introductory concepts: atomic and molecular orbitals

http://aimpro.ncl.ac.uk MMG Skills Lecture Series

Introductory concepts:Atomic and molecular orbitals

Jon Goss

MMG Skills Lecture Series2

Outline

Atomic orbitals (AOs)Linear combinations (LCAO):

HybridsMolecular orbitals (MOs)

One-electron vs. many-bodyCharge density and spin density


Atomic Orbitals: founding principles

Electrons are Fermions:The are indistinguishablespin-half particlesAnti-symmetric wave functionsObey the Pauli exclusion principle

(no two electrons can exist in the same quantum state)

The have mass and chargeThey move in the potential arising from the (point) nucleus and the other electrons in the atom.For the hydrogen atom the solutions may be obtained analytically.For other atoms, in general this is not (yet) possible.


Atomic orbitals

Electrons in atoms may be characterised by four quantum numbers

n: principal quantum numberl: orbital angular momentumms: spin magnetic angular momentum

ml: orbital magnetic quantum number[See for example, Atomic Spectra and Atomic Structure, Herzberg (Dover Press)]

In this lecture we are chiefly concerned with properties implied by the different values of n and l.


Atomic orbitals: l

The orbital angular momentum can take positive integer values, but they are commonly expressed using letters:

We can interpret the increase in l in terms of an increase in angular nodality.For a given value of l, ml can take any value from –l to l

E.g. l=1, ml can be -1, 0 and +1.

These are equal in energy: orbital degeneracyorbital degeneracy!

l 0 1 2 3 4 …

Term s p d f g …

ml 0 -1, 0, 1-2, -1, 0,

1, 2

-3, -2, -1, 0, 1,

2, 3

-4, -3, -2, -1, 0, 1, 2, 3,

4

…


Atomic orbitals: l, ml

There is a radial node, not shown


Atomic orbitals: n

The possible values of l are restricted by the principal quantum number, n.

l<n

Thus, for n=1, only l=0 (s) is allowed.For n=2, l can have values 0 and 1 (s and p).…and so on…Increasing n implies increasing radial nodality…


Atomic orbitals: n, l and ml


Atomic orbitals: mS

The final quantum number is the spin magnetic quantum number, which can take two values:

ms=+½ and ms=-½

“up” and “down” spinsThere is no real physical spin in the classical sense involved.


Pauli exclusion and the build up principles

The Pauli exclusion principle states that no two electrons may have the same set of quantum numbers…For atoms, therefore, we have definite groups of states (shells) that are incrementally occupied with increasing energy:

1s up, 1s down (there is only one value of ml)2s up, 2s down (2p, ml=-1, ms=+½), (2p, ml=0, ms=+½), (2p, ml=1, ms=+½), (2p, ml=-1, ms=-½), (2p, ml=0, ms=-½), (2p, ml=1, ms=-½)



H: 1s1

He: 1s2

Li: 1s22s1

Be: 1s22s2

B: 1s22s22p1

C: 1s22s22p2

N: 1s22s22p3

O: 1s22s22p4

F: 1s22s22p5

Ne: 1s22s22p6

Na: (Ne)3s1

Mg: (Ne)3s2

Al: (Ne)3s23p1

Si: (Ne)3s23p2

P: (Ne)3s23p3

S: (Ne)3s23p4

Cl: (Ne)3s23p5

Ar: (Ne)3s23p6



Nothing has been said about which ml states are involved, although we’ll touch on this in terms of the many-body effectsmany-body effects.It gets slightly more complicated with we move beyond Ar as we begin filling the 4s orbitals before the 3d…

You are referred to any reasonable inorganic chemistry text book


Linear combinations: Hybrids

In the presence of an applied field (typically as a consequence of nearby atoms) the atomic orbitals combined together to form “hybrids”.Some of the best known examples relate to carbon.

Graphite: sp2.Diamond: sp3.

In contrast the atomic orbitals, these are formed in weighted combinations…


Linear combinations: Hybrids

sp2:s+px+py

s+px-py

s-px-py

pz

sp3:s+px+py+pz

s-px-py-pz

s+px-py-pz

s-px+py-pz


Linear combinations: Molecular Orbitals

Both atomic orbitals and hybrids centred on different atoms combine to form covalent bonds.

σ-bonds (sigma-bonds) are made up from overlapping orbitals directed along the bond direction.π-bonds (pi-bonds) are made up from overlapping orbitals at an angle to the inter-nuclear direction:

πp bonds are combinations of p-orbitals perpendicular to the bond directionπp-d bonds are combinations of p- and d-orbitals but not precisely perpendicular to the bond-direction



σp-p anti-bonding combination



πp-bonding combination



πp* anti-bonding combination



πp-d bonding combination



πd-d bonding combination



πd-d anti-bonding combination



The bonds can be modelled by considering linear combinations of the atomic orbitals or atomic hybrids

This is only a simplification, as we shall see when we consider simple many-body concepts.

The combinations are dictated by the relative energies of the atomic orbitals.A prototypical example is the hydrogen molecule…



1sa 1sb

a1g

a1u

En

erg

y

Atom

Atom

Molecule



The same approach can be adopted for defects in solid solution:

The vacancy in diamond

Take out an atom and you generate four equivalent sp3 dangling-bonds.

We’ll label them a, b, c and d.

As in the H2 molecule, we form linear combinations of these orbitals to form the “molecular” orbitals for the four together:

(a+b+c+d) – the bonding combination(a+b+c-d;a+b-c+d;a-b+c+d) – a triply degenerate combinations involving some anti-bonding character.

Hood et al PRL 91, 076403 (2003).



What happens when the originating orbitals are inequivalent?


One-electron vs. many-body

Remember that electrons are indistinguishable particles, so the molecular and atomic orbitals are models for the electrons in real compound systemsA more precise description of the electronic states must be a function of the positions of all all the electrons in the system.the electrons in the system.This is the many-electron wave function of an atom, molecule, defect…As an example, lets look back at one of the atoms…


One-electron vs. many-body: Nitrogen Nitrogen

Nitrogen atoms have the electronic configurations 1s22s22p3

What does this mean in terms of the properties of the atom?

Note, for weak spin-orbit coupling, the electron spins combine to give the total effective spin, S.What is the spin state of a nitrogen atom?


One-electron vs. many-body: Nitrogen Nitrogen

The spins in the 1s and 2s are pairs with ms=+½ and ms=-½, yielding no net spin from these shells (S=0).We have three electrons in the n=2, l=1 states with ml=1,0,-1, ms=±½.

Which one combinations are involved?It can be shown that there are exactly three ways to three ways to combine the electronscombine the electrons:

Two have S=1/2, one has S=3/2.The combinations also yield effective orbital angular momenta, which in the many-body sense are labeled using upper case terms (S, P, D, F, G, …)

Do not confuse the total electron spin and the S orbital angular momentum term.

The two S=1/2 combinations are P and D, whereas S=3/2 yields S.We write 2P, 2D and 4S, where the leading numerical value indicates the multiplicity of the state and is given by (2S+1).


One-electron vs. many-body

Does this make any difference?Obviously the answer must be yes, otherwise I would not have tortured you with the preceding analysis

1. Spin selection rules for optical spectra (ΔS=0)

2. Spin state (magnetism, ESR, …)3. Orbital angular momentum selection rules in

optical spectra (|ΔL|=1)4. Jahn-Teller effects apply to many-body states


One-electron states vs. electron density

Experimentally observed properties may depend more or less on the many-body effects, with some accessible from the frontier orbitals alone:

Optical selection rules for orbitally non-degenerate one-electron states?orbitally degenerate one-electron states?

ESR?Bond strengths?Reactive sites?



Example of H/μ+ in diamond.There are two main forms of this centre

“Normal muonium” – a non-bonded site“Anomalous muonium” – residing in the centre of a carbon-carbon bond.

In the overall neutral charge state there are an odd number of electrons and therefore the net electron spin allows for access of these centres in ESR-like experiments.The interaction of the electron spin and the nuclear spin of H (or muonium) can be determined theoretically by analysis of the spin-densityspin-density at the nucleus.The question is, how well is the spin density represented by the unpaired one-electron state?



The answer is that qualitatively the correct kind of answer can be obtained for normal muonium, but not for anomalous muonium.

The unpaired electron (the state in the band-gap) is centred on the muon for the normal form, giving a large isotropic hyperfine interaction. There will also be a contribution from the polarisation of the valence states, but this is probably not the dominant term.The unpaired electron in the bond-centre is nodalnodal at the muonium, so there should be zero isotropic hyperfine interaction in this form, but this is not the case – for the bond-centre, the isotropic contribution to the hyperfine interaction arises purelypurely from the polarization of the valence density due to the unequal spin up and spin down populations,


Bond-centred muonium

sp3

sp3

1s



A

B

C



A+B+C

A+B-C

A-B

Note, we are choosing to ignore most of the electrons in the system

Ec

Ev



A+B+C

A+B-C

A-B


Ec

Ev


In this simplified model, the experiment can be qualitatively explained:

the isotropic part of the hyperfine comes from the small differences between spin up and spin down states “A+B+C”This is spin polarization in action!

For an accurate facsimile of the experiment, you must include the polarisation of all the electrons in your system.


Summary

Atomic and molecular orbital theory is a powerful tool for simplified, but highly illustrative explanations of a wide range of materials properties.However, it must always be remembered that it is only a simplified model!

introductory concepts: atomic and molecular orbitals

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