Answers to selected problems, Chapter 8

    Note: 1 the up and down arrows needed in several of these questions could not be made to reproduce in both Internet Explorer and Netscape. I chose Internet Explorer 5 because that is my default browser. If you don't have access to IE 5, I will give you the printed version of these answers.
    Note 2: For tutorial purposes, I have generally written the electron configurations in the order in which the subshells are filled. This practice differs from that of the book, which generally lists them by "position" instead.

Review questions

3. The principle at work in this question is that for the hydrogen atom, the energy levels of orbitals within a principal shell are all the same, whereas for atoms of other elements, the energy levels of all orbitals are different. (See point 1 in the middle of page 317 and Figure 8.1 on the same page.)
    (a) The three orbitals listed for this part (1s, 2s, 2p), are at different energy levels for atoms of any element because they come from different shells. Orbitals 2s and 2p are at identical levels for hydrogen, though.
    (b) The three orbitals listed here (3s, 3p, 3d) will be at the same energy levels for hydrogen but not for other elements.
    (c) The three orbitals listed here (3p_x, 3p_y, 3p_x) will be at the same energy levels for hydrogen and for other elements.

10. What subshell(s) is(are) being filled in these regions of the periodic table?
    (a) Groups 1A and 2A--the various s subshells 1s, 2s, etc.
    (b) Groups 3A through 7A--the p subshells 2p, 3p, etc.
    (c) the transition elements--the d subshells 3d, 4d, etc.,  but where 3d is being filled in row 4, 4d in row 5, etc.
    (d) the lanthanides and actinides--the f subshells 4f and 5f, but where 4f is being filled in row 6, etc.

18. Must all atoms with odd atomic numbers be paramagnetic? Yes. All paramagnetic elements have at least one odd electron.
    Must all atoms with even atomic numbers be diamagnetic? No, because an element with even atomic number may have two, four, etc. single electrons in different shells and thus be paramagnetic. An example is carbon: 1s²2s²2p_x¹2p_y¹.

Problems

29. Which orbital diagrams are possible and which not?
    (a) Not possible because one of the 2p pairs of electrons have the same (positive) spins.
    (b) Not possible because the last orbital contains three electrons (two is the maximum per orbital).
    (c) Possible--the orbital diagram of nitrogen.
    (d) Allowed, but the spins are by convention drawn upward first.
    (e) Not possible because the first three (unpaired) electrons in the 2p subshell must all have the same spin.
    (f) Possible--the orbital diagram of magnesium.

33. Explain why each of these electron configurations is wrong.
    (a) 1s²2s⁶3s²--The 2s orbital can have only two electrons, not the six indicated here.
    (b) 1s²2s²2p⁷3s¹--The 2p subshell can contain only six electrons, not the seven shown here.
    (c) 1s²2s²2p⁶2d³--There is no 2d subshell.

35. Use the spdf notation and the Aufbau principle (Figure 8.2, page 320) to write the electronic configuration for each of these elements:
    (a) Al (Z = 13) 1s²2s²2p⁶3s²3p¹ 
    (b) Cl (Z = 17) 1s²2s²2p⁶3s²3p⁵ 
    (c) Na (Z = 11) 1s²2s²2p⁶3s¹ 
    (d) B (Z = 5) 1s²2s²2p⁴ 
    (e) He (Z = 2) 1s² 
    (f) O (Z = 8) 1s²2s²2p⁴ 
    (g) C (Z = 6) 1s²2s²2p² 
    (h) Li (Z = 3) 1s²2s¹ 
    (i) Si (Z = 14) 1s²2s²2p⁶3s²3p² 

39. Orbital diagrams for the ground-state electronic configurations for atoms of the following elements.
    (a) C (Z = 6) 1s ↑↓ 2s ↑↓ 2p ↑ ↑
    (b) O (Z = 8)  1s ↑↓ 2s ↑↓ 2p ↑↓↑ ↑
    (c) K (Z = 19)  1s ↑↓ 2s ↑↓ 2p ↑↓↑↓↑↓ 3s ↑↓ 2p ↑↓↑↓↑↓ 4s ↑
    (d) Al (Z = 13) 1s ↑↓ 2s ↑↓ 2p ↑↓↑↓↑↓ 3s ↑↓ 2p ↑
    (e) S (Z = 16) 1s ↑↓ 2s ↑↓ 2p ↑↓↑↓↑↓ 3s ↑↓ 2p  ↑↓↑ ↑
    (f) Mg (Z = 12) 1s ↑↓ 2s ↑↓ 2p ↑↓↑↓↑↓ 3s ↑↓  

43. Orbital diagrams for these ions.
    (a) Br^-  The short answer is [Kr], because Br's extra electron gives it the same electronic configuration as Kr has.
        The long answer is [Ar] 3d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 4s ↑↓ 4p ↑↓↑↓↑↓. Be careful, though, because the electronic configurations, in the form that reproduces the order in which the subshells are filled (see discussion on page 324 of the two styles of writing electron configurations), would be [Ar] 4s²3d¹⁰4p⁶. It may be easier for you to start with the electronic configuration, for which you can use the Aufbau diagram in Figure 8.2, and then convert the order of the orbitals to that required for the orbital diagrams, which go by "position" rather than by energy. 
    (b) Ni²⁺   Careful! The electronic configuration of Ni is [Ar] 4s²3d⁸, which reflects the sequence of adding the electrons after the argon core. In terms of "position," they are written in the orbital diagram as [Ar] 3d ↑↓↑↓↑↓↑ ↑ 4s ↑↓ . (Note the blank spaces after the last two up-arrows in 3d.) The two electrons stripped to create Ni²⁺ come from the outermost shell, or 4s². Thus, Ni²⁺ is [Ar] 3d⁸, or [Ar] 3d ↑↓↑↓↑↓↑ ↑ . Whew!
    (c) Sb³⁺   The configuration of regular Sb (in sequence) is [Kr] 5s²4d¹⁰5p³. Giving it a triply positive charge means removing the last three electrons, which will give [Kr] 5s²4d¹⁰. The orbital diagram for this is [Kr] 4d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 5s ↑↓ . 
    (d) Te^2-   The configuration for Te is like that of Sb but with one more electron: [Kr] 5s²4d¹⁰5p⁴. Adding two electrons fills the 5p subshell, for [Kr] 4d¹⁰5s²5p⁶ (by position). That gives [Xe], or  [Kr] 4d ↑↓↑↓↑↓↑↓↑↓ 5s ↑↓ 5p ↑↓↑↓↑↓ .

47. (a) Bi has five valence electrons (6s²6p³). The 5d and 4f electrons in between are not in the valence shell (6).
    (b) The electrons in the fourth principal shell of Au are 4s²4p⁶4d¹⁰4f¹⁴ = 32 in number.
    (c) Elements with five valence-shell electrons are those in Group 5A, or N, P, As, Sb, Bi.
    (d) Any unpaired electron in Se will be found in its valence shell, whose electrons are 4s²4p⁴. By Hund's rule, the four electrons will enter the 4p subshell as follows: 4p_x4p_y4p_z4p_x, which will leave 4p_y and 4p_z unpaired.
    (e) There are ten transition elements in the fifth period: Y, Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag, and Cd. Note that these transition elements are in the fifth period but their transition electrons are 4d¹⁰.

49. To determine whether a given species is diamagnetic or paramagnetic, examine its orbital structure for unpaired electrons. If any are present, the species is paramagnetic. If not, it is diamagnetic.
    (a) S has two unpaired electrons, and so is paramagnetic: [Ne] 3s²3p_x²3p_y3p_z. 
    (b) Ba has no unpaired electrons, and so in diamagnetic: [Xe] 6s².
    (c) Careful! V²⁺ has three unpaired electrons, and so is paramagnetic. This problem is like 43b. The electronic structure of V is [Ar] 4s²3d³. The two electrons that are stripped to make the ion come from the outer shell, however, which leaves [Ar] 3d³. The 3d electrons enter in the sequence 3d_xy3d_xz3dyz, all unpaired (Hund's rule).
    (d) O^2- has the electronic configuration of [Ne], in which all electrons are paired: 1s²2s²2p⁶ . It is thus diamagnetic.
    (e) Ag's electronic configuration is [Kr] 5s²4d⁹. Ag is one of those transition elements whose electronic structure is exceptional. The nine 4d electrons should enter as 4d_xy²4d_xz²4d_yz²4d_x2-y2²4d_z, but it "borrows" one electron from the 5s² to fill the 4d subshell and give 5s¹4d¹⁰ (see page 326). One electron remains unpaired, though, as it must.
    Comment: Of course, one easy way to approach this problem is to recognize that any element with an odd atomic number must have at least one unpaired electron, and so will be paramagnetic. (The converse in not necessarily true, however, because even-numbered atoms can easily have two or four unpaired electrons--see S above.) S is even but with two unpaired electrons, Ba is even with no unpaired electrons, V2+ is odd minus two, or still odd, O2- is even plus two makes full shell, and Ag is odd.

53. Arranging atoms in order of increasing atomic radius:
    (a) Al, Mg, and Na are all in the third A-row of the periodic table. Since atomic radii of atoms decrease to the right along any A-row, the sequence of radii will be Al < Mg < Na.
    (b) Ca, Mg, and Sr are all in the second group. Atomic radii increase down a group. This gives the sequence of radii Mg < Ca < Sr.

57. (a) You can confidently predict that Ca is larger than Cl because or two factors acting in the same direction: moving leftward from Cl to Mg will increase the radii of the atoms, and moving downward from Mg to Ca will also increase radii.
    (b) Ill-defined; can't answer reliably. There are four steps to reasoning out this answer. 
        (1) K⁺ is [Ar] but with one more proton in the nucleus, so that K⁺ < [Ar].
        (2) F- is [Ne] but with one fewer proton in the nucleus, so that F- > [Ne].
        (3) [Ne] < [Ar].
        (4) So in comparing the sizes of K+ and F-, we are comparing < [Ar] with > [Ne], or < [Ar] with > (< [Ar]). Without knowing the magnitudes of the < and > factors, this can't be done.

59. (a) Because ionization energies decrease down the periodic table, Ba < Ca < Mg.
    (b) Because ionization energies decrease with increasing size and size increases leftward across the periodic table, ionization energies will increase rightward across the table. Thus, Al < P < Cl.
    (c) This problem just involves a lot of bookkeeping. Ne is a noble gas with a high first ionization energy. F is above Cl, so it will have an greater ionization energy. Cl is to the right of Na, so Cl will have a higher ionization energy. Fe, a transition metal, has about the same ionization energy as Ca, which is great than Na (Table 8.4) but less than Cl (deduction fro same table). Thus the order of ionization energies is Na < Fe < Cl < F < Ne.

65. Silicon has a greater electron affinity than P does because Si, with only 2 3p electrons, has one empty 3p orbital (3p_z) where the additional electron can go without interacting strongly with any existing 3p electrons. P, with one additional electron, has one in each of its 3p orbitals, so the next electron would have to enter an orbital that already contains one electron. Since there is less "resistance" to putting an electron into an empty orbital than into one that is half full, Si will have the greater electron affinity.

67. The graph of atomic volume (as volume per mole of atoms) is shown below. This curve resembles the curve for atomic radius versus atomic number, which is Figure 8.9 in the text.

71. This is an easy problem. Tl is lowest and farthest to the left, so it will have the lowest ionization energy. Ne, highest and farthest to the right, will have the highest. From Tl, next-lowest is Ge, above and to the right. Then comes Se, to its right, followed by S, which is directly above Se. From there, Ne is above and to the right, so it has higher ionization energy still. Thus the sequence of increasing ionization energies is Tl < Ge < Se < S < Ne.

75. (a) Completing the equation: N₂O₅(s) + H₂O(l) ® HNO₃(aq)
            Balancing the equation:  N₂O₅(s) + H₂O(l) ® 2 HNO₃(aq)
    (b) Completing the equation: MgO(s) + CH₃COOH(aq) ® Mg(CH₃COO)₂(aq) + H₂O(l)     
            Balancing the equation:  MgO(s) + 2 CH₃COOH(aq) ® Mg(CH₃COO)₂(aq) + H₂O(l)   
    (c) Completing the equation: Li₂O(s) + H₂O(l) ®  LiOH(aq)    
            Balancing the equation:  Li₂O(s) + H₂O(l) ®  2 LiOH(aq)    

77. The answers are shown in the periodic table below. 

81. 

83. 

89. 

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