1. 8 - 1 Sublevels and Orbitals Effective Nuclear Charge Inner (core) electrons act to shield outer (valence) electrons from the positive charge of the nucleus. Some orbitals penetrate to the nucleus more s < p < d < f. than others, therefore s < p < d < f. As a result, there are different energy levels for the different sublevels for any given principal quantum number.

2. 8 - 2 Sublevels energy 1s 2s 1s 2s 1s 2s 2p H Li F 2p

3. 8 - 3 Aufbau Principle The Aufbau Principal is used to write the electron configurations of atoms. For any element, the number of electrons in the neutral atom equals the atomic number. Start filling orbitals, from lowest to highest. If two or more orbitals exist at the same energy level, they are degenerate. Do not pair the electrons until you have to.

4. 8 - 4 Orbital Filling Order. spdf 11s 22s2p 33s3p3d 44s4p4d4f 55s5p5d5f 66s6p6d6f 77s7p7d7f

5. 8 - 5 Hund’s Rule When putting electrons into orbitals with the same energy, place one electron in each orbital before putting two in any one. In the orbital diagram, each electron must have opposite spins.

6. 8 - 6 Applying The Aufbau Principal 1s 2s 2p 1s 2s 2p 1s 2s 2p energy C O F

7. 8 - 7 Sublevels on the Periodic Table H Li Na Cs Rb K TlHgAuLaBa Fr PtIrOsReWTa He RnAtPoBiPb Be Mg Sr Ca CdAgYPdRhRuTcMoNb Ra ZnCu Hf Zr Ti ScNiCoFeMnCrV InXeITeSbSn GaKrBrSeAsGe AlArClSPSi BNeFONC Gd Cm Tb Bk Sm Pu Eu Am Nd U Pm Np Ce Th Pr Pa Yb No Ac Er Fm Tm Md Dy Cf Ho Es f f d d p p Lu Lr s s

8. 8 - 8 Using the Periodic Table to apply the Aufbau Principle Main Group Elements ns  Add electrons to the ns orbital as you move through s-block. np  Add electrons to the np orbital as you move through the p-block.

9. 8 - 9 Using the Periodic Table to apply the Aufbau Principle Transition Elements (n-1)d  Add electrons to the (n-1)d orbital as you move through d-block. (n-2)f  Add electrons to the (n-2)f orbital as you move through f-block.

10. 8 - 10 Writing Electron Configurations Expanded Format O1s 2 2s 2 2p 4 Ti1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2 Br1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5 Abbreviated Format O[He]2s 2 2p 4 Ti[Ar]4s 2 3d 2 Br[Ar]4s 2 3d 10 4p 5

11. 8 - 11 Electron Configurations for Ions Electron configurations can also be written for ions.  Start with the ground-state configuration for the atom.  For cations, remove the number of the outermost electrons equal to the charge.  For anions, add the number of outermost electrons equal to the charge.

12. 8 - 12 Electron Configurations for Anions Electron Configurations for Anions Example: Cl - (chloride) First, write the electron configuration for chlorine: Cl[Ne]3s 2 3p 5 Because the charge is 1-, add one electron. Cl - [Ne]3s 2 3p 6

13. 8 - 13 Electron Configurations for Cations Example: Ba 2+ (barium) First, write the electron configuration for barium. Ba[Xe]6s 2 Because the charge is 2+, remove two electrons. Ba 2+ [Xe]

14. 8 - 14 Isoelectronic Configurations Species having the same electron configurations. Each of the following has an electron configuration of 1s 2 2s 2 2p 6 O 2- F - Ne Na + Mg 2+ Al 3+

15. 8 - 15 Electron Dot Diagrams A simple way to show the valence electrons present in an atom. Valence electrons are those electrons found in the highest numbered principal energy level (PEL). Valence electrons are found only in the s and p sublevels and in most cases are the electrons responsible for bonding.

16. 8 - 16 Electron Dot Diagrams The chemical symbol represents the kernel of the atom. The kernel of an atom consists of the nucleus and the core electrons. Example X s px px py py pz pz Start with s and proceed cw!

17. 8 - 17 Electron Dot Diagrams Examples Na [Ne]3s 1 Br[Ar]4s 2 3d 10 4p 5 Cr[Ar]4s 1 3d 5.....

18. 8 - 18 Heisenberg Uncertainty Principle  In order to observe an electron, one would need to hit it with photons having a very short wavelength.  Short wavelength photons would have a high frequency and a great deal of energy.  If one were to hit an electron, it would cause the motion and the speed of the electron to change.  Lower energy photons would have a smaller effect but would not give precise data.

19. 8 - 19 Quantum Numbers Principal Quantum Number (n)  Tells the size of an orbital and determines its energy.  n = 1, 2, 3 … Angular Momentum (l)  The number of subshells that a principal level contains. It tells the shape of the orbitals.  l = 0 to n - 1

20. 8 - 20 Quantum Numbers Magnetic Quantum Number (m l )  Describes the orientation of the orbital.  m l = -l to +l (all integers, including zero)  For example, if l = 3, then m l would have values of -3, -2, -1, 0, 1, 2, 3

21. 8 - 21 Quantum Numbers. nlmlml SublevelOrbitals 1001s1 2002s1 1-1,0,12p3 3003s1 1-1,0,13p3 2-2,-1,0,1,23d5 4004s1 1-1,0,14p3 2-2,-1,0,1,24d5 3-3,-2,-1,0,1,2,34f7

22. 8 - 22 The s Orbital The s orbital is a sphere. Every PEL has one s orbital.

23. 8 - 23 The p Orbitals There are three p orbitals: p x, p y and p z

24. 8 - 24 Electron Spin Pauli added one additional quantum number that would allow only two electrons to be in an orbital. Spin quantum number (m s ).  It can have values of +1/2 and -1/2 Pauli Exclusion Principle  Pauli also proposed that no two electrons in an atom can have the same set of four quantum numbers.

25. 8 - 25 Waves Wavelength (l)  The distance measured from crest to crest or from trough to trough (m, cm, nm).Amplitude  The vertical distance from the node to the height of a wave (m, cm, nm).

26. 8 - 26 Frequency (f)  The number of cycles or complete vibrations that pass a point each second (s -1, vib/s).

27. 8 - 27 Electromagnetic Radiation A form of energy consisting of perpendicular electrical and magnetic fields that change, at the same time and in phase with time. The SI unit of frequency (f) is the hertz (Hz)  1 Hz = 1 s -1 = 1/s Wavelength and frequency are related  c = f λ c = 3.00 x10 8 m/s

28. 8 - 28 Energy Problems Calculate the frequency of a quantum of light, a photon, with a wavelength of 6.00 x 10 -7 m. λ = 6.00 × 10 -7 mc = 3.00 × 10 8 m/s C = f × λ f = = CλCλ = 5.00 × 10 14 /s 3.00 × 10 8 m/s 6.00 × 10 -7 m

29. 8 - 29 Energy Problems Calculate the energy of a photon, with a wavelength of 6.00 x 10 -7 m. λ = 6.00 × 10 -7 mc = 3.00 × 10 8 m/s h = 6.63 × 10 -34 Js E = h × fc = f × λ f = c/ λ E = h × c/ λ

30. 8 - 30 E = h × c/ λ E = 6.63 × 10 -34 Js × 3.00 × 10 8 m/s 6.0 × 10 -7 m E = 3.3 × 10 -19 J