Chemistry part 11, Julia Burdge,2e (2009) ppsx

240 CHAPTER 7 Electr on Configuration and the Periodic Table • Essential elements in the human body a ll biological molecules. Carbon is perhaps the mo st versatile element because it has the ability to form various types of chemical bond s. Carbon atom s can form bonds to each other, linking up to form an enormous variety of chains and ring structures. The metals play several different roles in living systems. As cations (Na +, K +, Ca 2 +, and Mg 2+), they serve to maintain the balance between intracellular and extracellular fluids, nerve transmissions, and other activities.1'hey are also needed for protein functions. For example, the Fe 2 + ion binds oxygen in hemoglobin molecules and Cu 2 +, Zn 2 +, and Mg2+ ions are essential for enzyme activit y. In addition, calcium in the form of Cas(P04MOH) and Ca3(P04)2 is an essential component of teeth and bones. 1 2 3 4 5 6 7 lA 1 H Na K 2A 2 3B Mg 3 Ca 4B 5B 6B 7B1s8B~ 1B 4 5 6 7 8 9 10 11 V Cr Mn Fe Co Ni Cu Mo Bu lk elements 3A 4A 5A 6A 7A 13 14 15 16 17 B C N 0 F 2B 12 Si P S Cl Zn Se I . Trace elements 8A 18 1 2 3 4 5 6 7 The periodic table shown highlights the essential elements in the human body. Of special interest are the trace elements, such as iron (Fe), copper (Cu), zinc (Zn), iodine (I), cobalt (Co), selenium (Se), and fluorine (F), which together make up about 0.1 percent of the body 's mass. Although the trace elements are present in very small amounts, they are crucial for our health. In many cases, however, their exact biological role is still not fully understood. These elements are necessary for biological functions such as growth, the transport of oxy- gen for metabolism, and defense against disease. There is a delicate balance in the amounts of these elements in our bodie s. Too much or too little over an extended period of time can lead to serious illness, retardation, or even death. Checkpoint 7.1 Development of the Periodic Table 7.1.1 Which of the following elements would yo u expect to ha ve chemical properties most similar to tho se of S? a) P b) CI c) Se d) Na e) Sr 7.1 .2 The Modern Periodic Table Which of the following elements would you expect to have properties simil ar to those ofBa? a) Sr b) Rb c) Na d) K e) B Figure 7.2 shows the modern periodic table together with the outermost ground-state electron con- figurations of the elements. (The electron configurations of the elements are also given in Figure 6.25.) Starting with hydrogen, the electronic subshells are filled in the order shown in Figure 6.23 [ ~~ Section 6.8] . • SECTION 7.2 The Modern Periodic Table 241 I 2 3 4 5 6 7 lA I 1 H ls i 3 Li 2s 1 11 Na 3s 1 19 K 4s 1 37 Rb 5s 1 55 Cs 6s 1 87 Fr 7s 1 2A 2 4 Be 2 S2 12 Mg 3s 2 20 Ca 4s 2 38 Sr 5s 2 56 Ba 6s 2 88 Ra 7s 2 3B 4B 3 4 21 22 Sc Ti 4s 2 3d 1 4s 2 3d 2 39 40 Y Zr 5s2 4d 1 5s24d 2 71 72 Lo Hf 6s 2 5d l 4l 6s 2 5d 2 103 104 Lr Rf 7s 2 5fl46d l 7s 2 6d 2 57 La 6s 2 5d 1 89 Ac 7s 2 6d l 5B 6B 7B I 5 6 7 8 23 24 25 26 V Cr Mn Fe 4s 2 3d 3 4s13d 5 ? • 4s-3d' 4s 2 3d 6 41 42 43 44 Nb Mo Tc Ru 5s 1 4d 4 5s 1 4d 5 5s2 4d 5 5s 1 4d 7 73 74 75 76 Ta W Re Os 6s 2 5d 3 6s 2 5d 4 ? . 6s-5d) 6s 2 5d 6 105 to6 107 108 Db Sg Bh Hs 7s 2 6d 3 7s26d 4 7s 2 6d 5 7s 2 6d 6 58 59 60 61 Ce Pr Nd Pm 6s 2 4f 1 sd l 6s 2 4]3 6s 2 4f4 6s 2 4 f5 90 91 92 93 Th Pa U Np 7s 2 6d 2 7s 2 Sf 2 6d l 7s2Sl6d 1 7 s2 Sf 4 6d 1 Figure 7.2 Va lence electron configurations of the elements. Classification of Elements 8B 9 27 Co 4s 23 d 7 45 Rb 5s 1 4d 8 77 Ir 6s 2 5d 7 109 Mt 7s 2 6d 7 62 Sm 6s 2 4f6 94 Po 7s 2 5f6 8A 18 2 3A 4A 5A 6A 7A He 1 13 14 15 16 17 ls 2 5 6 7 8 9 10 B C N 0 F Ne 2 2s22pl 2s 22p2 2s 2 2 p 3 2s 22 p4 2s 2 2 p 5 2s 2 2 p 6 13 14 15 16 17 18 1B 2B Al Si P S CI Ar 3 I 10 11 12 3s 2 3pl 3s 2 3p2 3s 2 3 p 3 3s 2 3p4 3s 2 3 p 5 3s 2 3 p 6 28 29 30 31 32 33 34 35 36 Ni Cu Zn Ga Ge As Se Br Kr 4 4s 2 3d 8 4s13dIO 3d 10 4s 2 4s24p I 4s 2 4p 2 4s 2 4 p 3 4s 2 4 p 4 4s 2 4 p 5 4s 2 4 p 6 46 47 48 49 50 51 52 53 54 Pd Ag Cd In Sn Sb Te I Xe 5 4d lo Ssl4d lO 5s 2 4d lO 5s 2 5 p l 5s 2 5p2 5s 2 5 p 3 5s 2 5 p 4 5s 2 5 p 5 5s 2 5 p 6 78 79 80 81 82 83 84 85 86 Pt Au Hg TI Pb Bi Po At Rn 6 6s 1 5d 9 6s 1 5dlO 6s 1 5d lO 6s 2 6pl 6s 2 6p2 6s 2 6 p 3 6s 2 6p4 6s 2 6 p 5 6s 2 6 p 6 110 111 ll2 1 1 3 114 115 116 (117) 118 Ds Rg - - - - - - 7 7s 2 6d 8 7s 2 6d 9 7s 2 6d lO 7s 2 7pl 7s 2 7p2 7s 2 7 p 3 7s 2 7 p 4 7s 2 7 p 6 63 64 65 66 67 68 69 70 Eo Gd Tb Dy Ho Er Tm Yb 6s 2 4f7 6s 2 4f 7 Sd i 6s 2 4f 9 6s 2 4flO 6s 24f 11 6s 2 4fl2 6s 2 4f l3 6s 2 4i 4 95 96 97 98 99 toO 101 102 Am Cm Bk Cf Es Fm Md No 75 2 5f 7 7s 2S l6 d l 7s 2 5f9 7s 2 5flO 7s 2 5f ll 7s 2 5fl2 7s 2 5f13 7s 2 5fl4 . . . , . . . . . Based on the type of subshell containing the outermost electron s, the elements can be divided into ategories the main group elements, the noble gases, the transition elements (or transition met- als), the lanthanides, and the actinides. The main group elements (also called the re pr esentative elements) are the elements in Groups lA through 7 A. With the exception of helium, each of the noble gases (the Group 8A elements) ha s a completely filled p subshell. The electron configura- tions are Ii for helium and ni np 6 for the other noble gases, where n is the principal quantum number for the outermost shell. The transition metals are the elements in Groups IB and 3B through 8B. Transition metals either have incompletely filled d subshells or readily produce cations with incompletely filled d . . . .,. ubshells. According to this defi nition, the elements of Group 2B are not transition metals. They are d-block elements, though, so they ge nerally are included in the discussion of transition metals. The lanthanides and actinides are sometimes calledf-block transition elements because they have incompletely filled f subshells. Figure 7.3 distinguishes the groups of elements discussed here. There is a distinct pattern to the electron configurations of the elements in a particular group. ee, for example, the electron configurations of Groups IA and 2A in Table 7.1. Each member of Group 1 A ha s a noble gas core plus one additional electron, giving each alkali metal the general electron configuration of [noble gas]nsl. Similarly, the Group 2A alkaline earth metals have a noble gas core and an outer electron configuration of ni. In this context, outermost electrons refe rs to those that are placed in orbitals last using the Aufbau principle [ ~ Section 6.8). The met a ls of Group 2B typically form + 2 ions, although they can also form + 1 i ons. In either case, the electron configuration includes a completed dsubshell [ ~~ Section 7.6]. Thus , strictly speaking, Zn, Cd, and Hg are not transition metals. ,i "" ~ Why Are There Two Different Sets of Numbers at the Top of the Periodic Table? The numbering of the transition metal groups 3B through 7B indicates the similarity between the outer electron configurations of these elements and those of the corresponding main group ele- ments. For example, scandium (Sc; Group 3B) and gallium (Ga; Group 3A) each have three outer electrons. However, because their outer electrons reside in different types of atomic orbitals (s and d orbitals in the case of Sc; s and p orbitals in the case of Ga), they belong in different groups. The elements of Groups IE and 2B have filled d subs hells [ ~~ Section 6.9] . Unlike the elements of Group 2B, elements of Group IB form cations with incompletely filled d subshells. Their group numbers correspond to the one and two electrons, respectively, that they have in s orbitals just like the main group elements in Groups lA and 2A. The elements iron (Fe), cobalt (Co), and nickel (Ni), and the Figure 7.3 Periodic table with color-coding of main group elements, noble gases, transition metals, group 2B metals, lanthanides, and actinides. 1 2 lA 1 H Li 2A 2 Be 3B 3 Na Mg 3 4 K Ca Sc S Rb Sr Y 6 Cs Ba Lu 7 Fr Ra Lr 6 7 elements that appear beneath them in the periodic table, cannot be classified in this way and are all placed in Group SB. The designation of A and B groups is not universal. In Europe, the practice is to use B for main group elements and A for transition or d-block elements, which is just the opposite of the American convention. The International Union of Pure and Applied Chemistry (IUPAC) has recommended eliminating ambiguity by numbering the columns sequentially with Arabic numerals 1 through IS (see Figure 7.2). The proposal has not been accepted universally, and many modern periodic tables retain the traditional group designations. Periodic tables in this text display both the IUPAC-recommended Arabic numerals and the traditional American numbering system. Discussions in the text will refer to the traditional American group numbers. 8A 18 3A 4A SA 6A 7A 13 14 15 16 17 He 1 B C N 0 F Ne 2 4B SB 6B 7B 1s8B~ IB 2B 4 S 678 91011 12 Al Si P S Cl Ar 3 Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 4 Zr Nb Mo Tc Ru Rb Pd Ag Cd In Sn Sb Te I Xe 5 Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 6 Rf Db Sg Bh Hs Mt Ds Rg 7 La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb 6 Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No 7 Although hydrogen's electron configuration is ls ' [ ~~ Section 6. 9, Figure 6.2S], it isa nonmetal and is not rea ll y a member of Group l A. . . . . . . Group 1A Group 2A • 242 Li [He]2s1 Na [Ne]3s 1 K Rb Cs Fr [Ar]4s1 [Kr]5s 1 [Xe]6s 1 [Rn]7s1 Be [He]2i Mg [Ne]3i Ca Sr Ba Ra [Ar]4s2 [Kr]5s 2 [Xe]6i [Rn]7i SECTION 7.2 The Modern Periodic Table 243 The outermost electrons of an atom are called valence electrons, which are the ones involved in the formation of chemical bonds between atom s. The similarity of the valence electron configu- rations (i.e., they have the same number and type of valence electrons) is what makes the elements in the same group resemble one another chemically. This observation holds true for the other main group elements as well. For instance, the halogens (Group 7 A) all have outer electron configura- tions of ni np 5, and they have similar properties. In predicting properties for Groups 3A through 7 A, we must take into account that each of these groups contains elements on both sides of the zigzag line that divides metals and nonmetals. For example, the elements in Group 4A all have the s ame outer electron configuration, ns 2 np2, but there is considerable variation in chemical properties among these elements because carbon is a nonmetal, silicon and germanium are metalloids, and tin and lead are metals. As a group, the noble gases behave very similarly. The noble gases are chemically inert because they all have completely filled outer ns and np subshells, a condition that imparts unusual stability. With the exception of krypton and xenon , they do not undergo chemical reactions. Although the outer electron configuration of the transition metals is not always the same within a group and there is often no regular pattern in the way the electron configuration changes from one metal to the next in the same period, all transition metals share many characteristics ( multiple oxidation states, richly colored compounds, magnetic properties, and so on) that se t them apalt from other elements. These properties are similar because all these metals have incom- pletely filled d subshells. Likewise, the lanthanide and actinide elements resemble one another because they have incompletely filled f subshells. Sample Problem 7.2 shows how to determine the electron configuration from the number of electrons in an atom. Without using a periodic table, give the ground-state electron configuration and block designation (s-, p-, d-, orf-block) of an atom with (a) 17 electrons, ( b) 37 electrons, and (c) 22 electrons. Classify each atom as a main group element or transition metal. Strategy Use the Aufbau principle discussed in Section 6.8. Start writing each electron configuration with principal quantum number n = 1, and then continue to assign electrons to orbitals in the order presented in Figure 6.23 until all the electrons have been accounted for. Setup According to Figure 6.23, orbitals fill in the following order: Is , 2s, 2p, 3s, 3p, 4s, 3d, 4p, Ss, 4d, Sp, 6s , and so on. Recall that an s subshell contains one orbital, a p subshell contains three orbitals, and a d subshell contains five orbitals. Remember, too, that each orbital can accommodate a maximum of two electrons. The block designation of an element corresponds to the type of subshell occupied by the last electrons added to the configuration according to the Aufbau principle. Solution (a) li2s 22p 6 3s 2 3i, p-block, main group (b) li2s22p 6 3s 2 3 p 6 4i3d104iSsl, s-block, main group ( c) li2s22p 6 3s 2 3 p 64i 3d 2 , d-block, transition metal Practice Problem A Without using a periodic table, give the ground-state electron configuration and block designation (s-, p-, d-, orf-block) of an atom with (a) 15 electrons, (b) 20 electrons, and (c) 35 electrons. Practice Problem B Identify the elements represented by (a) l S22s22i3s 2 3pl, (b) l s 22i 2p 6 3 i3i4s2 3d lO , and (c) lS22s2 2p63s23p64s23dl04isi. Representing Free Elements in Chemical Equations Ha ving classified the elements according to their ground-state electron configurations, we can now learn how chemists represent elements in chemical equations. etals Because metals do not exist in discrete molecular unit s but rather in complex, three- dim ensional networks of atom s, we always use their empirical formulas in chemical equations. Think About It Consult Figure 6.25 to confirm your answers. 244 CHAPTER 7 Electron Configuration and the Periodic Table The empirical formulas are the same as the symbols that represent the elements. For example, the empirical formula for iron is Fe, the same as the symbol for the element. Nonmetals There is no single rule regarding the representation of nonmetals in chemical equa- . . . , . . . . . . Recall that allotropes are different forms of the tions. Carbon, for example, exists in several allotropic forms. Regardless of the allotrope, we use same elem ent [ H~ Section 2.6]. its empirical formula C to represent elemental carbon in chemical equations. Often the symbol C will be followed by the specific allotrope in parentheses as in the equation representing the conver- sion of graphite to diamond, two of carbon's allotropic forms: Checkpoint 7.2 C(graphite) -_. C(diamond) For nonmetals that exist as polyatomic molecules, we generally use the molecular formula in equations: H 2 , N 2 , O?, F 2 , C1 2 , Brb 1 2 , and P 4, for example. In the case of sulfur, however, we usu- ally use the empirical formula S rather than the molecular formula S8' Thus, instead of writing the equation for the combustion of sulfur as we usually write although, technically, both ways are correct. Noble Gases All the noble gases exist as isolated atoms, so we use their symbols: He, Ne, Ar, Kr, Xe, and Rn. Metalloids The metalloids, like the metals, all have complex three-dimensional networks, so we also represent them with their empirical formulas that is, their symbol s: B, Si, Ge, and so on. The Modern Periodic Table 7.2.1 Which electron configuration is correct for a gennan ium ( Ge ) atom in the ground state? 7.2.2 Which of the following equations correctly repr ese nt the c hemical reaction in which graphite combines with sulfur to form carbon disulfide gas [CS 2 (g)]? (Select all that apply.) a) l i2i2p63s23l4i4p 2 b) li 2s 22 p 6 3i3 p 64s23d l 04p2 c) 1 i 2 s2 2p 6 3s 2 3p2 d) lS22s2 2p63i3p 64 s24 p24d lO e) ls 22i 2l3s 2 3p 2 3io Shielding is also known as scree ning. Core electrons are tho se in the completed inner s hell s. . a) C(graphite) + 2S(s) • CS 2 (g) b) C(graphite) + S 2( S) • CS 2 (g) c) C(graphite ) + Ss(S) -_. CSs(g) d) C(graphite) + l Ss(s) • CS 2 (g) e) 4C(graphite) + Ss(s) • 4CS 2 (g) Effective Nuclear Charge As we have seen, the electron configurations of the elements show a periodic variation with increasing atomic number. In this and the next few sections, we will examine how electron con- figuration explains the periodic variation of physical and chemical properties of the elements. We begin by introducing the concept of effective nuclear charge. Nuclear charge (Z) is simply the number of protons in the nucleus of an atom. Effective nuclear charge (ZerrJ is the actual magnitude of positive charge that is "experienced" by an elec- tron in the atom. The only atom in which the nuclear charge and effective nuclear charge are the same is hydrogen, which has only one electron. In all other atoms, the electrons are simultane- ously attracted to the nucleus and repelled by one another. This results in a phenomenon known as shielding. An electron in a many-electron atom is partially shielded from the positive charge of the nucleus by the other electrons in the atom. One way to illustrate how electrons in an atom shield one another is to consider the amounts of energy required to remove the two electrons from a helium atom, shown in Figure 7.4. Experi- ments show that it takes 3.94 X 10- 18 J to remove the first electron but 8.72 X 10 - 18 J to remove the second one. There is no shielding once the first electron is removed, so the second electron feels the full effect of the + 2 nuclear charge and is more difficult to remove. Although all the electrons in an atom shield one another to some extent, those that are most . . . . . . . effective at shielding are the core electrons. As a result, the value of Ze ff increases steadily from What Causes the Periodic Trends in Properties? Many of the periodic trends in properties of the elements can be explained using Coulomb's* law, which states that the force ( F) between two charged objects (QJ and Q2) is directly proportional to the product of the two charges and inversely proportional to the distance (d) between the objects squared: Force is inversely proportional to d 2 , whereas energy is inversely proportional to d [I ~~ Section 5.1]. The SI unit of force is the newton (N = m . kg/s2) and the SI unit of energy is the joule (J = m 2 • kg /s2) . When the charges have opposite signs, F is negative indicating an attractive force between the objects. When the charges have the same sign, F is positive- indicating a repulsive force. Table 7.2 shows how the magnitude of the attractive force between two oppositely charged objects at a fixed distance from each other varies with changes in the magnitudes of the charges. F et:. QJ X Q2 d 2 *Charles Au gustin de Co ul omb ( 17 36- 1806). French ph ys icis t. Coulomb did research in el ectrici ty and magnetism and applied Newton's inverse square law to electricity. He al so in ve nted a torsion balance. Attractive Cnaro-ed Fixed Eadi Q, Q2 Attractive force is proportional to +1 - 1 1 +2 - 2 4 +3 - 3 9 The product of a positive number and a negative number is a negative number . When we are s impl y co mparing the magnitudes of attractive forces, however, it is unnecessary to include the sign. left to right across a period of the periodic table because the number of core electrons remains the same (only the number of protons, Z, and the number of vale nc e electrons increases). As we move to the right across period 2, the nuclear charge increases by 1 with each new element, but the effective nuclear charge increases only by an average of 0.64. (If the valence elec- trons did not shield one another, the effective nuclear charge would also increase by I each time a proton was added to the nucleus.) Z Z e ff (felt by valence electrons) Li 3 1.28 Be 4 1.91 In general, the effective nuclear charge is given by Z eff = Z - IT B 5 2.42 C 6 3. 14 N 7 3.83 o 8 4.45 F 9 5.10 Equation 7.1 where IT is the shielding constant. The shielding constant is greater than zero but smaller than Z. The change in Z eff as we move from the top of a group to the bottom is generally less signifi- cant than the change as we move across a period. Although each step down a group represents a large increase in the nuclear charge, there is al so an additional shell of core electrons to shield the yalence electrons from the nucleus. Consequently, the eff ective nuclear charge changes less than rb e nuclear charge as we move down a column of the periodic table. +2 3.94 X 10- 18 ] to remove +2 8.72 X 10- 18 ] to remove Fig u re 7.4 Remov al of the first electron in He requires less energy than remov al of the second electron because of shielding. 245 246 CHAPTER 7 Electron Configuration and the Periodic Table (a) (b) Figure 7.5 (a) Atomic radius in metals is defined as half the distance between adjacent metal atoms. (b) Atomic radius in nonmetals is defined as half the distance between bonded identical atoms in a molecule. ~'!!- - Mul time d ia Periodic Table atomic radius . • Periodic Trends in Properties of Elements Several physical and chemical properties of the elements depend on effective nuclear charge. To understand the trends in these properties, it is helpful to visualize the electrons of an atom in shells. Recall that the value of the principal quantum number (n) increases as the distance from the nucleus increases [ ~~ Section 6.7]. If we take this statement literally, and picture all the electrons in a shell at the same distance from the nucleus, the result is a sphere of uniformly dis- tributed negative charge, with its distance from the nucleus depending on the value of n. With this as a starting point, we will examine the periodic trends in atomic radius, ionization energy, and electron affinity. Atomic Radius Intuitively, we think of the atomic radius as the distance between the nucleus of an atom and its valence shell (i.e., the outermost shell that is occupied by one or more electrons), because we usu- ally envision atoms as spheres with discrete boundaries. According to the quantum mechanical model of the atom, though, there is no specific distance from the nucleus beyond which an electron may not be found [ ~~ Section 6.7] . Therefore, the atomic radius requires a specific definition. There are two ways in which the atomic radius is commonly defined. One is the metallic radius, which is half the distance between the nuclei of two adjacent, identical metal atoms [Fig- ure 7.S(a) ]. The other is the covalent radius, which is halfthe distance between adjacent, identical nuclei in a molecule [Figure 7.S(b)]. Figure 7.6 shows the atomic radii of the main group elements according to their positions in the periodic table. There are two distinct trends. The atomic radius decreases as we move from left to right across a period and increases from top to bottom as we move down within a group. - Increasing atomic radius " ", .", - m =-""===:=:::= =~=::::;;;:::::: :=Il l en '" .~ "0 '" , u '8 0 ~ '" bll " .~ en '" !:l u " lA H 37 Li IS2 Na 186 K 227 Rb 24 8 Cs 265 2A Be 112 Mg 160 Ca 197 Sr 215 Ba 222 3A 4A B C 8S 77 Al Si 143 118 Ga Ge 135 123 In Sn 166 140 Ti Pb 171 175 Figure 7.6 Atomic radii of the elements (in picometers) . SA 6A N o 7S 73 P S 110 103 As Se 120 117 Sb Te 141 143 Bi Po 155 164 7A F 72 CI 99 Br 114 I 133 At 142 He Q 31 Ne 70 Ar 98 Kr 112 Xe 131 Rn 140 SECTION 7.4 Periodic Trends in Properties of Elements 247 The increase down a group is fairly easily explained. As we step down a column, the outermost occupied shell has an ever-increasing value of n, so it lies farther from the nucleus, making the radius bigger. Now let 's try to under s tand the decrease in radiu s from left to right across a period. Although this trend may at first see m counterintuitive, given that the number of vale nce electrons is increa sing with each new element, consider the shell model in which all the electrons in a shell form a uniform sphere of negative charge around the nucleu s at a di sta nce specifi ed by the value of n. As we move from left to right acro ss a period, the effective nuclear charge increa ses and each step to the right add s another electron to the valence shell. Coulomb's law dictates that there will be a more powerful attraction between the nucleus and the valence shell when the magnitude s of both charges increa se . The result is that as we step across a period the valence s hell is drawn closer to the nucleus, making . . . . . . . . . . . . . . . . . . . . . . . . the atomic radius smaller . Figure 7.7 shows how the effective nuclear charge, charge on the valence shell, and atomic radius vary across period 2. We can picture the valence shells in all the atoms as being initially at the sa me distance (determined by n) from the nuclei, but being pulled closer by a larger attractive force resulting from increase s in both Zeff and the number of valence electrons. Sample Problem 7.3 s hows how to use these trend s to compare the atomic radii of different elements. Referring only to a periodic table, arrange the elements P, S, a nd 0 in order of increasing atomic radius. Strategy Use the left-to-right (decreasing) and top-to-bottom (increasin g) trends to compare the atomic radii of two of the three elements at a time. Setup Sulfur is to the right of phosphorus in the third row, so sulfur should be smaller that phosphorus. Oxygen is abo ve sulfur in Group 6A, so oxygen should be smaller than sulfur. Solution: 0 < S < P. Practice Problem A Referring only to a periodic table, arrange the elemen ts Ge, Se, and F in order of increasing atomic radius. Practice Problem B For which of the following pairs of elements can the atomic radii not be compared using the periodic table alone: P and Se, Se and Cl, or P and O? ~ Ionization Energy Ionization energy (IE) is the minimum energy required to remove an electron from an atom in the gas pha se. Typically, we express ionization energy in kJ/mol, the number of kilojoules required to re move a mole of electron s from a mole of gaseous atoms. Sodium, for example, ha s an ionization energy of 495.8 kJ/mol, meaning that the energy input required to drive the process Na(g) • Na +(g) + e- is 495.8 kJ/mol. Specifically, this is the first ionization energy of so dium , lE I (Na), which corre- ponds to the removal of the mo st loo se ly held electron. Figure 7.8(a) s how s the first ionization Zeff Charge on valence shell Frx • • 0 Li Be B C N 0 F 1. 28 1.91 2.42 3. 14 3.83 4.45 5. 10 -I -2 -3 -4 - 5 - 6 - 7 (+1.28)(- 1) (+ 1.91) (-2) (+2 .42)(- 3) (+3.14)(-4) (+3.83)(-5) (+4.45)(-6) (+ 5.10)( -7) Although the overall trend in atomic size for transition elements is also to decrease from left to right and increase from top to bottom, the observed radii do not vary in as regular a way as do the main group elements. Think About It Consult Figure 7.6 to confirm the order. Note that there are circumstances under which the trends alone will be insufficient to compare the radii of two elements. Using only a periodic table, for example, it would not be possible to determine that chlorine (r = 99 pm) has a larger radius than oxygen (r = 73 pm). Figure 7.7 Atomic radius decreases from left to right across a period because of the increased electrostatic attraction between the effective nuclear charge a nd the charge on the valence shell. The dark circle shows the atomic size in each case. 248 CHAPTER 7 Electron Configuration and the Periodic Table IA 1 H 1312 Li 520 Na 496 K 419 Rb 403 Cs 376 2A 3A 2 13 Be B 899 800 Mg Al 738 577 Ca Ga 590 579 Sr In 549 558 Ba TI 503 589 4A SA 14 15 C N 1086 1402 Si P 786 1012 Ge As 761 947 Sn Sb 708 834 Pb Bi 715 703 6A 16 0 1314 S 999 Se 941 Te 869 Po 813 7A 17 F 1681 CI 1256 Br 1143 I 1009 At (92 6) 8A 18 He 2372 Ne 2080 Ar 1520 Kr 1351 Xe 1170 Rn 1037 2500 He 2000 Ne ~ - 0 E ~ , ~ ~ Ar >. 1500 OJ) Kr ~ " c Q) c H Xe Hg 0 ~ co Rn N 1000 c I 0 I ~ I ~ Y ~ - IE) values for main group elements. (kl lmol) (a) 500 Li Na K Rb Cs o 10 20 30 40 50 60 70 80 90 Atomic number (Z) (b) Figure 7.8 (a) First ionization energies (in kJ /mol) of the main group elements. (b) First ionization energy as a function of atomic number. As with atomic radius, ionization energy changes in a similar but somewhat less regular way among the transition elements. Harder to remove E 1 ' ' np ns Group 2A energies of the main group elements according to their positions in the periodic table. Figure . . . "" . 7.8(b) shows a graph of lE\ as a function of atomic number. In general, as effective nuclear charge increases, ionization energy also increases. Thus, lE[ increases from left to right across a period. Despite this trend, the graph in Figure 7.8(b) shows that lE [ for a Group 3A element is smaller than that for the corresponding Group 2A element. Likewise, lE [ for a Group 6A element is smaller than that for the corresponding Group SA ele- ment. Both of these interruptions of the upward trend in lE \ can be explained by using electron configuration. Recall that the energy of an electron in a many-electron system depends not only on the principal quantum number ( .e ), but also on the angular momentum quantum number (.e) [~~ Sec- tion 6.8, Figure 6.23] . Within a given shell, electrons with the higher value of.e have a higher energy (are less tightly held by the nucleus) and are therefore easier to remove. Figure 7.9(a) shows the relative energies of an s subshell (.e = 0) and a p subshell (.e = 1). Ionization of an ele- ment in Group 2A requires the removal of an electron from an s orbital, whereas ionization of an element in Group 3A requires the removal of an electron from a p orbital; therefore, the element in Group 3A has a lower ionization energy than the element in Group 2A. As for the decrease in ionization energy in elements of Group 6A compared to those in Group SA, both ionizations involve the removal of a p electron, but the ionization of an atom in Group 6A involves the removal of a paired electron. The repulsive force between two electrons in the same orbital makes it easier to remove one of them, making the ionization energy for the Group 6A element actually lower than that for the Group SA element. [See Figure 7.9(b).] The first ionization energy l Ei decreases as we move from top to bottom within a group due to the increasing atomic radius. Although the effective nuclear charge does not change signifi- cantly as we step down a group, the atomic radius increases because the value of n for the valence Easier to remove Harder to remove Easier to remove / " / I ' 1 1 I' 1 1 1 ElliJ np E[ll] np ElliJ np ns ns ns Group 3A Group SA Group 6A (a) (b) Figure 7.9 (a) It is harder to remove an electron from an s orbital than it is to remove an electron from a p orbital with the same principal quantum number. (b) Within a p subshell, it is easier to remove an electron from a doubly occupied orbital than from a si ngly occupied orbital. SECTION 7.4 Periodic Trends in Properties of Elements 249 kJ/mol) lements 3 Z IE1 IE2 IE3 IE4 IE5 IE6 IE7 Li 3 520 7,298 11,815 Be 4 899 1,757 14,848 21,007 21,007 B 5 800 2,427 3,660 25,026 32,827 C 6 1,086 2,353 4,621 6,223 37,831 47,277 N 7 1,402 2,856 4,578 7,475 9,445 53,267 64 ,3 60 0 8 1,314 3,388 5,301 7,469 10,990 l3,327 71,330 F 9 1,681 3,374 6,050 8,408 11 ,023 15,164 l7,868 Ne 10 2,080 3,952 6,122 9,371 12,177 15,238 19,999 Na 11 496 4,562 6,910 9,543 l3,354 16,6l3 20,117 *Cells shaded with blue represent the removal of core electrons. shell increases. According to Coulomb 's law, the attractive force between a valence electron and the effective nuclear charge gets weaker as the distance between them increases. This makes it easier to remove an electron, and so lEI decreases. It is possible to remove additional electrons in subsequent ionizations, giving IE z, IE 3 , and so o n. The second and third ionizations of sodium, for example, can be represented, respectively, as However, the removal of successive electrons requires ever-increasing amounts of energy because it is harder to remove an electron from a cation than from an atom (and it gets even harder as the charge on the cation increases). Table 7.3 li sts the ionization energies of the elements in period 2 and of sodium. These data show that it takes much more energy to remove core electrons than [0 remove valence electrons. There are two reasons for this. First, core electrons are closer to the nucleus, and second, core electrons experience a greater effective nuclear charge because there are fewer filled shells shielding them from the nucleus. Both of these factors contribute to a greater attractive force between the electrons and the nucleus, which must be overcome in order to remove me electrons. Sample Problem 7.4 shows how to use these trends to compare first ionization energies, and ubsequent ionization energies, of specific atoms. Would you expect Na or Mg to have the greater first ionization energy (lEI)? Which should have the greater second ionization energy (IE 2 )? Strategy Consider effective nuclear charge and electron configuration to compare the ionization energies. Effective nuclear charge increases from left to right in a period (thus increasing IE), and it is more difficult to remove a paired core electron than an unpaired valence electron. Setup Na is in Group lA, and Mg is beside it in Group 2A. Na has one valence electron, and Mg has two valence electrons. Solution lEI (Mg) > lEI (Na) becau se Mg is to the right of Na in the periodic table (i.e., Mg has the greater effective nuclear charge, so it is more difficult to remove its electron). lE z(Na) > IE 2 (Mg) because the second ionization of Mg removes a valence electron, whereas the second ionization of Na removes a core electron. Practice Problem A Which element, Mg or AI, will have the higher first ionization energy and which will have the higher third ionization energy? Practice Problem B Explain why Rb has a lower fEI than Sr, but Sr has a lower fE z than Rb. ~ IE8 lEg IE 10 84,078 92,038 106,434 23,069 115,380 l31,432 25,496 28,932 141,362 Think About It The first ionization energies of Na and Mg are 496 and 738 kJ/ mol , respectively. The second ionization energies of Na and Mg are 4562 and 1451 kJ/ mol , respectively. [...]... 2HzO(I) + 2NaOH(aq) + 2HzCg) CaHz(s) + 2H zO(I) • Ca(OH)z(aq) + 2H 2(g) The most important compound of hydrogen is water, which forms when hydrogen burns in air: In many ways the chemistry of Li resembles that of Mg and the chemistry of Be resembles that of AI This behavior is called the diagonal relationship 258 CHAPTER 7 Figure 7.14 Electron Configuration and the Periodic Table I Group lA elements... them valuable in the manufacture of coins and jewelry For this reason, these metals are also known as "coinage metals." The differences in the chemistry of the elements in Group 2A from that of the elements in Group 2B can be explained in a similar way Bringing Chemistry to life Rad ioactive Bone One of the consequences of two chemical species having similar properties is that human physiology sometimes... the case of Fe, that would be the 4s subshell Fe: [Ar]4i3d 6 • Fe2+ : [Ar]3d 6 ::-on can also form the Fe 3 + ion, in which case the third electron is removed from the 3d subshell • • • This explains in part why many of the transition metals can form ions with a + 2 charge 254 CHAPTER 7 Electron Configuration and the Periodic Table In general, when a d-block element becomes an ion, it loses electrons... atoms Then add electrons (for anions) or remove • electrons (for cations) to account for the charge The electrons removed from a d-block element must come first from the outermost s subshell, not the partially filled d subshell • • • · • • • • • • • • Think About It Be sure to add electrons to form an anion and remove electrons to form a cation Also, double-check to make sure that electrons removed... example, depends on the relative sizes of its cations and anions SECTION 7.6 • Comparing Ionic Radius with Atomic Radius When an atom loses an electron and becomes a cation, its radius decreases due in part to a reduction in electron-electron repulsions (and consequently a reduction in shielding) in the valence shell A significant decrease in radius occurs when all of an atom's valence electrons are... configuration to compare the electron affinities The effective nuclear charge increases from left to right in a period (thus generally increasing EA), and it is more difficult to add an electron to a partially occupied orbital than to an empty one Writing out orbital diagrams for the valence electrons is helpful for this type of problem Setup (a) Al is in Group 3A and Si is beside it in Group 4A AI... prepared from xenon and krypton by exposing them to very strong oxidizing agents such as fluorine and oxygen Some of the compounds that have been prepared are XeF4 , Xe03, XeOF4 , and KrF2 Although the chemistry of the noble gases is interesting, their compounds are not involved in any natural biological processes and they currently have no major commercial applications Comparison of Group 1A and Group... (Si) > EA] (P) because although P is to the right of Si in the third period of the periodic table (giving P the larger Zeff), adding an electron to a P atom requires placing it in a 3p orbital that is partially occupied The energy cost of pairing electrons outweighs the energy advantage of adding an electron to an atom with a larger effective nuclear charge Practice Problem A Would you expect Mg or... facilities Strontium-90 released into the atmosphere will eventually settle on land and water, and it can reach our bodies through ingestion especially of vegetation-and through the inhalation of airborne particles Because calcium and strontium are chemically similar, SrH ions are mistaken by the body for Ca?+ ions and are incorporated into the bones Constant exposure to the radiation emitted by strontium-90 . properti es of alkali and alkaline earth metals . In many ways the chemistry of Li rese mb les that of Mg and the chemistry of Be re sembl es that of AI. Th is behav io r is called. just the opposite of the American convention. The International Union of Pure and Applied Chemistry (IUPAC) has recommended eliminating ambiguity by numbering the columns sequentially with. because they all have completely filled outer ns and np subshells, a condition that imparts unusual stability. With the exception of krypton and xenon , they do not undergo