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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, 15 Multi-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

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

1s2s2p Ne:1s 2 2s 2 2p 6 1s2s2p F:1s 2 2s 2 2p 5 1s2s2p O:1s 2 2s 2 2p 4 1s2s2p N:1s 2 2s 2 2p 3 Nitrogen Oxygen Fluorine Neon THE ELECTRON CONFIGURATIONS FOR NITROGEN TO NEON

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!

Sodium has the electron configuration Na: 1s 2 2s 2 2p 6 3s 1 abbreviate electron configurations by... 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 And Rb is: dividing the electrons into…….. valence electrons and core electrons. What does this mean?????

Example Na: 1s 2 2s 2 2p 6 3s 1 VALENCE ELECTRONS  Noble gases have fully-filled valence shells Example Ar: 1s 2 2s 2 2p 6 3s 2 3p 6 are the electrons in the outermost principle quantum level of an atom. highest value of n. Valence electrons are the ones involved in bonding. CORE ELECTRONS 

can be abbreviated as a noble gas core and the valence electrons. The configuration for sodium is: Na: 1s 2 2s 2 2p 6 3s 1 orNa: [Ne]3s 1 ELECTRON CONFIGURATIONS Now we can easily write the electron configurations….. QUESTION…..

1s1s E 2s2s 2p2p 3s3s 3p3p 3d3d 4s4s 4p4p 5s5s 4d4d   POTASSIUM ELECTRONIC CONFIGURATION 1s22s22p62p23s23p64s11s22s22p62p23s23p64s1 or [Ar]4s 1 Q: why does 3d lie above 4s??

SCREENING AND PENETRATION PENETRATIONIs to get close to the nucleus SCREENINGIs to block the view of other electrons of the nucleus Consider the excited state of helium 1s 1 3p 1 LOOK AT RADIAL PROBABILITY DISTRIBUTIONS

SCREENING AND PENETRATION THE 1s close to the nucleus PENETRATES WELLSEES A CHARGE OF Z=2 EFFECTIVE NUCLEAR CHARGE…... Is 3p THE 3p DOES NOT SCREEN THE NUCLEUS

SCREENING AND PENETRATION THE 1s close to the nucleus EFFECTIVE NUCLEAR CHARGE Z eff = 2 ENERGY MORE NEGATIVE THAN IN H-ATOM Is 3p E2.178 x n  Z n eff  2

SCREENING AND PENETRATION THE 3p far from the nucleus PENETRATES POORLY 1s SCREENS 3p WELL SEES A CHARGE OF Z eff = Z – S = 2-1 = 1 Is 3p

THE 3s orbital penetrates better than 3p orbital 3s 3p 3d The 3p orbital penetrates better than 3d orbital Z eff (s) > Z eff (p) > Z eff (d)

So, in a multi-electron atomZ eff (s) > Z eff (p) > Z eff (d) E 4s < E 3d ENERGY LEVELS REVERSE!!

The first transition element is scandium, Sc (Z=21) 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 KCa Scandium has the electron configuration: Sc: [Ar]4s 2 3d 1 Sc TRANSITION METALS. Because…..

1s1s E 2s2s 2p2p 3s3s 3p3p 3d3d 4s4s 4p4p 5s5s 4d4d   ENERGY LEVEL DIAGRAM FOR A MULTI- ELECTRON ATOM SCANDIUM ELECTRONIC CONFIGURATION Scandium is the first TRANSITION ELEMENT  Sc: [Ar]4s 2 3d 1

the metals that fill the d orbitals in their valence shell. TRANSITION METALS HUND’S RULE OBEYED FOR ALL EXCEPT Cr and Cu Cr:[Ar] 4s 2 3d 4 EXPECTED OBSERVED Cr:[Ar] 4s 1 3d 5 …. The d-shell is ½ filled this way; all spin up WHEN n=3 FOR COPPER

the metals that fill the d orbitals in their valence shell. TRANSITION METALS HUND’S RULE OBEYED FOR ALL EXCEPT Cr and Cu Cu: [Ar] 4s 2 3d 9 EXPECTED OBSERVED Cu:[Ar] 4s 1 3d 10 …. The d-shell is filled this way – extra stable WHEN n=3 IONS…...