1. Quantum Theory and the Electronic Structure of Atoms Chapter 7 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Recall: electrons are outside the nucleus. Electronic structure of atoms: arrangements of electrons in an atom Z: atomic number = no. of protons in the nucleus of an atom In atoms; no. of protons = no. of electrons Elements in a group exhibit similar chemical properties IA metals  cations of +1 charge IIA metals  cations of +2 charge Halogens (VIIA)  monoatomic anions of -1 charge (halides) Noble Gases (VIIIA) are not reactive or inert All of the above can be explained based on comparisons of electronic structure. Electronic structure models were obtained from the interaction of light with elements.

3. Properties of Waves Wavelength ( ) is the distance between two adjacent peaks or two adjacent troughs or the length of 1 cycle. (SI unit = m) Frequency ( ) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = x  distance in m travelled in 1 s Amplitude is the vertical distance from the midline of a wave to the peak or trough. 7.1 Wave: oscillation; A cycle of a wave = 1 peak + 1 trough

4. Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 10 8 m/s All electromagnetic radiation x  c 7.1

6. 1. A laser used to fuse detached retinas produces light of frequency 4.69 x 10 14 s -1. Calculate the wavelength of this light in nm. x  c  x 4.69 x 10 14 s -1 = 3.00 x 10 8 m s -1  3.00 x 10 8 m s -1 /4.69 x 10 14 s -1 = 6.40 x 10 -7 m= 640 nm 2. The yellow light given off by a Na vapor lamp has a wavelength of 589 nm. What is the frequency of this light? x  c 589 x 10 -9 m x  3.00 x 10 8 m s -1  3.00 x 10 8 m s -1 /589 x 10 -9 m = 5.09 x 10 14 s -1

7. x = c = c/ = 3.00 x 10 8 m/s / 6.0 x 10 4 Hz = 5.0 x 10 3 m Radio wave A light has a frequency of 6.0 x 10 4 Hz (= s -1 ). Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? = 5.0 x 10 12 nm 7.1

8. Quantum Theory of Max Planck (1900) Energy (light) is emitted or absorbed in discrete units called a quantum. That is light behaves as if it is made up of packets of energy. Each packet of energy is called a photon (an imaginary particle of light). The energy, E, of a photon is called a quantum. E = h x Planck’s constant (h) h = 6.63 x 10 -34 J s 7.1

9. 1.A laser used to fuse detached retinas produces light of frequency 4.69 x 10 14 s -1. (a) Calculate the energy of 1 photon of this light. (b) Calculate the energy of 1 mole of photons of this light. 2.The yellow light given off by a Na vapor lamp has a wavelength of 589 nm. If the total energy given off is 15.0J, how many photons were present? 1.(a) Energy of 1 photon, E = h  = 6.63 x 10 -34 Js × 4.69 x 10 14 s -1 = 3.10 × 10 -19 J. (b) (3.10 × 10 -19 J/photon) × (6.022 × 10 23 photons/mole) = 1.87 × 10 5 J/mole 2.Energy of 1 photon, E = hc  = =6.63 x 10 -34 Js × 3.00 x 10 8 ms -1 /(589 ×10 -9 m) = 3.38 ×10 -19 J Assume there are n photons. The energy of n photons = n×E = 15.0J. 15.0 J = n× 3.38 ×10 -19 J, Therefore n= number of photons = 15.0J/ 3.38 ×10 -19 J = 4.44 ×10 19 photons

10. h = KE + BE “ Photoelectric Effect”( Einstein in 1905) KE = h - BE h KE e - 7.2 BE: binding energy = measure of how tightly the electron is ‘bound’ in atom KE: kinetic energy of ejected electron =energy of photon - BE Photoelectric effect is the ejection of electrons from a metal surface when light of frequency above a certain minimum value strikes the metal surface. It established that light has both wave nature and particle nature. This is called the wave-particle duality of light. Photon is an imaginary “particle” of light

11. E = h x E = 6.63 x 10 -34 (J s) x 3.00 x 10 8 (m/s) / 0.154 x 10 -9 (m) E = 1.29 x 10 -15 J E = h x c /  7.2 When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.

12. 7.3 Line Emission Spectrum of Hydrogen Atoms The emission spectrum of an element consists of the lights of different frequencis given off by the hot vapor of that element.

13. 7.3

14. The lines in the emission spectrum of an element correspond to lights of different frequencies and hence different energies. Emission lines are due to the movement of electrons or electron transitions from regions of higher energy to regions of lower energy. When an electron moves from a higher energy region to a lower energy region, the difference in energy is given off as a photon of that energy.  E = difference in energy = h = energy of the photon emitted Therefore an accurate model of the electronic structure has to be able to account for the lines in the emission spectrum by accurately describing the allowed locations of the electrons. Niels Bohr came up with the first model of electronic structure that could account for the line spectrum of H. But the Bohr model failed for all other atoms. The Bohr model is only applicable to 1 electron species: H, He +, Li +2, Be +3..

15. Bohr Model 1.The electron in a H atom moves in fixed circular paths called orbits. It cannot be in any region between orbits. 2.The orbits are numbered starting with the one closest to the nucleus in whole number values. These numbers are called principal quantum numbers, n. The orbit closest to the nucleus is n=1 3.The energy of electron, E n, in an orbit n is fixed; 4.R H = Rydberg constant = 2.18 x 10 -18 J 5.There is no upper limit to the value of n. When n is very large (infinity), E = 0. That is, the electron is not bound to the atom. It is free and we have H +. 6.Orbits with larger n have higher energy and larger radii. 7.When an electron moves from a higher energy orbit to a lower energy orbit, it gives off the excess energy,  E, as a photon of that energy. 8.When an electron moves from a lower energy orbit to a higher energy orbit, it absorbs the additional energy,  E, as a photon of that energy.

16. 1.e - can only have specific (quantized) energy values 2.light is emitted as e - moves from one energy level to a lower energy level Bohr’s Model of the Atom (1913) E n = -R H ( ) 1 n2n2 n (principal quantum number) = 1,2,3,… R H (Rydberg constant) = 2.18 x 10 -18 J 7.3

17. E = h 7.3

18. E photon =  E = E f - E i E f = -R H ( ) 1 n2n2 f E i = -R H ( ) 1 n2n2 i i f  E = R H ( ) 1 n2n2 1 n2n2 n f = 1 n i = 2 n f = 1 n i = 3 n f = 2 n i = 3 7.3

19. Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 2 state.

20. E photon = 2.18 x 10 -18 J x (1/25 - 1/9) E photon =  E = -1.55 x 10 -19 J = 6.63 x 10 -34 (Js) x 3.00 x 10 8 (m/s)/1.55 x 10 -19 J = 1280 nm Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. E photon = h x c /  = h x c / E photon i f  E = R H ( ) 1 n2n2 1 n2n2 E photon = 7.3

21. Light has dual wave-particle nature. Matter also has dual wave particle nature. Wave nature of matter was quantified by Louis de Broglie. de Broglie equation, the wavelength of a matter wave, = h/(mv) h= Planck’s constant; m= mass of the matter in kg; v= velocity of the matter in m/s. For very tiny particles such as electrons, protons, neutrons, the wave nature cannot be ignored. The Bohr model did not incorporate the wave nature of the electron and hence it failed. The Bohr model also does not account for the electron-electron repulsions when there are more than 1 electron in the atom. What is still correct about the Bohr model is that the energy of the electron can take only certain fixed values – that is, the energy of the electron is quantized. Since the energy of the electron can take only certain fixed values, the wavelengths of the electron wave can also take only certain fixed values – they are quantized

22. De Broglie (1924) reasoned that e - is both particle and wave. Why is e - energy quantized? 7.4 u = velocity of e- m = mass of e- 2  r = N = h mu

23. What is the de Broglie wavelength (in nm) of an electron (mass = 9.109×10 −28 g) travelling at a speed of 15.0 m/s = h/(mv) m = 9.109×10 −28 g = 9.109×10 −31 kg =6.63 x 10 -34 Js/(9.109×10 −31 kg x 15.0m/s) = 4.85 x 10 -5 m = 4.85 x 10 4 nm What is the de Broglie wavelength (in nm) of a ping-pong ball(mass = 2.5 g) travelling at a speed of 15.0 m/s = h/(mv) m = 2.5 g = 2.5×10 -3 kg =6.63 x 10 -34 Js/(2.5×10 −3 kg x 15.0m/s) = 1.77 x 10 -32 m = 1.77 x 10 -23 nm

24. = h/mu = 6.63 x 10 -34 / (2.5 x 10 -3 x 15.6) = 1.7 x 10 -32 m = 1.7 x 10 -23 nm What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s? m in kgh in J su in (m/s) 7.4

25. Chemistry in Action: Electron Microscopy STM image of iron atoms on copper surface e = 0.004 nm What was the speed of the electrons used?

26. Erwin Schrodinger formulated the current model of electronic structure: quantum mechanical model or wave mechanical model or the Schrodinger model. His model incorporates the wave and particle nature of the electron. It incorporates Planck’s quantum theory. That is, it accounts for the energy and the wavelengths of the electron having only certain fixed values Schrodinger solved an equation to obtain expressions for the electron waves for various allowed values of the energy. These expressions are called wavefunctions or ORBITALS and describe the allowed electron waves.

27. Schrodinger Wave Equation In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e - Wave function (  ) describes: 1. energy of e - with a given  2. probability of finding e - in a volume of space Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems. 7.5

28. 4 quantum numbers are required to describe an electron wave, . These 4 quantum numbers are: 1.The principal quantum number, n 2.The angular momentum or azimuthal quantum number, l 3.The magnetic moment quantum number m l 4.The spin quantum number, m s The energy of the electron and its distance from the nucleus is determined by the principal quantum number, n. The shape of the region of space around the nucleus where the electron is likely to be found is determined by the angular momentum or azimuthal quantum number, l. The orientation (direction) of this region in 3D is determined by the magnetic moment quantum number m l The electron can spin in two possible directions (clockwise or anticlockwise). The spin direction is specified by the two possible values of the spin quantum number, m s.

30. Schrodinger Wave Equation  fn(n, l, m l, m s ) principal quantum number n n = 1, 2, 3, 4, …. n=1 n=2 n=3 7.6 distance of e - from the nucleus

31. Allowed values of n for an electron wave = 1, 2, ….. In nature, we see n values of 1, 2, 3, 4, 5, 6, 7 Allowed values of l depend on the value of n. For every allowed value of n, there are n allowed values of l ranging from 0 to n-1. Each l value suggests a different shaped region. To describe all the atoms in the periodic table we need up to 4 l values although the model does not have that restriction. n l 10 20, 1 30, 1,2 40, 1, 2, 3 50, 1, 2, 3, 4 60, 1, 2, 3, 4, 5 70, 1, 2, 3, 4, 5, 6

32. Each of the 4 different l values is assigned a letter. All orbitals with the same value of l have the same shape. All s orbitals are spherical All p orbitals have two lobes with the nucleus at center d orbitals have 4 lobes with the nucleus at center f orbitals have 6 lobes with the nucleus at center. For H atom, the electron’s energy only depends on the value of n, E=-R H /n 2 For all other atoms, the energy of electron depends both on the n value and the l value. As the value of n increases energy increases. For a given n, as the value of l increases energy increases. Orbital’s l value Letter 0s 1p 2d 3f

33.  = fn(n, l, m l, m s ) angular momentum quantum number l for a given value of n, l = 0, 1, 2, 3, … n-1 n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 Shape of the “volume” of space that the e - occupies l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital Schrodinger Wave Equation 7.6

34. For every l value there are 2 l +1 directions. Each direction corresponds to a different orbital. There is only one s orbital ( l =0) for any n There are three p orbitals ( l =1) for any n > 1 There are five d orbitals ( l =2) for any n > 2 There are seven f orbitals ( l =3) for any n > 3 Orbitals that differ in just their direction are assigned different values of the magnetic quantum number, ml. For every l value, values of m l range from – l to + l for a total of 2 l + 1 values. All orbitals with the same value of n and l have the same energy. All orbitals with the same value of n constitute a SHELL. All orbitals with the same value of n and l constitute a subshell. A shell n has n 2 orbitals. Each of these orbitals differs in at least their l or m l value. A subshell (n, l ) has 2 l + 1 orbitals that differ in the value of m l. To describe an orbital we need 3 quantum numbers: (n, l, m l ) To describe the spin of the electron along with its orbital: 4 quantum numbers: (n, l, m l, m s )

35. l = 0 (s orbitals) l = 1 (p orbitals) 7.6

36. l = 2 (d orbitals) 7.6

37.  = fn(n, l, m l, m s ) magnetic quantum number m l for a given value of l m l = -l, …., 0, …. +l orientation of the orbital in space if l = 1 (p orbital), m l = -1, 0, or 1 if l = 2 (d orbital), m l = -2, -1, 0, 1, or 2 Schrodinger Wave Equation 7.6

38. m l = -1m l = 0m l = 1 m l = -2m l = -1m l = 0m l = 1m l = 2 7.6

39.  = fn(n, l, m l, m s ) spin quantum number m s m s = +½ or -½ Schrodinger Wave Equation m s = -½m s = +½ 7.6

40. Existence (and energy) of electron in atom is described by its unique wave function . Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. Schrodinger Wave Equation  = fn(n, l, m l, m s ) 7.6 Consequence: Any orbital can accommodate a maximum of 2 electrons, one with m s = ½ and the other with m s = -½.

41. 7.6

42. Schrodinger Wave Equation  = fn(n, l, m l, m s ) Shell – electrons with the same value of n Subshell – electrons with the same values of n and l Orbital – electrons with the same values of n, l, and m l How many electrons can an orbital hold? If n, l, and m l are fixed, then m s = ½ or - ½  = (n, l, m l, ½ ) or  = (n, l, m l, - ½ ) An orbital can hold 2 electrons 7.6

43. How many 2p orbitals are there in an atom? 2p n=2 l = 1 If l = 1, then m l = -1, 0, or +1 3 orbitals How many electrons can be placed in the 3d subshell? 3d n=3 l = 2 If l = 2, then m l = -2, -1, 0, +1, or +2 5 orbitals which can hold a total of 10 e - 7.6

44. Energy of orbitals in a single electron atom Energy only depends on principal quantum number n E n = -R H ( ) 1 n2n2 n=1 n=2 n=3 7.7

45. Energy of orbitals in a multi-electron atom Energy depends on n and l n=1 l = 0 n=2 l = 0 n=2 l = 1 n=3 l = 0 n=3 l = 1 n=3 l = 2 7.7

46. Ground state: lowest energy arrangement Excited States: Higher energy arrangements Rules for achieving Ground state: “Fill up” electrons in lowest energy orbitals (Aufbau principle) Orbitals filled following both i. Pauli exclusion Principle: Maximum of 2 electrons in any orbital ii. Hund’s rule of maximum multiplicity: Orbitals in a subshell are filled 1 electron at a time before pairing up Electrons represented by arrows; arrow pointing up: ms =1/2; arrow pointing down: ms = -1/2 Orbitals represented as squares; s subshell : 1 square; p subshell: 3 squares connected together; d subshell: 5 squares connected together; f subshell: 7 squares connected Orbital diagram: diagram showing the distribution of electrons in the various orbitals (squares) with electrons (arrows) 7.9

47. Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s 7.7

48. 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

49. Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. 1s 1 principal quantum number n angular momentum quantum number l number of electrons in the orbital or subshell Orbital diagram H 1s 1 7.8

50. Core electrons: electrons in inner shells Valence electrons : electrons in outer shells For any element the number of core electrons = number of electrons in the NOBLE GAS in PREVIOUS PERIOD. Valence electrons: remaining electrons (outside the core) Ground state electron configuration: lowest energy arrangement of electrons (aufbau principle; pauli exclusion principle; Hund’s rule) Arrangement of core electrons = ground state electron configuration of NOBLE GAS in PREVIOUS PERIOD. The ground state electron configuration can be condensed by replacing the core electron configuration by following: [Symbol of NOBLE GAS in PREVIOUS PERIOD]

51. What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 2 + 2 + 6 + 2 = 12 electrons 7.8 Abbreviated as [Ne]3s 2 [Ne] 1s 2 2s 2 2p 6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 3p 5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n = 3l = 1m l = -1, 0, or +1m s = ½ or -½

52. Outermost subshell being filled with electrons 7.8

54. Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p 7.8

55. Diamagnetic Elements in ground state: alkaline earth metals, noble gases, Zn, Cd, Hg: repelled slightly by an external magnetic field Paramagnetic elements: all the others: attracted by an external magnetic field To determine how many unpaired electrons, draw the orbital diagram for the VALENCE ELECTRONS ALONE.

56. Chemistry Mystery: Discovery of Helium In 1868, Pierre Janssen detected a new dark line in the solar emission spectrum that did not match known emission lines In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay). Mystery element was named Helium

57. 1. The maximum number of electrons that can have the following set of quantum numbers n=3, l=2, ml=0 is 1B. 2C. 10D. 18 2. Which one of the following sets of quantum numbers corresponds to an electron in a 4f orbital?: 3. The principle that states that no two electrons in an atom can have the same set of 4 quantum numbers is A. Heisenberg Uncertainity Principle B. Hund’s Rule of Maximum Multiplicity C. Pauli Exclusion Principle D. Planck’s Theory

58. 1. Which set of quantum numbers is permissible? A. n=7, l=4, ml=-3, ms = -1/2B. n=2, l=0, ml=-2, ms=-1/2 C. n=4, l=4, ml=0, ms = 1D. n=3, l=0, ml=0, ms=0 2. For the element Sc, the core electron configuration corresponds to the electron configuration of : A.Ca B. Kr C. Ar D. K 3. For an electron in a 5s subshell which of the following sets of quantum numbers is the only one possible : 4. Orbitals that have the same energy are said to be A. equivalentB. degenerateC. ground state D. excited state

59. 1. The orbital designation when n= 6, l = 3, ml= 2 A. 6dB. 3dC. 6fD. 3p 6d: n= 6; l= 2; 3d: n=3; l=2;3p: n=3; l=1 2. List the values of n, l, m l and ms for electrons in the 5p subshell. 5p: n=5, l=1 mlms 1/2 -1/2 01/2 0-1/2 11/2 1-1/2