1. Chapter 7 Periodic Properties of the Elements
Lecture Presentation
Chapter 7 Periodic Properties of the Elements
John D. Bookstaver
St. Charles Community College
Cottleville, MO

2. Development of Periodic Table
Dmitri Mendeleev and Lothar Meyer independently came to the same conclusion about how elements should be grouped.

3. Development of Periodic Table
Mendeleev, for instance, predicted the discovery of germanium (which he called eka-silicon) as an element with an atomic weight between that of zinc and arsenic, but with chemical properties similar to those of silicon.

4. Periodic Trends
In this chapter, we will rationalize observed trends in
Sizes of atoms and ions.
Ionization energy.
Electron affinity.

5. Effective Nuclear Charge
In a many-electron atom, electrons are both attracted to the nucleus and repelled by other electrons.
The nuclear charge that an electron experiences depends on both factors.

6. Effective Nuclear Charge
The effective nuclear charge, Zeff, is found this way:
Zeff = Z − S
where Z is the atomic number and S is a screening constant, usually close to the number of inner electrons.

7. What Is the Size of an Atom?
The bonding atomic radius is defined as one-half of the distance between covalently bonded nuclei.

8. Sizes of Atoms The bonding atomic radius tends to
— Decrease from left to right across a row
(due to increasing Zeff).
— Increase from top to bottom of a column
(due to the increasing value of n).

9. Sample Exercise 7.1 Bond Lengths in a Molecule
Natural gas used in home heating and cooking is odorless. Because natural gas leaks pose the danger of explosion or suffocation, various smelly substances are added to the gas to allow detection of a leak. One such substance is methyl mercaptan, CH3SH. Use Figure 7.6 to predict the lengths of the C—S, C—H, and S—H bonds in this molecule.
Solution
Analyze and Plan We are given three bonds and told to use Figure 7.6 for bonding atomic radii. We will assume that each bond length is the sum of the bonding atomic radii of the two atoms involved.
Solve
C—S bond length = bonding atomic radius of C + bonding atomic radius of S
= 0.77Å Å = 1.79Å
C—H bond length = 0.77Å Å = 1.14Å
S—H bond length = 1.02Å = 0.37Å = 1.39Å
Check The experimentally determined bond lengths are C—S = 1.82Å, C—H = 1.10Å and S—H = 1.33Å. (In general, the lengths of bonds involving hydrogen show larger deviations from the values predicted from bonding atomic radii than do bonds involving larger atoms.)
FIGURE 7.6 Trends in bonding atomic radii for periods 1 through 5.

10. Sample Exercise 7.1 Bond Lengths in a Molecule
Continued
Comment Notice that our estimated bond lengths are close but not exact matches to the measured bond lengths. Bonding atomic radii must be used with some caution in estimating bond lengths.
Practice Exercise
Using Figure 7.6, predict which is longer, the bond in PBr3 or the As—Cl bond in AsCl3.
Answer: P—Br

11. Sample Exercise 7.2 Atomic Radii
Referring to a periodic table, arrange (as much as possible) the atoms 15P, 16S, 33As, and 34Se in order of increasing size. (Atomic numbers are given to help you locate the atoms quickly in the table.)
Solution
Analyze and Plan We are given the chemical symbols for four elements and told to use their relative positions in the periodic table to predict the relative size of their atomic radii. We can use the two periodic trends just described to help with this problem.
Solve P and S are in the same period, with S to the right of P. Therefore, we expect the radius of S to be smaller than that of P because radii decrease as we move across a period. Likewise, the radius of Se is expected to be smaller than that of As. As is directly below P, and Se is directly below S. We expect, therefore, the radius of P to be smaller than that of As and the radius of S to be smaller than that of Se. Thus, so far we can say S < P,
P < As, S < Se, Se < As. We can therefore conclude that S has the smallest radius and As has the largest radius and so can write S < ? < ? < As.
Our two periodic trends for atomic size do not supply enough information to allow us to determine whether P or Se (represented by the two question marks) has the larger radius, however. Going from P to Se in the periodic table, we move down (radius tends to increase) and to the right (radius tends to decrease). In Figure 7.6 we see that the radius of Se is greater than that of P. If you examine the figure carefully, you will discover that for the s- and p-block elements the increase in radius moving down a column tends to be the greater effect. There are exceptions, however.

12. Answer: Be < Mg < Na
Sample Exercise 7.2 Atomic Radii
Continued
Solution
Check From Figure 7.6, we have S(1.02Å) < (1.06Å) < Se(1.16Å) < As(1.19Å).
Comment Note that the trends we have just discussed are for the s- and p-block elements. Figure 7.6 shows that the transition elements do not show a regular decrease moving across a period.
Practice Exercise
Arrange 11Na, 4Be, and 12Mg in order of increasing atomic radius.
FIGURE 7.6 Trends in bonding atomic radii for periods 1 through 5.
Answer: Be < Mg < Na

13. Sizes of Ions Ionic size depends upon The nuclear charge.
The number of electrons.
The orbitals in which electrons reside.

14. Sizes of Ions Cations are smaller than their parent atoms:
The outermost electron is removed and repulsions between electrons are reduced.

15. Sizes of Ions Anions are larger than their parent atoms”
Electrons are added and repulsions between electrons are increased.

16. Sizes of Ions Ions increase in size as you go down a column:
This increase in size is due to the increasing value of n.

17. Sample Exercise 7.3 Atomic and Ionic Radii
Arrange Mg2+, Ca2+, and Ca in order of decreasing radius..
Solution
Cations are smaller than their parent atoms, and so Ca2+ < Ca. Because Ca is below Mg in group 2A, Ca2+ is larger than Mg2+. Consequently, Ca > Ca2+ > Mg2+.
Practice Exercise
Which of the following atoms and ions is largest: S2–, S, O2–?
Answer: S2– 17

18. Sizes of Ions
In an isoelectronic series, ions have the same number of electrons.
Ionic size decreases with an increasing nuclear charge.

19. Sample Exercise 7.4 Ionic Radii in an Isoelectronic Series
Arrange the ions K+, Cl–, Ca2+, and S2– in order of decreasing size.
Solution
This is an isoelectronic series, with all ions having 18 electrons. In such a series, size decreases as nuclear charge (atomic number) increases. The atomic numbers of the ions are S 16, Cl 17, K 19, Ca 20. Thus, the ions decrease in size in the order S2– > Cl– > K+ > Ca2+.
Practice Exercise
In the isoelectronic series Rb+, Sr2+, Y3+, which ion is largest?
Answer: Rb+ 19

20. Ionization Energy
The ionization energy is the amount of energy required to remove an electron from the ground state of a gaseous atom or ion.
The first ionization energy is that energy required to remove the first electron.
The second ionization energy is that energy required to remove the second electron, etc.

21. Ionization Energy
It requires more energy to remove each successive electron.
When all valence electrons have been removed, the ionization energy takes a quantum leap.

22. Sample Exercise 7.5 Trends in Ionization Energy
Three elements are indicated in the periodic table in the margin. Which one has the largest second ionization energy?
Solution
Analyze and Plan The locations of the elements in the periodic table allow us to predict the electron configurations. The greatest ionization energies involve removal of core electrons. Thus, we should look first for an element with only one electron in the outermost occupied shell.
Solve The red box represents Na, which has one valence electron. The second ionization energy of this element is associated, therefore, with the removal of a core electron. The other elements indicated, S (green) and Ca (blue), have two or more valence electrons. Thus, Na should have the largest second ionization energy.
Check A chemistry handbook gives these I2 values: Ca 1145 kJ/mol, S 2252 kJ/mol, Na 4562 kJ/mol.
Practice Exercise
Which has the greater third ionization energy, Ca or S?
Answer: Ca

23. Trends in First Ionization Energies
As one goes down a column, less energy is required to remove the first electron.
For atoms in the same group, Zeff is essentially the same, but the valence electrons are farther from the nucleus.

24. Trends in First Ionization Energies
Generally, as one goes across a row, it gets harder to remove an electron.
As you go from left to right, Zeff increases.

25. Trends in First Ionization Energies
However, there are two apparent discontinuities in this trend.

26. Trends in First Ionization Energies
The first occurs between Groups IIA and IIIA.
In this case the electron is removed from a p orbital rather than an s orbital.
The electron removed is farther from the nucleus.
There is also a small amount of repulsion by the s electrons.

27. Trends in First Ionization Energies
The second discontinuity occurs between Groups VA and VIA.
The electron removed comes from a doubly occupied orbital.
Repulsion from the other electron in the orbital aids in its removal.

28. Sample Exercise 7.6 Periodic Trends in Ionization Energy
Referring to a periodic table, arrange the atoms Ne, Na, P, Ar, K in order of increasing first ionization energy.
Solution
Analyze and Plan We are given the chemical symbols for five elements. To rank them according to increasing first ionization energy, we need to locate each element in the periodic table. We can then use their relative positions and the trends in first ionization energies to predict their order.
Solve Ionization energy increases as we move left to right across a period and decreases as we move down a group. Because Na, P, and Ar are in the same period, we expect I1 to vary in the order Na < P < Ar. Because Ne is above Ar in group 8A, we expect Ar < Ne. Similarly, K is directly below Na in group 1A, and so we expect K < Na.
From these observations, we conclude that the ionization energies follow the order
K < Na < P < Ar < Ne
Check The values shown in Figure 7.9 confirm this prediction.
Practice Exercise
FIGURE 7.9 Trends in first ionization energies of the elements
Which has the lowest first ionization energy, B, Al, C, or Si? Which has the highest?
Answer: Al lowest, C highest

29. Answers: (a) [Ar]3d10, (b) [Ar]3d3, (c) [Ar]3d104s24p6 = [Kr]
Sample Exercise 7.7 Electron Configurations of Ions
Write the electron configuration for (a) Ca2+, (b) Co3+, and (c) S2–.
Solution
Analyze and Plan We are asked to write electron configurations for three ions. To do so, we first write the electron configuration of each parent atom, then remove or add electrons to form the ions. Electrons are first removed from the orbitals having the highest value of n. They are added to the empty or partially filled orbitals having the lowest value of n.
Solve (a) Calcium (atomic number 20) has the electron configuration [Ar]4s2. To form a 2+ ion, the two outer electrons must be removed, giving an ion that is isoelectronic with Ar: Ca2:[Ar]
(b) Cobalt (atomic number 27) has the electron configuration [Ar]3d74s2. To form a 3+ ion, three electrons must be removed. As discussed in the text, the 4s electrons are removed before the 3d electrons. Consequently, the electron configuration for Co3+:[Ar]3d6
(c) Sulfur (atomic number 16) has the electron configuration [Ne]3s23p4. To form a 2– ion, two electrons must be added. There is room for two additional electrons in the 3p orbitals. Thus, the S2– electron configuration is S2–:[Ne]3s23p6 = [Ar]
Comment Remember that many of the common ions of the s- and p-block elements, such as Ca2+ and S2–, have the same number of electrons as the closest noble gas. •(Section 2.7)
Practice Exercise
Write the electron configuration for (a) Ga3+, (b) Cr3+, and (c) Br–.
Answers: (a) [Ar]3d10, (b) [Ar]3d3, (c) [Ar]3d104s24p6 = [Kr]

30. Electron Affinity
Electron affinity is the energy change accompanying the addition of an electron to a gaseous atom:
Cl + e−  Cl−

31. Trends in Electron Affinity
In general, electron affinity becomes more exothermic as you go from left to right across a row.

32. Trends in Electron Affinity
There are again, however, two discontinuities in this trend.

33. Trends in Electron Affinity
The first occurs between Groups IA and IIA.
The added electron must go in a p orbital, not an s orbital.
The electron is farther from the nucleus and feels repulsion from the s electrons.

34. Trends in Electron Affinity
The second discontinuity occurs between Groups IVA and VA.
Group VA has no empty orbitals.
The extra electron must go into an already occupied orbital, creating repulsion.

35. Properties of Metal, Nonmetals, and Metalloids

36. Metals versus Nonmetals
Differences between metals and nonmetals tend to revolve around these properties.

37. Metals versus Nonmetals
Metals tend to form cations.
Nonmetals tend to form anions.

38. Metals
Metals tend to be lustrous, malleable, ductile, and good conductors of heat and electricity.

39. Metals Compounds formed between metals and nonmetals tend to be ionic.
Metal oxides tend to be basic.

40. Sample Exercise 7.8 Metal Oxides
(a) Would you expect scandium oxide to be a solid, liquid, or gas at room temperature?
(b) Write the balanced chemical equation for the reaction of scandium oxide with nitric acid.
Solution
Analyze and Plan We are asked about one physical property of scandium oxide—its state at room temperature—and one chemical property—how it reacts with nitric acid.
Solve
(a) Because scandium oxide is the oxide of a metal, we expect it to be an ionic solid. Indeed it is, with the very high melting point of 2485ºC.
(b) In compounds, scandium has a 3+ charge, Sc3+, and the oxide ion is O2–. Consequently, the formula of scandium oxide is Sc2O3. Metal oxides tend to be basic and, therefore, to react with acids to form a salt plus water. In this case the salt is scandium nitrate, Sc(NO3)3:
Sc2O3(s) + 6 HNO3(aq) → 2 Sc(NO3)3(aq) + 3 H2O(l)
Practice Exercise
Write the balanced chemical equation for the reaction between copper(II) oxide and sulfuric acid.
Answer: CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l)

41. Nonmetals
Nonmetals are dull, brittle substances that are poor conductors of heat and electricity.
They tend to gain electrons in reactions with metals to acquire a noble-gas configuration.

42. Nonmetals
Substances containing only nonmetals are molecular compounds.
Most nonmetal oxides are acidic.

43. Sample Exercise 7.9 Nonmetal Oxides
Write the balanced chemical equation for the reaction of solid selenium dioxide, SeO2(s), with (a) water, (b) aqueous sodium hydroxide.
Solution
Analyze and Plan We note that selenium is a nonmetal. We therefore need to write chemical equations for the reaction of a nonmetal oxide with water and with a base, NaOH. Nonmetal oxides are acidic, reacting with water to form an acid and with bases to form a salt and water.
Solve (a) The reaction between selenium dioxide and water is like that between carbon dioxide and water (Equation 7.13):
SeO2(s) + H2O(l) → H2SeO3(aq)
(It does not matter that SeO2 is a solid and CO2 is a gas under ambient conditions; the point is that both are water-soluble nonmetal oxides.)
(b) The reaction with sodium hydroxide is like the reaction in Equation 7.15:
SeO2(s) + 2 NaOH(aq) → Na2SeO3(aq) + H2O(l)
Practice Exercise
Write the balanced chemical equation for the reaction of solid tetraphosphorus hexoxide with water.
Answer: P4O6(s) + 6 H2O(l) → 4 H3PO3(aq)

44. Metalloids
Metalloids have some characteristics of metals and some of nonmetals.
For instance, silicon looks shiny, but is brittle and a fairly poor conductor.

45. Group Trends

46. Alkali Metals Alkali metals are soft, metallic solids.
The name comes from the Arabic word for ashes.

47. Alkali Metals
They are found only in compounds in nature, not in their elemental forms.
They have low densities and melting points.
They also have low ionization energies.

48. Alkali Metals
Their reactions with water are famously exothermic.

49. Alkali Metals
Alkali metals (except Li) react with oxygen to form peroxides.
K, Rb, and Cs also form superoxides:
K + O2  KO2
They produce bright colors when placed in a flame.

50. Answer: 2 K(s) + S(s) → K2S(s)
Sample Exercise 7.10 Reactions of an Alkali Metal
Write a balanced equation for the reaction of cesium metal with (a) Cl2(g), (b) H2O(l), (c) H2(g).
Solution
Analyze and Plan Because cesium is an alkali metal, we expect its chemistry to be dominated by oxidation
of the metal to Cs+ ions. Further, we recognize that Cs is far down the periodic table, which means it is among the most active of all metals and probably reacts with all three substances.
Solve The reaction between Cs and Cl2 is a simple combination reaction between a metal and a nonmetal, forming the ionic compound CsCl: Cs(s) + Cl2(g) → 2 CsCl(s)
From Equations 7.18 and 7.16, we predict the reactions of cesium with water and hydrogen to proceed as follows:
2Cs(s) + 2 H2O(l) → 2 CsOH(aq) + H2(g)
2 Cs(s) + H2(g) → 2 CsH(s)
All three reactions are redox reactions where cesium forms a Cs+ ion. The Cl–, OH–, and H– are all ions,
which means the products have 1:1 stoichiometry with Cs+.
Practice Exercise
Write a balanced equation for the reaction between potassium metal and elemental sulfur.
Answer: 2 K(s) + S(s) → K2S(s)

51. Alkaline Earth Metals
Alkaline earth metals have higher densities and melting points than alkali metals.
Their ionization energies are low, but not as low as those of alkali metals.

52. Alkaline Earth Metals
Beryllium does not react with water, and magnesium reacts only with steam, but the other alkaline earth metals react readily with water.
Reactivity tends to increase as you go down the group.