Why do bonds form?

E

r

Energy of two separate H atoms

Lennard-Jones potential energy diagram for the hydrogen molecule.

Forces involved:

We understand that the electrons are shared between the to H nuclei. How do we describe their positions? How do we describe electron position in atoms?

Chemical bonds...

... result from the sharing of electrons between atoms.

H H

electrons

protons

represented in a Lewis structure as:

The Schrödinger equation – the solution for the H atom

o Austrian physicist Erwin Schrödinger used de Broglie’s insight to calculate what electron waves might act like, using a branch of mathematics known as wave mechanics

o The result is encapsulated in a complex formula known as the Schrödinger equation:

o This equation was know to belong to a special class known as an eigenvector equation: an operator acts on a function (ψ) and generates a scalar times the same function

o Ψ is known as the wavefunction of the electron: there are an infinite number of such wavefunctions, each of which is characterized by a precise energy En, where n is an integer from 1,2,3,4….∞. V in the equation is a constant value: the potential energy from attraction to the nucleus. The remaining terms are all fundamental constants

o A wave function that satisfies the Schrödinger equation is often called an orbital. Orbitals are named for the orbits of the Bohr theory, but are fundamentally different entities

o An orbital is a wave function

o An orbital is a region of space in which an electron is most likely to be found

o The square of the amplitude of the wavefunction, ψ2, expresses the probability of finding the electron within a given region of space, which is called the electron density. Wavefunctions do not have a precise size, since they represent a distribution of possible locations of the electron, but like most distributions, they do have a maximum value.

2 2 2 2

2 2 2 2

8( )

mE V

x y z b

∂ ∂ ∂ π∂ ∂ ∂Ψ Ψ Ψ

+ + =− − Ψ

Atomic Orbitals

1s

2s

2pThe size and energy of orbitals depend on …

Orbitals are wavefunctions and have wave properties:

These 2s orbitals are the same in all respectsexcept that they have opposite phase.

This is a single 2p orbital. It has two lobes of different phase and a node (an angular node) at the nucleus.

We have a picture of where electrons are found around atoms: in atomic orbitals of various types and energies.

E

Atoms to Molecules

• Recall that the Schroedinger solution works for the H atom but becomes extremely complex for multi-electron atoms.• We begin then by examining the simplest possible molecule: H2+

Forces involved:

• This is a three-body problem which can be simplified by assuming that the motion of the nuclei is small compared to the motion of the electron (a fixed internuclear distance).• The result is an approximation of what one obtains by overlapping the atomic orbitals at the optimal bonding distance. This is the LCAO approach: Linear Combination of Atomic Orbitals.

What happens when two standing waves interact?

= =

In phase overlap:Out of phase overlap:

In orbital terms...

r2Ψ21s r2Ψ2

1s

two overlapping 1s H atomic orbitals

• In H2+, the electron is found in an orbital that is the sum of the overlap of the atomic orbitals • The electron is distributed equally between both nuclei• This is, therefore, a molecular orbital (MO) – an orbital for the H2+ molecule• This type of MO has radial symmetry and electron density directly between the two nuclei. These are classified as σ bonding orbitals.

MOs

• are orbitals just like atomic orbitals• can span the entire molecule• can house up to two electrons

But if you can form MOs by combining AOs, then electron occupancy must be conserved. Two AOs can hold four electrons, so their combination must also hold four electrons. There must be a second MO… In all cases, the number of MOs that exist for a molecule must equal the number of AOs in its constituent atoms.

Out of phase:

=

σ1s bonding MO

This orbital has low electron density between the two nuclei and destabilizes the attraction between them. This is referred to as …

Electron energies…Orbital occupancy…Orbital labels…MO electron configuration…

We use Correlation Diagrams (MO Diagrams) to show the MO electron configuration and show from which AOs the MOs are derived.

AOs appear on the outer parts of the diagram at their corresponding energy level…… and MOs in the middle.

The MO diagram for the H2+ cation…

Bond order = ½(#bonding electrons - #antibonding electrons)

The Hydrogen Molecule – all that changes is the number of electrons…

MO electron configuration: (1σ1s)2

300 pm

250 pm220 pm

200 pm150 pm

100 pm

73 pm

We construct a picture of the electronic structure of molecules using correlation diagrams, which indicate the energy and type of molecular orbitals that are occupied and vacant.

E

1s 1s

1σ1s

2σ∗1s

The AOs appear on the left (and for diatomics, right) side of the diagram at their corresponding energy level and the MOs that result from the AOs are in the middle at their corresponding energy level.

An appropriate no. of valence electrons are added to the diagram following Hund's Rule and the Pauli exclusion principle to determine which orbitals are occupied.

Irradiation of H2

Ground state BO:

hυ

Excited state BO:

1s 1s

1σ1s

2σ∗1s

1s 1s

1σ1s

2σ∗1s

First Row Diatomics

1s 1s

1σ1s

2σ∗1s

1s 1s

1σ1s

2σ∗1s

1s 1s

1σ1s

2σ∗1s

1s 1s

1σ1s

2σ∗1s

Second Row Homonuclear Diatomics: Li2 – Ne2

We now have access to four new orbitals: 2s, 2px, 2py and 2pz. These are the valence AOs.

The 1s orbitals are too small to overlap at the optimal bond distances so they can be ignored.

Li Li1s 1s

σ2s

Constructing MO Diagrams … for the second row diatomics.

Principles:1) We have four valence AOs per atom, therefore there must be ___________ MOs.

2) Only orbitals of compatible symmetry (both σ or both π) can be combined.

OK

Not OK

3) The energy of an MO is closely related to the energy of the AOs from which it is derived.4) The best (i.e. most energetically favourable) MOs are formed from AOs of similar

energy.

The 2s AOs are lowest in energy and equal in energy…

2s 2s

2σ

2σ*

2s 2s

1σ2s

2σ∗2s

The 2pz AOs overlap to form a σ bond…Using two 2pz orbitals…

Out of phase

In phase

σ2pz

σ∗2pz

2s 2s

1σ2s

2σ∗2s

2p 2p

3σ2p

4σ∗2p

The remaining 2p orbitals?Once the 2pz AO has been used to form a σ bond, the remaining 2p orbitals cannot overlap head on because these AOs are at right angles to the 2pz and are therefore…

Out of phase

In phase

π2px

π∗2px

=

=

The π Bond

The 2py AOs overlap in exactly the same way to form π2py and π2py orbitals with the same shape and same energy but different…

As easy as…

Where do these fit?

O2

Bond orderBDE 498 kJ

Correlation diagram for homonuclear diatomics, Z = 8 and above (O2-Ne2)

MOEC:

E

2s 2s

1σ2s

2σ∗2s

3σ2pz

4σ∗2pz

1π2px

2π∗2px

1π2py

2π∗2py

2p 2p

Rules for filling MO diagram:

• The number of electrons added to theMO diagram must equal the total number of valence electrons• Orbitals of lowest energy are filled first.• The Pauli exclusion principle applies.• Hund’s Rule must be obeyed.

F2

Bond orderBDE 155 kJ

MOEC:

E

2s 2s

1σ2s

2σ∗2s

3σ2pz

4σ∗2pz

1π2px

2π∗2px

1π2py

2π∗2py

2p 2p

Ne2

Bond orderBDE

MOEC:

E

2s 2s

1σ2s

2σ∗2s

3σ2pz

4σ∗2pz

1π2px

2π∗2px

1π2py

2π∗2py

2p 2p

21

Figure courtesy of Prof. Marc Roussel

Which p orbital combines with which other p orbital is symmetry-determined. Since orthogonal orbitals on different atoms don’t combine, you won’t see combination of a px orbital on one atom and a py orbital on its neighbour, for example.

s orbitals *can* combine with p orbitals when makingσ MOs *if* the orbitals areclose enough in energy. The figure at the right shows 2s and 2p atomic orbital energies for the elements in period 2. We can see that there will be little mixing between 2s and 2p orbitals for the heavier elements in period 2.

MO Diagrams for Z = 3-7 differ from 8-10

Because Li, Be, B, C and N have smaller energy gaps between their 2s and 2p orbitals, some mixing is observed when forming the σ and σ* orbitals, primarily σ*

2s and σ2pz: Mixing in some “p character” lowers the energy of the σ*

2s MO

Mixing in some “s character” raises the energy of the σ2pz MO

If this effect is strong enough, the 3σ orbital can end up higher in energy than the 1π orbital, giving the MO diagram on the next page. This is the case in Li2, Be2, B2, C2 and N2.

instead of

instead of

The MO Diagram for Li2-N2

3σ2pz

2s 2s

1σ2s

2σ∗2s

2π∗2p

2p 2p

4σ∗2p

2π∗2p

1π2p 1π2p

Li2 and Be2

Be2

BO:

Li2

MOEC:

BO:Bond energy: 106 kJ/mol

2s 2s

1σ2s

2σ∗2s

2s 2s

1σ2s

2σ∗2s

Correlation diagram for homonuclear diatomics, Z up to 7 (Li2-N2)

B2?

Bond order =

BDE = 290 kJ

MOEC:

2s 2s

1σ2s

2σ∗2s

2π∗2p

2p 2p

4σ∗2p

2π∗2p

1π2p 1π2p

C2

Bond orderBDE 620 kJ

MOEC:

2s 2s

1σ2s

2σ∗2s

2π∗2p

2p 2p

4σ∗2p

2π∗2p

1π2p 1π2p

N2

Bond orderBDE 945 kJ

MOEC:

2s 2s

1σ2s

2σ∗2s

2π∗2p

2p 2p

4σ∗2p

2π∗2p

1π2p 1π2p

Recap