VII Atomic Structure and Quantum Theory B

1. VII Atomic Structure and Quantum Theory B Atomic Models Quantum Numbers and Bonding

2. Atomic Models • Thomson Model (Early 1900’s) • “plum pudding” model • Atoms is cloud of positive charge with nuggets of negative charge embedded in it.

3. Atomic Models • Rutherford Model (1909) • Devised experiment to test Thomson Model • Stream of alpha particles () (He2+) at metal foil - if Thomson correct - expected to pass through with minimum dispersion. But saw major deflection, some even coming right back at the source. • Rutherford concluded: 1) mass of atom is concentrated in a small region in the center of the atom (nucleus) and has a positive charge. origin of nuclear atom 2) nucleus surrounded by large void containing electrons moving at random.

4. Bohr Model (1913) • Was attempting to explain line spectra • Used a combination of classical and quantum physics • Treated only H atom (one electron system)

5. Bohr Model (1913) Assumptions 1) Only certain set of allowable circular orbits for an electron in an atom 2) An electron can only move from one orbit to another. It can not stop in between. So discrete quanta of energy involved in the transition in accord with Planck (E = h) 3) Allowable orbits have unique properties particularly that the angular momentum is quantized.

6. Bohr Model (1913) r3 • Equations derived from Bohr’s Assumption • Radius of the orbit r2 h = Planck’s constant m = mass of electron e = charge on electron r1 n = orbit number Z = atomic number n=1 n=2 n=3

7. Bohr Model (1913) Called Bohr radius For H: For He+ (also 1 electron) Smaller value for the radius. This makes sense because of the larger charge in the center For H and any 1 electron system: n = 1 called ground state n = 2 called first excited state n = 3 called second excited state etc.

8. Bohr Model (1913) Which of the following has the smallest radius? • First excited state of H • Second excited state of He+ • First excited state of Li+2 • Ground state of Li+2 • Second excited state of H

9. Bohr Model (1913) Which of the following has the smallest radius? • First excited state of H • Second excited state of He+ • First excited state of Li+2 • Ground state of Li+2 • Second excited state of H

10. Bohr Model (1913) For H, A = 2.18 x 10-18 J • Energy of the Orbit What’s happening to the energy of the orbit as the orbit number increases? Energy is becoming less negative, therefore it is increasing. The value approaches 0. Completely removed the electron from the atom.

11. Bohr Model (1913) • Energy differences Between Orbits • For H: electron moves from n = 1 to n = 2 Is energy absorbed or emitted? How much energy is needed? the amount of energy needed is the difference between the energy of orbit 1 and orbit 2.

12. Bohr Model (1913) + sign shows that energy was absorbed. xJ) = 1.64 x 10-18 J What is E when electron moves from n = 2 to n = 1? xJ) = - 1.64 x 10-18 J

13. What is the wavelength of the photon needed to move an electron from n = 1 to n = 2 in the H atom? recall: h = 6.62 x 10-34J sec,  = c • 1.64 x 10-18 m • 2.47 x 1015 m • 6.62 x 10-34 m • 1.21 x 10-7 m • 2.18 x 10-18 m

14. What is the wavelength of the photon needed to move an electron from n = 1 to n = 2 in the H atom? So, light with this wavelength is absorbed When the electron goes back to n = 1, light with the same energy and wavelength will be emitted.

15. So: Ephoton = |E|transition = h = h(c/) h = Planck’s constant = 6.62 x 10-34 J sec c = speed of light = 3.00 x 108 m/sec When E is positive, the photon is absorbed When E is negative, the photon is emitted

16. Ionization Energy energy needed to remove the outermost electron from an atom in its ground state. For H:electron moves from n = 1 to n = ∞ E∞= 0, E1 = - 2.18 x 10-18 J Therefore, the ionization energy for H is 2.18 x 10-18 J

17. DeBroglie Postulate (1924) Said if light can behave as matter, i.e. as a particle, then matter can behave as a wave. That is, it moves in wavelike motion. So, every moving mass has a wavelength () associated with it. where h = Planck’s constant v = velocity m = mass

18. What is the in nm associated with a ping pong ball (m = 2.5 g) traveling at 35.0 mph. • 1.69 x 10-32 B) 1.7 x 10-32 C) 1.69 x 10-23 D) 1.7 x 10-23

19. Heisenberg Uncertainty Principle To explain the problem of trying to locate a subatomic particle (electron) that behaves as a wave Anything that you do to locate the particle, changes the wave properties He said: It is impossible to know simultaneously both the momentum(p) and the position(x) of a particle with certainty

20. Mathematically: uncertainty So: the better you know the position, the speed is less well known and vice versa, and anything done to find one changes the other

21. Schrodinger Model(1926) developed theory that treated electrons as a matter waves [new field: wave mechanics = quantum mechanics] model combined certain allowable quantities of energy that corresponds to certain allowable 3D wave patterns of an electron in an atom

22. H = Hamiltonian operator (set of math operations)  = Wave function E = Energy  describes a particle with a wave pattern, when H is carried out on the function (the solution) you get an allowable energy value, so each solution is associated with a given atomic orbital

23.  = wave function = atomic orbital gives the probability of finding an electron in a certain volume from the nucleus Look at  probability density. How likely are we to “find” an electron at a particular distance from the nucleus. So when solve the wave equation and get probability, yields 3 quantum numbers used to describe the atomic orbital: size(n), shape(l), location (ml)

24. Quantum Numbers and Atomic Orbitals n = principal quantum number (shell) describes orbital size specifies primary energy level greater n, higher E (like Bohr’s E) l = angular momentum quantum number (subshell)(azimuthal quantum number) describes orbital shape l = 0 = s orbital l = 1 = p orbital inc E l = 2 = d orbital within l = 3 = f orbital n

25. Section 20-26 Section 27-29,50 - 52 Section 53-59

26. ml = magnetic quantum number describes orbital orientation the direction is space also tells the number of orbitals of same energy (degenerate orbitals) So, the combination of n, l, and ml, completely describes a specific orbital, its size , shape and orientation. To make a complete picture, a 4th quantum number added, ms. ms = spin quantum number - describes the electron spin in the orbital

27. m lquantum numbers gives the number of each type of orbital • So: for an s orbital, l = 0 and m l = 0, so there is only 1 s type orbital for each n level which can hold 2 electrons • For a p orbital, l = 1, so m l = -1,0,or +1 thus there are 3 distinct p orbitals for each n level which can hold a total of 6 electrons • For a d orbital, l = 2, so m l =-2, -1,0,or +1,+2 thus there are 5 distinct d orbitals for each n level which can hold a total of 10 electrons

28. Shapes of Atomic Orbitals p orbitals s orbital d orbitals

29. Pauli Exclusion Principle • No two electrons in the same atom can have the same four quantum numbers • Each electron has a unique address • Designated by a set of quantum numbers (n, l ,m l,ms) • So: (1,0,0, +1/2) indicates an electron in a 1s orbital • While (3,1, -1, +1/2) would be a 3p electron

30. Which of the following does not represent a 3d electrons? • (3,2,1,+1/2) • (3,2,-1,+1/2) • (3,1,1,+1/2) • (3,2,2,-1/2) • (3,2,0,+1/2)

31. Energy Levels For H atom All other atoms 4s_4p_ _ _ 4d_ _ _ _ _4f _ _ _ _ _ _ _ 4s _ 3d _ _ _ _ _ 3p _ _ _ 3s _ 3p _ _ _ 3d _ _ _ _ _ 3s _ 2p _ _ _ (degenerate orbitals) 2s _ 2p _ _ _(degenerate orbitals) 2s _ 1s _ 1s _ Inc E Inc E

32. Aufbau Principle • For an atom in its ground state ( the lowest energy • configuration) fill the lowest energy orbital first • then go up in energy until all the electrons are used. 1s2s2p3s3p3d4s4p4d…..

33. Spectroscopic NotationElectron Configuration H(1e) 1s1 He(2e) 1s2 Li(3e) 1s22s1 Be(4e) 1s22s2 B(5e) 1s22s22p1 C(6e) 1s22s22p2 N(7e) 1s22s22p3 Ne(10e) 1s22s22p6 Na(11e) 1s22s22p63s1 Si(14) 1s22s22p63s23p2 Ar(18) 1s22s22p63s23p6 K(19) 1s22s22p63s23p6 4s1

34. Predicting Orbital Filling Diagonal Rule 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 6f 7s 7p 7d 7f

35. Using The Periodic Table To Predict Orbital Filling s1 p1 p5 1 2 d6 3 d1 4 5 6 7 p d s 4f 5f f

36. Exam 2 Range = 10 - 120 Average = 58.7 Median = 57

37. [Ar]4s23d1 = [Ar]3d14s2 Sc(21) [Ar]4s23d3 V(23) Zn(30) [Ar]4s23d10 As(33) [Ar]4s23d104p3 [Ar]4s23d104p6 Kr(36) [Kr]5s24d1 Y(39)

38. What would be the correct electron configuration for P? • [Ne]3s5 • [Ar]3s23p3 • [Ar]4s23p3 • [Ne]4s23p3 • [Ne]3s23p3

39. Electron Configuration of Ions Add extra electrons in lowest available orbital Anions O 1s22s22p4 O -2 1s22s22p6 Br [Ar]4s23d104p5 Br - [Ar]4s23d104p6

40. Chloride ion would have the same electron configuration as? • Ne • Ar • Kr • S • P

41. Cations form by losing e from the highest n level First. Not necessarily the last e added. Cations Mg 1s22s22p63s2 1s22s22p6 Mg+2 [Ar]4s23d6 Fe Fe+2 [Ar]4s03d6 = [Ar]3d6 Fe+3 [Ar]3d5 Sn [Kr]5s24d105p2 Sn+2 [Kr]5s24d10 [Kr] 4d10 Sn+4

42. Which of the following would have the same electron configuration as Sn4+? • Cd • Cd2+ • Ge4+ • Ag+ • Sb3-

43. [Kr]5s24d10 Which of the following would have the same electron configuration as Sn4+? • Cd • Cd2+ • Ge4+ • Ag+ • Sb3- [Kr]5s04d10 [Kr] 4d10 Sn+4