1 The Periodic Properties of the Elements David A. Katz Department of Chemistry Pima Community College Tucson, AZ, USA
2 Electron Configurations
3 Lewis Dot Symbols Show the outermost electrons only
4 1 IA Oxidation States 1 H IIA The most common oxidation state(s) is shown in red IIIA IVA VA VIA 3 Li Na K +1 4 Be Mg Ca +2 3 IIIB 21 Sc +3 4 IVB 22 Ti VB 23 V Rb Sr Y Zr Nb Cs Fr Ba Ra La Ac Ce Lu Th Lr 72 Hf Rf 73 Ta Db 6 VIB 24 Cr Mo W +6 7 VIIB 25 Mn Tc Re Sg Bh 5 B Al +3 6 C Si N P O S VIII IB 12 IIB Co Ni Cu Zn Ga Ge As Se Fe 44 Ru Rh Pd Ag Cd In Sn Sb Os Ir Pt Au Hg Tl Pb Bi Hs 109 Mt 110 Ds 111 Rg 112 Uub 113 Uut 114 Uuq 115 Uup 52 Te Po Uuh 17 VII A 9 F 1 17 Cl Br I At Uus Noble Gas 2 He 10 Ne 18 Ar 36 Kr 54 Xe 86 Rn 118 Uuo
5 Melting Points of the Elements Symbol Atomic number Melting point, C Symbol Atomic number Melting point, C Symbol Atomic number Melting point, C Symbol Atomic number Melting point, C H Co I Au He Ni Xe Hg Li Cu Cs Tl Be Zn Ba Pb B Ga La Bi C Ge Ce Po N As Pr At O Se Nd Rn F Br 35 7 Pm Fr Ne Kr Sm Ra Na Rb Eu Ac Mg Sr Gd Th Al Y Tb Pa Si Zr Dy U P Nb Ho Np S Mo Er Pu Cl Tc Tm Am Ar Ru Yb Cm K Rh Lu Bk Ca Pd Hf Cf Sc Ag Ta Es Ti Cd W Fm V In Re Md 101 Cr Sn Os No Mn Sb Ir Lr Fe Te Pt
6 Melting Points of the Elements ing point, C Melt Series Atomic Number
8 Boiling Points of the Elements Symbol Atomic Boiling Atomic Boiling Atomic Boiling Atomic Boiling Symbol Symbol Symbol number point, C number point, C number point, C number point, C H Co I Au He Ni Xe Hg Li Cu Cs Tl Be Zn Ba Pb B Ga La Bi C Ge Ce Po N As Pr At O Se Nd Rn F Br Pm Fr Ne Kr Sm Ra Na Rb Eu Ac Mg Sr Gd Th Al Y Tb Pa 91 Si Zr Dy U P Nb Ho Np S Mo Er Pu Cl Tc Tm Am Ar Ru Yb Cm 96 K Rh Lu Bk 97 Ca Pd Hf Cf 98 Sc Ag Ta Es 99 Ti Cd W Fm 100 V In Re Md 101 Cr Sn Os No 102 Mn Sb Ir Lr 103 Fe Te Pt
9 Boiling Points of the Elements Boiling point, C Series Atomic Number
11 Symbol Atomic number Density g/ml or g/l (gas) Densities of the Elements Symbol Atomic number Density g/ml or g/l (gas) Symbol Atomic number Density g/ml or g/l (gas) Symbol Atomic number H Co I Au He Ni Xe Hg Li Cu Cs Tl Be Zn Ba Pb B Ga La Bi C Ge Ce Po N As Pr At 85 O Se Nd Rn F Br Pm Fr 87 Ne Kr Sm Ra Na Rb Eu Ac Mg Sr Gd Th Al Y Tb Pa Si Zr Dy U P Nb Ho Np S Mo Er Pu Cl Tc Tm Am Ar Ru Yb Cm K Rh Lu Bk Ca Pd Hf Cf Sc Ag Ta Es 99 Ti Cd W Fm 100 V In Re Md 101 Cr Sn Os No 102 Mn Sb Ir Lr 103 Fe Te Pt Density g/ml or g/l (gas)
12 Densities of the Elements 25 Dens sity, g/ml (so olids) and g/ /L (gases) Series Atomic Number
14 Atomic Rdii Radii The atomic radius is defined as one half of the distance between covalently bonded atoms.
15 Metallic and covalent radii Not every element will form covalent bonds with itself
17 Atomic Radii of the main group atoms in pm
18 Atomic Radii Size decreases across a period owing to increase innuclearchargenuclear charge. Each added electron feels a greater + charge. Large Small Increase in Z*
19 Sizes of Transition Elements
20 Periodicity of atomic radius
21 Atomic radius decreases from left to right across a period increases from top to bottom down a group or family Rb has the largest size He has the smallest size
22 Atomic radii vs Atomic radii vs Metallic radii
23 Atomic radii vs Metallic radii for 2 nd Period elements
24 Atomic radii vs Metallic radii for 4 th Period elements
25 Ionization Energy Amount of energy required to remove an electron from the ground state of a gaseous atom or ion. First ionization energy is that energy required to remove first electron. Second ionization energy is that energy required to remove esecond electron, eecto,etc.
26 First Ionization Energies of the Elements Symbol Atomic Ionization Atomic Ionization Atomic Ionization Atomic Ionization Symbol Symbol Symbol number energy, ev number energy, ev number energy, ev number energy, ev H Co I Au He Ni Xe Hg Li Cu Cs Tl Be Zn Ba Pb B Ga La Bi C Ge Ce Po N As Pr At O Se Nd Rn F Br Pm Fr Ne Kr Sm Ra Na Rb Eu Ac Mg Sr Gd Th Al Y Tb Pa Si Zr Dy U P Nb Ho Np S Mo Er Pu Cl Tc Tm Am Ar Ru Yb Cm K Rh Lu Bk Ca Pd Hf Cf Sc Ag Ta Es Ti Cd W Fm V In Re Md Cr Sn Os No Mn Sb Ir Lr Fe Te Pt
27 First Ionization Energies of the Elements, ev in ev 15 Series1 10 Ionization energy Atomic Number
28 Periodicity of first ionization energy, kj/mol
29 Trends in Ionization Energy st Ionization energy (kj/mol) He Ne 1500 Ar Kr H Li Na K Atomic Number
30 First ionization energies of the main group elements
31 Ranking Elements by First Ionization Energy PROBLEM: Using the periodic table only, rank the elements in each of the following sets in order of decreasing IE 1 : (a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs RECALL: IE decreases as you proceed down in a group IE increases as you go across a period SOLUTION: (a) He > Ar > Kr Group 8A(18) IE decreases down a group. (b) Te > Sb > Sn (c) Ca > K > Rb (d) Xe > I > Cs Period 5 elements IE increases across a period. Ca is to the right of K; Rb is below K. I is to the left of Xe; Cs is furtther to the left and down one period.
32 Trends in First Ionization Energies Thetrend in ionization The trend in ionization energies is not linear, there are discontinuities in this trend.
33 Trends in First Ionization Energies The first exception occurs between Groups IIA and IIIA. Electron removed from p orbital rather than s orbital Only one electron in the p orbital Electron farther from nucleus Small amount of repulsionby s electrons Filled s orbital
34 Trends in First Ionization Energies The second exception occurs between Groups VA and VIA. Hlffill Half filled rule Electron removed comes from doubly occupied orbital. Repulsion from other electron in orbital helps in its removal.
35 Factors Affecting Atomic Orbital Energies The Effect of Nuclear Charge (Z effective ) Higher nuclear charge lowers orbital energy (stabilizes the system) by increasing nucleus electron attractions. The Effect of Electron Repulsions (Shielding) Additional electron in the same orbital An additional electron raises the orbital energy through electron electron repulsions. Additional electrons in inner orbitals Inner electrons shield outer electrons more effectively than do electrons in the same sublevel. l
36 Effective Nuclear Charge In a many electron atom, electrons are both attracted to the nucleus and repelled by other electrons. The nuclear charge that anelectron experiences depends on both factors.
37 Effective Nuclear Charge The effective nuclear charge, Z eff, is found by: Z eff = Z S where Z is the atomic number and S is a screening constant, usually close to the number of inner electrons. Note: For simple estimation, we may use the inner number of electrons
38 Effective Nuclear Charge, g, Z* Z* is the nuclear charge experienced by the outermost electrons. Explains why E 2s < E 2p Z* increases across a period owing to incomplete shielding by inner electrons. Estimate Z* = [ Z (no. inner electrons) ] Where Z = total number of electrons Examples: Charge felt by 2s e in Li Z* = 3 2 = 1 Be Z* = 4 2 = 2 B Z* = 5 2 = 3
39 Effective Nuclear Charge Z* is the nuclear charge experienced by the outermost electrons. Electron cloud for 1s electrons
40 Effective Nuclear Charge Z* The 2s electron PENETRATES the region occupied by the 1s electron. 2s electron experiences a higher positive charge than expected.
41 Effective Nuclear Charge, Z* Atom Li Be B C N O F Z* Experienced by Electrons in Valence Orbitals [Values calculated using Slater s Rules] Increase in Z* across a period
42 Slater s Rules In 1930, J.S. Slater formulated the following set of empirical rules for determining the values of the shielding constant σ. Slater's Rules Write out the electronic configuration of the element and group the orbitals in the following order: (1s)(2s, 2p)(3s, 3p)(3d)(4s, 4p)(4d)(4f)(5s, 5p)... To establish the screening constant for any electron, sum up the following contributions: 1. Electrons in groups outside (to the right) of the one being considered do not contribute to the shielding. 2. Electrons in the same group contribute 0.35 to the shielding (except the 1s group, where a contribution of 0.30 is used 3. For s or p electrons being observed, each electron in the (n 1) shell contributes 0.85 to the shielding and each electron in the (n 2), (n 3),... shells contribute 1.00 to the shielding 4. For d or f electrons being observed, each electron in an underlying group contributes 1.00 to the shielding.
43 Slater s Rules Example: Calculate Z* for a 4s and a 3d electron in Zn Determine the electron configuration for Zn (1s) 2 (2s, 2p) 8 (3s, 3p) 8 (3d) 10 (4s) 2 For a 4s electron: Establish the screening constant for the 4s electron σ = (1 x 0.35) + (18 x 0.85) + (10 x 1.00) = Calculate the effective nuclear charge Z*= Z σ = = 4.35 For a 3d electron: Establish the screening constant for the 3d electron σ = (9 x 0.35) + (18 x 1.00) = Calculate the effective nuclear charge Z*= Z σ = = 8.85 From this example, you can see that the 3d electrons experience a much greater positivecharge than the 4s electron and would be held more tightly. Thus, the 4s electrons will be the first removed when Zn is ionized.
44 Orbital Energies Orbital energies drop as Z* increases
45 Theeffect effect of another electron in the same orbital: He vs He +
46 Theeffect effect of other electrons in inner orbitals: Li vs Li 2+
47 The effect of orbital shape
48 Redox Reactions A redox (oxidation reduction) reaction is one in which one eee element e loses electrons and another element gains electrons This is recognized by a change in oxidation number of the elements involved Why do metals lose electrons in their reactions? Why does Mg form Mg 2+ ions and not Mg 3+? Why do nonmetals take on electrons?
49 Ionization Energy IE = energy required to remove an electron from an atom in the gas phase. Mg (g) kj Mg + (g) + e 2 nd IE = energy required to remove a second electron from an atom Mg + (g) kj Mg 2+ (g) + e Mg + has 12 protons and only 11 electrons. Therefore, IE for Mg + > Mg.
50 Ionization Energy Mg (g) kj Mg + (g) + e (1 st IE) Mg (g) 1451 kj Mg (g) + e (2 nd IE) 3 rd IE = energy required to remove a third electron from an atom Mg 2+ (g) kj Mg 3+ (g) + e Mg 2+ has a noble gas electron configuration. Energy cost is very high h to dip into a shell of lower n. This is why ox. no. = Group no.
51 Li Ratio: 2nd IE / 1st IE Na K B Al
52 The first three ionization energies of beryllium (in MJ/mol)
53 Ionization Energies, IE 1 IE 10 It requires more energy to remove each successive electron. Whenall valence electrons have been removed, and the ion is isoelectronic with the nearest noble gas, the ionization energy shows a very large increase
54 Ionization Energies, IE IE 1 7
55 Atomic Electron Ionization Energy, kj mol 1 Number Symbol Config 1st 2nd 3rd 4th 5th 6th 7th 8th 1 H 1s He 1s Li 2s Be 2s B 2s 2 2p C 2s 2 2p N 2s 2 2p O 2s 2 2p F 2s 2 2p Ne 2s 2 2p Na 3s Mg 3s Al 3s 2 3p Si 3s 2 3p P 3s 2 3p S 3s 2 3p CI 3s 2 3p Ar 3s 2 3p K 4s Ca 4s Sc 3d 1 4s Ti 3d 2 4s V 3d 3 4s Cr 3d 5 4s Mn 3d 5 4s Fe 3d 6 4s Co 3d 7 4s Ni 3d 8 4s Cu 3d 10 4s Zn 3d 10 4s Ga 3d 10 4s 2 4p
56 Identifying an Element from Successive Ionization Energies PROBLEM: Name the Period 3 element with the following ionization energies (in kj/mol) and write its electron configuration: IE 1 IE 2 IE 3 IE 4 IE 5 IE ,230 PLAN: Look for a large increase in energy which indicates that all of the valence electrons have been removed. SOLUTION: There is a large increase after removing 3 electrons The largest increase occurs after IE 5, that is, after the 5th valence electron has been removed. Three electrons and five electrons would mean that the valence configuration is 3s 2 3p 3 and the element must be phosphorous, P (Z = 15). The electron configuration is 1s 2 2s 2 2p 6 3s 2 3p 3 OR [Ne] 3s 2 3p 3
57 Ion Sizes: Formation of a cation Li, 152 pm 3e and 3p + + Li +, 78 pm 2e and 3 p + CATIONS are SMALLER than the atoms from which they are formed The proton/electron attraction has increased and size DECREASES.
58 Ion Sizes: Formation of an anion F, 71 pm 9e and 9p + F, 133 pm 10e and 9p + ANIONS are LARGER than the atoms from which they are formed The proton/electron attraction has decreased and size INCREASES. Trends in ion sizes are the same as atom sizes.
59 Ionic Radius Ionic radius is a measure of an ion size in the crystal lattice of an ionic compound. The values are determined by x-ray crystallography Vl fi i dii Values of ionic radii vary with the ionic compound and with the method of measurement
60 Ionic vs. atomic radius Sizes in pm This is an isoelectronic series of the ions Metal atoms are blue Metal atoms are blue Non-metal atoms are red Ions are grey
61 Isoelectronic Series of Main Group Atoms in Groups 6A to 2A
62 Sizes of Ions (Ionic Radii) Ionic radius decreases from left to right across a period for similar charged ions Due to increasing nuclear charge. In an isoelectronic series, ions have the same number of electrons. These ions are isoelectronic with He These ions are isoelectronic with Ne
63 Sizes of Ions (Ionic Radii) Ionic radius increases as you go down a group Due to increasing value of n.
64 Trends in Ion Sizes
65 Arranging Ions by Size PROBLEM: Rank each set of ions in order of decreasing size, and explain your ranking: (a) Ca 2+, Sr 2+, Mg 2+ (b) K +, S 2, Cl (c) Au +, Au 3+ PLAN: Compare positions in the periodic table, formation of positive and negative ions and changes in size due to gain or loss of electrons. SOLUTION: (a) Sr 2+ > Ca 2+ > Mg 2+ These are members of the same Group (2A/2) and therefore decrease in size going up the group. These ions are isoelectronic; (b) S 2 > Cl > K + S 2 is an anion with the smallest Z eff and is the largest K + is a cation with a large Z eff and is the smallest. (c) Au + > Au 3+ The higher the + charge, the smaller the ion.
66 Electron Affinity Energy released when an electron is added to a neutral gaseous atom: Cl + e Cl kj/mol
67 Trends in Electron Affinity In general: Electron affinity increases (becomes more exothermic) going from left to right across a period Electron affinity decreased going down a group Note: Energy released is indicated by a negative sign
68 Trends in Electron Affinity There are two There are two discontinuities in this trend.
69 Trends in Electron Affinity The first occurs between Groups IA and IIA. Added electron must go in a p orbital, not an s orbital. Electron is farther from nucleus and feels repulsion from s electrons.
70 Trends in Electron Affinity The second occurs between Groups IVA and VA. Group VA has no empty orbitals (half filled filled rule) Extra electron must pair up in an occupied orbital, creating repulsion.
71 Writing Electron Configurations of Main Group Ions PROBLEM: Using condensed electron configurations, write reactions for the formation of the common ions of the following elements: (a) Iodine (Z = 53) (b) Potassium (Z = 19) (c) Indium (Z = 49) RECALL: Ions of elements in Groups 1A(1), 2A(2), 6A(16), and 7A(17) are usually isoelectronic with the nearest noble gas. Metals in Groups 3A(13) to 5A(15) lose their outermost p and/or s electrons. SOLUTION: () (a) Iodine (Z = 53) is in Group 7A(17) and will gain one electron to be isoelectronic with Xe: ([Kr]5s 2 4d 10 5p 5 ) + e I ([Kr]5s 2 4d 10 5p 6 ) (b) Potassium (Z = 19) is in Group 1A(1) and will lose one electron to be isoelectronic with Ar: K ([Ar]4s 1 ) K + ([Ar]) + e (c) Indium (Z = 49) is in Group 3A(13) and can lose either one electron or three electrons: I ([K ]5 2 4d ) I + ([K ]5 2 4d 10 ) + In ([Kr]5s 2 4d 10 5p 1 ) In + ([Kr]5s 2 4d 10 ) + e + In ([Kr]5s 2 4d 10 5p 1 ) In 3+ ([Kr] 4d 10 ) + 3e
72 The Period 4 The Period 4 crossover in sublevel energies
73 Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions PROBLEM: Use condensed electron configurations to write the reaction for the formation of each transition metal ion, and predict whether the ion is paramagnetic. (a) Mn 2+ (Z = 25) (b) Cr 3+ (Z = 24) (c) Hg 2+ (Z = 80) PLAN: SOLUTION: Write the electron configuration and remove electrons starting with ns to match the charge on the ion. If the remaining configuration has unpaired electrons, it is paramagnetic. (a) Mn 2+ (Z = 25) Mn([Ar]4s 2 3d 5 ) Mn 2+ ([Ar] 3d 5 ) + 2e paramagnetic (b) Cr 3+ (Z = 24) Cr([Ar]4s 2 3d 6 ) Cr 3+ ([Ar] 3d 5 ) + 3e paramagnetic (c) Hg 2+ (Z = 80) Hg([Xe]6s 2 4f 14 5d 10 ) Hg 2+ ([Xe] 4f 14 5d 10 ) + 2e diamagnetic
74 General Applications of the Periodic Law Electronic structure is periodic and the following properties have a dependence on electronic structure: Atomic volume Coefficient of expansion Compressibility Electrical conductivity (or resistance) Hardness Heats of solution, fusion, vaporization, and sublimation Ion mobility Magnetic behavior Malleability Optical spectrum Refractive index Standard reduction potential Thermal conductivity
75 General Applications of the Periodic Law As illustrated by Mendeleev, it is possible to predict many of the physical and chemical properties of an element based on information of the elements surrounding it on the periodic table. See: Mendeleev s predictions See: Seaborg s paper on expanding the periodic table
76 General Applications of the Periodic Law Prediction of formulas of inorganic compounds If you know the formula of a compound, you can substitute another element from the same group into the formula and get a correct formula for the new compound Compound Compound Compound NaCl CaSO 4 HClO 3 LiCl MgSO 4 HBrO 3 KCl SrSO 4 HIO 3 RbCl BaSO 4 KI MgSeO 4 KBr CaTeO 4 LiF
77 Anomalies of the Periodic Classification Position of Hydrogen is not determined Prediction of formulas sometimes results in writing of formulas of compoundsthat are unstable or do not exist Properties of elements in the second period differ from the properties of the elements below them in the same family