1. Chapter 5.Periodicity and the Periodic Table

2. Many properties of the elements follow a regular pattern. In this chapter, we will look at theory that has been developed to explain this periodicity

4. Much of what we have learned about atomic and molecular structure, has come from our understanding of how matter interacts with light. What is light? The interaction of light with matter forms the foundation of our understanding of atomic structure, molecular structure, and the structure of the universe! IR tutor

7. The interaction of light with matter forms the foundation of our understanding of molecular structure. How so? Wave properties of light c = where c = 3x10 10 cm/s speed of light in vacuum = wavelength of light and = frequency of light Particle properties of light E = h where E is the energy of a photon of light and h = 6.626 x 10 -34 J. s

8. Heisenberg uncertainty principle: (uncertainty in position)(uncertainty in momentum (mv))  h/4 

9. What happens if we take a hydrogen atom and heat it up, or for that matter, any element?

10. The emission spectrum of H and sodium atoms in the visible region of the spectrum The significance of observing discrete line may not be immediately apparent but these atomic species when heated do not give off all wavelengths of light but only discrete wavelengths

11. http://jersey.uoregon.edu/vlab/elements/Elements.html

12. The interpretation of these observations is that upon heating H atoms, the hydrogen does not emit any light until a certain amount of energy is put into the atom. Since an atom of hydrogen consists of only a proton and an electron, it is believed that the emission of light by the hydrogen is due to excitation of the electron. Unexcited hydrogen Excited hydrogen energy

13. The interpretation of these observations is that upon heating H atoms, the hydrogen does not emit any light until a certain amount of energy is put into the atom. Since an atom of hydrogen consists of only a proton and an electron, it is believed that the emission of light by the hydrogen is due to excitation of the electron. Alternatively, we can’t excite hydrogen electronically unless we put in the correct amount of energy. Unexcited hydrogen Excited hydrogen energy h given off

14. Balmer equation 1/ = R[1/2 2 -1/n 2 ] where R = 1.097x10 -2 nm -1 and n is some integer > 2 The energy of the light observed in the visible region is only a portion of the light emitted by a hydrogen atom

15. ionization potential I.E.

16. Balmer Rydberg equation 1/ = R[1/m 2 -1/n 2 ] where R = 1.097x10 -2 nm -1 and n is some integer > m. This equation accounts for the lines observed for hydrogen both in the visible region and elsewhere. Why are these other lines also included?

17. Why are the orbital energies of hydrogen written as 1s, 2s, 2p, 3s, 3p, 3d….? Also why the difference in energy between the 2s and 2p level, for example, in a multi-electron atom? Many more emission lines are observed in multi electron atoms. These terms are used to describes the levels an electron can occupy

18. How does the emission spectrum of multi-electron atoms look like? http://jersey.uoregon.edu/vlab/elements/Elements.html

19. 3d 3p 3s 2p 2s 1s multi-electron atom 3d 3p 3s 2p 2s 1s multi-electron states in a magnetic field Energy Absorption lines were observed to increase in magnetic field States needed to explain emission lines in:

20. What is observed in the spectra of multielectron atoms are multiple lines closely spaced followed by big gaps. The number of lines observed with other atoms are numerous and beyond our concern. We will be interested in summarizing the theory that has been developed to explain these emision lines.

21. Attempts to explain the emission (and absorption spectra) of atomic hydrogen and the other atoms, resulted in discovery/development of a mathematical equation with properties that mimicked the observed spectra of atoms. Schroedinger Equation is a differential equation: Properties of a differential equation: 1.the equation may have more than one solution. 2.any combination of solutions (sum or difference) is also a solution Solutions to this equation are found only when certain terms in the equation have unique values: these terms have been called quantum numbers and have been given the symbols: n, l, m, and s.

22. The quantum numbers have names and also must have certain relationships between each other, otherwise the equation vanishes (has no solution) n = principle quantum number, must have integer values of 1, 2, 3, … L = is called the angular momentum quantum number and must have integer values from –(n-1), -(n-2), …0 m = is called the magnetic quantum number and can have values from –L, (L-1),..0,..+L ( cap L used because lowercase l looks like the number 1) s = is called the spin quantum number, must be +1/2 or –1/2 Each electron in an atom is assigned 4 quantum numbers; no two electrons can have the same 4 quantum numbers or the solution vanishes?

24. What do the solutions to the Schroedinger Equation look like and what information do they provide. The solutions are mathematical equations often described in spherical coordinates. What are spherical coordinates? What are Cartesian coordinates?

25. Cartesian Coordinates x y z. (x 1,y 1,z 1 )

26. Spherical Coordinates x y z. (r, ,  )  

27. Some solutions to the Schroedinger Equation Solution to this equation are called  (psi) What do they look like:  1s = (1/  a 3 )(2.718) r/a where a is a constant 5.29*10 -9 cm and r is the distance of the particle from the origin (n=1, l =0)  2s = 1/4(1/2  a 3 ).5 (2-r/a)(2.718) r/2a (n = 2, l=0)  2p = 1/4(1/2  a 3 ).5 (r/a)(2.718) r/2a cos  (n = 2, l=1) … What is the physical interpretation of the information they provide? The functions  (psi) are amplitude functions, when squared and multiplied by an element of volume, they provide the probability of finding an electron at some location ((r, ,  ) in space. What do these functions look like?

28. 1s n = 1, l =0 2s n = 2, l = 0 3s n = 3, l = 0  1s  2s  3s a node is a region where the function = 0 + -+-+ + - +

29. What do the p orbital look like? How do they compare in energy to s orbitals?

30. P orbitals n = 2, l = 1, m = -1n = 2, l = 1, m =0 n = 2, l = 1, m = 1  2p - ++ - -+-+

31. What do the d orbitals look like and how many are there? How do they compare in energy to p orbitals?

32.  3d  4f

33. Why are these orbitals significant: These orbitals are solutions to the Schroedinger Equation for the hydrogen atom. However they are very useful because they provide a model to mimic the behavior observed for the remaining element in the periodic table. Rules for predicting the electronic properties of the remaining elements of the periodic table: 1.Electrons want to occupy orbitals with the lowest energy possible 2.No two electrons can have the same four quantum numbers 3.Electrons repel each other and prefer to go in orbitals of equal energy that are unoccupied; they prefer to go in with the same spin (Hund’s rule) 4.A maximum of 2 electron are possible in any given orbital

34. H 1 proton and 1 electron Designation: 1s 1

35. Remember, if we excite hydrogen, we can excite it to a 2s level, 3s level, 4s level, and then it can decay from any one of these leves to a lower level by emitting a specific wavelength of light. This model explains the observed spectra of hydrogen, both emission (light given off from an exited state to one of lower energy) or absorption (light aborbed in going from the ground state to an excited state)

36. He has 2 protons and 2 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 2 protons Designation: 1s 2 Also note that this fills the 1s level; the next level is much higher in energy

37. Li has 3 protons and 3 electrons; note that the orbital energy scale will change again because each electron will be attracted to a nucleus that has 3 protons Designation: 1s 2 2s 1

38. Be has 4 protons and 4 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 4 protons Designation: 1s 2 2s 2

39. B has 5 protons and 5 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 5 protons Designation: 1s 2 2s 2 2p 1

40. C has 6 protons and 6 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 6 protons Designation: 1s 2 2s 2 2p 2 Note Hund’s rule: electrons occupy different orbitals with the same spin

41. N 7 has protons and 7 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 7 protons Designation: 1s 2 2s 2 2p 3

42. O has 8 protons and 8 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 8 protons Designation: 1s 2 2s 2 2p 4

43. F has 9 protons and 9 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 9 protons Designation: 1s 2 2s 2 2p 5

44. Ne has 10 protons and 10 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 9 protons Designation: 1s 2 2s 2 2p 6 Also note that this fills this level

45. Na has 11 protons and 11 electrons; note that the orbital energy scale will change because each electron will be attracted to a nucleus that has 11 protons Designation: 1s 2 2s 2 2p 6 3s 1

46. Name the element with the following electronic configurations 1s 2 2s 2 2p 6 3s 1 (Ne 3S 1 ) Na 1s 2 2s 2 2p 6 3s 2 Mg 1s 2 2s 2 2p 6 3s 2 3p 6 Ar 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 K 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 5 Mn 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 3p 3 As

47. In a multi-electron atom, which orbital shape do you think shields the nucleus best to an electron further out in space? s orbital p orbital d orbital f orbital

48. Some anomalous electron configurations Stability associated with half filled and fully filled shells Cr [Ar] 4s 2 3d 4  [Ar] 4s 1 3d 5 Cu [Ar] 4s 2 3d 9  [Ar] 4s 1 3d 10

49. Which of the following combination of quantum numbers can refer to any electron in a ground state Co atom (Z =27)? 1.n = 3, l = 0, m l = 2 2.n = 4, l = 2 m l = -2 3.n = 3, l =1, m l = 0 Using the periodic table to assist in determining electric configurations ground state Co 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 7 n = 3, l =0 is a s orbital, m l = 0 n= 4, l =2 is a 4d orbital, m l = -2, -1, 0, 1, 2 n= 3, l =1 3p orbital, m l = -1, 0,1

50. Which of the following electron configurations refer to an excited state of V? [Ne]3s 2 3p 6 4s 2 3d 3 [Ne]3s 2 3p 6 4s 2 3d 2 3f 1 [Ne]3s 2 3p 6 4s 2 3d 2 4p 1 ground state V 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 3 [Ne]3s 2 3p 6 4s 2 3d 2 4p 1

51. Which of the following electron configurations refer to an excited state of V +3 ? [Ne]3s 2 3p 6 4s 2 [Ne]3s 2 3p 6 4s 2 3d 2 4f 1 [Ne]3s 2 3p 6 4s 1 3d 0 4p 1 ground state V +3 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 0 [Ne]3s 2 3p 6 4s 1 3d 0 4p 1

52. What is the electronic configuration of Se? 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4

53. How many unpaired electrons are there in a.K? b.Cr? c.Fe?