Chapter 7
The Structure of Atoms and Periodic Trends
Chapter 7
The Structure of Atoms and Periodic Trends
INSTRUCTOR’S NOTES
As noted in the previous chapter, about nine lectures are devoted to the material in Chapters 6 & 7.
Challenges for teaching this material include:

Nodes in probability (“How does the electron get to the other side of the nucleus?”)

Subshell filling order (Figure 7.2)

Exceptions to the usual filling order of electrons in the transition metals

Electron configurations of transition metal ions

The indistinctness of atomic size

Exceptions to periodic properties.
Note: The definition of electron affinity used is the energy involved when an atom in the gas phase acquires an
electron. Some texts use a different definition. Therefore, for example, the electron affinity for F is –328 kJ/mol. In
class, we emphasize that the affinity for an addition electron beyond the number in the neutral atom increases
(becomes more negative) moving across a period.
SUGGESTED DEMONSTRATIONS
1.
Electron Configurations

Garofalo, A. “Housing Electrons: Relating Quantum Numbers, Energy Levels, and Electron
Configurations,” Journal of Chemical Education 1997, 74, 709.
3.
Periodic Properties of the Elements

The Periodic Table Live! CD-ROM from JCE: Software is most valuable with this chapter. (It is also still
available on videodisc!) See the information in the Preface to this Manual and the following article: Kotz,
J. Journal of Chemical Education 1989, 66, 750. The Periodic Table Live! CD-ROM can be used to
illustrate periodic trends in reactions. The CD is essentially an image database that contains images and
movies of the elements reacting with air, water, acids, and a base. In addition, it contains information on the
history and uses of the elements. We use it, for example, to illustrate the trend in reactivity of the alkali and
alkaline earth metals with air and water. In addition, we carry out the reactions of some of the alkali and
alkaline earth metals with water in petri dishes on an overhead projector.
137
Chapter 7
The Structure of Atoms and Periodic Trends
SOLUTIONS TO STUDY QUESTIONS
7.1
P
1s22s22p63s23p3


  

1s
2s
2p
3s



3p
Phosphorus is in Group 5A and has five electrons in its outer shell.
Cl 1s22s22p63s23p5


  

  
1s
2s
2p
3s
3p
Chlorine is in Group 7A and has seven electrons in its outer shell.
7.2
Mg 1s22s22p63s2


  

1s
2s
2p
3s
Magnesium is in group 2A and has two electrons in its outer shell.
Ar 1s22s22p63s23p6


  

  
1s
2s
2p
3s
3p
Argon is in group 8A and has eight electrons in its outer shell.
7.3
Cr 1s22s22p63s23p63d 54s1
Fe 1s22s22p63s23p63d 64s2
1s22s22p63s23p63d 34s2
7.4
V
7.5
(a) As
[Ar]3d 104s24p3
(b) Kr
[Ar]3d 104s24p6
(a) Sr
[Kr]5s2
(b) Zr
[Kr]4d 25s2
(c) Rh
[Kr]4d 75s2
(d) Sn
[Kr]4d 105s25p2
(a) Ta
[Xe]4f 145d 36s2
(b) Pt
[Xe]4f 145d 86s2
(actual configuration [Xe]4f 145d 96s1)
(a) Sm
[Xe]4f 55d 16s2
(actual configuration [Xe]4f 66s2)
(b) Yb
[Xe]4f 135d 16s2
(actual configuration [Xe]4f 146s2)
7.6
7.7
7.8
[Rn]5f 77s2
7.9
Am
7.10
(a) Pu
[Rn]5f 56d 17s2
(b) Cm
[Rn]5f 76d 17s2
138
(actual configuration [Kr]4d 85s1)
(actual configuration [Rn]5f 67s2)
Chapter 7
7.11
The Structure of Atoms and Periodic Trends
(a) 2 electrons
(b) 1 electron
(c) none; the maximum value of  is (n – 1)
7.12
(a) 18 electrons
(b) 10 electrons
(c) 1 electron
(d) none; when  = 0, m can only have a value of 0
7.13
Mg
[Ne]

3s
n = 3,  = 0, m = 0, ms = +1/2
n = 3,  = 0, m = 0, ms = –1/2
7.14
P
[Ne]


3s


3p
n = 3,  = 0, m = 0, ms = +1/2
n = 3,  = 0, m = 0, ms = –1/2
n = 3,  = 1, m = –1, ms = +1/2
n = 3,  = 1, m = 0, ms = +1/2
n = 3,  = 1, m = +1, ms = +1/2
7.15
Ga
[Ar]
    

3d
4s

4p
n = 4,  = 1, m = 0, ms = +1/2
7.16
Ti
[Ar]



3d
4s
n = 3,  = 2, m = 2, ms = +1/2
n = 3,  = 2, m = 1, ms = +1/2
n = 4,  = 0, m = 0, ms = +1/2
n = 4,  = 0, m = 0, ms = –1/2
139
Chapter 7
7.17
The Structure of Atoms and Periodic Trends
(a) Mg2+
(b) K+
(c) Cl–
(d) O2–
7.18
(a) Na+
(b) Al3+
(c) Ge2+
(d) F–
7.19
(a) V
(b) V2+
(c) V5+


  
1s
2s
2p


  

  
1s
2s
2p
3s
3p


  

  
1s
2s
2p
3s
3p


  
1s
2s
2p


  
1s
2s
2p


  
1s
2s
2p


  

  
    

1s
2s
2p
3s
3p
3d
4s


  
1s
2s
2p
[Ar]
[Ar]






3d
4s

3d
4s
3d
4s
[Ar]
The V and V2+ ion are paramagnetic with three unpaired electrons
140
Chapter 7
7.20
The Structure of Atoms and Periodic Trends
(a) Ti

[Ar]
(b) Ti2+

[Ar]
(c) Ti4+


3d
4s
3d
4s
3d
4s

[Ar]
The Ti2+ ion is paramagnetic with two unpaired electrons.
7.21
(a) Mn
[Ar]






3d
(b) Mn4+
[Ar] 

4s

3d
4s
(c) Yes, the +4 ion is paramagnetic.
(d) The ion has three unpaired electrons.
7.22
Ni2+
[Ar]
   

3d
Ni3+
[Ar]
  
Paramagnetic
4s


3d
Paramagnetic
4s
7.23
C < B < Al < Na < K
7.24
P < Ge < Ca < Sr < Rb
7.25
(a) Cl– > Cl
(b) Al > O
(c) In > I
7.26
(a) Cs > Rb
(b) O2– > O
(c) As > Br
7.27
(c) Li < Si < C < Ne
7.28
K < Li < C < N
7.29
(a) Na
(b) O
(c) Na < Mg < P < O
7.30
(a) Al
(b) Al
(c) C (based on periodic trends)
Si (according to experimental data)
(d) Al < B < C
141
Chapter 7
7.31
The Structure of Atoms and Periodic Trends
(a) S < O < F
The trend is to increase to the right and decrease down the periodic table.
(b) O
The trend is to decrease down the periodic table.
(c) Cl
The trend is to be more negative to the right and less negative down the periodic table
(d) O2– Ions are larger than neutral atoms. O2– and F– are isoelectronic, but the O2– ion has only 8
protons in its nucleus to attract the 10 electrons whereas the F– ion has 9.
7.32
(a) F < O < S
The trend is to increase to the left and down the periodic table.
(b) Based on a knowledge of first-order periodic trends, we would predict that S should have the largest
IE of the group P, Si, S, and Se. However, recall that the O atom IE is smaller than that of N, so it is
not surprising that the same effect carries over into the third period. That is, the order of ionization
energies is Si < Se < S < P.
(c) F– < O2– < N3–
These ions are isoelectronic and size increases as the number of protons in the
nucleus available to attract the electrons decreases.
(d) Cs < Ba < Sr The trend is to increase to the right and decrease down the periodic table.
7.33
U
U4+
[Rn]
[Rn]






5f
6d
7s
5f
6d
7s

Both uranium and the uranium(IV) ion are paramagnetic.
7.34
(a) Ce
[Xe]
Ce3+ [Xe]
(b) Ho
[Xe]


4f
5d
6s
4f
5d
6s

    


4f
Ho3+ [Xe]
   


(a) atomic number = (2 + 8 + 8 + 2) = 20
(b) total number of s electrons = (2 + 2 + 2 + 2) = 8
(c) total number of p electrons = (6 + 6) = 12
(d) total number of d electrons = 0
(e) The element is Ca, calcium, a metal
7.36
142
Mt [Rn]5f 146d 77s2
Co, Rh, Ir

5d
6s
5d
6s

4f
7.35

Chapter 7
7.37
The Structure of Atoms and Periodic Trends
(a) valid; Li or Be
(b) the maximum value of  is (n – 1)
(c) valid; B, C, N, O, F, or Ne
(d) valid; Y, Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag, or Cd
7.38
n = 4,  = 1, m = –1, ms = +1⁄2
n = 4,  = 1, m = –1, ms = –1⁄2
n = 4,  = 1, m = 0, ms = +1⁄2
n = 4,  = 1, m = 0, ms = –1⁄2
n = 4,  = 1, m = +1, ms = +1⁄2
n = 4,  = 1, m = +1, ms = –1⁄2
7.39
(a) Nd [Xe]4f 46s2

[Xe]




4f
5d
6s
Fe [Ar]3d 64s2
[Ar]
 




3d
B
4s
[He]2s22p1
[He]


2s
2p
(b) All three elements are paramagnetic.
(c) Nd3+ [Xe]



4f
5d
6s
paramagnetic
Fe3+
[Ar]



3d


4s
paramagnetic
7.40
(a) P, phosphorus
(b) Be, beryllium
(c) N, nitrogen
(d) Tc, technetium
(e) Cl, chlorine
(f) Zn, zinc
7.41
K < Ca < Si < P
143
Chapter 7
7.42
The Structure of Atoms and Periodic Trends
Cl– < Cl < Ca2+
Ca2+ and Cl– are isoelectronic. Ca2+ has a larger IE than Cl– because the calcium ion has a 2+ charge and
removing another electron from that ion would take more energy. Removing an electron from the chloride
ion would require less energy than the removal of an electron from the Cl atom due to electron repulsion
forces.
7.43
(a) metal
(b) B
(c) A
(d) A
(e) A2B or Rb2Se
7.44
(a) alkaline earth metal
(b) nonmetal (halogen)
(c) B
(d) B
7.45
In4+
Indium has three outer shell electrons, so it is unlikely to form a 4+ ion
Fe6+
Ions with charges greater than +3 or +4 are unlikely to form
Sn5+
Tin has four outer shell electrons, so it is unlikely to form a 5+ ion
7.46
S2– > Cl– > K+ > Ca2+
7.47
(a) Se
(b) Br–
(c) Na
(d) N
(e) N3–
7.48
(a) Ca2+ < K+ < Cl–
(b) Cl– < K+ < Ca2+
(c) Cl– < K+ < Ca2+
7.49
(a) Na
(b) C
(c) Na < Al < B < C
7.50
Tc and Rh
7.51
(a) Co, cobalt
(b) paramagnetic (three unpaired electrons)
(c) The 3+ ion would be formed by the loss of both 4s electrons and one d electron leaving four unpaired
electrons
144
Chapter 7
7.52
The Structure of Atoms and Periodic Trends
(a) V, vanadium
(b) Group 5B, Period 4
(c) Transition element
(d) Paramagnetic, three unpaired electrons
(e) n = 3,  = 2, m = –2, ms = +1⁄2
n = 3,  = 2, m = –1, ms = +1⁄2
n = 3,  = 2, m = 0, ms = +1⁄2
n = 4,  = 0, m = 0, ms = +1⁄2
n = 4,  = 0, m = 0, ms = –1⁄2
(f) V2+ [Ar]3d 3 The two 4s electrons are removed and the resulting ion is paramagnetic
7.53
(a) 0.500 g Ni(CH3CO2)2 ·
(b) Ni2+
1mol Ni(CH3 CO2 )2
1mol Ni(HCO2 )2
148.7 g
·
·
= 0.421 g
176.8 g
1mol Ni(CH3CO2 )2 1mol Ni(HCO2 )2
[Ar]    

3d
(c) 253 mg Ni(HCO2)2 ·
Ni
4s
58.69 g
1g
1mol Ni(HCO2 )2
1mol Ni
·
·
·
= 0.0999 g Ni
3
1 mol Ni(HCO2 )2 1 mol Ni
148.7 g
10 mg
[Ar]    

3d
7.54
(a) Fe2+
Paramagnetic, 2 unpaired electrons

Paramagnetic, 2 unpaired electrons
4s
 
[Ar]



3d
Fe3+
4 unpaired electrons



4s
[Ar]


3d
4s
5 unpaired electrons
(b) Co2+
[Ar]   


3d
Al3+
Sn2+
Co3+
4s


  
1s
2s
2p
[Kr]
Paramagnetic, 3 unpaired electrons
Diamagnetic
    

4d
5s
[Ar]
Diamagnetic
5p
Paramagnetic, 4 unpaired
electrons
145
Chapter 7
7.55
The Structure of Atoms and Periodic Trends
For Li+, there is one less electron to be attracted by the same nuclear charge, but the outer electrons are in
the n = 1 shell, closer to the nucleus. For F–, there is one additional electron in the 2p orbital. This adds
additional electron–electron repulsions while the nuclear charge remains the same.
7.56
K2+
Potassium has one outer shell electron, so it is unlikely to form a 2+ ion
Al4+
Aluminum has three outer shell electrons, so it is unlikely to form a 4+ ion
F2–
Fluorine has seven outer shell electrons, so adding one electron forms an anion with the same
number of electrons as the nearest noble gas. Adding a second electron would increase
electron–electron repulsions and require a great amount of energy.
7.57
Element 1 comes from Group 4A. The first two IEs correspond to removing electrons from a p subshell.
With the third IE, there is a fairly large jump in IE corresponding to removing an electron from an s
subshell. The fourth electron removed comes from the same s subshell and therefore does not increase the
IE by as much. None of the IEs are large enough to correspond to removing an electron from a lower
energy level. Element 2 comes from Group 3A. There is a large change in IE between the third and fourth
IEs. The first three IEs correspond to removing electrons from the same energy level. The large jump at the
fourth IE corresponds to having to remove the electron from a lower energy level.
7.58
 22 
Assumption 1: IE = –1312  2  = –5248 kJ/mol
1 
 12 
Assumption 2: IE = –1312  2  = –1312 kJ/mol
1 
The actual IE value is between the two extreme limits. This suggests that in He one electron probably
screens the other from the full nuclear charge to some extent, but not completely.
7.59
Most stable: (d) The two electrons are in separate orbitals, following Hund’s rule, and are of the same spin.
Least stable: (a) In this case the electrons violate both Hund’s rule and the Pauli exclusion principle.
7.60
7.61
Sn—Cl: Calculated bond distance = 141 pm + ½(200 pm) = 241 pm
Actual bond distance = 233 pm
Sn—Br: Calculated bond distance = 141 pm + ½(228 pm) = 255 pm
Actual bond distance = 250 pm
Sn—I: Calculated bond distance = 141 pm + ½(266 pm) = 274 pm
Actual bond distance = 270 pm
K(g)

1s22s22p63s23p64s1
K+(g)
1s22s22p63s23p6
K+(g)
1s22s22p63s23p6

K2+(g)
1s22s22p63s23p5
The first ionization is for the removal of an electron from the outermost valence shell of electrons. The
146
Chapter 7
The Structure of Atoms and Periodic Trends
second electron, however, is removed from the 3p subshell. This subshell is significantly lower in energy
than the 4s subshell and considerably more energy is required to remove this second electron.
7.62
For s- and p-block elements, first ionization energy generally decreases down a group because the electron
removed is increasingly farther from the nucleus, thus reducing the nucleus-electron attractive force.
7.63
(a) In going from one element to the next across a period, the effective nuclear charge increases slightly
and the attraction between the nucleus and electrons increases.
(b) The slight decrease in atomic radius of the transition metals is a result of increased repulsions of
(n–1)d electrons for ns electrons. This repulsion reduces the effects of the increasing nuclear charge
across a period.
7.64
Li, lithium Of the four elements shown, only Li is in Group 1A. The loss of the first electron results in an
ion with a filled outer shell. Removal of a second electron (from a filled electron shell) would require a
much larger amount of energy.
7.65
There are many arguments for an Mg/O compound being composed of Mg2+ and O2– ions. A few are:
(1) The groups characteristically form ions with noble gas configurations; Group 2A loses two electrons
and Group 6A gains two electrons.
(2) Look at other ionic compounds from the same groups; MgS and CaO.
(3) Calculate and compare fHº values for the two compounds Mg+/O– and Mg2+/O2–. The Mg2+/O2–
combination is more favorable thermodynamically.
Some possible experiments:
(1) Determine the first and second ionization energy values for Mg and the first and second electron
affinity values for O.
(2) Determine the melting point of MgO and compare it to the +1/–1 compound NaF (990 ºC) and
the +2/–2 compound CaO (2580 ºC).
7.66
Ca is smaller than K, so we would expect the first IE of Ca to be greater than that of K. Once K has lost
one electron, it has a noble gas (Ar) configuration. Removal of a second electron (from a filled electron
shell) requires much additional energy. Ca, on the other hand, can lose a “second” electron to obtain a
noble gas configuration with a much smaller amount of energy (smaller IE).
7.67
(a) The effective nuclear charge increases, causing the valence orbital energies to become more negative
on moving across the period.
(b) As the valence orbital energies become more negative, it is increasingly difficult to remove an electron
from the atom, and the IE increases. Toward the end of the period, the orbital energies have become
so negative that removing an electron requires significant energy. Instead, the effective nuclear charge
has reached the point that it is energetically more favorable for the atom to gain an electron.
147
Chapter 7
The Structure of Atoms and Periodic Trends
(c) The valence orbital energies are in the order:
Li (–520.0 kJ) < Be (–899.3 kJ) > B (–800.8 kJ) < C (–1029 kJ)
This means it is more difficult to remove an electron from Be than from either Li or B. The energy is
more negative for C than for B, so it is more difficult to remove an electron from C than from B.
7.68
Generally the increasing effective nuclear charge across a period causes ionization energy to also increase.
For sulfur, however, two of its four 2p electrons are paired in the same orbital. The greater repulsion
experienced by these electrons makes it easier to remove one of them, and the ionization energy of sulfur is
lower than expected.
7.69
Atomic radius decreases from potassium to vanadium. Because the mass of these elements increases from
K to V as the radius decreases, the density is expected to increase.
7.70
The fifth and sixth period transition metals have similar atomic radii. The sixth period transition metals
have a higher mass resulting in higher density for the sixth period transition metals.
7.71
(a) Element 113 [Rn]5f 146d 107s27p1
Element 115 [Rn]5f 146d 107s27p3
(b) Element 113: Group 3A, B, Al, Ga, In, Tl
Element 115: Group 5A, N, P, As, Sb, Bi
(c) Argon, Ar
7.72
To form CaF3, calcium would have to form a 3+ cation. Since calcium (in Group 2A) normally forms 2+
cations (with noble gas configuration), the formation of the 3+ cation is highly unlikely.
7.73
(a) S


  

1s
2s
2p
3s
 

3p
(b) n = 3,  = 1, m = 1, ms = –1/2
(c) smallest ionization energy: S
smallest radius: O
(d) S < S2–
(e) 675 g SOCl2 ·
(f) 10.0 g SO3 ·
10.0 g SCl2 ·
1 mol SOCl2
1 mol SCl2
103.0 g
·
·
= 584 g SCl2
119.0 g
1 mol SOCl2 1 mol SCl2
1 mol SO3 1 mol SOCl2
119.0 g
·
·
= 14.9 g SOCl2
80.06 g
1 mol SO3
1 mol SOCl2
1 mol SCl2 1 mol SOCl2
119.0 g
·
·
= 11.6 g SOCl2
103.0 g
1 mol SCl2
1 mol SOCl2
The theoretical yield of SOCl2 is 11.6 g
(g) rHº = fHº[SO2(g)] + fHº[SOCl2(g)] – (fHº[SO3(g)] + fHº[SCl2(g)])
–96.0 kJ = 1 mol (–296.84 kJ/mol) + 1 mol (–212.5 kJ/mol)
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Chapter 7
The Structure of Atoms and Periodic Trends
– [1 mol (–395.77 kJ/mol) + fHº[SCl2(g)])
fHº[SCl2(g)] = –17.6 kJ/mol
7.74
(a) The reducing agent is Na. The low ionization energy of sodium plays a major role in making it a
good reducing agent.
(b) The oxidizing agent is Cl2. Among other properties, the element has a high electron affinity.
(c) Na2Cl would have a Cl2– ion. Adding a second electron to Cl– means placing an electron in a higher
energy electron shell. Conversely, NaCl2 would have a Na2+ ion. Here one would have to remove the
second electron from the atom’s core.
7.75
(a) F: (1s2)(2s22p5)
Ne:
(1s2)(2s22p6)
Z* = 9 – [(2  .85) + (6  0.35)] = 5.2
Z* = 10 – [(2  .85) + (7  0.35)] = 5.85
Z* increases from O to F to Ne. As the effective nuclear charge increases, the atomic radius decreases
and the first ionization energy increases.
(b) Mn: (1s2)(2s22p6)(3s23p6)(3d5)(4s2)
Z*(3d) = 25 – [(18  1.00) + (4  0.35)] = 5.6
Z*(4s) = 25 – [(10  1.00) + (13  0.85) + (1  0.35)] = 3.6
The Z* experienced by a 4s electron is much smaller than that experienced by a 3d electron. A 4s
electron in Mn is thus more easily removed than a 3d electron.
SOLUTIONS TO APPLYING CHEMICAL PRINCIPLES: THE NOT SO RARE EARTHS
1.
(a) Sm3+ [Xe]4f5
(b) 4 Sm(s) + 3 O2(g)  2 Sm2O3(s)
2.
Y has a [Kr]4d15s2 configuration, while La has a [Xe]5d16s2 configuration and Lu has a [Xe]4f145d16s2
configuration. Based on electronic structure, both La and Lu are appropriately located under yttrium. All
149
Chapter 7
The Structure of Atoms and Periodic Trends
three atoms have electronic structures that finish with a filled outermost s-orbital and one electron in the
outermost d-orbital.
3.
(a) [Xe]
↑
↑
↑
↑
↑
↑
↑
↑
4f
↑↓
5d
6s
(b) The most common oxidation state is +3. The electronic configuration is [Xe]4f7.
4.
(a) The electron configuration of the outermost shell of La and Lu is 6s2. Lutetium’s 14 additional electrons
are located in the 4f orbitals. These electrons are poorly shielding so that the outermost electrons feel a large
portion of the charge from the additional 14 protons in the nucleus.
(b) The elements in the 5d block have similar radii to the atoms directly above them in the 4d block.
Lanthanum and yttrium have similar radii, so it might be considered a more appropriate fit below yttrium on
the periodic table.
5.
 =
E=
6.
c

=
hc

2.998  108 m/s
= 4.90  1014 s -1
-9
612  10 m
 6.626  10
=
-34
Js  2.998  108 m/s 
612  10 m
-9
= 3.25  10-19 J
Molar mass of Nd2Fe14B = 1081.12 g/mol
% Nd = (2 mol)(144.24 g/mol Nd)/(1081.12 g) x 100 % = 26.683 %
150