C H E M 1 CHEM 101-GENERAL CHEMISTRY CHAPTER 6 THE PERIODIC TABLE & ATOMIC STRUCTURE INSTR : FİLİZ ALSHANABLEH

Transcription

1 C H E M 1 CHEM 101-GENERAL CHEMISTRY CHAPTER 6 THE PERIODIC TABLE & ATOMIC STRUCTURE 0 1 INSTR : FİLİZ ALSHANABLEH

2 CHAPTER 6 THE PERIODIC TABLE & ATOMIC STRUCTURE The Electromagnetic Spectrum The Wave Nature of Light The Particulate Nature of Light Model of the Atom Quantum Numbers The Pauli Exclusion Principle and Electron Configurations The Periodic Table and Electron Configurations Periodic Trends in Atomic Properties 2

3 The Electromagnetic Spectrum Visible light is a small portion of the electromagnetic radiation spectrum detected by our eyes. Electromagnetic radiation includes radio waves, microwaves and X-rays. Described as a wave traveling through space. There are two components to electromagnetic radiation, an electric field and magnetic field. 3

4 The Wave Nature of Light Wavelength, λ, is the distance between two corresponding points on a wave. Amplitude is the size or height of a wave. Frequency, ν, is the number of cycles of the wave passing a given point per second, usually expressed in Hz. The fourth variable of light is velocity. All light has the same speed in a vacuum. c = x 10 8 m/s The product of the frequency and wavelength is the speed of light. c = λν 4

5 The Wave Nature of Light Refraction is the bending of light when it passes from one medium to another of different density. Speed of light changes. Light bends at an angle depending on its wavelength. Light separates into its component colors. 5

6 The Wave Nature of Light Electromagnetic radiation can be categorized in terms of wavelength or frequency. Visible light is a small portion of the entire electromagnetic spectrum. 6

7 Example Problem 6.1 Neon lights emit an orange-red colored glow. This light has a wavelength of 670 nm. What is the frequency of this light? Answer: c = λν 1 nm = 1x 10-9 m ν = c / λ = (3.00 m/s) / (670x 10-9 m) ν = 4.5 x s -1 7

8 The Particulate Nature of Light Photoelectric effect: light striking a metal surface generates electrons. The light s energy is transferred to electrons in metal. With sufficient energy, electrons break free of the metal. Electrons given more energy move faster (have higher kinetic energy) when they leave the metal. Particles of light are called photons. 8

9 The Photoelectric Effect The energy of a photon (E) is proportional to the frequency (ν). and is inversely proportional to the wavelength (λ). h = Planck s constant = x J s Binding Energy - energy holding an electron to a metal. If frequency of incoming photon sufficiently high, extra energy is imparted to the electrons as kinetic energy. E = hν = hc λ E photon = Binding E + Kinetic E This explains the photoelectric effect. 9

10 Problem 6.18 For photons with the following energies, calculate the wavelength and identify the region of the spectrum they are from: (a) 6.0x10-19 J, (b) 8.7x10-22 J, (c) 3.2x10-24 J, (d) 1.9x10-28 J Answers: λ = hc/e h = 6.63x10-34 Js (a) E = 6.0x10-19 J λ = (6.63x10-34 Js)(3.0x10 8 m/s)/ (6.0x10-19 J) λ = 3.32 x 10-7 m...( UV ) (b) E =8.7x10-22 J, λ = 2.29 x 10-4 m...( IR ) (c) E =3.2x10-24 J, λ = 6.21 x 10-2 m...( Microwaves) (b) E =1.9x10-28 J, λ = 1.05 x 10 3 m...( AM Radiowaves) 10

11 Problem 6.23 When light with a wavelength of 58.5 nm strikes the surface of tin metal, electrons are ejected with the maximum kinetic energy of 2.69 x J. What is the electron binding energy of this electrons to the metal? Answer: E photon = Binding E + Kinetic E E photon = hc / λ = [(6.63x10-34 Js)(3.0x10 8 m/s)]/ (58.5x10-9 m) E photon = 3.4 x J Kinetic E = E photon Binding E Kinetic E = 3.4 x J 2.69 x J Kinetic E = 7.1 x J 11

12 Atomic Spectra Atomic Spectra: the particular pattern of wavelengths absorbed and emitted by an element. Wavelengths are well separated or discrete. Wavelengths vary from one element to the next. Atoms can only exist in a few states with very specific energies. When light is emitted, the atom goes from a higher energy state to a lower energy state. Electrical current dissociates molecular H 2 into excited atoms which emit light with 4 wavelengths. 12

13 The Bohr s Atom Model Bohr model - electrons orbit the nucleus in stable orbits. Electrons move from lower energy to higher energy orbits by absorbing energy. Electrons move from high energy to lower energy orbits by emitting energy as light (photons). Ground state: where the electrons at the lowest possible energy state. Excited state: electrons in a state that is higher then the lowest possible energy state. 13

14 The Quantum Mechanical Model of the Atom Quantum mechanical model replaced the Bohr s atom model. Bohr model depicted electrons as particles in circular orbits of fixed radius. Quantum mechanical model depicts electrons as waves spread through a region of space (delocalized) called an orbital. The energy of the orbitals is quantized like the Bohr model. Electrons exhibit wave-like behavior. Wave behavior described using a wave function, the Schrödinger equation. H is an operator, E is energy and ψ is the wave function. Hψ = Eψ Total energy for electrons includes potential and kinetic energies. 14

15 Orbitals ψ can be written in terms of two components. Radial component, depends on the distance from the nucleus. Angular component, depends on the direction or orientation of electron with respect to the nucleus. Square of the wave function, ψ 2, is always positive and gives probability of finding an electron at any particular point. Each solution of the wave function defines an orbital. Each solution labeled by a letter and number combination: 1s, 2s, 2p, 3s, 3p, 3d, etc. An orbital in quantum mechanical terms is actually a region of space rather than a particular point. 15

16 Quantum Numbers Quantum numbers - solutions to the functions used to solve the wave equation. Quantum numbers used to name atomic orbitals. Vibrating string fixed at both ends can be used to illustrate a function of the wave equation. When solving the Schrödinger equation, three quantum numbers are used. Principal quantum number, n (n = 1, 2, 3, 4, 5, ) Secondary quantum number, l Magnetic quantum number, m l 16

17 Principal Quantum Number (n) The principal quantum number, n, defines the shell in which a particular orbital is found. n must be a positive integer n = 1 is the first shell, n = 2 is the second shell, etc. Each shell has different energies. So n defines the main energy levels. 17

18 Secondary Quantum Number (l ) The secondary quantum number, l, indexes energy differences between orbitals in the same shell in an atom. l has integral values from 0 to n-1. l specifies subshell Each shell contains as many l values as its value of n. 18

19 Secondary Quantum Number (l ) When n = 1 (first energy level) l = 0 (only 1 sublevel) n = 2 ( 2nd energy level) l = 0, 1 ( 2 sublevels) n = 3 (3rd energy level) l = 0,1,2 (3 sublevels) The energies of orbitals are specified completely using only the n and l quantum numbers. 1s (n = 1, l = 0 ) n l 2s (n = 2, l = 0 ) 2p (n = 2, l = 1 ) 19

20 Magnetic Quantum Number (m l ) In magnetic fields, some emission lines split into three, five, or seven components. A third quantum number describes splitting. The third quantum number is the magnetic quantum number, m l. Orientation of the orbital in space relative to the other orbitals in the atom. m l has integer values. m l defines number of orbitals. For l m l = - l,( -l +1),..., 0,..., + l For l = 0 ( s sublevel) m l = 0 ( only 1 orbital) For l = 1 (p sublevel) m l = -1,0,+1 ( 3 orbitals) For l = 2 (d sublevel) m l = -2, -1,0,+1,+2 ( 5 orbitals) 20

21 Quantum Numbers Note the relationship between number of orbitals within s, p, d, and f and m l. 21

22 Exercises Example : What is the number of electrons in the third energy level (n=3)? n=3 l = 0,1,2 l = 0 m l = 0 ( 1 orbital = 2e - ) l = 1 m l = -1,0,+1 ( 3 orbitals= 6e - ) l = 2 m l = -2, -1,0,+1,+2 ( 5 orbitals= 10e - ) Total # of orbitals= 9, Total # of electrons= 18e - # of orbitals in energy level= n 2 # of electrons in energy level= 2n 2 Example : What is the number of orbitals in 4d sublevel? 4d n = 4 l = 2 m l = -2, -1,0,+1,+2 ( 5 orbitals= 10e - ) Example : (Example problem 6.5)Write all of the allowed sets of quantum numbers (n, l, and m l ) for a 3p orbital. 22

23 Visualizing Orbitals s orbitals are spherical p orbitals have two lobes separated by a nodal plane. A nodal plane is a plane where the probability of finding an electron is zero (here the yz plane). d orbitals have more complicated shapes due to the presence of two nodal planes. 23

24 Visualizing Orbitals Cross sectional views of the 1s, 2s, and 3s orbitals. 24

25 Visualizing Orbitals Electron density plots for 3s, 3p, and 3d orbitals showing the nodes for each orbital. 25

26 The Pauli Exclusion Principle and Electron Configurations The spin quantum number, m s, determines the number of electrons that can occupy an orbital. m s = ±1/2 Electrons described as spin up or spin down. An electron is specified by a set of four quantum numbers. Pauli Exclusion Principle - no two electrons in an atom may have the same set of four quantum numbers. Two electrons can have the same values of n, l, and m l, but different values of m s. Two electrons maximum per orbital. Two electrons occupying the same orbital are spin paired. 26

27 Problem 6.38 & Problem 6.41 Prob Which of the following represent valid sets of quantum numbers? For a set that is invalid, explain briefly why it is not correct. (a) n=3, l = 3, m l = 0 (b) n=2, l = 1, m l = 0 (c) n=6, l = 5, m l = -1 (d) n=4, l = 3, m l = -4 Prob What is the maximum number of electrons in an atom that can have the following quantum numbers? (a) n=2 (2n 2 = 2(2) 2 = 8 e - ) (b) n=3, l = 1 (l = 1 (p sublevel) = 6 e - ) (c) n=3, l = 1, m l =0 (m l =0 (only one orbital) = 2 e - ) (d) n=3, l = 1, m l = -1, m s = -1/2 ( 1 e - ) 27

28 Orbital Energies and Electron Configurations Electrons in smaller orbitals are held more tightly and have lower energies. So electrons fill first the lowest permitted energy levels. The energy ordering for atomic orbitals is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p. An orbital s size and penetration when treated quantitatively produces the order of filling represented. Electronic configurations are written in order of energy for atomic orbitals. 28

29 The Aufbau Principle & Hund s Rule Aufbau principle - when filling orbitals, start with the lowest energy and proceed to the next highest energy level. Hund s rule - within a subshell, electrons occupy the maximum number of orbitals possible. Electron configurations are sometimes depicted using boxes to represent orbitals. This depiction shows paired and unpaired electrons explicitly. 29

30 Electron Configurations & Orbital Diagrams A simplified depiction uses superscripts to indicate the number of electrons in an orbital set. 1s 2 2s 2 2p 2 is the electronic configuration for carbon ( 6 C). is the orbital diagram of C Noble gas electronic configurations are used as a shorthand for writing electronic configurations. Electrons in outermost occupied orbitals give rise to chemical reactivity of an element. [He] 2s 2 2p 2 is the shorthand for carbon 30

31 Hund s Rule and the Aufbau Principle The inner electrons, which lie closer to the nucleus, are referred to as core electrons. 6C = [He] 2s 2 2p 2 Core electrons can be represented by the noble gas with the same electronic configuration. (# of core e - for 6 C = 2 e - ) The outer electrons are usually referred to as valence electrons. Valence electrons are shown explicitly when a noble gas shorthand is used to write electronic configurations. (# of valence e - for 6 C = 4 e - ) 31

32 Exercise Write the electron configuration for helium, flourine, neon, sulfur, sodium, zinc Helium, 2 He : 1s 2 = [He] Flourine, 9 F : 1s 2 2s 2 2p 5 = [He] 2s 2 2p 5 Neon, 10Ne : 1s 2 2s 2 2p 6 = [He] 2s 2 2p 6 = [Ne] Sodium 11 Na : 1s 2 2s 2 2p 6 3s 1 = [Ne] 3s 1 Sulfur 16S : 1s 2 2s 2 2p 6 3s 2 3p 4 = [Ne] 3s 2 3p 4 Zinc 30Zn: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 = [Ar] 4s 2 3d 10 32

33 Problem 6.55 & Problem 6.56 Prob Which of these electron configurations are for atoms in the ground state? In excited state? Which are impossible? (a) 1s 2 2s 1 (the ground state electron conf. of 3 Li ) (b) 1s 2 2s 2 2p 3 (the ground state electron conf. of 7 N ) (c) [Ne] 3s 2 3p 3 4s 1 (the excited state electron conf. of 16 S ) (d) [Ne] 3s 2 3p 6 4s 3 3d 2 (impossible) (e) [Ne] 3s 2 3p 6 4f 4 (the excited state electron conf. of 22 Ti ) (f ) 1s 2 2s 2 2p 4 3s 1 (the excited state electron conf. of 9 F ) Prob From the list of atoms and ions given, identify any pairs that have the same electron configurations and write that configuration. Na +, S 2, Ne, Ca 2+, Fe 2+, Kr, I 11Na +, 16 S 2, 10 Ne, 20 Ca 2+, 26 Fe 2+, 36 Kr, 53 I Na +, Ne = 10 e - = 1s 2 2s 2 2p 6 Ca 2+, S 2- = 18 e - = 1s 2 2s 2 2p 6 3s 2 3p 6 33

34 Exercise Write the four quantum number of each electron for oxygen atom. Oxygen, 8 O : 1s 2 2s 2 2p 4 = [He] 2s 2 2p 4 1s 2 2s 2 2p 4 e - no. n l m l m s / /2 1s / /2 2s / / /2 2s /2 34

35 The Periodic Table and Electron Configurations The periodic table and the electronic configurations predicted by quantum mechanics are related. The periodic table is broken into s, p, d, and f blocks. Elements in each block have the same subshell for the highest electron. Structure of periodic table can be used to predict electronic configurations. 35

36 The Periodic Table and Electron Configurations The shape of the periodic table can be broken down into blocks according to the type of orbital occupied by the highest energy electron in the ground state. 36

37 Periodic Trends in Atomic Properties Using the understanding of orbitals and atomic structure, it is possible to explain some periodic properties. Atomic size Ionization energy Electron affinity 37

38 Atomic Size The shell in which the valence electrons are found affects atomic size. The size of the valence orbitals increases with n, so size must increase from top to bottom for a group. The strength of the interaction between the nucleus and the valence electrons affects atomic size. The effective nuclear charge increases from left to right across a period, therefore the valence electrons are drawn closer to the nucleus, decreasing the size of the atom. 38

39 Atomic Size 39

40 Example Problem 6.9 Using only the periodic table, rank the following elements in order of increasing size: Fe, K, Rb, S, and Se. S K Fe Se Rb S < Se < Fe < K < Rb 40

41 Problem 6.67 Using only the periodic table, arrange each of the following series of atoms in order of increasing size. (a) Na, Be, Li (b) P, N, F, (c) I, O, Sn (a) Li Be Be < Li < Na Na (b) N F F < N < P P (c) O O < I < Sn Sn I 41

42 Ionization Energy Ionization energy - the energy required to remove an electron from a gaseous atom, forming a cation. Formation of X + is the first ionization energy, X 2+ would be the second ionization energy, etc. X(g) X + (g) + e Effective nuclear charge increases left to right across a period. The more strongly held an electron is, the higher the ionization energy must be. As valence electrons move further from the nucleus, they become easier to remove and the first ionization energy becomes smaller. 42

43 Ionization Energy Graph of the first ionization energy (in kj/mol) vs. atomic number for the first 38 elements. 43

44 Example Problem 6.10 Using only the periodic table, rank the following elements in order of increasing ionization energy: Br, F, Ga, K, and Se. F K Ga Se Br K < Ga < Se < Br < F 44

45 Electron Affinity Electron affinity - energy required to place an electron on a gaseous atom, forming an anion. X(g) + e X (g) Electron affinities may have positive or negative values. Negative values - energy released Positive values - energy absorbed Electron affinities increase (numerical value becomes more negative) from left to right for a period and bottom to top for a group. The greater (more negative) the electron affinity, the more stable the anion will be. 45

46 Electron Affinity Graph of electron affinity (in kj/mol) vs. atomic number for the first 20 elements. 46

47 Problem 6.74 Indicate which species in each pair has the more favorable (more negative) electron affinity. (a) Cl or S, (b) S or P, (c) Br or As P S Cl As Br (a) S < Cl (b) P < S (c) As < Br 47

48 Problem 6.75 Compare the elements Na, B, Al and C witht regard the following properties (a) Which has the largest atomic radius (b) Which has the most negative electron affinity. (c) Place the elements in order of increasing ionization energy B C Na Al (a) The largest radius Na (b) The most negative electron affinity C (c) The order of ionization energy Na < Al < B < C 48