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CHAPTER 7 | Electrons in Atoms and Periodic Properties

7.85. Collect and Organize / Analyze We are asked what is meant by a degenerate orbital.

SolveDegenerate orbitals have the same energy and are indistinguishable from each other.

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Think about ItIn 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.

AnalyzeMendeleev 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.

SolveThe 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 R2+, which then combines with O2–.

Think about ItNotice 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.

AnalyzeThe 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.

SolveAs 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 ItAlso, 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.

SolveThe 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 ItThe 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

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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 quantum numbers in order of increasing energy.

AnalyzeThe 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.

SolveThe orbitals described are: (a) 3d, for n = 3, = 2; (b) 5g, for n = 5, = 4; (c) 3s, for n = 3, = 0; and (d) 4p for n = 4, = 1, = 1.In increasing order of energy: (c) 3s < (a) 3d < (d) 4p < (b) 5g.

Think about ItTo 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 quantum numbers in order of increasing energy.

AnalyzeThe 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.

SolveThe orbitals described are: (a) 2p, for n = 2, = 1; (b) 5f, for n = 5, = 3; (c) 3d, for n = 3, = 2; and (d) 4f, for n = 4, = 3.In increasing order of energy: (a) 2p < (c) 3d < (d) 4f < (b) 5f.

Think about ItTo determine the energy of an orbital, first look to the n quantum number, then to

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.

AnalyzeWhen 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.

SolveLi: [He]2s1 Na+: [Ne] or [He]2s22p6

Li+: [He] or 1s2 Mg2+: [Ne] or [He]2s22p6

Ca: [Ar]4s2 Al3+: [Ne] or [He]2s22p6

F–: [He]2s22p6 or [Ne]

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Think about ItBecause 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.

AnalyzeNeon’s electronic configuration is [He]2s22p6 with a total of 10 e–.

SolveThe 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 ItThe 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.

AnalyzeWhen 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.

SolveK: [Ar]4s1 Ba: [Xe]6s2

K+: [Ar] Ti4+: [Ar] or [Ne]3s23p6

S2–: [Ne]3s23p6 or [Ar] Al: [Ne]3s23p1

N: [He]2s22p3

Think about ItNotice 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.

AnalyzeThe electron configurations are

H: 1s1 K: [Ar]4s1

Li: [He]2s1 Rb: [Kr]5s1

Na: [Ne]3s1 Cs: [Xe]6s1

SolveAll of these elements (the alkali metals) have an ns1 configuration after a noble gas core.

Think about ItThe 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.


AnalyzeWhen 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.

SolveNa: [Ne]3s1 Mn: [Ar]3d

54s2

Cl: [Ne]3s23p5 Mn2+: [Ar]3d5

Think about ItTo 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.

AnalyzeWhen 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.

SolveC: [Ne]2s22p2 Ti: [Ar]3d

24s2

S: [Ne]3s23p4 Ti4+: [Ar]

Think about ItCarbon could obtain a noble gas configuration by either gaining four electrons to become C4– 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.

AnalyzeIf 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 3 unpaired e–

(b) O: [He]2s22p4 2 unpaired e–

(c) P3–: [Ne]3s23p6 0 unpaired e–

(d) Na+: [Ne] or [He]2s22p6 0 unpaired e–

Think about ItNotice 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.

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AnalyzeIf 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 1 unpaired e–

(b) Ag+: [Kr]4d

10 0 unpaired e–

(d) Cd2+: [Kr]4d

10 0 unpaired e–

(e) Zr4+: [Kr] or [Ar]3d

104s24p6 0 unpaired e–

Think about ItRemember, 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.

AnalyzeThis atom has no charge so we do not have to account for additional or lost electrons.

SolveThe 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 ItAlthough 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.

AnalyzeThis atom has no charge, so we do not have to account for additional or lost electrons.

SolveThe 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 ItThe 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.


AnalyzeBecause the atom has an extra electron, to form the monatomic anion, the neutral atom would have an electron configuration of one less electron.

SolveIon’s electron configuration: [Ne]3s23p6 = X– Atom’s electron configuration: [Ne]3s23p5 = XThis 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 ItWhen 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.

AnalyzeBecause the atom has lost an electron to form the X+ cation, the neutral atom would have an electron configuration of one more electron.

SolveIon’s electron configuration: [Kr]4d

105s2 = X+

Atom’s electron configuration: [Kr]4d

105s25p1 = XThis 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 ItWhen 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.

AnalyzeThe 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.

SolveAl 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 ItRecall 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.

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AnalyzeThe 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.

SolveS gains two electrons to become S2–, with the electron configuration of Ar.P gains three electrons to become P3–, 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 ItThe 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.

AnalyzeThe 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 ItIf 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.

AnalyzeThe 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.


(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 ItIf 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.

AnalyzeThe electron configuration for iodine is [Kr]4d

105s25p5. The difference between 131I and 127I is that 131I has 4 additional neutrons in its nucleus.

SolveThe 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 131I and 127I (which are based on total number of electrons in the atom) are the same.

Think about ItElectron 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.

AnalyzeFor 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).

SolveFor 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 ItBecause 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?

AnalyzeWhen 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.

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SolveIf 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 ItThe 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?

AnalyzeAs 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.

SolveWhen 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 ItEven 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.

AnalyzeThe sizes of atoms increase down a group because electrons have been added to higher n levels.

SolveRb is the largest atom.

Think about ItThe 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.

AnalyzeCompared 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.

SolveI– is the largest monatomic anion.

Think about ItThis 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.


AnalyzeThe 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 ItIonization 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.

AnalyzeIn Figure 7.34, we see that ionization energy comparisons between the pairs are

He > Li, Li < Be, Be > B, N > OFor 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 ItGenerally, 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.

AnalyzeBoth fluorine and boron have 2s22pn configurations and the ionized electron is removed from the 2p orbital.

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SolveFluorine, with a higher nuclear charge, exerts a higher Zeff on the 2p electrons than boron, resulting in higher ionization energy.

Think about ItThe 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.

AnalyzeWith one extra electron on the atoms in X– there will be higher electron–electron repulsions in X–

compared to those in X.

SolveThe higher repulsion of the electrons in X– means that the ionization energy of X– is lower than that of X.

Think about ItThe 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 (IE2).

AnalyzeElement Number of

Protons in Nucleus

Electron Configuration Cation (X+) Electron Configuration

Br 35 [Ar]3d

104s24p5 [Ar]3d

104s24p4

Kr 36 [Kr] [Ar]3d

104s24p5

Rb 37 [Kr]5s1 [Kr]Sr 38 [Kr]5s2 [Kr]5s1

Y 39 [Kr]4d

15s2 [Kr]4d

15s1

SolveRb+, 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 IE2 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 ItIn 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.


AnalyzeAs 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.

SolveThe 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 ItOther 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.

AnalyzeThe energy difference between two n levels in the hydrogen atom is given by the equation

The wavelength of light associated with a particular energy is = hc/E

Solve

(a)

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.

(b)

(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 ItOur 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.

AnalyzeThe 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.

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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 ItEach 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.

AnalyzeBoth 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:

E31 = hc/31

E31, 21 = hc/32 + hc/21

These energies are equal so

Multiplying both sides by 31 gives

(c) True. Because the energies are additive, the frequencies are also additive:E31 = h31

E32, 21 = h32 + h21

Because these energies are equalh31 = h32 + h21

31 = 3 2 + 21

(d) True. Using Equation 7.19 in the textbook

For He+:

For H+:


Think about ItBe 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.

AnalyzeThe 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 ItThe 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 IE1 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.

SolveThe 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+

Ga [Ar]3d

104s24p1 [Ar]3d

104s2

Ge [Ar]3d

104s24p2 [Ar]3d

104s24p1

As [Ar]3d

104s24p3 [Ar]3d

104s24p2

Se [Ar]3d

104s24p4 [Ar]3d

104s24p3

Br [Ar]3d

104s24p5 [Ar]3d

104s24p4

Kr [Ar]3d

104s24p6 [Ar]3d

104s24p5

(a) For the first ionization energy, the IEs increase in the following order:Ga < Ge < Se < As < Br < Kr

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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 < KrAgain, 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 ItIn 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 ItExtending 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 Sn2+ and Mg2+, and which 2+ ion is isoelectronic with Sn4+.


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 ItWhen 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 O5+ 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:

(d) Because energy is inversely proportional to wavelength through = hc/E we can calculate through E (the fifth ionization energy) expressed in J/atom.

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.

(d)

Think about ItIf a shorter wavelength of light is used (<12.7 nm) to ionize O4+, the extra energy would be converted to kinetic energy of the ionized electron.

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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 ItNotice 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.

AnalyzeThe 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.

SolveSodium 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 ItSince 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.

AnalyzeWhen 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.


SolveWhen 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 ItRemember 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.

AnalyzeThe 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).

SolveEinstein 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 ItIn 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?

AnalyzeIn 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.

SolveThe 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 ItWe 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?

AnalyzeA 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.

SolveFor 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.

363 | Chapter 7

Think about ItIf 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.

AnalyzeHelium’s light mass gives helium atoms high velocities at normal temperatures as expressed by the root-mean-square speed equation.

SolveThe 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 ItHelium 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.

AnalyzeSodium has one more proton in its nucleus than neon.

SolveThe 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 ItContinuing the trend, the third ionization energy of Mg would be higher still.

Chapter 6 | Thermochemistry - Page Not Found - …web.mnstate.edu/marasing/Chem150/EOC150/Solutions... · Web viewWe are to explain why the transition metals often form ... The

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