CHAPTER 7 | Electrons in Atoms and Periodic Properties
7.85. Collect and Organize / Analyze
We are asked what is meant by a degenerate orbital.
Solve
Degenerate orbitals have the same energy and are indistinguishable from each other.
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Electrons in Atoms and Periodic Properties | 345
Think about It
In the hydrogen atom, all the orbitals in a given n level are degenerate. This means that in hydrogen
the 3s, 3p, and 3d orbitals, for example, all have the same energy. In multielectron atoms, however,
these orbitals split in energy and are no longer degenerate.
7.86. Collect and Organize
The electron configuration of group 2 elements is ns2. Mendeleev’s periodic table shows these
elements in the second column. We are to link the position in the periodic table to the electron
configuration.
Analyze
Mendeleev based his periodic table on similar chemical and physical properties, not electron
configurations. Each element in the second column of his periodic table combined with oxygen to
give RO, the element in 1:1 molar ratio with oxygen.
Solve
The electron configuration of [core]ns2 that all the group 2 elements share indicates the same
reactivity with oxygen to form RO. The two ns2 electrons can be lost relatively easily to form R 2+,
which then combines with O2–.
Think about It
Notice that Zn and Cd were also included in Mendeleev’s periodic table. These elements also have an
ns2 configuration at the highest n level: Zn is [Ar]3d 104s2, Cd is [Kr]4d 105s2. These elements are
chemically similar to Ca and Sr.
7.87. Collect and Organize
In the filling of atomic orbitals, the 4s level fills before the 3d. We are asked how this is evident in the
periodic table.
Analyze
The two leftmost columns in the periodic table correspond to the s block while columns 3–12 starting
in period 4 correspond to the d block.
Solve
As we start from an argon core of electrons, we move to potassium and calcium, which are located in
the s block on the periodic table. It is not until Sc, Ti, V, etc., that we begin to fill electrons into the 3d
shell.
Think about It
Also, notice that the 6s orbitals fill (Cs and Ba) followed by the 4f orbitals (Ce–Yb) then the 5d
orbitals (La–Hg).
7.88. Collect and Organize
We are to explain why the transition metals often form ions with a 2+ charge. The metals have an
electron configuration of [noble gas core](n – 1)d xns2.
Solve
The electrons in the ns subshell of transition metals are lost to form the 2+ ion, leaving the (n – 1)d x
electrons in the outermost shell.
Think about It
The loss of the s-orbital electrons “first” in forming transition metal compounds seems to contradict
the fact that the s orbitals (being lower in energy) are filled before electrons are placed into the d
orbitals. We know, however, by experiment, that transition metal cations (particularly of 2+ charge or
greater) have indeed lost both s electrons. This is because of a change in orbital energies (4s versus
346 | Chapter 7
3d) upon formation of a transition metal cation. Once electrons are lost, the d orbitals become lower
in energy than the s orbitals, so the s-orbital electrons are those that are “lost” to form the cation.
7.89. Collect and Organize
For multielectron atoms, we are to list a set of orbitals defined by their n and
order of increasing energy.
quantum numbers in
Analyze
The higher the energy of an orbital, the farther the electron is from the nucleus. This means that for
differing n values the order of energies is 1 < 2 < 3, etc. For orbitals in the same n shell, the orbitals
increase in energy, that is, s < p < d < f for multielectron atoms.
Solve
The orbitals described are: (a) 3d, for n = 3, = 2; (b) 5g, for n = 5,
and (d) 4p for n = 4, = 1, m = 1.
In increasing order of energy: (c) 3s < (a) 3d < (d) 4p < (b) 5g.
= 4; (c) 3s, for n = 3,
Think about It
To determine the energy of an orbital, first look to the n quantum number, then to the
7.90. Collect and Organize
For multielectron atoms, we are to list a set of orbitals defined by their n and
order of increasing energy.
= 0;
.
quantum numbers in
Analyze
The higher the energy of an orbital, the farther the electron is from the nucleus. This means that for
differing n values the order of energies is 1 < 2 < 3, etc. For orbitals in the same n shell, the orbitals
increase in energy, that is, s < p < d < f for multielectron atoms.
Solve
The orbitals described are: (a) 2p, for n = 2, = 1; (b) 5f, for n = 5,
and (d) 4f, for n = 4, = 3.
In increasing order of energy: (a) 2p < (c) 3d < (d) 4f < (b) 5f.
= 3; (c) 3d, for n = 3,
= 2;
Think about It
To determine the energy of an orbital, first look to the n quantum number, then to m .
7.91. Collect and Organize
We can use the periodic table and Figure 7.29 to write the electron configurations for several
elemental species, including anions and cations.
Analyze
When a cation is formed, electrons are removed from the highest energy orbital. None of the species
are transition metals, so we remove the electrons from the orbitals last filled in building the electron
configuration of the element. To form an anion, we need to add electrons to the highest energy orbital
or the next orbital up in energy. We use the previous noble gas configuration as the “core” to write the
condensed form of the configurations.
Solve
Li: [He]2s1
Li+: [He] or 1s2
Ca: [Ar]4s2
F–: [He]2s22p6 or [Ne]
Na+: [Ne] or [He]2s22p6
Mg2+: [Ne] or [He]2s22p6
Al3+: [Ne] or [He]2s22p6
Electrons in Atoms and Periodic Properties | 347
Think about It
Because F–, Na+, Mg2+, and Al3+ all have the same electron configurations and, thus, the same number
of electrons, they are isoelectronic with each other.
7.92. Collect and Organize
From the species in Problem 7.91 (Li, Li +, Ca, F–, Na+, Mg2+, Al3+) we are to choose which species
have the same number of electrons as (are isoelectronic with) Ne.
Analyze
Neon’s electronic configuration is [He]2s22p6 with a total of 10 e–.
Solve
The species that also have 10 e– are F–, Na+, Mg2+, and Al3+. These all have the electron configuration
[Ne] or [He]2s22p6 and are, therefore, isoelectronic with neon.
Think about It
The atomic species found as ions in compounds tend to show charges that reflect the loss or gain of
electrons so as to attain a noble gas configuration.
7.93. Collect and Organize
We are to write the condensed electron configurations (using the noble gas core configuration in
brackets) for several species including cationic and anionic species. Figure 7.29 is helpful here.
Analyze
When a cation is formed, electrons are removed from the highest energy orbital. To form an anion, we
need to add electrons to the highest energy orbital or the next orbital up in energy.
Solve
K: [Ar]4s1
K+: [Ar]
S2–: [Ne]3s23p6 or [Ar]
N: [He]2s22p3
Ba: [Xe]6s2
Ti4+: [Ar] or [Ne]3s23p6
Al: [Ne]3s23p1
Think about It
Notice that K+, S2–, and Ti4+ are isoelectronic with each other and with Ar.
7.94. Collect and Organize
After writing the condensed electron configurations for H, Li, Na, K, Rb, and Cs, we can describe
how they are similar.
Analyze
The electron configurations are
H: 1s1
Li: [He]2s1
Na: [Ne]3s1
K: [Ar]4s1
Rb: [Kr]5s1
Cs: [Xe]6s1
Solve
All of these elements (the alkali metals) have an ns1 configuration after a noble gas core.
Think about It
The ns1 electron in these metals is relatively easy to lose, so the elements are typically found as 1+
cations in compounds.
7.95. Collect and Organize
We are to write the condensed electron configurations (using the noble gas core configuration in
brackets) for several species including a cationic species. Figure 7.29 is helpful here.
348 | Chapter 7
Analyze
When a cation is formed, electrons are removed from the highest energy orbital. When electrons are
removed from a transition metal, ns electrons are removed first to form 2+ ions and then additional
electrons are removed from the (n – 1) orbital to form higher charged ions.
Solve
Na: [Ne]3s1
Cl: [Ne]3s23p5
Mn: [Ar]3d 54s2
Mn2+: [Ar]3d5
Think about It
To obtain a noble gas configuration, Na would lose one electron to become Na+ whereas Cl would
gain an e– to become Cl–.
7.96. Collect and Organize
We are to write the condensed electron configurations (using the noble gas core configuration in
brackets) for several species including a cationic species. Figure 7.26 is helpful here.
Analyze
When a cation is formed, electrons are removed from the highest energy orbital. When electrons are
removed from a transition metal, ns electrons are removed first to form 2+ ions and then additional
electrons are removed from the (n – 1) orbital to form higher charged ions.
Solve
C: [Ne]2s22p2
S: [Ne]3s23p4
Ti: [Ar]3d 24s2
Ti4+: [Ar]
Think about It
Carbon could obtain a noble gas configuration by either gaining four electrons to become C 4– or
losing four electrons to become C4+.
7.97. Collect and Organize
To determine the number of unpaired electrons in the ground-state atoms and ions, we have to first
write the electron configuration for each species and then detail how the electrons are distributed
among the highest energy orbitals.
Analyze
If the highest energy orbital (s, p, d, or f ) is either empty or completely filled, the species have no
unpaired electrons. If the highest energy orbital is partially full, Hund’s rule states that electrons
singly occupy the degenerate orbitals at that level before pairing up in those orbitals.
Solve
(a) N: [He]2s22p3
(b) O: [He]2s22p4
(c) P3–: [Ne]3s23p6
(d) Na+: [Ne] or [He]2s22p6
3 unpaired e–
2 unpaired e–
0 unpaired e–
0 unpaired e–
Think about It
Notice that the ground-state configuration of these elements fills the s orbital completely first then
places electrons into the p orbitals. This is because for a multielectron atom, s < p in terms of energy
for a given principal quantum level.
7.98. Collect and Organize
To determine the number of unpaired electrons in the ground-state atoms and ions, we have to first
write the electron configuration for each species and then detail how the electrons are distributed
among the highest energy orbitals.
Electrons in Atoms and Periodic Properties | 349
Analyze
If the highest energy orbital (s, p, d, or f) is either empty or completely filled, the species have no
unpaired electrons. If the highest energy orbital is partially full, Hund’s rule states that electrons
singly occupy the degenerate orbitals at that level before pairing up in those orbitals.
Solve
(a) Sc: [Ar]3d 14s2
(b) Ag+: [Kr]4d 10
(d) Cd2+: [Kr]4d 10
(e) Zr4+: [Kr] or [Ar]3d 104s24p6
1 unpaired e–
0 unpaired e–
0 unpaired e–
0 unpaired e–
Think about It
Remember, when forming a cation of a transition metal ion the electrons are removed from the ns
orbital before the (n – 1) d orbitals. For Ag+ you might think that this would result in an electron
configuration of [Kr]4d 95s1, but like Cu+, the s electron is lower in energy when placed in the 4d
orbital to complete the 4d shell.
7.99. Collect and Organize
An atom with the electron configuration [Ar]3d 24s2 is in the fourth period in the periodic table and is
among the transition metals.
Analyze
This atom has no charge so we do not have to account for additional or lost electrons.
Solve
The 4s orbital is filled for the element Ca. Two additional electrons are present in the 3d orbitals for
the second transition metal of the fourth period: titanium, Ti. The electron filling orbital box diagram
shows 2 unpaired electrons.
4s
3d
4p
Think about It
Although we write the electron configuration so that 3d comes before 4s, remember that the 4s
orbital fills before the 3d in building up electron configurations.
7.100. Collect and Organize
An atom with the electron configuration [Ne]3s23p3 is in the third period in the periodic table and is
among the p-block elements.
Analyze
This atom has no charge, so we do not have to account for additional or lost electrons.
Solve
The 3s orbitals are filled with the element Mg. Three electrons will be in the 3p orbitals for
phosphorus, P. The electron filling orbital box diagram shows three unpaired electrons.
Think about It
The elements that are in the same group as phosphorus (N, As, Sb, Bi) also have an ns2np3
configuration.
7.101. Collect and Organize
We are to name the monatomic anion which has a filled-shell configuration of [Ne]3s23p6 or [Ar] and
determine the number of unpaired electrons in the ion in its ground state.
350 | Chapter 7
Analyze
Because the atom has an extra electron, to form the monatomic anion, the neutral atom would have
an electron configuration of one less electron.
Solve
Ion’s electron configuration: [Ne]3s23p6 = X–
Atom’s electron configuration: [Ne]3s23p5 = X
This atom is chlorine and the monatomic anion is chloride, Cl –. Because electrons fill the s and p
orbitals, Cl– has no unpaired electrons in its ground state.
Think about It
When identifying elements with the electron configurations of anions, remove the electrons
associated with the anionic charge to obtain the electron configuration of the neutral atom.
7.102. Collect and Organize
We are to name the monatomic cation, with a 1+ charge, that has an electron configuration of
[Kr]4d105s2, and determine the number of unpaired electrons in the ground state of the ion.
Analyze
Because the atom has lost an electron to form the X + cation, the neutral atom would have an electron
configuration of one more electron.
Solve
Ion’s electron configuration: [Kr]4d 105s2 = X+
Atom’s electron configuration: [Kr]4d 105s25p1 = X
This atom is indium and the monatomic cation is In+. The s and d orbitals are completely filled so
there are no unpaired electrons for In+ in its ground state.
Think about It
When identifying elements with the electron configuration of cations, add the electrons associated
with the amount of charge to obtain the electron configuration of the neutral atom.
7.103. Collect and Organize
For Al, N, Mg, and Cs we can use the electron configurations and the positions of the elements in the
periodic table to predict the charge on these elements as monatomic ions.
Analyze
The elements lose electrons to reach a noble gas configuration if they are located among the metals.
If the elements are nonmetals, they gain electrons so as to reach a noble gas configuration.
Solve
Al loses three electrons to become Al3+, with the electron configuration of Ne.
N could either gain three electrons to become N3–, having the electron configuration of Ne, (more
likely) or lose five electrons to become N5+, with the electron configuration of He (less likely).
Mg loses two electrons to become Mg2+, with the electron configuration of Ne.
Cs loses one electron to become Cs+, having the electron configuration of Xe.
Think about It
Recall earlier in the textbook that nitrogen can have varying oxidation states from negative to
positive as in NO2, NO, and NH3. This is reflected in nitrogen’s middle position between two noble
gases.
7.104. Collect and Organize
For S, P, Zn, and I, we can use the electron configurations and the positions of the elements in the
periodic table to predict the charge in these elements as monatomic ions.
Electrons in Atoms and Periodic Properties | 351
Analyze
The elements lose electrons to reach a noble gas configuration if they are located among the metals.
If the elements are nonmetals, they gain electrons so as to reach a noble gas configuration.
Solve
S gains two electrons to become S2–, with the electron configuration of Ar.
P gains three electrons to become P 3–, with the electron configuration of Ar. It may also lose five
electrons to become P5+, having the electron configuration of Ne.
Zn loses two electrons from its 4s shell to become Zn2+. This leaves an electron configuration of
[Ar]3d10.
I gains one electron to become I–, having the electron configuration of Xe.
Think about It
The filled 3d orbital shell in Zn2+ can be considered to be “stable” and therefore, the 3d electrons are
not lost (usually) to form higher charged cations of zinc.
7.105. Collect and Organize
An electronic excited state is when an electron has been placed into a higher energy orbital than
would be predicted using the filling rules shown by the periodic table.
Analyze
The order of filling for the orbitals is as follows:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p
Solve
(a) Because the 2s orbital is lower in energy than the 2p orbital, the lowest energy configuration for
this atom is [He]2s22p4, so the configuration [He]2s12p5 represents an excited state.
(b) The order of filling of orbitals for atoms after krypton is 5s < 4d < 5p. This atom has a total of 13
electrons in its outer shell: 2 fill the 5s orbital, 10 fill the 4d orbitals and one is placed in a 5p orbital.
This configuration, [Kr]4d 105s25p1, does not represent an excited state.
(c) The order of filling of orbitals for atoms after argon is 4s < 3d < 4p. This atom has a total of 17
electrons in its outer shell, 2 fill the 4s orbital, 10 fill the 3d orbitals, and 5 are placed in the 4p
orbitals. This configuration, [Ar]3d 104s24p5, does not represent an excited state.
(d) Because the 3p orbital is lower in energy than the 4s orbital, the lowest energy configuration for
this atom is [Ne]3s23p3, so the configuration [Ne]3s23p24s1 represents an excited state.
Think about It
If each of these configurations are for neutral atoms, we can assign the elements: (a) excited-state O,
(b) ground-state In, (c) ground-state Br, and (d) excited-state P.
7.106. Collect and Organize
An electronic excited state is when an electron has been placed into a higher energy orbital than
would be predicted using the filling rules shown by the periodic table.
Analyze
The order of filling for the orbitals is as follows:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p
Solve
(a) The order of filling of orbitals for atoms after neon is 3s < 3p. This atom has a total of 3 electrons
in its outer shell: 2 fill the 3s orbital, and one is placed in a 3p orbital. This configuration, [Ne]3s23p1,
does not represent an excited state.
(b) Because the 4s orbital is lower in energy than the 3d orbital and the 3d orbital is lower in energy
than the 4p orbital, the lowest energy configuration for this atom is [Ar]3d 104s24p1, so the
configuration [Ar]3d 104s14p2 represents an excited state.
352 | Chapter 7
(c) Because the 5s orbital is lower in energy than the 4d orbital and the 4d orbital is lower in energy
than the 5p orbital, the lowest energy configuration for this atom is [Kr]4d 105s2, so the configuration
[Kr]4d 105s15p1 represents an excited state.
(d) The order of filling of orbitals for atoms after neon is 3s < 3p < 4s. This atom has a total of 9
electrons in its outer shell: 2 fill the 3s orbital, 6 fill the 3p orbitals and one is placed in the 4s orbital.
This configuration, [Ne]3s23p64s1, does not represent an excited state.
Think about It
If each of these configurations are for neutral atoms, we can assign the elements: (a) ground-state Al,
(b) excited-state Ga, (c) excited-state Cd, and (d) ground-state K.
7.107. Collect and Organize
Iodine-131 has 53 protons, 78 neutrons, and 53 electrons as a neutral atom. We are to identify the
subshell containing the highest-energy electrons and compare the electron configuration of 131I to 127I.
Analyze
The electron configuration for iodine is [Kr]4d 105s25p5. The difference between
131
I has 4 additional neutrons in its nucleus.
131
I and
127
I is that
Solve
The electron configuration of iodine shows that the highest-energy electrons are in the 5p subshell.
Because the difference in isotopes is the number of neutrons present, the electron configurations of
131
I and 127I (which are based on total number of electrons in the atom) are the same.
Think about It
Electron configurations, however, do change if the atom gains or loses electrons to become either
anionic or cationic, respectively.
7.108. Collect and Organize
The g orbitals lie higher in energy because they will only be present for very large atoms. We are
asked what is the minimum atomic number of an element having g orbitals in its ground state atom.
Analyze
For the g orbitals = 4 so the first n level for which the g orbitals will be available is n = 5 (because
= n – 1, n – 2, . . . , 0).
Solve
For the hypothetical element 118, the electron configuration is [Rn]7s25f 146d 107p6. As we fill
electrons for elements 119 and 120, we would expect those electrons to fill the 8s orbital. Next in
energy could be the 5g orbitals (for element 121) or the 6f orbitals. If 6f orbitals are lower in energy
than 5g, then the first 5g orbital would fill for element 135. Therefore, the lowest possible atomic
number for the 5g orbitals would be 121.
Think about It
Because for g orbitals, ml = –4, –3, –2, –1, 0, 1, 2, 3, 4 there are nine g orbitals, which could hold 18
electrons total.
7.109. Collect and Organize
Sodium and chlorine atoms are neutral in charge, but the sodium atom in NaCl has a charge of 1+
and the chlorine atom has a charge of 1–. These changes in charge also come with a change in size.
Why?
Analyze
When we remove an electron from an atom, we reduce the repulsion for the remaining electrons in
the atom. When we add electrons, we increase e–– e– repulsion.
Electrons in Atoms and Periodic Properties | 353
Solve
If electrons do not repel each other as much in Na + as they do in Na, they will have lower energy and
be, on average, closer to the nucleus, resulting in a smaller size. When electrons are added to an atom
(Cl), the e–– e– repulsion increases, so the electrons have higher energy and they will be, on average,
farther from the nucleus, thereby creating a larger size species (Cl –).
Think about It
The change in size upon forming a cation or anion can be dramatic, as seen in Figure 7.30.
7.110. Collect and Organize
As we proceed across a row in the periodic table we are adding both electrons to the orbitals and
protons to the nucleus. The sizes of the atoms across this series decrease. Why?
Analyze
As we add electrons to an atom we would expect the atoms to become larger. Adding protons,
however, would draw the electrons closer to the nucleus.
Solve
When we add electrons for elements in the same period, we add them to the same shell; for example,
for period 3, electrons fill the 3s and then the 3p orbitals. The electrons in the same shell do not
shield each other well, so with added protons in the nucleus, the electrons feel greater effective
nuclear charge (Zeff) and are therefore pulled in closer to the nucleus, resulting in smaller atoms.
Think about It
Even though Zeff also increases as you go down a family (which would mean smaller atoms), the
electrons are placed at higher n values (which leads to larger atoms). The effect of increasing n down
a group dominates so the atoms are larger as you go down a family.
7.111. Collect and Organize
Of the group 1 elements (Li, Na, K, Rb) we are to predict the largest and explain our selection.
Analyze
The sizes of atoms increase down a group because electrons have been added to higher n levels.
Solve
Rb is the largest atom.
Think about It
The largest atoms are those situated to the lower left in the periodic table.
7.112. Collect and Organize
From among F–, Cl–, Br–, and I–, we are to predict which is the largest and explain why.
Analyze
Compared to the neutral atoms, all the monatomic anions will be larger. As we go down a group, the
atoms become larger due to placement of electrons at higher n levels.
Solve
I– is the largest monatomic anion.
Think about It
This trend will also be true for cationic monatomic species. For example, the 2+ cations of group 2
increase in size in the order Be2+ < Mg2+ < Ca2+ < Sr2+ < Ba2+.
7.113. Collect and Organize
Ionization energy is the energy required to remove an electron from a gaseous atom.
X(g)  X–(g) + e–
We are to state the trends in ionization energies down and across the periodic table.
354 | Chapter 7
Analyze
The ionization energy will change with effective nuclear charge (the higher the Zeff, the greater the
ionization energy) and with size (an electron farther away from the nucleus requires less energy to
remove).
Solve
(a) As the atomic number increases down a group, electrons are added to higher n levels, leading to a
decrease in ionization energy.
(b) As the atomic number increases across a period, the effective nuclear charge increases. This
means that the ionization energy increases across a period of elements.
Think about It
Ionization energy trends follow atomic size trends; smaller atoms require more energy to ionize than
larger atoms.
7.114. Collect and Organize
Between pairs of elements that are adjacent to each other in the periodic table, we are asked to
explain the differences in ionization energies (IE) using Figure 7.34.
Analyze
In Figure 7.34, we see that ionization energy comparisons between the pairs are
He > Li, Li < Be, Be > B, N > O
For each of these we can compare electron configurations:
He = 1s2, Li = [He]2s1, Be = [He]2s2, B = [He]2s22p1, N = [He]2s22p3, O [He]2s22p4
and consider the atom’s effective nuclear charge (Zeff).
Solve
(a) He has a higher IE than Li because the effective nuclear charge on the electrons in the closed shell
(1s) is very high. The Li 2s electron, however, is shielded by the two 1s electrons, and therefore feels
less effective nuclear charge and is more easily ionized.
(b) Lithium has a lower IE than Be because its 2s electron feels a lower Zeff than the 2s electrons on
Be do. This is because of the increased nuclear charge on Be (4 protons) versus Li (3 protons).
(c) Beryllium has a higher ionization energy than B even though there are more protons in boron’s
nucleus, which would cause us to reason that the electron on B would feel a higher Zeff. However, the
electron that is ionized in B is in the p orbital, which has a higher orbital energy than the s orbital,
making it easier to remove.
(d) Nitrogen has a higher ionization energy than O because oxygen’s ionized electron comes from a p
orbital in which 2 electrons are paired. Placing two electrons in the same orbital destabilizes them,
making O easier to ionize.
Think about It
Generally, across a period, ionization energy increases due to increased Zeff. However, where
electrons are first placed in higher energy orbitals or when electrons are first paired in an orbital set,
we get a slight decrease in IE.
7.115. Collect and Organize
Fluorine and boron are located in the same period of the periodic table (period 2). Fluorine has 9
protons in its nucleus while boron has 5. We are to explain why F is more difficult to ionize than B.
Analyze
Both fluorine and boron have 2s22pn configurations and the ionized electron is removed from the 2p
orbital.
Electrons in Atoms and Periodic Properties | 355
Solve
Fluorine, with a higher nuclear charge, exerts a higher Zeff on the 2p electrons than boron, resulting in
higher ionization energy.
Think about It
The general trend across a period for ionization energies follows the trend for effective nuclear
charge. As effective nuclear charge increases, so does ionization energy.
7.116. Collect and Organize
We are to compare the relative ionization energies of the group 17 anions, X–, with that of the neutral
atoms, X.
Analyze
With one extra electron on the atoms in X – there will be higher electron–electron repulsions in X–
compared to those in X.
Solve
The higher repulsion of the electrons in X– means that the ionization energy of X– is lower than that
of X.
Think about It
The ionization energy of X+, on the other hand, is greater than that of either X or X –.
7.117. Collect and Organize
We have to consider the electron configurations of the cations of Br, Kr, Rb, Sr, and Y to determine
which of the neutral atoms would have the smallest second ionization energy (IE 2).
Analyze
Element
Br
Kr
Rb
Sr
Y
Number of
Protons in
Nucleus
35
36
37
38
39
Electron Configuration
Cation (X+) Electron
Configuration
[Ar]3d 104s24p5
[Kr]
[Kr]5s1
[Kr]5s2
[Kr]4d 15s2
[Ar]3d 104s24p4
[Ar]3d 104s24p5
[Kr]
[Kr]5s1
[Kr]4d 15s1
Solve
Rb+, with the noble gas configuration of Kr as Rb +, has the highest IE2. Both Br+ and Kr+ lose the
second electron from a 4p orbital, which is lower in energy (harder to remove) than the removal of a
5s electron (higher in energy, easier to remove). Therefore, the IE 2 for Br+ and Kr+ is expected to be
higher than that for Sr+ or Y+. Sr+ with fewer protons in the nucleus holds onto the 5s electron less
tightly than Y+. Therefore, Sr is expected to have the smallest IE2.
Think about It
In determining relative orders for second, third, etc., IEs, we have to be sure to consider the electron
configuration of the cation that will lose the electron for that particular ionization step.
7.118. Collect and Organize
Aluminum’s first ionization energy is lower than both of its neighboring elements in the periodic
table. To answer why, we must consider both the effects of effective nuclear charge (Zeff) and relative
orbital energies of the s orbital versus the p orbital.
356 | Chapter 7
Analyze
As atomic number increases across a period, Zeff increases, so we expect IE to steadily increase across
a period. In terms of orbital energies, the p orbital within a shell lies higher in energy than the s orbital.
Solve
The electron configurations for the elements Mg, Al, and Si are
Mg: [Ne]3s2
Al: [Ne]3s23p1
Si: [Ne]3s23p2
When aluminum is ionized, it loses the electron from the higher-lying p orbital, so its IE is slightly
lower than magnesium’s. Silicon, with its increased atomic number, increases the Zeff on the p
electron so it is harder to ionize than aluminum.
Think about It
Other group 13 elements (B, Ga) also show this lowered IE with respect to their immediate element
neighbors in the periodic table.
7.119. Collect and Organize
We consider an electron dropping from n = 732 to n = 731 in a hydrogen atom. We are to calculate
the energy of this transition along with its wavelength and say what kind of telescope could detect
such radiation.
Analyze
The energy difference between two n levels in the hydrogen atom is given by the equation
 1
1
E  –2.18  10–18 J  2  2 
 n f ni 
The wavelength of light associated with a particular energy is
 = hc/E
Solve
 1
1 
(a) E  –2.18  10 –18 J 

 –1.11  10 –26 J
2
 731 7322 
Because this energy represents a loss of energy as the electron drops from a higher energy level to a
lower energy level, this process is exothermic, so the sign of ∆E is negative.
6.626  10 –34 J  s  3.00  108 m/s
 17.9 m
(b)  
1.11  10 –26 J
(c) This long wavelength occurs in the radio portion of the electromagnetic spectrum, so we would
need a radio telescope to detect this transition.
Think about It
Our result makes sense. As n increases in the hydrogen atom, the energy levels get closer and closer
together in energy and a transition between any two adjacent n levels where n is high would emit
very little energy (long wavelength).
7.120. Collect and Organize
In this problem we are to consider how the relative energy of the 2p and 2s orbitals is affected by
increasing atomic number across a period and down a column in the periodic table.
Analyze
The 2s orbital is lower in energy than the 2p orbital. This means that the 2s orbital is closer to the
nucleus than the 2p and when the charge on the nucleus increases the s orbital is pulled in (lowered
in energy) more than the p orbital is. Thus, as the number of protons increases, the energy separation
between the 2s and 2p orbital becomes greater.
Electrons in Atoms and Periodic Properties | 357
Solve
(a) As we move across the fourth period (K to Kr), the energy separation between the 2s and 2p
orbitals would increase. Because energy is inversely proportional to wavelength, the wavelength for
the 2p  2s transition would decrease.
(b) As we move down a given column in the periodic table, the energy separation between the 2s and
2p orbitals would again increase, so the wavelength for the 2p  2s transition would decrease.
Think about It
Each element has a distinctive 2p  2s transition energy.
7.121. Collect and Organize
We consider the emission of energy from an He + ion from n = 3 to n = 1 compared to a stepwise
relaxation of an He+ ion to the ground state (n = 3 to n = 2 then n = 2 to n = 1). We are to state which
statements given are true.
Analyze
Both He+ ions have the same nuclear charge (2+) and the same energies for n = 3, n = 2, and n = 1.
Solve
(a) True. Because the energies of n = 1, 2, and 3 do not depend on how the electron relaxes to the
ground state, the total energy of n = 3 to n = 1 is equal to the sum of the energy of n = 3 to n = 2 and
the energy of n = 2 to n = 1.
(b) False. Although the energies are additive, the wavelengths are not:
E31 = hc/31
E31, 21 = hc/32 + hc/21
These energies are equal so
hc
hc
hc


31
1
31
 3 2

1
 3 2
 21

1
 21

 21
3 2

3 2  21  213 2
Multiplying both sides by 31 gives
1
 2131 3 2 31  2131  3 2 31


3 2  21  213 2
3 2  21
3 2  21   2131  3 2 31  31   21  3 2 
3 2  21
 31
 21  3 2
(c) True. Because the energies are additive, the frequencies are also additive:
E31 = h31
E32, 21 = h32 + h21
Because these energies are equal
h31 = h32 + h21
31 = 3 2 + 21
(d) True. Using Equation 7.19 in the textbook
1 1
For He+: E  –(2.18  10 –18 J)(2) 2  2 – 2   –7.75  10–18 J
1 3 
hc 6.626  10 –34 J  s  3.00  108 m/s

 2.56  10 –8 m or 25.6 nm
E
7.75  10 –18 J
1 1
For H+: E  –(2.18  10 –18 J)(1) 2  2 – 2   –1.94  10 –18 J
1 3 


hc 6.626  10 –34 J  s  3.00  108 m/s

 1.02  10 –7 m or 102 nm
E
1.94  10 –18 J
358 | Chapter 7
Think about It
Be careful in jumping to the conclusion that the wavelengths of transitions are additive. It is only
their energies and frequencies that can be added in steps to get to the overall energy.
7.122. Collect and Organize
Using electron configurations we are asked to explain why silver’s typical ion is Ag +, why the
heavier group 13 elements form both 1+ and 3+ ions, and why the heavier group 14 and group 4
elements form both 2+ and 4+ ions.
Analyze
The electron configuration for Ag is [Kr]4d 105s1 (completely filled d orbitals). The electron
configuration for a group 13 element is [core](n – 1)d 10ns2np1. The electron configuration for a
group 14 element is [core](n – 1)d 10ns2np2. The electron configuration for a group 4 element is [core]
(n – 1)d 10ns2(n – 1)d 2.
Solve
(a) Silver forms a 1+ ion through the loss of a high-lying 5s electron. Palladium ([Kr]4d 85s2) and
cadmium ([Kr]4d 105s2) each lose two 5s electrons to form 2+ cations.
(b) The heavier group 13 elements may form 1+ cations through the loss of the np electron and form
3+ cations through the loss of np electrons and the two ns electrons.
(c) The heavier group 14 elements may form 2+ cations through the loss of two np electrons
and form 4+ cations through the loss of both np electrons and the two ns electrons. The group 4
elements may lose the two ns electrons to form 2+ cations and may lose both the ns electrons and the two
(n – 1)d electrons to form 4+ cations.
Think about It
The formation of ions that are two less than typical for the group (as in the heavier group 13 and 14
elements) is sometimes called the “inert pair effect.”
7.123. Collect and Organize
Ionization energy (IE1) is correlated with electronic structure. In this problem we examine the trends
in first and second ionization energies for elements 31–36 (Ga through Kr) and then compare the
second ionization energies of Kr and Rb.
Analyze
(a) The general trend is for increasing IE1 as atomic number (Z) increases across a period. However,
electronic structure (configuration) plays a role. In particular, the IE 1 and IE2 for Ga through Kr
depend on whether the electron is being removed from an s or p orbital.
(b) When comparing the IE1 of Kr to that of Rb, we have to be aware of the orbital level from which
the electron is being removed.
Solve
The electron configurations for Ga through Kr for both neutral atoms (X) and singly charged cations
(X+) are as follows:
Element
Electron Configuration X
Electron Configuration X+
10 2
1
Ga
[Ar]3d 4s 4p
[Ar]3d 104s2
10 2
2
Ge
[Ar]3d 4s 4p
[Ar]3d 104s24p1
10 2
3
As
[Ar]3d 4s 4p
[Ar]3d 104s24p2
10 2
4
Se
[Ar]3d 4s 4p
[Ar]3d 104s24p3
10 2
5
Br
[Ar]3d 4s 4p
[Ar]3d 104s24p4
10 2
6
Kr
[Ar]3d 4s 4p
[Ar]3d 104s24p5
(a) For the first ionization energy, the IEs increase in the following order:
Ga < Ge < Se < As < Br < Kr
Electrons in Atoms and Periodic Properties | 359
In this series, as Z increases, IE1 generally increases. The IE1 of Se is less than that of As because the
electron pairing (4p4) in one of the p orbitals for Se lowers Se’s IE1 slightly.
For the second ionization, the IE2 values increase in the following order:
Ge < Ga < As < Br < Se < Kr
Again, it is generally observed that as Z increases, so does the IE2. However, Ge’s second IE2 is
lower than Ga’s because to ionize the second electron in Ga, we need to remove an electron from a
lower energy 4s orbital. Also, Br’s IE2 is lower than Se’s because the electron pairing (4p4) in one of
the p orbitals for the Br+ ion lowers its IE2 slightly.
(b) Rubidium’s second ionization would occur from the electron configuration [Ar]3d 104s24p6 while
krypton’s would occur from [Ar]3d 104s24p5. Both would have an electron lost from the 4p orbital.
Since Rb has a higher Z, it exerts a higher Zeff on the 4p electron being lost, so it has the higher IE2
compared to krypton.
Think about It
In comparing the first and second ionization energies for Ga through Kr notice that the reversal of the
general trend at As–Se in IE1 occurs one pair to the right (Se–Br) in IE2.
7.124. Collect and Organize
We consider the effect of replacing Cl– in photo-gray sunglasses with Br– after defining an excited
state and writing the electron configurations of Ag +, Ag, Cl, and Cl–.
Analyze
(a) The electron configuration of Cl– results in a closed-shell configuration. For the Ag atom a 5s
electron is placed into the 4d shell to complete that subshell. For the Ag+ ion, the electron in Ag is
removed from the 5s orbital.
(b) In a ground state, all electrons are in their lowest energy orbital according to the Aufbau
principle. An excited state occurs when an electron absorbs light to be placed in a higher energy
orbital.
(c) The ionization energies of species decrease as we descend a group in the periodic table so the IE
of Br– is less than that of Cl–.
(d) According to part c, less energy is needed to ionize Br –. Energy is inversely proportional to
wavelength.
Solve
(a) Cl–: [Ne]3s23p6 or [Ar]
Cl: [Ne]3s23p5
Ag: [Kr]4d 105s1
Ag+: [Kr]4d 10
(b) An excited state occurs when an electron occupies a higher energy orbital. The electron is not in
its lowest energy state.
(c) More energy is needed to remove an electron from Cl – compared to Br– because the electron
removed from Cl– is at a lower n (principal quantum number) level and is held more tightly by the
nucleus.
(d) If AgBr were used in place of AgCl, longer wavelength (lower energy) light would remove the
electron. The AgBr sunglasses would darken perhaps in the infrared region.
Think about It
Extending the periodic trend in this question, AgF would be sensitive to shorter wavelengths of light.
7.125. Collect and Organize
We are to determine which neutral atoms are isoelectronic with Sn 2+ and Mg2+, and which 2+ ion is
isoelectronic with Sn4+.
360 | Chapter 7
Analyze
(a) The ground-state electron configurations for the neutral atoms Sn and Mg are Sn = [Kr]4d 105s25p2
and Mg = [Ne]3s2. To form Sn2+, remove the two 5p electrons; to form Sn4+, remove the two 5p
electrons and the two 5s electrons. To form Mg2+, remove the two 3s electrons.
(b) The neutral atom that has the same electron configuration as Sn2+ would have to have two 5s
electrons and a filled 4d shell. The neutral atom that has the same electron configuration as Mg2+
would have to have a filled n = 2 shell (two 2s electrons and six 2p electrons).
(c) Isoelectronic species are those that have the same number of electrons. The 2+ cation that would
be isoelectronic with Sn4+ would have to have no 5s or 5p electrons but would have a filled 4d shell.
Solve
(a) Sn2+: [Kr]4d 105s2
Sn4+: [Kr]4d 10
Mg2+: [Ne] or [He]2s22p6
(b) Cadmium has the same electron configuration as Sn2+ and neon has the same electron
configuration as Mg2+.
(c) Cd2+ is isoelectronic with Sn4+.
Think about It
When writing electron configurations for ionic species, start with the neutral atom and add or remove
electrons to form the ions.
7.126. Collect and Organize
We consider the unusual species O5+. We are asked to write the electron configuration of O 5+ and
consider the ionization energy required to produce this species.
Analyze
(a and b) The electron configuration for a neutral oxygen atom is [He]2s22p4. To form O5+, we need
to remove the four p electrons and one of the s electrons.
(c) Separating charge requires energy as described by Coulomb’s law:
QQ 
E  2.31  1019 J  nm  1 2 
 d 
(d) Because energy is inversely proportional to wavelength through  = hc/E we can calculate 
through E (the fifth ionization energy) expressed in J/atom.
9391kJ 1000J
1 mol


 1.559  10–17 J/atom
mol
kJ
6.022  1023 atoms
Solve
(a) O5+: [He]2s1
(b) One 2s electron and all four 2p electrons are removed from oxygen to make O5+.
(c) As electrons are removed from the oxygen atom, the negative electron has to be separated from
increasingly positively charged ions. Separation of a 1– charge from a 5+ charge will require more
energy than separating a 1– charge from a 4+ or lower charge. Also, the small amount of electron
shielding from the other electrons in the same shell has decreased, contributing to higher and higher
successive ionization energies.
6.626  10 –34 J  s  3.00  108 m/s
 1.27  10 –8 m or 12.7 nm
(d)  
1.559  10 –17 J
Think about It
If a shorter wavelength of light is used (<12.7 nm) to ionize O 4+, the extra energy would be converted
to kinetic energy of the ionized electron.
Electrons in Atoms and Periodic Properties | 361
7.127. Collect and Organize
Using the equation Zeff = Z – , where Z is the atomic number and  is the shielding parameter, we
are to compare the Zeff (effective nuclear charge) for the outermost electron in neon and argon.
Analyze
(a) In the effective nuclear charge equation given, use Z = 10 and  = 4.24 for Ne and Z = 18 and  =
11.24 for Ar.
(b) Shielding depends on the number of electrons that are lower in energy than the electron of
interest.
Solve
(a) Ne: Zeff = 10 – 4.24 = 5.76
Ar: Zeff = 18 – 11.24 = 6.76
(b) The outermost electron in argon is a 3p electron which is mostly shielded by the electrons in the
n = 2 level (10 electrons) and the n = 1 level (2 electrons), whereas the outermost electron in neon is
a 2p electron which is shielded only by the electrons in the n = 1 level (2 electrons).
Think about It
Notice that Zeff is greater for the outermost electron in Ar compared to that of Ne. The ionization
energy of Ar, however, is lower than the ionization energy for Ne. The effective nuclear charge
equation therefore, doesn’t seem to predict the trend in decreasing ionization energy as we descend a
group in the periodic table. The effective nuclear charge equation here does not take into account the
n level from which the electron is removed (ionized) to form the cation. Remember that the farther
away the electron is from the nucleus, the lower the energy that is required to remove it.
7.128. Collect and Organize
We compare the light that is emitted from sodium atoms versus light that might be emitted from
sodium ions.
Analyze
The sodium ion has the electron configuration of neutral neon, a closed shell configuration, while the
sodium atom has an electron configuration of [Ne]3s1. The 3s electrons in sodium atoms are excited
in sodium lamps which, when the electron returns to the ground state, emit yellow-orange light
(589 nm). Sodium ions lack this 3s electron.
Solve
Sodium ions do not emit the same orange-yellow light because the electron that would be excited
would be a 2p electron, not the 3s electron. When an excited 2p electron returns to the ground state, it
returns to the lower-lying 2p orbital, not to the higher-lying 3s orbital and so Na+ will emote a
different wavelength of light.
Think about It
Since the 2p orbital is lower lying in energy than the 3s, an electron returning to the 2p orbital from a
particular excited state will emit a shorter wavelength of light than an electron returning to the 3s
orbital for sodium atoms.
7.129. Collect and Organize
The p orbital has two lobes of different phase with a node between the lobes. We are asked how an
electron gets from one lobe to the other without going through the node between them.
Analyze
When we think of an orbital, we should think of the electron not as a particle (which in this case
would have to move through the node, a region of zero probability), but as a wave.
362 | Chapter 7
Solve
When we think of the electron as a wave, we can envision the node between the two lobes as a wave
of zero amplitude and the p orbital as a standing wave.
Think about It
Remember that an orbital describes the wave function for the electron and does not specifically
locate the electron as a particle.
7.130. Collect and Organize
We are asked what Einstein and Bohr meant in their statements about the uncertainty principle.
Einstein was concerned about the uncertainty principle’s implications of probability and Bohr’s
answer showed that he was comfortable with Heisenberg’s conclusions.
Analyze
The uncertainty principle states that we cannot know with great accuracy both the location and the
momentum of the electron in an atom (or other small particle).
Solve
Einstein meant that he thought nature (God) must be definitively knowable, that the randomness of
quantum mechanics must not be a fundamental property. Bohr’s reply was meant to remind Einstein
that a human could not presume to tell nature (God) how the universe should be constructed.
Think about It
In the century since quantum mechanics and the uncertainty principle were first developed, these
theories have withstood many experimental tests and become part of chemists’ and physicists’
picture of matter.
7.131. Collect and Organize
The heavier noble gases can form compounds with oxygen and fluorine, but the light noble gases do
not. Why?
Analyze
In order to form compounds, electrons must be either exchanged or shared between two atoms. In
compounds, we can assign oxidation numbers to the atoms. The oxidation number for oxygen is
typically 2– and for fluorine, it is 1–. This means that the noble gases would take on a positive
oxidation number in compounds with fluorine and oxygen.
Solve
The heavier noble gases are easier to ionize (IE decreases down a group in the periodic table) and
therefore can combine with oxygen and fluorine.
Think about It
We will learn later that fluorine and oxygen are the most electronegative elements in the periodic
table and therefore combine with most of the elements to form both covalent and ionic compounds.
7.132. Collect and Organize
Large jumps in successive ionization energies occur for elements other than the noble gases, which
show a smooth increase in their ionization energies. Why?
Analyze
A large jump in successive ionization energy occurs when electrons have been completely removed
from the outermost shell of the atom and the next electron must be removed from a lower n level.
Solve
For the noble gases, the ionization energies show a smooth increase because the large jump in
ionization energy would not occur until the eighth ionization energy for Ar and Ne, for example.
Electrons in Atoms and Periodic Properties | 363
Think about It
If we could remove 10 electrons from argon, we would observe a large jump in ionization energy
from the seventh to the eighth ionization energy. These high successive ionization energies are
difficult to measure, however.
7.133. Collect and Organize
Helium’s name derives from helios, Greek for sun, where it was first discovered. Helium, as
evidenced by its use in party balloons, is lighter than air. We are asked why helium was discovered
extraterrestrially before being found on Earth.
Analyze
Helium’s light mass gives helium atoms high velocities at normal temperatures as expressed by the
root-mean-square speed equation.
Solve
The high velocity of helium atoms means that once helium atoms are released into the atmosphere,
they can escape Earth’s gravitational pull. Therefore, Earth’s atmosphere contains very little helium.
Think about It
Helium on Earth is found trapped with natural gas underground and is a result of radioactive decay
( particles) of heavier elements. The United States is the world’s largest supplier of helium.
7.134. Collect and Organize
We are asked why it requires more than twice the energy to remove an electron from Na + than from
Ne. Both species have a closed-shell electronic configuration of [He]2s22p6.
Analyze
Sodium has one more proton in its nucleus than neon.
Solve
The higher nuclear charge on sodium increases the effective nuclear charge that the 2p electrons feel
in Na+. Its ionization energy, therefore, is greater than that of neutral Ne. Also, in Na +, the electron is
being removed from an already positively charged ion, which requires more energy than the removal
of an electron from a neutral Ne atom.
Think about It
Continuing the trend, the third ionization energy of Mg would be higher still.