the radial distribution function);

||2 = probability density

(2) Wavefunctions

Different systems have different wavefunctions. If the systemis in a stationary state, then its probability density does notchange in time. The 1s state of the hydrogen atom is such astate. This means that if I prepare a lot of identical 1s H atomsand wait a while before measuring the position of theirelectrons, then the probability distribution I get doesn't dependon how long I wait. If the H atoms are in a superposition ofthese stationary states, then the distribution will change with time in a complicated way.

So what are these stationary states?

energies: E 1 / n2

principal quantum number: n = 1, 2, 3...

angular momentum quantum number: l = 0, 1, 2, ..., n-1l=0 (s); l=1 (p); l=2 (d); l=3 (f); ...

magnetic quantum number: m = -l, -l+1, ..., 0, ..., l-1, l- essentially give different orientations of similar orbitals

- for example, for n=2, there are three possible solutions:

(n, l, m)

(2, 0, 0) = 2s

(2, 1, -1), (2, 1, 0), (2, 1, 1):linear combinations are the 2px, 2py, and 2pz

E

En = -13.6 eV / n2

n=1-13.6 eV

r0

all positive energies allowed

1,0,01s

n,l,m

n=22,0,0

2s2,1,-1 2,1,0 2,1,1

2p

n=3..........n=4 and up...........

3s 3p 3d

Chem 106E. Kwan Lecture 1: Bonding and HybridizationHydrogenlike Atoms(3) Quantum Numbers

As it turns out, these stationary states can't be just anything. Infact, they are labelled by three quantum numbers, which haveto be integers. Because these numbers can only take oncertain discrete values, the stationary states are said to bequantized.

(4) Energies

Notice that the energy depends only on n. That means the 2sand 2p wavefunctions (any one electron wavefunction is calledan orbital) are of the same energy. The 3s, 3p, and 3d statesare also of the same energy (but don't have the same energyas the 2s/2p).

(for Z=1)

(n, l, m)

Actually, not all the energies are quantized. Zero energy isdefined as the electron and the proton at an infinite distancewhere they don't interact. If the electron has more than zeroenergy, then you can interpret that as an ionized atom: theelectron has escaped the potential of the hydrogen atom, andis free to be anywhere it wants, with any energy (more or less).

Notice that the levels get closer and closer together as nincreases, but the s/p/d/f energies remain degenerate oneach level. This is only true for hydrogenlike atoms.

credit: The OrbitronUniv. of Sheffield

Chem 106E. Kwan Lecture 1: Bonding and HybridizationHydrogenlike Atoms(5) Orthogonality

A key idea is that of orthogonality. The Pauli Exclusion Principlesays that no two electrons can occupy the same state. Moreprecisely, the electrons have to be in orthogonal orbitals, whichmeans their overlap integral

has to be zero. For example, 2s orbitals are orthogonal to2p ones:

Even though the 2s and 2p orbitals occupy roughly the samespace, in that there's a reasonable probability of finding theelectron within a common set of coordinates, they are stillorthogonal because the 2p orbital has an angular node wherethe wavefunction changes sign.

Although the hydrogenlike atom only has one electron, theidea is that atoms with more electrons will look similar, withelectrons piling on top of each other in successively higherenergy levels.

So how do the 1s and 2s orbitals remain orthogonal? This time,there is a radial node in the 2s orbital:

12 1 2S d

Thus, even though the 1s and 2s orbitals "occupy the samespace," they still have zero overlap:

In general, there are:

n-1 total nodes,l angular nodes (m in the xy-plane; l-m on the z-axis), andn-l-1 radial nodes.

Multielectron Atoms

2s 2p

Once there is more than one electron, interelectronicrepulsions makes energy depend on both n and l:

2s

2p

hydrogen(Z=1)

helium (Z=2)

The ordering of the levels is very complicated, as shown onthe following page. Nonetheless, theory and experimentagree very well on the energies of atoms.

Chem 106E. Kwan Lecture 1: Bonding and HybridizationMultielectron AtomsThis chart (Levine, pg 312-315) shows the intricateordering of energies. Eventually, if the atomic numberZ gets high enough, the energies once again onlydepend on n (and not l) as nucleus-electron attractionsoverwhelm electron-electron repulsions. These energiesare for neutral, isolated atoms only; "real" energies andorbital shapes depend on the specific chemicalenvironment of the atom.

1 Hydrogen H 2 Helium He 3 Lithium Li 4 Beryllium Be 5 Boron B 6 Carbon C 7 Nitrogen N 8 Oxygen O 9 Fluorine F 10 Neon Ne 11 Sodium Na 12 Magnesium Mg 13 Aluminium Al 14 Silicon Si 15 Phosphorus P 16 Sulfur S 17 Chlorine Cl 18 Argon Ar 19 Potassium K 20 Calcium Ca

21 Scandium Sc 22 Titanium Ti 23 Vanadium V 24 Chromium Cr 25 Manganese Mn 26 Iron Fe 27 Cobalt Co 28 Nickel Ni 29 Copper Cu 30 Zinc Zn 31 Gallium Ga 32 Germanium Ge 33 Arsenic As 34 Selenium Se 35 Bromine Br 36 Krypton Kr 37 Rubidium Rb 38 Strontium Sr 39 Yttrium Y 40 Zirconium Zr

atomic number

hydrogen atom

at Z=7 (N), the4s becomes lowerin energy than the 3d

Chem 106E. Kwan Lecture 1: Bonding and HybridizationThe Hydrogen Molecule

H H2 e

Proof: Levine, pg 208-209.

The hydrogen molecule has two protons and two electrons:

What we want is to obtain the energy levels and wavefunctionsfor the hydrogen molecule.

Unfortunately, the interelectronic repulsions make it impossibleto do this analytically, so we need to resort to approximatemethods. Despite the fact that they are not "analytic" in nature,they are still good enough to describe chemistry!

The Variational Theorem

If a system has a ground state energy of E1, and is anormalized, well-behaved function of the system's, then

E1 is no higher than the variational integral W:

1*H d E W =

a "guess" for the wavefunction; has to be"reasonable" (satisfy the boundary conditionsof the problem being considered)

* the complex conjugate of ( need not be real)

the Hamiltonian operator, which you can think ofas a widget which gives the energy of the guess H

d a notation that means "integrate over all space"

W the variational energy

(Note that has nothing to do with the spherical coordinate.)

So, for any guess , the ground state energy of the system E1is no higher than W. This suggests we should try to vary , andseek a form of it that minimizes W:

= c11 + c221 = 1s orbital on hydrogen A2 = 1s orbital on hydrogen B

H H

A B

Let's try this for the hydrogen molecule. A reasonable guessfor might be 1s orbitals of the hydrogen atom. One mightguess that the wavefunctions are linear combinations of theform:

Exact solutionunknown.

Make a reasonableguess,

The real ground state energyis below the value of the integral.

Translation:

Compute theintegral W.

Refine andtry again.

The Linear Combination of Atomic Orbitals (LCAO)

Thus, we seek values of c1 and c2 that minimize W. Moreprecisely:

This is advantageous because it's hard to vary a function,but it's easy to vary the coefficients c1 and c2 in a linearcombination.

1

0Wc

2

0Wc

Chem 106E. Kwan Lecture 1: Bonding and HybridizationThe Secular EquationsAs it turns out, if you vary c1 and c2 to minimize W, then c1 andc2 have to satisfy the secular equations:

For details, please see Levine, 220-223.

*12 1 2 1 2H H H d where Hamiltonian

matrix element/resonance integral

1 11 11 2 12 12

1 21 21 2 22 22

0

0

c H S W c H S W

c H S W c H S W

*12 1 2 1 2S d overlap

integral

Note: ij jiS S*

ij jiH H

(overlap of i with j = overlap of j with i)

(the Hamiltonian is Hermitian)

1 2

energy

H11

W1

W2

(The factors N are to normalize the wavefunctions. Thisamounts to saying that the electron has a 100% chance ofbeing somewhere.)

For the moment, don't worry too much about theseequations.

Instead, let us return to the hydrogen atom. A feature of thesolution to these equations is that if n atomic orbitals i go intothe calculation, then n molecular orbitals j come out. Byconvention, the molecular orbital j has energy Ej. For example,the ground state wavefunction is 1 and has energy E1.

This is the idea behind the Hartree-Fock method.

If this seems too abstract, let's try the hydrogen moleculeas an example. Here is the energy diagram:

ground state,bonding

first excited state,anti-bonding

1sorbital

(1) We are not saying that each hydrogen atom has a 1s orbital on it. Clearly, it doesn't, since a 1s orbital is what an isolated hydrogen atom looks like, not a hydrogen in the hydrogen molecule. Rather, we are just using the 1s orbitals in the hydrogen atom as a convenient guess.

(2) [A technical, perhaps controversial detail]: In this diagram, it looks like the anti-bonding level goes up by more than the bonding orbital goes down. It turns out this is true only if S12 > 0. But for orthogonal orbitals, S12=0!


What about something like hydrogen fluoride (HF), wherethe atoms are of different electronegativities? As you can seefrom slide 8, the energy of the 1s orbital declines as Zincreases. This isn't too surprising, since, there are moreprotons in the nucleus to attract the 1s electrons, which lowersthem in energy. Thus, we get a diagram that looks like:

Unequal Energy Levels

1

2

c11c21

energy

H11

W1

W2

H22

c12c22

more electronegativeatom, lower in energy

lesselectronegative

atom

This time, we need four coefficients to describe the molecularorbitals. For example, c12 means the coefficient of atomicorbital 1 in molecular orbital 2.

Basis Sets A natural question is: what do people use for these atomicorbitals i in actual computations? The set {i} is called thebasis set: the collection of all the atomic orbitals. A reallycommon basis set type was developed by Boys and Pople.Instead of using hydrogenlike orbitals as above, they useGaussian functions of the Cartesian form:

2expi j kijkg Nx y z r

normalizationconstant a Gaussian

indicatesangular

momentum

i + j + k = 0: s-type Gaussiani + j + k = 1: p-type Gaussiani + j + k = 2: d-type Gaussian

Each gijk is a primitive Gaussian. Special linear combinationsof primitives are selected to look like hydrogenlike orbitals.These are called contracted Gaussian-type orbitals (CGTO).These orbitals are used because integrating Gaussians isvery easy computationally.

The 6-31g(d) Basis SetThis is used in many, many computational papers (it's alsocalled 6-31G*). Let's look at what's inside it. Each atom hasdifferent basis orbitals associated with it. For example, eachhydrogen in a molecule gets two 1s-type CGTOS:


hydrogens:

carbons:

2x1s

2x2s 6x2p 5x3d

Carbons, however, have a lot more electrons, and so needmany more basis orbitals:

2x1s

CORE VALENCE

But wait! How come carbon has "d orbitals" on it? All sortsof chemical evidence suggests that carbon does not use 3dorbitals for bonding. So if that's the case, then should weexpect to find that adding 3d "polarization" functions doeslittle to improve the variational energy?

Let's try it for acetone. Here are the results of a Hartree-Fockcalculation on its equilibrium geometry with different basissets. The "*" notation denotes the presence of d orbitals;as you go down the list, bigger basis sets (with correspondingly larger and larger shells of atomic orbitals) are shown.

wavefunction.org

Clearly, the polarization functions are necessary!

Does this mean that the carbon in acetone is using d orbitals?No, because the GTOs are not real orbitals.

In fact, if you try to ask "how many electrons are in GTO X,"you can actually get numbers that are less than 0 or greaterthan 2!

So what's the problem?

Basis Set Description

Number of Functions

Rel. Energy(kcal/mol)

STO-3G 26 1577.03-21G 48 728.76-31G 48 108.96-31G* 72 54.76-311G* 90 28.86-311+G* 106 26.26-311++G** 130 20.46-311++G(2df,2pd) 226 11.46-311++G(3df,3pd) 264 10.1cc-pVTZ 204 9.3cc-pVQZ 400 0.7aug-cc-pCVQZ 712 0.0

Physical Issues with CGTOsNo Nodes

According to the Pauli Principle, two electrons of the samespin cannot occupy the same space.

"1s" and "2s" CGTOs for the neutral lithium atom:

The real 1s and 2s orbitals (again for a neutral Li atom)do have nodes (for the precise meaning of "real," see below):

The CGTOs have no nodes, which is unphysical.

Chem 106E. Kwan Lecture 1: Bonding and HybridizationConstant Orbital Size

CGTOs: for the same element, the same size, regardless of nuclear environment

reality: more positively charged atoms attract their electrons more, and have larger orbitals

2s (Li anion)

2s (neutral Li) 2s (Li cation)

Chem 106E. Kwan Lecture 1: Bonding and HybridizationSymmetry and Molecular Orbitals

In general, there will be an interaction between any twoorbitals with a non-zero Hamiltonian off-diagonal element:

i interacts with j if Hij is nonzero

an integral

high symmetryenvironment

many integralsare zero

reduce amountwork

grouptheory

A criticism: Just because Hij can be nonzero, does not meanthat it actually is nonzero. The requirement for symmetryadaptation can lead to spurious results. Consider H2:

1 2

energy

H11

W1

W2

Regardless of the distance between the hydrogen atoms,this symmetry-adapted representation applies.

H H

bonding and antibondingMOs are reasonable

descriptors

r = 1 Å

H H

these are isolated H atoms,not bonded to each other

r = 1 km

HOMO

LUMO

In the low-symmetry environment of organic chemistry,this approach is less useful. With fast modern computers,we are free to consider all possible combinations of Hij andlet the form of H take care of the symmetry automatically.

Another criticism:Methane and ethane are generally thought of as quite similarchemically, but their HOMOs look quite different:

So MOs are not always transferable from one molecule toanother. How can we recover the idea of a chemical bondfrom this approach?

Chem 106E. Kwan Lecture 1: Bonding and HybridizationDelocalized vs. Localized Orbitals

The NBO Method

PNAOs(intraatomically

orthogonal)

NAOs(intra- and inter-

atomically orthogonal

One question you might ask is: what do the atomic orbitalslook like in molecules? The NBO method does this in twosteps:

atomic orbitals(CGTOs or

other basis set)

calculation method(Hückel/by hand,

Hartree-Fock, DFT, etc.)

canonical molecularorbitals (CMOs)

(expressed as a LCAOin the GTO basis)

input

output

The results of calculations are accurate, in that they predictmolecular gemetries, vibrational frequencies, reaction barrierheights, etc. So chemically useful information is containedwithin the wavefunctions {i,} at least when considered as awhole.

But to talk about bonds, we need to ask questions about theproperties of one-electron subsystems of {i,}.

One approach that seems to accomplish this in a sensibleway is the natural bond orbital (NBO) analysis methoddeveloped by Weinhold and coworkers. The idea is to convertthe unphysical CGTO basis into one which corresponds toregular chemical ideas of bonds, lone pairs, and so forth. Thedetails of how this is done are complicated and are well beyondthe scope of this course. Suffice it to say, we seek new basisorbitals which are maximally occupied; i.e., have as closeto two electrons as possible. The NBO algorithimgenerates orthogonal orbitals which satisfy the PauliPrinciple.

For details, see: Weinhold, F. Encyclopedia of ComputationalChemistry, 1998, 1792-1811.

Natural Atomic Orbitals

We take the wavefunction, and search for new basis orbitals,each of which is centered on one atom ("one-center orbitals")which are maximally occupied. This generates pre-orthogonalnatural atomic orbitals (PNAOs):

CGTO basis PNAO basis

1s and 2soverlap

1s and 2sorthogonal

Here, the 1s- and 2s-type orbitals in the CGTO basis areshown. They are clearly not orthogonal! In the PNAObasis, they are properly orthogonal and satisfy the Pauliprinciple.

Reasoning: Can’t have one core pair of electrons occupying the same space as another, so nodes required.

2s PNAOon C

2s NAOon C

1s PNAOon H

C H

CH

1s PNAOon H 3.000.78 2 2 0.62 1CH C C Hs p s

NHO basis NBO basis

Thus, in a second step, these orbitals are converted to a fullyintra- and inter-atomically orthogonal basis. This is thenatural atomic orbital (NAO) basis. The increased curvatureand nodal behavior of these NAOs reflects the importanteffects of atomic confinement.

However, there is no orthogonality between electrons ondifferent atoms in PNAOs. For example, in methane, the2s core electrons on carbon and the 1s electrons on hydrogenclearly overlap:

Chem 106E. Kwan Lecture 1: Bonding and HybridizationNatural Atomic Orbitals

HC

H HH

methanemolecule

localized (C-H)NBOs

H3C H

sp3 (carbon) + s (hydrogen)

= +

Example (Methane):

NAOs

Here's what it looks like:

Natural Bond Orbitals

2-center basis functions: bonds and antibonds

1-center basis functions: lone pairs and core electrons

If we want to talk about bonds, we search for one- and two-center orbitals of maximum occupancy instead:

Each NBO bond and antibond can be thought of as beingcomposed of two natural hybrid orbitals (NHOs).(The span of the basis is completed by largely unoccupiedRydberg orbitals. For example, 3d orbitals on carbon.)

linearcombination

of NHOs1s 2p

Chem 106E. Kwan Lecture 1: Bonding and HybridizationReview: Conceptual Framework

http://www.chem.wisc.edu/~nbo5/

natural atomic orbitals (NAOs)1-center orbitals

(1s, 2s, 2p...)

standard global approach:

The standard approach is to use computations to predict bondlengths, transition state energies, vibrational frequencies, etc.These calculations correlate well with experimentalobservations. Therefore, it is reasonable to think that otheruseful information might be contained in the wavefunctionsproduced by these calculations. However, the canonicalmolecular orbitals (CMOs) are relatively delocalizedentities that are hard to interpret chemically.

new localized approach:

If we want to talk about "normal" chemical ideas, like bonds orlone pairs, then we need to ask questions about localized one-electron subsystems. The natural bond orbital (NBO) approachis a very reasonable way to do this.

The NBO Hierarchy

Example: Hydrogen Molecule

HB HA

energy

1s 1s

bond

* antibond

NBO picture: and * bonds arelinear combinations

of natural atomic orbitals

MO picture:bonds are "delocalizedblobs": accurate, but

confusing

natural bond orbitals (NBOs)1 or 2-center orbitals

(bonds, antibonds, lone pairs)

natural hybrid orbitals (NHOs)linear combinations of NAOs(eg., sp3 hybrids on carbon)

canonical molecular orbitals (CMOs)fully delocalized orbitals

(HOMO, LUMO)

The NBO algorithim converts the wavefunctions/CMOs intoother basis sets that are mathematically equivalent (span thesame space), but easier to understand:

contracted Gaussian-type orbitals (CGTOs)linear combinations of Gaussian chosento look like orbitals in the hydrogen atom;

no physical significance

The various natural orbitals are constructed such that everyoccupied orbital has as close to two electorns as possible("maximum occupancy").

Chem 106E. Kwan Lecture 1: Bonding and HybridizationOrthogonality

The Pauli Principle requires that no two electrons can occupythe same state, and therefore, any reasonable orbitals mustbe orthogonal. The NAOs, NHOs, and NBOs are allorthogonal. But orthogonal means overlap is zero!

However, we can recover the idea of "overlap" but consideringthe "pre-orthogonal" versions of each:

PNBOs: overlap corresponds to donor-acceptor interactions (e.g., hyperconjugation from a lone pair into an antibond)

PNHOs/PNAOs: overlap corresponds to strength of chemical bonding

Q: How do you think about the difference between thepre-orthogonal orbitals and the fully orthogonal ones?

A: Here is an approximate picture. Imagine having an electronin a molecule where the nuclei are frozen in place. Now takeaway all the other electrons. The one remaining electron sitsin orbitals whose shapes reflect the chemical environment ofthe nuclei around it. However, these orbitals don't have to haveas many nodes as they would if other electrons were around(because of the Pauli principle).

Thus, it not surprising that NAOs don't look like hydrogenlikeorbitals! (Electrons in molecules are largely under the influenceof the local nuclear potential, so their orbitals look kind ofhydrogenlike, but are also influenced by the other electronsaround them.)

NAO/NBO Terminology

e.g. neon

1s

2s

2p

3sRydberg orbitals (RY*):empty, and of littlephysical significance

By convention, orbitals are given abbreviations to indicate theirtype. Asterisks (*) mean the orbital is mostly empty. You canthink of these unoccupied orbitals, such as antibonds, as amolecule's capacity to respond to change, like irradiation bylight or the approach of a reagent.

valence orbitals (VAL)these are the onesinvolved in bonding

lone pairs (LP):one-center valenceelectrons (none here)

core electrons (CR)

e.g. methane

1s

C-H

3s

C-H

RY*

antibonds (BD*)

bonds (BD)

CR


Example: Hydrogen Molecule

HB HA

energy

net stabilization

1s 1s

bond

* antibond

Let's go back to the hydrogen molecule:

LCAO-MO: AB2 = 1sA

+ 1sB (fully occupied)

*AB0 = 1sA

0 - 1sB0 (fully unoccupied)

NBO/NHO: In a more sophisticated treatment, we can considera combination of hybrids (NHOs):

hA = (1 + 2)-1/2 (1sA + 2pA) sp hybridizedhB = (1 + 2)-1/2 (1sA + 2pA)

is the hybridization parameter; it tells you how muchp-character is in the hybrid.

One finds that is very small, which means p-orbitals are hardlybeing used here (not really a surprise).

Q: What is the driving force for covalent bond formation?A: Consider what hydrogen looks like when the atoms are very far apart:

This is essentially a diradical. Each hydrogen atom hasan either 100% or spin electron on it. However, asthey get closer together, the 1sA() electron starts todelocalize into the empty 1sB(a) spin-orbital:

HAHB

1sA() 1sA() 1sB() 1sB()

HAHB

1sA() 1sA() 1sB() 1sB()

By the time the H2 equilibrium distance of 1.4 A isreached, the orbital populations are equalized andthe diradical character is entirely quenched:

HAHB

1sA() 1sA() 1sB() 1sB()

There is still only one electron on each hydrogen, buteach has half and half character.

two localized radicals

a sigma bond

Why is dihydrogen stable?Spin-Orbital Occupancy in Dihydrogen

This qualitative analysis is borne out in a natural populationanalysis. Thus, when two moieties interact in an equal,bidirectional way, strong bonding results.

In the NBO picture, we can write these complementarydonor-acceptor interactions as:

AB = (1 + 2)-1/2 (hA + hB)

AB = (1 + '2)-1/2 (hB + ' hA)

where is a parameter which measures the strength of thebi-directional donor-acceptor interaction.

Interpretation: Nodes and the Virial Theorem

At their core, all bonds are electrostatic in nature, withinternuclear and interelectronic Coulombic repulsions (as wellas interelectronic exchange repulsions) preventingcomplete collapse to the united atom limit. But why is thein-phase bonding mode more stable than the out of phaseantibonding mode?

One explanation is that the antibond incurs a node,which increases the kinetic energy through the secondderivative of the wavefunction:

HB HA

energy

net stabilization

1s 1s

bond

* antibond

2 2

2

( ) ( ) ( ) ( )2

d x V x x E xm dx

kineticenergy

It is therefore reasonable to ask whether the stabilization ofbonding in diatomic molecules is purely related to kineticenergy. Using the virial theorem (see Levine, pg 466-469),one can draw some general conclusions about the depend-ence of kinetic energy (T), potential energy (V), and overallenergy (U) on internuclear separation.


Common Misconceptions About Covalent Bonding

E = <Tel> + <V>U: total energy<Tel>: mean kinetic energy of e-

<V>: mean potential energy

2 dEV E RdR

0eqR

dEdR


Unfortunately, many textbooks get it wrong.

Rioux, F. "The Covalent Bond Examined Using the VirialTheorem." Chem. Educator 2003, 8, 10-12.

Here are some common misconcpetions:

(1) the covalent bond is purely electrostatic (the total energy is a sum of both potential (electrostatic) and kinetic energy, so you can never understand bonding from just an electrostatic perspective

(2) electrons go between the atoms because the potential energy is low there (as they are attracted to both positively charged nuclei equally) (the electron-nuclear nuclear potential energy is actualy higher in the internuclear region; on the basis of electrostatics alone, the electrons would rather be in the nucleus)

(3) electrons occupy the region between the nuclei (the amount of electron density transfered to the bonding region is not as much as usually drawn)

The Virial Intepretation (Diatomics)Consider a diatomic molecule joined by a single covalent bond:

A B2 e

The system has some total energy E which varies as a functionof internuclear separation (R). (This can be approximated bythe Morse function. E has potential and kinetic parts:

Using the virial theorem, we can relate these quantities:

For a bond to exist, E(R) must reach a substantial minimum:

R: distance between A and B

One finds that the U, T, and V curves for dihydrogen look like:

Let us dissect this graph in detail. (Figure copied from Riouxreference.)

(1) the energy at infinite distance is defined as zero

(4) if you try to push thenuclei very close together, the two positive charges repel each other a lot

(2) the initial overlap of orbitals actually increases the potential energy because the electrons are getting farther away from the positive charge on the nuclei

(5) the kinetic energy decreases initially because the electrons occupy a bigger volume of space than they do in the isolated atoms (K.E. is related to the second derivative of position)

(6) eventually, the K.E. goes up because the electrons are getting too close together, and have to form nodes to remain orthogonal

(3) the potential energy is going down here because the electrons are getting closer to the nuclei

Req

(7)

(7) This is the equilibrium bonddistance, which reflects the tradeoff between V and T. The immediate cause of the well is the increase in T, not V. The delocalization of electron densityhere is only 16%!

These are curves for dihydrogen (H2).

Chem 106E. Kwan Lecture 1: Bonding and HybridizationThe Virial Intepretation (Diatomics)

Bonding in Formaldehyde

Chem 106E. Kwan, D.A. Evans Lecture 1: Bonding and Hybridization

"Ultraviolet Spectra and Excited States of Formaldehyde."Moule, D.C.; Walsh, A.D. Chem. Rev. 1975, 75, 67-84.

In virtually every textbook, formaldehyde is represented assomething like:

H H

O

H H

O

This implies that the lone pairs on oxygen are equivalent.However, photoelectron spectroscopy shows they areactually different in energy!

Bonding AnalysisLet us place the formaldehyde molecule along the y-axisas shown here:

x

y

z

C O

atomic orbitals on oxygen

(1) Formaldehyde has one -bond, one -bond, and two lone pairs in its valence shell.

(2) Oxygen has 2s, 2px, 2py, and 2pz atomic orbitals on it.

(3) Orbitals only hybridize in response to bonding interactions.

If we don't hybridize the orbitals, then simplistically, we candraw this diagram (let us ignore the orbitals on carbon):

y

z

C O

- there is no additional ligand in the y-direction either, so the py is non-bonding as well

y

z

C O

- use a 2pz orbital on oxygen to form a -bond with carbon

y

z

C O

- there is no ligand in the x-direction, so the px is non-bonding

x

x

x

y

z

C O

x

- use the 2s orbital to form a -bond with carbon

s-richlone pair: 60% s

exclusivelyp lone pair 0% s

Bonding in Formaldehyde

Chem 106E. Kwan, D.A. Evans Lecture 1: Bonding and Hybridization

y

z

C O

x

- oxygen directs an spy-hybrid towards carbon to form a better -bond

However, mixing the 2s and 2py orbitals can result in the energyof the -bond going down. This is s-p hybridization. The energygoes down because the s-orbitals are not very directional innature, and adding some p-character allows them to point inthe direction of the carbon more efficiently:

y

z

C O

x

- to conserve the total number of orbitals, the lone pair must now also have some s-character

(1) the spy hybrid on oxygen in the negative y-direction is used for a s-bond; the spy hybrid on oxygen in the positive y-direction is nonbonding--this is a lone pair

(2) s-p hybridization will not lower the energy of the -bond, so oxygen uses a pure pz hybrid for that

(3) there is no ligand for px, so it remains purely unhybridized; this represents the other lone pair

(4) s orbitals are lower in energy than p orbitals

(5) the electrons in the C-O bond spend most, but not all, of their time on oxygen, while the electrons in the lone pair on the y-axis spend all of their time on oxygen

(6) therefore, it is advantangeous to emphasize low energy s-character in the hybrid directed towards the lone pair:

y

z

C O

x

y

z

C O

x

Thus, there is an s-rich lone pair, and and a purely px lone pair:

more p-character more s-character

This tendency to direct hybrids of increased s-charactertowards more electropositive substituents is known asBent's Rule.

O-B bond:uses an sp5.4 hybrid on oxygen (16% s)

O lone pair:is an sp1.3 hybrid on oxygen (56% s)

- the C=O-B bond angle is: 119.6°- the difference in s-character reflects Bent’s Rule- calculated at B3LYP/6-31g(d)

- the C=O-H bond angle is: 99°

- this lone pairis 100% p

- this lone pairis 59% s

Bonding in FormaldehydeChem 106E. Kwan, D.A. Evans Lecture 1: Bonding and Hybridization

Of course, this does not agree with the picture in organicchemistry textbooks. However, it is widely accepted in thephysical chemistry community.

An important follow-up question, then, is: how do carbonyls bindto Lewis acids?

Strong Lewis Acids

With a strong Lewis acid like BF3, the complex is roughly sp2

hybridized, and in fact, there is quite a strong penalty forhaving a linear geometry:

Weak Lewis Acids

Water, which is quite a weak Lewis acid, also prefers an sp2-like geometry, but the lone pairs remain largely unperturbed:

Summary:

Strong Carbonyl-Lewis Acid Complexes: one sp2 lone pairWeak Carbonyl-Lewis Acid Complexes: one pure p lone pair one s-rich lone pair

Other weak interactions like hyperconjugation fall into thesecond category.

Hybridization occurs to strengthen bondinginteractions.

100 110 120 130 140 150 160 170 1800

20

40

60

80

C=O-B angle (degrees)

C-C

=O-B

dih

edra

l ang

le (d

egre

es)

0.1

1.8

3.4

5.1

6.7

8.4

10.0

acetone-BF3 complex

5 2X/6 311 (d )

electronic energy(kcal/mol)

separated reactants:+17.4 kcal/mol

global min.: -0.4 kcal/mol angle: 124.3°

dihedral: 0.3°global minimum (acetone-BF3):

80 100 120 140 160 1800

20

40

60

80

C=O...H bond angle (degrees)C

-C=O

...H

dih

edra

l ang

le (d

egre

es)

0.1

1.8

3.4

5.1

6.7

8.4

10.0

electronic energy(kcal/mol)

acetone-MeOH complex

separated reactants:+7.4 kcal/mol

global min.: -0.11 kcal angle: 115.7° dihedral: 4.4°

global minimum(acetone-MeOH):

Bonding in AcetoneChem 106E. Kwan, D.A. Evans Lecture 1: Bonding and Hybridization

Calculations on acetone-BF3 and acetone-MeOH at M05-2X/6-311+g(d,p) confirm this finding (our unpublished work).

The graphs are to the same scale. The penalty for moving the hydrogen bond around the carbonyl group is very low compared tothe penalty for moving the Lewis acid. There is no evidence of VSEPR-type lone pairs for the weak interaction.

SummaryChem 106E. Kwan Lecture 1: Bonding and Hybridization

(1) Multielectron atoms - The orbitals must be orthogonal.

- s,p,d,f orbitals are equal in energy for hydrogen, but not for multielectron atoms.

(2) Molecules - LCAO method: use the variational theorem to put upper bounds on the molecular orbital energies based on a guess for the wavefunction

- the guess is a linear combination of orbitals; the component orbitals don't have to be orthogonal; the forms of these components are not generally meaningful

- bonding can be viewed as complementary donor-acceptor interactions; it is not purely electrostatic

(3) NBO Method - computations produce accurate energies, geometries, and wavefunctions - the molecular orbitals are not readily transferable between molecules

- NBO: rotate guess orbital basis to a new, fully orthogonal, maximally occupied basis set of natural orbitals

(4) Lone Pairs - orbitals only hybridize in response to bonding

- s-orbitals are lower in energy than p-orbitals, but will hybridize to acquire more directionality

- in strongly bound complexes, the orbitals in the isolated starting materials and the orbitals in the complex are not the same

natural atomic orbitals (NAOs)1-center orbitals

(1s, 2s, 2p...)

natural bond orbitals (NBOs)1 or 2-center orbitals

(bonds, antibonds, lone pairs)

natural hybrid orbitals (NHOs)linear combinations of NAOs(eg., sp3 hybrids on carbon)

canonical molecular orbitals (CMOs)fully delocalized orbitals

(HOMO, LUMO)

contracted Gaussian-type orbitals (CGTOs)linear combinations of Gaussian chosento look like orbitals in the hydrogen atom;

no physical significance

The NBO Hierarchy

Harvard Chem 106 Lecture 1

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