Why do bonds form? 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?

The Schrödinger equation – the solution for the H atom oAustrian 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 oThe result is encapsulated in a complex formula known as the Schrödinger equation: oThis 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 oΨ 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 E n, 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 oA 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 oAn orbital is a wave function oAn orbital is a region of space in which an electron is most likely to be found oThe square of the amplitude of the wavefunction, ψ 2, 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.

Atomic Orbitals The size and energy of orbitals depend on … 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: H 2 + 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.

In orbital terms... two overlapping 1s H atomic orbitals In H 2 +, 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 H 2 + 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. σ 1s 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 H 2 + cation… Bond order = ½(#bonding electrons - #antibonding electrons)

The Hydrogen Molecule – all that changes is the number of electrons… MO electron configuration: (1σ 1s ) pm 250 pm 220 pm 200 pm 150 pm 100 pm 73 pm

Irradiation of H 2 Ground state BO: hυ Excited state BO:

First Row Diatomics

Second Row Homonuclear Diatomics: Li 2 – Ne 2 We now have access to four new orbitals: 2s, 2p x, 2p y and 2p z. These are the valence AOs. The 1s orbitals are too small to overlap at the optimal bond distances so they can be ignored.

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… 2s2s2s2s 2σ2σ 2σ*2σ*

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

The remaining 2p orbitals? Once the 2p z 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 2p z and are therefore… The 2p y AOs overlap in exactly the same way to form  2py and π  2py orbitals with the same shape and same energy but different…

As easy as… Where do these fit?

O 2 Bond order BDE 498 kJ Correlation diagram for homonuclear diatomics, Z = 8 and above (O 2 -Ne 2 ) MOEC: Rules for filling MO diagram: The number of electrons added to the MO 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.

F 2 Bond order BDE 155 kJ MOEC:

Ne 2 Bond order BDE MOEC:

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 p x orbital on one atom and a p y orbital on its neighbour, for example. s orbitals *can* combine with p orbitals when making  MOs *if* the orbitals are close 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 σ 2pz : Mixing in some “p character” lowers the energy of the σ * 2s MO Mixing in some “s character” raises the energy of the σ 2pz 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 Li 2, Be 2, B 2, C 2 and N 2. instead of

The MO Diagram for Li 2 -N 2 3σ 2pz

Li 2 and Be 2 Be 2 BO: Li 2 MOEC: BO: Bond energy: 106 kJ/mol

Correlation diagram for homonuclear diatomics, Z up to 7 (Li 2 -N 2 ) B2?B2? Bond order = BDE = 290 kJ MOEC:

C 2 Bond order BDE 620 kJ MOEC:

N 2 Bond order BDE 945 kJ MOEC:

Recap