1. Warmup – 5 minutes 1. A quantum of energy absorbed by an electron is 4.7 x 10 -19 J. What is the wavelength of this quantum? What type of radiation is this quantum? E = hc/λ 6.626 x 10 -34 J·s)( (4.7 x 10 -19 J)λ= (6.626 x 10 -34 J·s)(3.00 x 10 8 m/s) 4.22 x 10 -7 m (visible purplish-blue light) 2. In each of these elements, identify the number of TOTAL e- and the number of VALENCE e- : P K U Pb VCu Find your periodic table!

2. Electron Configurations and Orbitals In other words……the system that quantum physicists use to describe the specific location of an electron. WARNING! Today is very conceptual. Just go with it.

3. 1 st Quantum Number (N) –Energy Level 2 nd Quantum Number ( l ) –orbital 3 rd Quantum Number (m l ) –orientation/shape of orbital 4 th Quantum Number (m s ) –Spin of Electron Which floor? Cool Weird Analogy! A ‘friend’ turns out to be a stalker! He/she walks you to your building and asks “where can I find you in the middle of the night?” In order to kill you, he/she needs to know some specifics: Which wing of the building? Which room? Which bed?

4. 1. Calculate the maximum number of electrons that can exist on the energy levels 1 through 4 n = number of electrons = 1 2n 2 2(1) 2 2 2 2n 2 2(2) 2 8 3 2n 2 2(3) 2 18 4 2n 2 2(4) 2 32 Principal (1 st ) Quantum Number There are no more than 2N 2 electrons per energy level(N)

5. On each energy level (n), electrons are paired up into orbitals Atomic orbitals are regions where there is a high probability (90 %) of finding an electron. sometimes called sublevels ( l, the 2 nd quantum number) The atom is found inside a blurry “electron cloud”

6. one s orbital for every energy level spherical shaped Each s shape can hold 2 electrons Called the 1s, 2s, 3s, etc.. orbitals. S orbitals

7. P orbitals Start at the 2nd energy level 3 different orientations & shapes Each shape holds 2 electrons Dumbbell

8. D orbitals Start at the 3rd energy level 5 different shapes Each can hold 2 electrons 4-Lobed

9. F orbitals Start at the 4 th energy level Have 7 different shapes 2 e- per shape 6-8 Lobed

11. Principal energy level # diff orbitals Type of sublevel (number of shapes per orbital) Total # of sublevels (n 2 ) Total # of electrons (2n 2 ) n=12 n=28 n=318 n=432 1s (one shape only) 2s (1 shape) 2p (3 shapes) 3s (1 shape) 3p (3 shapes) 3d (5 shapes) 4s (1 shape) 4p (3 shapes) 4d (5 shapes) 4f (7 shapes) 12341234 1 4 9 16

12. Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f 2. Draw the orbital-filling diagram and determine the electron configuration for phosphorus. Pauli Exclusion Principle: no 2 e- (in the same atom) can have the same set of 4 quantum numbers. Electrons in the same shape must have opposite spins

13. Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f Sublevel 1 fills up first. Aufbau Principle: electrons occupy the lowest energy orbitals first

14. Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f

15. Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f

16. Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f

17. Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f The last three e- go into the 3p orbitals, but they each go into separate shapes! electron configuration: 1s 2 2s 2 2p 6 3s 2 3p 3 Hund’s Rule: one electron will occupy each orientation in an orbital before a second electron moved in

18. Potassium’s last electron is put in the 4 th energy level and not in the 3 rd level 3d sublevel 3. Write the electron configuration for potassium 1s 2 2s 2 Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f 3p 6 4s 1 2p 6 3s 2

19. 6d 7 Energy Level sublevel # of e - in sub level 4. The e- config. of which element would end in 6d 7 ? Mt Meitnerium

20. The easy way to remember 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p 1s 2 2 electrons

21. Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p 1s 2 2s 2 4 electrons

22. Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p 1s 2 2s 2 2p 6 3s 2 12 electrons

23. Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 20 electrons

24. Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 38 electrons

25. Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 56 electrons

26. Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p 88 electrons 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 6 7s 2

27. Fill from the bottom up following the arrows 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 6 7s 2 5f 14 6d 10 7p 6 108 electrons

28. 5. Write the full e- configuration for lead 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 2 Abbreviated: [Xe]6s 2 4f 14 5d 10 6p 2 Pb Write the noble gas preceding the desired element [brackets], then proceed

29. # 6) Write the full e- configuration for chromium. [Ar]4s 2 3d 4 is expected But this is wrong!! Chromium is actually: [ Ar]4s 1 3d 5 Why? A half-filled (5 e-) or full (10 e-) d shell is more stable and lower in energy than a full s subshell of the next n value

30. Increasing energy 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 7p 6d 4f 5f This gives Cr two half filled orbitals (s and d)

31. 7. Copper also behaves contrary to the norm. Predict the e- config of copper. we expect: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 9 But the actual configuration is: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10

32. Isoelectronic Ions 8. What is the e- configuration of the aluminum ion? Al +3 It WAS neutral: 1s 2 2s 2 2p 6 3s 2 3p 1 But Al loses 3 electrons to form an ion So Al +3 is 1s 2 2s 2 2p 6 9. What is the electron configuration of the fluoride ion? F WAS neutral: 1s 2 2s 2 2p 5 then gained one e- to form F - : 1s 2 2s 2 2p 6 Mg +2 Na +1 Ne O -2 and N -3 are also 1s 2 2s 2 2p 6