1. Chapter 7 Periodic Properties of the Elements
Lecture Presentation
Chapter 7 Periodic Properties of the Elements
(With chapter 6)
6 out of 60
FRQ almost every year
James F. Kirby
Quinnipiac University
Hamden, CT

2. 7.1 Development of the Periodic Table

3. 7.1 Development of the Periodic Table
Dmitri Mendeleev and Lothar Meyer independently came to the same conclusion about how elements should be grouped.
Elements were grouped by similarities in chemical properties and increasing atomic weight.

4. Mendeleev and the Periodic Table

5. Atomic Number
Mendeleev’s table was based on atomic masses. It was the most fundamental property of elements known at the time.
About 35 years later, the nuclear atom was discovered by Ernest Rutherford.
Henry Moseley developed the concept of atomic number experimentally. The number of protons was considered the basis for the periodic property of elements.

6. Development of Periodic Table
* Henry Moseley – 1913
Rearranged the table by increasing atomic number ( # protons)

7. Periodicity (Periodic Trends)
Periodicity is the repetitive pattern of a property for elements based on atomic number.
The following properties are discussed in this chapter:
Sizes of atoms and ions
Ionization energy
Metallic character
Some group chemical property trends
First, we will discuss a fundamental property that leads to may of the trends, effective nuclear charge.
Don’t confuse the explanation of the trend with the trend.

8. 7.2 Effective Nuclear Charge
Many properties depend on attractions between valence electrons and the nucleus.
Electrons are both attracted to the nucleus and repelled by other electrons.
The forces an electron experiences depend on both factors.

9. Effective nuclear charge is the net positive charge experienced by an electron on a many electron atom.
The electron is attracted to the nucleus, but repelled by the inner-shell electrons that shield or screen it from the full nuclear charge. This shielding is called the screening (shielding) effect.

10. We don’t do the calculation. Just understand the concept.

12. Zeff Trends Effective nuclear charge is a periodic property:
Zeff increases across a period.
-Adding valence e’s does little to screen the effect of the increase in nuclear charge
Zeff not relevant down a group.
- Zeff is not as important as the increased the distance from the nucleus as n increases down a group
By Coulomb’s Law, force of attractions decreases as distance increases!

15. Why is the 2s lower in energy than 2p ?

16. Why is the 2s lower in energy than 2p ?
Many students believe the 2s e’s are closer to the nucleus and are therefore attracted more by the nucleus and have a lower energy.
This is a common misconception!
The 2s e’s actually have a slightly larger radial probability!

17. Why is the 2s lower in energy than 2p ?
Because of the small peak
close to the nucleus, the 2s
e’s penetrate closer to the
nucleus and are less
effectively screened which
increases the zeff.
The greater attraction
lowers the energy of the 2s
orbital.
This makes orbital energies:
ns < np < nd < nf
More effective nuclear stregnth

18. 7.3 Size of Atoms and Ions
The bonding atomic radius is defined as one-half of the distance between covalently bonded nuclei.
The nonbonding atomic radius or van der Waals radius is half of the shortest distance separating two nuclei during a collision of atoms.

19. Sizes of Atoms The bonding atomic radius tends to
— decrease from left to right across a period (Zeff ↑).
— increase from top to bottom of a group (n ↑).

20. 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 7.7 to predict the lengths of the C—S, C—H, and S—H bonds in this molecule.

21. Sample Exercise 7.1 Bond Lengths in a Molecule
Continued
Solution
Analyze and Plan We are given three bonds and told to use Figure 7.7 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
= 0.76 Å Å = 1.81 Å
C—H bond length = 0.76 Å Å = 1.07 Å
S—H bond length = 1.05 Å Å = 1.36 Å
Check The experimentally determined bond lengths are C—S = 1.82 Å, C—H = 1.10 Å, and S—H = 1.33 Å. (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.)
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.

22. Sample Exercise 7.1 Bond Lengths in a Molecule
Continued
Practice Exercise 1
Hypothetical elements X and Y form a molecule XY2, in which both Y atoms are bonded to atom X (and not to one another). The X—X distance in the elemental form of X is 2.04 Å, and the Y—Y distance in elemental Y is 1.68 Å. What would you predict for the X—Y distance in the XY2 molecule?
(a) 0.84 Å (b) 1.02 Å (c) 1.86 Å (d) 2.70 Å (e) 3.72 Å
Practice Exercise 2
Using Figure 7.7, predict which is longer, the P—Br bond in PBr3 or the As—Cl bond in AsCl3.

23. Sample Exercise 7.2 Predicting Relative Sizes of Atomic Radii
Referring to the periodic table, arrange (as much as possible) the atoms B, C, Al, and Si in order of increasing size.
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 C and B are in the same period, with C to the right of B. Therefore, we expect the radius of C to be smaller than that of B because radii usually decrease as we move across a period. Likewise, the radius of Si is expected to be smaller than that of Al. Al is directly below B, and Si is directly below C. We expect, therefore, the radius of B to be smaller than that of Al and the radius of C to be smaller than that of Si. Thus, so far we can say C < B, B < Al, C < Si, and Si < Al. We can therefore conclude that C has the smallest radius and Al has the largest radius, and so can write C < ? < ? < Al.
Our two periodic trends in atomic size do not supply enough information to allow us to determine, however, whether B or Si (represented by the two question marks) has the larger radius. Going from B to Si in the periodic table, we move down (radius tends to increase) and to the right (radius tends to decrease). It is only because Figure 7.7 provides numerical values for each atomic radius that we know that the radius of Si is greater than that of B. 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.

24. Sample Exercise 7.2 Predicting Relative Sizes of Atomic Radii
Continued
Check From Figure 7.7, we have
C(0.76 Å) < B(0.84 Å) < Si(1.11 Å) < Al(1.21 Å).
Comment Note that the trends we have just discussed are for the s- and p-block elements. Figure 7.7 shows that the transition elements do not show a regular decrease moving across a period.
Practice Exercise 1
By referring to the periodic table, but not to Figure 7.7, place the following atoms in order of increasing bonding atomic radius: N, O, P, Ge.
(a) N < O < P < Ge (b) P < N < O < Ge (c) O < N < Ge < P (d) O < N < P < Ge
(e) N < P < Ge < O
Practice Exercise 2
Arrange Be, C, K, and Ca in order of increasing atomic radius.

25. Sizes of Ions Determined by interatomic distances in ionic compounds
Ionic size depends on
the nuclear charge.
the number of electrons.
the orbitals in which electrons reside.

26. Sizes of Ions Cations are smaller than their parent atoms:
The outermost electron is removed (lower n) and repulsions between electrons are reduced.
Anions are larger than their parent atoms:
Electrons are added and repulsions between electrons are increased.

28. Sample Exercise 7.3 Predicting Relative Sizes of 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 1
Arrange the following atoms and ions in order of increasing ionic radius: F, S2–, Cl, and Se2–.
(a) F < S2– < Cl < Se2– (b) F < Cl < S2– < Se2– (c) F < S2– < Se2– < Cl
(d) Cl < F < Se2– < S2– (e) S2– < F < Se2– < Cl
Practice Exercise 2
Which of the following atoms and ions is largest: S2–, S, O2–?

29. Size of Ions - Isoelectronic Series
In an isoelectronic series, ions have the same number of electrons.
Ionic size decreases with an increasing nuclear charge.
An Isoelectronic Series (10 electrons)
Note increasing nuclear charge with decreasing ionic radius as atomic number increases
O2– F–
Na+
Mg2+
Al3+
1.26 Å
1.19 Å
1.16 Å
0.86 Å
0.68 Å

30. 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 1
Arrange the following atoms and ions in order of increasing ionic radius: Br–, Rb+, Se2–, Sr2+, Te2–.
(a) Sr2+ < Rb+ < Br– < Se2– < Te2–
(b) Br– < Sr2+ < Se2– < Te2– < Rb+
(c) Rb+ < Sr2+ < Se2– < Te2– < Br–
(d) Rb+ < Br– < Sr2+ < Se2– < Te2–
(e) Sr2+ < Rb+ < Br– < Te2– < Se2–
Practice Exercise 2
In the isoelectronic series Ca2+, Cs+, Y3+, which ion is largest?

31. 7.4 Ionization Energy (I)
The ionization energy is the minimum 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.
Note: the higher the ionization energy, the more difficult it is to remove an electron!

32. Ionization Energy
It requires more energy to remove each successive electron.
When all valence electrons have been removed the ionization energy increases dramatically.
Inner level p electrons are closer to the nucleus, this gives them a much higher Zeff and are much harder to remove.

39. 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.

40. Sample Exercise 7.5 Trends in Ionization Energy
Continued
Practice Exercise 1
The third ionization energy of bromine is the energy required for which of the following processes?
(a) Br(g) → Br+(g) + e – (b) Br+(g) → Br2+(g) + e – (c) Br(g) → Br2+(g) + 2e –
(d) Br(g)→ Br3+(g) + 3e– (e) Br2+(g) → Br3+(g) + e –
Practice Exercise 2
Which has the greater third ionization energy, Ca or S?

41. Periodic Trends in First Ionization Energy (I1)
I1 generally increases across a period.
I1 generally decreases down a group. Zeff doesn’t change much, but the valence e’s are farther from the nucleus (greater n) and easier to remove (coulomb’s law)
The s- and p-block elements show a larger range of values for I1. (The d-block generally increases slowly across the period; the f-block elements show only small variations.)
2A-3A, 5A-6A

42. Factors that Influence Ionization Energy
Smaller atoms have higher I values.
I values depend on effective nuclear charge and average distance of the electron from the nucleus.

43. Irregularities in the General Trend
The trend is not followed when the added valence electron in the next element
enters a new sublevel (higher energy sublevel);
is the first electron to pair in one orbital of the sublevel (electron repulsions lower energy). Wh

44. K < Na < P < Ar < Ne
Sample Exercise 7.6 Periodic Trends in Ionization Energy
Referring to the 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.
Sample
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

45. Sample Exercise 7.6 Periodic Trends in Ionization Energy
Continued
Check The values shown in Figure 7.10 confirm this prediction.

46. Sample Exercise 7.6 Periodic Trends in Ionization Energy
Continued
Practice Exercise 1
Consider the following statements about first ionization energies:
(i) Because the effective nuclear charge for Mg is greater than that for Be, the first ionization energy of Mg is greater than that of Be.
(ii) The first ionization energy of O is less than that of N because in O we must pair electrons in the 2p orbitals.
(iii) The first ionization energy of Ar is less than that of Ne because a 3p electron in Ar is farther from the nucleus than a 2p electron in Ne.
Which of the statements (i), (ii), and (iii) is or are true?
(a) Only one of the statements is true.
(b) Statements (i) and (ii) are true.
(c) Statements (i) and (iii) are true.
(d) Statements (ii) and (iii) are true.
(e) All three statements are true.
Practice Exercise 2
Which has the lowest first ionization energy, B, Al, C, or Si? Which has the highest?

47. Electron Configurations of Ions
Cations: The electrons are lost from the highest energy level (n value).
Li+ is 1s2 (losing a 2s electron).
Fe2+ is 1s22s22p63s23p63d6 (losing two 4s electrons).
Anions: The electron configurations are filled to ns2np6; e.g., F– is 1s22s22p6 (gaining one electron in 2p).

48. Sample Exercise 7.7 Electron Configurations of Ions
Write the electron configurations 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, and 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 4s electrons must be removed, giving an ion that is isoelectronic with Ar:
Ca2+: [Ar]
(b) Cobalt (atomic number 27) has the electron configuration [Ar]4s23d7. 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, we remove the two 4s electrons and one of the 3d electrons, and the electron configuration for Co3+ is
Co3+: [Ar]3d6

49. Sample Exercise 7.7 Electron Configurations of Ions
Continued
(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 1
The ground electron configuration of a Tc atom is [Kr]5s24d5.
What is the electron configuration of a Tc3+ ion?
(a) [Kr]4d4 (b) [Kr]5s24d2 (c) [Kr]5s14d3 (d) [Kr]5s24d8 (e) [Kr]4d10
Practice Exercise 2
Write the electron configurations for (a) Ga3+, (b) Cr3+, and (c) Br–.

50. 7.5 Electron Affinity
Electron affinity is the energy change accompanying the addition of an electron to a gaseous atom:
Cl + e−  Cl−
It is typically exothermic, so, for most elements, it is negative!

51. General Trend in Electron Affinity
Not much change in a group.
Across a period, it generally increases. Three notable exceptions include the following:
Group 2A: s sublevel is full!
Group 5A: p sublevel is half-full!
Group 8A: p sublevel is full!
Note: the electron affinity for many of these elements is positive (X– is unstable).

52. 7.6 Metal, Nonmetals, and Metalloids
Metallic character refers to the extent to which element exhibits the physical and chemical properties of metals.
- In other words, how easily it loses electrons.

53. 7.6 Metal, Nonmetals, and Metalloids

54. Metals vs. Nonmetals
Differences between metals, and nonmetals tend to revolve around these properties.

55. Metals Differ from Nonmetals
Metals tend to form cations.
Nonmetals tend to form anions.

56. Metals Most of the elements in nature are metals.
Properties of metals:
Shiny luster
Conduct heat and electricity
Malleable and ductile
Solids at room temperature (except mercury)
Low ionization energies/form cations easily
Tend to be oxidized (loses electrons!) when they react.

57. Metal Chemistry
Compounds formed between metals and nonmetals tend to be ionic.
Metal oxides tend to be basic.
- Elements forms of each will combine in a synthesis reaction.
2K + S  K2S
2Al + 3Cl2  2AlCl3
2Na + H2  2NaH
Compound formulas are based on charges (hydrides are 1-)
Elemental formulas are based on if it is diatomic or monoatomic.

58. Sc2O3(s) + 6 HNO3(aq) → 2 Sc(NO3)3(aq) + 3 H2O(l)
Sample Exercise 7.8 Properties of 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 °C.
(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)

59. Sample Exercise 7.8 Properties of Metal Oxides
Continued
Practice Exercise 1
Suppose that a metal oxide of formula M2O3 were soluble in water. What would be the major product or products of dissolving the substance in water?
(a) MH3(aq) + O2(g) (b) M(s) + H2(g) + O2(g) (c) M3+ (aq) + H2O2(aq)
(d) M(OH)2(aq) (e) M(OH)3(aq)
Practice Exercise 2
Write the balanced chemical equation for the reaction between copper(II) oxide and sulfuric acid.

60. Nonmetals
Nonmetals are found on the right hand side of the periodic table.
Properties of nonmetals include the following:
Solids are dull, brittle, poor conductors
Large negative electronegativity/form anions readily
Tend to gain electrons in reactions with metals to acquire a noble gas configuration.

61. H2 , N2, O2, F2, Cl2 (g) , Br2(l) and I2(s)
Seven nonmetallic elements exist as diatomic molecules under ordinary conditions:
H2 , N2, O2, F2, Cl2 (g) , Br2(l) and I2(s)
Substances containing only nonmetals are usually molecular compound.

62. Nonmetal Chemistry
Substances containing only nonmetals are molecular compounds.
Most nonmetal oxides are acidic.

63. Recap of a Comparison of the Properties of Metals and Nonmetals

64. Sample Exercise 7.9 Reactions of Nonmetal Oxides
Write a 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 that in Equation 7.15:
SeO2(s) + 2 NaOH(aq) → Na2SeO3(aq) + H2O(l)

65. Sample Exercise 7.9 Reactions of Nonmetal Oxides
Continued
Practice Exercise 1
Consider the following oxides: SO2, Y2O3, MgO, Cl2O, N2O5. How many are expected to form acidic solutions in water?
(a) 1 (b) 2 (c) 3 (d) 4 (e) 5
Practice Exercise 2
Write a balanced chemical equation for the reaction of solid tetraphosphorus hexoxide with water.

66. Metalloids
Metalloids have some characteristics of metals and some of nonmetals.
Several metalloids are electrical semiconductors (computer chips). For instance, silicon looks shiny, but is brittle and fairly poor conductor.

67. 7.7 Group Trends Elements in a group have similar properties.
Trends also exist within groups.
Groups Compared:
Group 1A: The Alkali Metals
Group 2A: The Alkaline Earth Metals
Group 6A: The Oxygen Group
Group 7A: The Halogens
Group 8A: The Noble Gases

68. Alkali Metals – Group 1A Alkali metals are soft, metallic solids.
The name comes from the Arabic word for ashes.
They are found only in compounds in nature, not in their elemental forms.
Typical metallic properties (luster, conductivity) are seen in them.

69. Alkali Metal Properties
They are found only in compounds in nature, not in their elemental form.
They have low densities and melting points.
They also have low ionization energies.
Reactivity increases as we move down the group.
Lower ionization energies as n increases!

70. Alkali Metal Chemistry
Their reactions with water are famously exothermic.
2M + 2H2O  2MOH + H2

71. Flame Tests
Qualitative tests for alkali metals include their characteristic colors in flames.
They produce bright colors when placed in a flame.

72. Differences in Alkali Metal Chemistry
Lithium reacts with oxygen to make an oxide:
4 Li + O2  2 Li2O
Sodium reacts with oxygen to form a peroxide:
2 Na + O2  Na2O2
K, Rb, and Cs also form superoxides:
M + O2  MO2

73. 2 Cs(s) + 2 H2O(l) → 2 CsOH(aq) + H2(g)
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:
2 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:
2 Cs(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 1– ions, which means the products have 1:1 stoichiometry with Cs+.

74. Sample Exercise 7.10 Reactions of an Alkali Metal
Continued
Practice Exercise 1
Consider the following three statements about the reactivity of an alkali metal M with oxygen gas:
(i) Based on their positions in the periodic table, the expected product is the ionic oxide M2O.
(ii) Some of the alkali metals produce metal peroxides or metal superoxides when they react with oxygen.
(iii) When dissolved in water, an alkali metal oxide produces a basic solution.
Which of the statements (i), (ii), and (iii) is or are true?
(a) Only one of the statements is true.
(b) Statements (i) and (ii) are true.
(c) Statements (i) and (iii) are true.
(d) Statements (ii) and (iii) are true.
(e) All three statements are true.
Practice Exercise 2
Write a balanced equation for the expected reaction between potassium metal and elemental sulfur, S(s).

75. Alkaline Earth Metals—Compare to Alkali 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.
Reactivity tends to increase as you go down the group.
Same reason as with alkali earth metals – easier to remove electrons.

76. 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.

77. 7.8 Group 6A—Increasing in Metallic Character down the Group
As we move down the group the metallic character increases
Oxygen, sulfur, and selenium are nonmetals.
Tellurium is a metalloid.
The radioactive polonium is a metal.

78. Group 7A—Halogens The halogens are prototypical nonmetals.
The name comes from the Greek words halos and gennao: “salt formers”.
High electronegativities.
Great oxidizers (easily reduced).
They react directly with metals to form metal halides.
Chlorine is added to water supplies to serve as a disinfectant.

79. Group 8A—Noble Gases
The noble gases have very large ionization energies.
Therefore, they are relatively unreactive.
They are found as nonmetallic monatomic gases.

80. Noble Gases become slightly more reactive as you go down the group because of the decrease in Ionization energy cause
Xe forms 3 compounds
- XeF2, XeF4, XeF6
Kr forms only one stable compound
- KrF2
The unstable HArF was synthesized in 2000.

81. Sample Integrative Exercise Putting Concepts Together
The element bismuth (Bi, atomic number 83) is the heaviest member of group 5A. A salt of the element, bismuth subsalicylate, is the active ingredient in Pepto-Bismol®, an over-the-counter medication for gastric distress.
(a) Based on values presented in Figure 7.7 and Tables 7.5 and 7.6, what might you expect for the bonding atomic radius of bismuth?
(b) What accounts for the general increase in atomic radius going down the group 5A elements?
(c) Another major use of bismuth has been as an ingredient in low-melting metal alloys, such as those used in fire sprinkler systems and in typesetting. The element itself is a brittle white crystalline solid. How do these characteristics fit with the fact that bismuth is in the same periodic group with such nonmetallic elements as nitrogen and phosphorus?
(d) Bi2O3 is a basic oxide. Write a balanced chemical equation for its reaction with dilute nitric acid. If 6.77 g of Bi2O3 is dissolved in dilute acidic solution to make L of solution, what is the molarity of the solution of Bi3+ ion?
(e) 209Bi is the heaviest stable isotope of any element. How many protons and neutrons are present in this nucleus?
(f) The density of Bi at 25 °C is g ⁄ cm3. How many Bi atoms are present in a cube of the element that is 5.00 cm on each edge? How many moles of the element are present?

82. Sample Integrative Exercise Putting Concepts Together
Continued
Solution
(a) Bismuth is directly below antimony, Sb, in group 5A. Based on the observation that atomic radii increase as we go down a column, we would expect the radius of Bi to be greater than that of Sb, which is 1.39 Å. We also know that atomic radii generally decrease as we proceed from left to right in a period. Tables 7.5 and 7.6 each give an element in the same period, namely Ba and Po. We would therefore expect that the radius of Bi is smaller than that of Ba (2.15 Å) and larger than that of Po (1.40 Å). We also see that in other periods, the difference in radius between the neighboring group 5A and group 6A elements is relatively small. We might therefore expect that the radius of Bi is slightly larger than that of Po—much closer to the radius of Po than to the radius of Ba. The tabulated value for the atomic radius on Bi is 1.48 Å, in accord with our expectations.
(b) The general increase in radius with increasing atomic number in the group 5A elements occurs because additional shells of electrons are being added, with corresponding increases in nuclear charge. The core electrons in each case largely screen the outermost electrons from the nucleus, so the effective nuclear charge does not vary greatly as we go to higher atomic numbers. However, the principal quantum number, n, of the outermost electrons steadily increases, with a corresponding increase in orbital radius.
(c) The contrast between the properties of bismuth and those of nitrogen and phosphorus illustrates the general rule that there is a trend toward increased metallic character as we move down in a given group. Bismuth, in fact, is a metal. The increased metallic character occurs because the outermost electrons are more readily lost in bonding, a trend that is consistent with its lower ionization energy.

83. Sample Integrative Exercise Putting Concepts Together
Continued
(d) Following the procedures described in Section 4.2 for writing molecular and net ionic equations, we have the following:
Molecular equation: Bi2O3(s) + 6 HNO3(aq) → 2 Bi(NO3)3(aq) + 3 H2O(l)
Net ionic equation: Bi2O3(s) + 6 H+(aq) → 2 Bi3 + (aq) + 3 H2O(l)
In the net ionic equation, nitric acid is a strong acid and Bi(NO3)3 is a soluble salt, so we need to show only the reaction of the solid with the hydrogen ion forming the Bi3+(aq) ion and water. To calculate the concentration of the solution, we proceed as follows (Section 4.5):
(e) Recall that the atomic number of any element is the number of protons and electrons in a neutral atom of the element (Section 2.3) Bismuth is element 83; there are therefore 83 protons in the nucleus. Because the atomic mass number is 209, there are 209 – 83 = 126 neutrons in the nucleus.

84. Sample Integrative Exercise Putting Concepts Together
Continued
(f) We can use the density and the atomic weight to determine the number of moles of Bi, and then use Avogadro’s number to convert the result to the number of atoms (Sections 1.4 and 3.4) The volume of the cube is (5.00)3 cm3 = 125 cm3. Then we have