What’s coming up??? Oct 25The atmosphere, part 1Ch. 8 Oct 27Midterm … No lecture Oct 29The atmosphere, part 2Ch. 8 Nov 1Light, blackbodies, BohrCh. 9 Nov 3,5Postulates of QM, p-in-a-boxCh. 9 Nov 8, 10Hydrogen and multi – e atoms Ch. 9 Nov 12,15Multi-electron atomsCh.9,10 Nov 17Periodic propertiesCh. 10 Nov 19Periodic propertiesCh. 10 Nov 22Valence-bond; Lewis structuresCh. 11 Nov 24Hybrid orbitals; VSEPRCh. 11, 12 Nov 26VSEPRCh. 12 Nov 29MO theoryCh. 12 Dec 1MO theoryCh. 12 Dec 2Review for exam

We can split the hydrogen wavefunction into two:  (x,y,z)   (r,  ) = R(r) x Y(  ) Depends on r only Depends on angular variables

The solutions have the same features we have seen already: –Energy is quantized E n =  R Z 2 / n 2 =  x Z 2 / n 2 J [ n = 1,2,3 …] –Wavefunctions have shapes which depend on the quantum numbers –There are (n-1) nodes in the wavefunctions

Because we have 3 spatial dimensions, we end up with 3 quantum numbers: n, l, m l n = 1,2,3, …; l = 0,1,2 … (n  1); m l =  l,  l+1, …0…l  1, l n is the principal quantum number – gives energy and level l is the orbital angular momentum quantum number – it gives the shape of the wavefunction m l is the magnetic quantum number – it distinguishes the various degenerate wavefunctions with the same n and l

E n =  R Z 2 / n 2 =  x Z 2 / n 2 J [ n = 1,2,3 …] … degenerate

Probability Distribution for the 1s wavefunction: 0 -r/a e 3/2          a  Maximum probability at nucleus

A more interesting way to look at things is by using the radial probability distribution, which gives probabilities of finding the electron within an annulus at distance r (think of onion skins) max. away from nucleus

nodes 90% boundary: Inside this lies 90% of the probability

P-orbitals Node at nucleus

The result (after a lot of math!) Node at  = 2!! Nodes at  = 0 !!

The Boundary Surface Representations of All Three 2p Orbitals

A Comparison of the Radial Probability Distributions of the 2s and 2p Orbitals

A Cross Section of the Electron Probability Distribution for a 3p Orbital Spatial nodes and Angular nodes!

Nodes at  = 0 and 4 Nodes at  = 0 !!

The Boundary Surfaces of All of the 3d Orbitals

Representation of the 4f Orbitals in Terms of Their Boundary Surfaces

The Radial Probability Distribution for the 3s, 3p, and 3d Orbitals

Another quantum number! Electrons are influenced by a magnetic field as though they were spinning charges. They are not really, but we think of them as having “spin up” or “spin down” levels. These are labeled by the 4 th quantum number: m s, which can take 2 values.

In silver (and many other atoms) there is one more “spin up” electron than “spin down” or vice versa. This means that an atom of silver can interact with a magnetic field and be deflected up or down, depending on which type of spin is in excess. This 2-valued electron spin can be shown in an experiment

In multi-electron atoms things change, because of the influence of electrons which are already there The degenerate energy levels are changed somewhat to become Remember the energies are <0 3s3p3d2s2p1s3s3p3d2s2p1s E

1s1s E 2s2s 2p2p 3s3s 3p3p 3d3d 4s4s 4p4p 5s5s 4d4d THE MULTI-ELECTRON ATOM ENERGY LEVEL DIAGRAM

THE PAULI PRINCIPLE An orbital is described by three quantum numbers, each orbital may contain a maximum of two electrons, and they must have opposite spins. No two electrons in the same atom can have the same set of four quantum numbers ( n, l, m l, m s ). Then each electron in a given orbital must have a different m s HOW MANY ELECTRONS IN AN ORBITAL?

THE BUILDING-UP PRINCIPLE. assign electrons to orbitals one at a time lowest energy electronic configuration Electrons go into the available orbital of lowest energy. Electrons are placed in orbitals according to the Pauli Principle. A maximum of two electrons per orbital. ELECTRONIC CONFIGURATIONS GROUND STATE

THE AUFBAU (BUILDING-UP) PRINCIPLE: The electron configuration of any atom or ion….... electrons are added to hydrogen-like atomic orbitals in order of increasing energy can be represented by an orbital diagram 1s1s E 2s2s 2p2p 3s3s 3p3p 3d3d 4s4s 4p4p 5s5s 4d4d

1s2s2p He: Hydrogen has its one electron in the 1s orbital: 1s2s2pH:1s2s2pH: 1s2s2p He:  1s11s1 ORBITAL DIAGRAM with opposite spins: both occupy the 1s orbital Pauli principle Helium has two electrons: 1s21s2 

Hydrogen has its one electron in the 1s orbital: 1s2s2pH:1s2s2pH: 1s2s2p He: 1s2s2p He: helium ground state Helium has two electrons:     1s11s1 1s12s11s12s1 1s21s2 ORBITAL DIAGRAM Now onto the next atoms Helium can also exist in an excited state such as: with opposite spins: both occupy the 1s orbital Pauli principle

Lithium has three electrons, so it must use the 2s orbital: Beryllium has four electrons, which fill both the 1s and 2s orbitals: Boron’s five electrons fill the 1s and 2s orbitals, and begin to fill the 2p orbitals. Since all three are degenerate, the order in which they are filled does not matter. 1s2s2p Li:1s 2 2s 1 1s2s2p Be:1s 2 2s 2 1s2s2p B:1s 2 2s 2 2p 1

1s2s2p C:1s 2 2s 2 2p 2 1s2s2p C:1s 2 2s 2 2p 2 How can we decide????? A CHOICE OR CARBONZ=6

HUND’S RULE FOR THE GROUND STATE ELECTRONS OCCUPY DEGENERATE ORBITALS SEPARATELY THE SPINS ARE PARALLEL SO FOR CARBON THE GROUND STATE IS 1s2s2p C:1s 2 2s 2 2p 2

1s1s E 2s2s 2p2p 3s3s 3p3p 3d3d 4s4s 4p4p 5s5s 4d4d   ENERGY LEVEL DIAGRAM FOR A MULTI- ELECTRON ATOM BROMINE ELECTRONIC CONFIGURATION  [Ar] 4s 2 3d 10 4p 5

The idea of penetration explains why the 3d orbitals lie higher in energy than the 4s.

The valence electron configuration of the elements in the periodic table repeat periodically! H He 1s 1 1s 2 Li Be B C N O F Ne 2s 1 2s 2 2p 1 2p 2 2p 3 2p 4 2p 5 2p 6 Na Mg Al Si P S Cl Ar 3s 1 3s 2 3p 1 3p 2 3p 3 3p 4 3p 5 3p 6 Every element in a group has the same valence electron configuration!