Chemistry 121
Mines, Fall 2015
Answer Key, Problem Set 10 – Full
1. MP; 2. NT1; 3. 8.59; 4. 8.58; 5. MP; 6. NT2; 7. NT3; 8. MP; 9. MP; 10. MP; 11. NT4; 12. NT5; 13. MP; 14.
NT6; 15. MP; 16. MP; 17. NT7; 18. NT8; 22. MP; 23. 8.70; 24. 8.72; 19. MP; 20. MP; 21. NT9; 25. NT10; 26.
NT11; 27. 8.86; 28. MP; 29. NT12; 30. NT13; 31. NT14; 32. NT15; 33. NT16; 34. NT17
-------------------------------------------------------
Effective Nuclear Charge, the Shell Model, and Trends in Atomic Radii and First Ionization Energy
1. MP. No answer in key for this problem, but see answer to NT1 below for Zeff description.
2. NT1.
(a) Define Zeffective, Zactual, and S in the equation Zeffective = Zactual - S, and then "translate" briefly in words what this
equation "means". (b) For our purposes (simplified model), how have I defined a "shielding electron"? (c) Tro
problem 8.57
(a) Zeffective = the effective or “apparent” positive charge that a given electron in an atom “sees” coming
from the center of the atom
Zactual = the actual charge on the nucleus of a given atom; it is equal to the number of protons (the
atomic number)
S = the amount of shielding, or the number of shielding electrons
The equation says that the apparent positive charge that a given electron in an atom “sees” is
equal to the actual nuclear charge minus however much of that positive charge is shielded (“from
view”), or “cancelled out” by shielding electrons.
In other words, the apparent positive charge that a given electron “sees” is generally less than the
actual nuclear charge by an amount equal to the number of shielding electrons.
(b) shielding electron (in our simplified model): any electron in an atom that is in an energy level
closer to the nucleus (value of n is smaller) than the energy level of the electron being
considered.
Note that if the electron being considered is a valence electron, then every core electron is a
shielding electron. Note also that in this simplified model, electrons in the same energy level do
not shield one another (at all).
(c)
8.57. Which electrons experience a greater effective nuclear charge, the valence electrons in Be, or the valence
electrons in N? Why?
Answer: the ones in N. In each case, the number of shielding electrons is 2, but in Be, Zactual is
only +4 whereas Zactual in N is three units greater at +7.
3. 8.59.
(paraphrasing) Calculate the effective nuclear charge for the following using the simplified model discussed in
class and in NT1 (which is the same as the model described in this problem)
Answers: (a) +1 (b) +2 (c) +6 (d) +4
Complete Reasoning / Work:
Electron
Configuration
(a) K:
Zactual
Core electrons bolded
(atomic # of
element)
1s2 2s2 2p6 3s2 3p6 4s1
+19
(b) Ca
1s2 2s2 2p6 3s2 3p6 4s2
+20
(c) O
(d) C
1s2 2s2 2p4
+8
+6
1s2
2s2
2p2
# core
electrons*
2+2+6+2+6=
18
2+2+6+2+6=
18
2
2
S
(#
shielding
electrons)*
Zeff
18
+19 – 18 = +1
18
+20 – 18 = +2
2
2
+8 – 2 = +6
+6 – 2 = +4
* Since we are assumed to be finding the Zeff of the outermost electrons, all core electrons are shielding
electrons (in the simple model discussed in class). As such, the number of shielding electrons also ends up
PS10-1
Answer Key, Problem Set 10
equaling the (total) number of electrons in the preceding noble gas (which is probably easier to assess if you
are looking at a periodic table).
4. 8.58 No answer in key for this problem, but see answers to previous problems (where Zeff is discussed).
5. MP. No answer in key for this problem, but you can watch the video for a good description of the trends.
You can also see my answer to Problem 8.64 at the end of this key (was not assigned).
Please note that “describing” a trend or “using a trend to make predictions” is not an explanation of
that trend or prediction. This problem only requires the former, but you are responsible for the
latter (see NT2 and NT3 below).
6. NT2.
If the following drawings are meant to very simplistically represent the distribution of electrons in two different
atoms, predict which atom would be the smaller of the two, and provide your reasoning. (Hint: use the concept of Zeffective
--calculate it for each and go from there. State explicitly the number of shielding electrons in each case.)
+15
+17
2 e-
2 e-
8 e5 e-
Atom A
8 e-
Zeff = +15 – 2 – 8 = +5
7 e-
Atom B
Zeff = +17 – 2 – 8 = +7
Answer: Atom B would be smaller.
Reasoning: The outer electrons are in the same energy level (the third), so there’s no difference there.
However, the effective nuclear charge “seen by” the outer electrons in Atom B is BIGGER than for the
outer electrons in Atom A. Why? Because the number of shielding electrons in both cases is 10 (the 2
in the 1st energy level and the 8 in the 2nd energy level), but the number of protons in the nucleus is
larger for Atom B. The net effect is that the amount of nuclear charge “seen” by each of the (seven)
outer electrons in Atom B is +17 – 10 = +7, whereas the amount of nuclear charge “seen” by each of
the (five) outer electrons in Atom A is +15 – 10 = +5. So the electrons in Atom B will be pulled in with
greater force and thus they will be pulled in closer, leading to a smaller atom. (Note that for the same
reason, the ionization energy would be predicted to be larger for Atom B.)
7. NT3.
Use the concepts in Coulomb's Law (force of attraction depends on "distance" and "magnitude of charge"...), as
well as related ideas (such as effective nuclear charge, shielding, etc.) to provide explanations for the following
observations:
(a) It is easier (requires less energy; smaller IE1) to remove the outermost electron from a neutral sodium atom than to
remove one of the outermost electrons in a neutral chlorine atom.
The brief answer: The Na valence electron experiences a smaller Zeff than each of the outermost
electrons in a Cl atom. At a similar base distance (n = 3 in both atoms), a smaller Zeff (“magnitude of
charge”) leads to a smaller force of attraction according to Coulomb’s Law, and a smaller force leads
to it being easier to remove the electron (takes less energy).
The detailed explanation. The valence electron configuration for Na is 3s1; that for Cl is 3s23p5. The
core electron configuration for both atoms is 1s22s22p6 ([Ne]), which means there are 10 core
electrons in each. But there are more protons in the nucleus of Cl, so the Zeff ends up larger for Cl:
Zeff(Na v. e-) = +11 – 10 = +1;
Zeff(Cl v. e-‘s) = +17 – 10 = +7
Since the force of attraction between oppositely charged particles depends on the magnitude of the
charge, a +7 charge will attract an electron much more strongly than a +1 charge, if distance is the
same. If the force is stronger, then it will be harder to pull an electron away. So the electron in Cl is
much harder to remove (takes more energy) than the electron in Na.
PS10-2
Answer Key, Problem Set 10
Note: although the average distances are not identical, since the valence electrons are in the same
energy level (n = 3), this difference is small compared to the charge difference (a factor of 7-fold!)
(b) It is easier (requires less energy;
smaller IE1) to remove the outermost electron from a neutral potassium atom
than to remove the outermost electron from a neutral sodium atom.
The brief answer: The outer electron in a K atom “sees” about the same magnitude of positive
charge coming from its nucleus (same Zeff) as the outer electron in a Na atom, but is farther away
(on average) from the nucleus. By Coulomb’s Law, the force of attraction is weaker if the distance
between two charged particles is greater (magnitudes of charge being equal).
The detailed answer: The outer electrons in both Na and K have an “s1” configuration. That
means that in both cases, there is only one electron that is farther away from the nucleus than all of
the others. The result is that the Zeff “seen” by the outer electron in both cases is about +1. The
specific calculations are (noting the full configuration of K is 1s22s22p63s23p64s1):
Zeff(Na v. e-) = +11 – 10 = +1;
Zeff(K v. e-) = +19 – 10 – 8 = +1
That means the “magnitude of charge effect” is a non-issue here (unlike in part (a)). The big
difference here is that the outer electron in Na is in the 3s orbital but the outer electron in K is in the
4s orbital. That means that the one in K is significantly farther away from the nucleus (in the
ground state). Since the force of attraction between oppositely charged particles decreases as
distance increases, the electron farther away will experience a weaker force of attraction and thus
be easier to remove.
(c) An atom of F has a smaller radius than an atom of C.
Since F is to the right of C in the periodic table, the nucleus of F has more protons in it (9) than the
nucleus of C (6). But the number of shielding electrons is the same in both (just the two 1s
electrons). This means that the outer electrons in F will see a greater amount of positive charge
(Zeff  +9 – 2  +7) from the nucleus than the outer electrons in C (Zeff  +6 – 2  +4.) Thus they will
experience a stronger force of attraction toward the nucleus and thus get pulled in closer.
(Coulomb’s Law: “magnitude of charge” effect. Note: the outer electrons are in the second energy
level in both atoms, so there is no “base” distance effect here. The size difference comes
essentially as a result of the greater Zeff and force of attraction experienced by the outer electrons.)
8. MP. No answer in key for this problem, but see notes or text for the definition of “ionization energy” and
then make sure you know how to “translate” that into a chemical equation representing a process.
9. MP.
No answer in key for this problem, but see my answer to 8.74 at the end of this key (extra
practice). You may also view the video, my Powerpoint (or your notes), or the text for the
ionization energy trends. As noted above for the radius trends, please note that “knowing the
trend” (and using it) is different than explaining the trend (or why one is larger than or smaller than
another).
10. MP.
No answer in key for this problem, but Part A is like the prior problem. Part B is why I asked this
question. I want you to think about how the radius trends and ionization energy trends are
related. Since both are fundamentally dependent on the Coulomb’s Law (electrostatic) force of
attraction between each valence electron and the nucleus, their trends are related to one
another. (You may find reviewing NT3’s answer key entry helpful for this problem.)
Electron Configurations of Ions, Para- and Dia- magnetism, Effective Nuclear Charge for Ions
11. NT4. 8.66(abc only [add f, g]). Write electron configurations for each ion.
Reasoning: Remember way back from PS3 that one can view ions as “atoms (or molecules) with a
charge”, and that a positive charge means electrons have been removed from the
PS10-3
Answer Key, Problem Set 10
neutral atom (or molecule), while a negative charge means electrons have been added
to the neutral atom (or molecule). So to get the electron configuration for a main group
monatomic ion, you start with the neutral atom’s configuration and then add or remove
as many electrons as needed.
12. NT5.
“Parent” Atom Electron
Configuration
Addition or Removal of
electrons
Configuration of the ion
(a) Cl-
Cl: 1s2 2s2 2p6 3s2 3p5
Add 1 e- →
1s2 2s2 2p6 3s2 3p6
(b) P3-
P: 1s2 2s2 2p6 3s2 3p3
Add 3 e- →
1s2 2s2 2p6 3s2 3p6
(c) K+
K: 1s2 2s2 2p6 3s2 3p6 4s1
Remove 1 e- →
1s2 2s2 2p6 3s2 3p6
(f) P+
P: 1s2 2s2 2p6 3s2 3p3
Remove 1 e- →
1s2 2s2 2p6 3s2 3p2
(g) Ar 
Ar: 1s2 2s2 2p6 3s2 3p6
Add 1 e- →
1s2 2s2 2p6 3s2 3p6 4s1
(probably has never been
experimentally observed)
For each ion in the prior problem (my modified 8.66(abcfg)), state the number of unpaired electrons and indicate
whether it is diamagnetic (no unpaired electrons) or paramagnetic (at least one unpaired electron).
(a), (b), and (c) are isoelectronic and thus have
the same electron configuration (see 8.66),
same orbital diagram (see right), and same
number of unpaired electrons (none)
1s
2s
2p
3s
3p
1s
2s
2p
3s
3p
 diamagnetic
(f) P+ has two unpaired electrons (see right)
 paramagnetic
(g) Ar- (if it were stable) would have one unpaired
electron (see right)  paramagnetic
1s
2s
2p
3s
3p
4s
13. MP. No answer in key for this problem, but see prior problem and the information in the problem (as
well as the comment I left for you before this particular problem in Mastering regarding electron
configurations of transition metal cations) for help here.
14. NT6. For each ion in NT4 (my modified 8.66(abcfg), calculate (show work!) the effective nuclear
charge (Zeff).
Note how the effective nuclear charge for each of these ions is NOT equal numerically to the
number of valence electrons! It just “accidentally” works out that way for many neutral atoms.
Don’t ever connect the two, please!!
(a)
Electron Configuration
Zactual
(See NT4)
Core electrons bolded
(atomic # of
element)
Cl-
1s2 2s2 2p6 3s2 3p6
+17
P3-
1s2 2s2 2p6 3s2 3p6
(b)
(c) K+
(f) P+
1s2 2s2 2p6 3s2 3p6
1s2 2s2 2p6 3s2 3p2
(g) Ar  1s2 2s2 2p6 3s2 3p6 4s1
# core
electrons*
(bolded)
2+2+6=
S
(# shielding
electrons)*
Zeff **
10
+17 – 10 = +7
+15
+19
+15
10
10
10
10
10
10
10
+15 – 10 = +5
+19 – 10 = +9
+15 – 10 = +5
+18
2+2+6+2+
6=
18
+18 – 18 = 0
18
PS10-4
Answer Key, Problem Set 10
* Since we are assumed to be finding the Zeff of the outermost electrons, all core electrons are shielding
electrons (in the simple model discussed in class).
** Technically, this calculation does not tell the full story, because it does not take into account the difference
in electron-electron repulsions relative to the neutral atom. For example, P3- and P+ both show the same Zeff
with this calculation, but clearly P3- will experience a greater amount of overall electron-electron repulsion
since there are three extra electrons in the species relative to its being neutral (which will push the electrons
farther out and make it easier to remove one). And clearly P+ will go the other way: it will experience a
lesser amount of overall e-e repulsion since there is one fewer electron, making all electrons held more
tightly to the nucleus).
Higher Ionization Energies, Explaining the Charges on (Certain) Cations and the “Big Jump”
15. MP. No answer in key for this problem, but see notes or text for the definition of “nth ionization
energy” and then make sure you know how to “translate” that into a chemical equation
representing a process. I guess I can say here that the “nth” ionization energy is the energy
needed to remove the nth electron from a neutral gaseous atom, which means that you are
removing it after (n-1) electrons have already been removed. You may also find the
discussions in my answers to Q17 (NT7) and Q18 (NT8) helpful here (see below).
16. MP. No answer in key for this problem, but see the paragraph entitled “Successive Ionization
Energies” for guidance (and embedded hints) on this. NOTE: I rewrote much of that paragraph
so that it would better help you. You may also find the discussions in my answers to Q17
(NT7) and Q18 (NT8) helpful here (see below).
17. NT7. (a)
Use the shell model and related concepts (see NT4) to explain why the fourth ionization energy (IE4) of
Al is MUCH greater than the third ionization energy (IE3). (As a part of your answer, write out the chemical equation
that represents these processes [see p. 339], as well as the electron configuration of the reactant and product ions in
both cases).
Third ionization:
Al2+

1s22s22p63s1
Al3+
+
E = IE3
e -;
1s22s22p6
Here, the e- comes from the 3rd energy
level (the valence shell) and the effective
nuclear charge is moderate:
Zeff = +13 – (2 + 2 + 6) = +3 (a low
estimate since e-- e- repulsion is low in the
starting cation)
Fourth ionization:
Al3+
1s22s22p6

Al4+
+
E = IE 4
e -;
1s22s22p5
Here, the e- comes from the 2nd energy
level (part of the original core) and the
effective nuclear charge is huge!
Zeff = +13 – 2 = +11 (a low estimate since e- e- repulsion is even lower in the starting cation)
Answer: Since there are 3 valence electrons in Al, the third ionization of Al involves removing the last
electron from the valence shell, but the fourth ionization involves removing an electron from (what
was originally) the core. The Zeff of core electrons is always MUCH GREATER than the Zeff of any
valence electron, and so the force of attraction of core electrons toward the nucleus is enormous by
Coulomb’s Law. Thus, they are much harder to remove. On top of that, there is an additional
distance effect here which makes the force even greater! The 4th electron is coming from an
energy level that is closer to the nucleus (n = 2 vs n = 3 level). Again, by Coulomb’s Law, the force
will be greater for the core electron vs. the valence electron.
(b) Tro, 8.78.
Consider this set of successive ionization energies:
IE1
IE2
IE3
IE4
578 kJ/mol
1820 kJ/mol
2750 kJ/mol
11,600 kJ/mol
To which third period element do these ionization values belong?
Answer: Al
PS10-5
Answer Key, Problem Set 10
Reasoning: The “big jump” happens when the fourth electron is to be removed (IE4 >> IE3). That
indicates that the fourth electron is the first core electron, which means there should be three
valence electrons in the original atom. Al is the only 3rd period element with 3 v. e-‘s [See answer
to Part (a) for detailed analysis of why the first core electron is so much harder to remove than any of the valence
electrons.]
18. NT8.
Predict the ordering of these elements from smallest to largest fourth ionization energy (IE4): Na, Mg, Al, Si, P.
You must provide detailed reasoning which involves the use of the “shell model” and related ideas! This is not
as straightforward as it might initially appear.
Hint: Write out the e- configuration for the species whose electron is being removed in each case and consider Zeff!
Answer: Smallest IE4: __Si__ < ___P__ < __ Na___ < __ Mg___ < __ Al___ Largest IE 4
Reasoning: The fourth ionization involves removing an electron from the X3+ species in each case
(see NT5). The configuration, Zactual, S, and Zeff for each such ion in this problem are:
Electron Configuration of
“Parent” Atom
(being ionized)
Electron
Configuration*
Zactual
S
Zeff
Na: 1s2 2s2 2p6 3s1
Na3+
1s2 2s2 2p4
+11
2
+9
Mg: 1s2 2s2 2p6 3s2
Mg3+
1s2 2s2 2p5
+12
2
+10
Al: 1s2 2s2 2p6 3s2 3p1
Al3+
1s2 2s2 2p6
+13
2
+11
Si: 1s 2s 2p 3s 3p
2
Si3+
1s2 2s2 2p6
3s1
+14
10
+4
P: 1s2 2s2 2p6 3s2 3p3
P3+
1s2 2s2 2p6 3s2
+15
10
+5
2
2
6
2
Species
*Remove 3 electrons from the “parent” atom’s configuration to get the configuration of the +3 cation
Fundamentally, Si has the smallest 4th ionization energy because Si3+ has the smallest Zeff. Al has
the largest IE4 because Al3+ has the greatest Zeff. Note that the Zeff of the neutral atoms is
irrelevant here. One must look at the Zeff of the species being ionized (here, the +3 cations).
Distance also contributes in that the electrons in Na3+, Mg3+, and Al3+ are all coming from the n = 2 level where they are
closer to the nucleus than the n = 3 electrons in Si3+ and P3+. So among the three species with the electron closest to
the nucleus, Al3+ ‘s greater Zeff makes it have the greatest force of attraction for the nucleus and thus the greatest (4th)
ionization energy. Among the two species with the electron farthest from the nucleus, Si3+’s smaller Zeff makes it have
the weakest force of attraction for the nucleus and thus the smallest (4th) ionization energy.
Notice that the effective nuclear charge does not end up steadily increasing across the row here
(as it does for the neutral atoms, which are the species being ionized when the first ionization
energy occurs). It starts off increasing (from Na to Mg to Al, where it is quite large in all of them
because the fourth electron is in what was the core of the neutral atom), but then drops from +11 to
+4 at Si because Si has its fourth electron still in the valence shell. Then it continues its upward
trend after that as the Zactual increases (while S remains the same).
Electron Affinity (EA), EA Trends for Neutral Atoms
19. MP. No answer in key for this problem, but watch the video and read the information provided in the
problem!
20. MP. No answer in key for this problem, but see the information in the prior problem!
21. NT9. 8.80(ab only). Choose the element with the more negative (more exothermic) electron affinity from each
pair. (Use the “shell model” along with a description of what process is occurring during the process associated with
“electron affinity” to explain your choices in all parts)
(a) Mg or S (greater Zeff)
(b) K or Cs (closer to nucleus)
Answers: (a) S
(b) K
(e) O or Ne (added electron in Ne would have Zeff ~ 0 b/c no room left in p sublevel)
PS10-6
(c) O
Answer Key, Problem Set 10
Full Reasoning:
A negative electron affinity means “energy-lowering to add an electron”, which means “the added
electron is attracted to the nucleus in the anion formed” (remember, “when things that attract one
another come together, energy is released”). Thus, the more negative electron affinity in each
pair should be for the atom that more strongly attracts an added electron.
(a) Mg and S are both in the 3rd row, but S has more protons in the nucleus. Thus, S has a
greater Zeff and a valence shell at a similar (base) distance. Therefore, as long as the added
electron gets to go into the valence shell in both (which it does), it will have a stronger force
of attraction for the S nucleus (greater magnitude of charge).
(b) K and Cs are both in the 1st column, so their Zeff’s will be essentially the same. But the
electron added to K will go into the 4th energy level (closer to the nucleus) whereas it will go
into the 6th level in Cs (farther away). Thus, the force of attraction will be stronger for the
electron added to K.
(e) O and Ne are both in the 2nd row, but Ne has more protons in the nucleus. Thus, Ne has a
greater Zeff and a valence shell at a similar (base) distance. Therefore, if the added electron
were able to go into the valence shell in both, it would have a stronger force of attraction for
the Ne nucleus (greater magnitude of charge). But unlike in (a), the p sublevel in Ne is full,
so an added electron would end up going into the next energy level (3s orbital). From that
energy level, the Zeff (for the added electron) is essentially zero (because it would be
“completely” shielded from the nucleus), so the force of attraction for an added electron is
much stronger in O (where there is still space in the 2p sublevel for the added electron).
(Relative) Sizes of Ions
22. MP.
No answer in key for this problem, but see my answer to 8.70 below for some similar
problems. Also, I covered the topics associated with Parts B and C pretty thoroughly starting
near the middle of PowerPoint 21. Lastly, for Part A, since ions of the same charge for
elements down a column in the periodic table will have analogous valence electron
configurations (just like their neutral atom counterparts do), the radius and ionization energy
trends will be the same for the monatomic ions as they are for the neutral atoms (for the same
reasons, of course!).
23. 8.70.
Which is the larger species in each pair? [“Larger” here means “bigger volume”  larger radius]
(a) Sr or Sr2+ Cations have less e--e- repulsion than their neutral atom counterparts (same # protons
in the nucleus), so the electrons get pulled in closer to the nucleus, making the radius
smaller. Thus the neutral atom Sr is larger than Sr2+.
(b) N or N3-
Anions have more e--e- repulsion than their neutral atom counterparts (same #
protons in the nucleus), so the electrons get pushed farther out from the nucleus,
making the radius larger. Thus the anion N3- is larger than the neutral atom N.
(c) Ni or Ni2+
Same reasoning as in (a).
(d) S2- or Ca2+ These ions are isoelectronic, each having the electron configuration of Ar, Element
18. So they have the exact same amount of e--e- repulsion. The difference here is
what’s in the nucleus: Ca2+ has 20 protons (i.e., a nuclear charge of +20) while S2has only 16 protons (i.e., a nuclear charge of +16). So all the electrons in Ca2+ are
pulled in more tightly (vs. S2-), making the radius smaller. Thus S2- is larger.
24. 8.72. No answer in key, but this problem is like the prior two problems, so see my
comments for these problems.
Reactivities of Alkali Metals and Halogens
25. NT10. Short answer now. This problem is similar to the following problem, but in the “other
direction”. Halogens have an s2p5 configuration. That means there is still a spot left for an
PS10-7
Answer Key, Problem Set 10
electron in the p sublevel of the valence shell. That coupled with the fact that halogens have
high effective nuclear charge values means that adding an electron will be fairly exothermic
(“favorable”).
26. NT11. 8.92.
Potassium is a highly reactive metal while argon is an inert gas. Explain the difference based on their
electron configurations.
Answer:
K: 1s2 2s2 2p6 3s2 3p6 4s1
Ar: 1s2 2s2 2p6 3s2 3p6
K has one electron in the valence (4th) energy level, and the other 18 are in the core, shielding
that outer one from the nuclear charge (+19). Thus Zeff is small (+1) and the force of attraction
for it to the nucleus is small. That means it is not that hard to remove it (small IE1). That makes
K a very good reducing agent (gives electrons away readily), and makes the element “reactive”.
NOTE: It is still endothermic to remove the electron even from an alkali metal. However, the cations and anions that
result from a redox reaction will come together to lower energy [forming ionic “bonds”]. If the ionization energy of
the metal is low, that tends to make the overall process more exothermic or “energy lowering”. See Figure 9.4, step
3, in Chapter 9.
On the other hand, Ar has 8 electrons in its valence (3rd) energy level, with only 10 electrons in
the core shielding the outer ones from the nuclear charge (+18). Thus Zeff is quite large, making
it difficult (takes a lot of energy) to give electrons away. That makes Ar a very poor reducing
agent. On the other hand, with such a large Zeff you might think Ar would be a good oxidizing
agent, like the halogens. However, it is not because it does not have any room left in its 3p
sublevel for an electron, so an added electron would go into the next higher energy orbital, the
4s, and would be shielded much more from the nucleus (see ____).
27. 8.86.
Based on the ionization energies of the alkali metals, which alkali metal would you expect to undergo the most
exothermic reaction with chlorine gas? Write a balanced chemical equation for the reaction.
Answer: Cs (Cesium) [I wouldn’t mark it wrong if you said Fr, Francium, although that element is radioactive with a fairly
short half life, so there is never any significant amount of it on Earth. I did not find its ionization energy in the text so
I assumed the authors intended the answer to be Cs.]
Reasoning: Cs has the lowest (first) ionization energy (376 kJ/mol; Figure 8.16), and so one would
expect it to contribute the least amount to the positive “part” of the overall H than other alkali metals
(with larger (positive) ionization energies). (Actually, the overall reaction can be thought of as comprising 5 distinct
steps (again, see Fig. 9.4 in Chapter 9 to see all of these), but the key one involving the metal with a positive H is the
ionization energy). The balanced equation for the reaction is:
2 Cs(s) + Cl2(g) → 2 CsCl(s)
Lattice Energy (and Prediction of Relative Magnitudes of)
28. MP. No answer in key for this problem, but see notes or text for the definition of “lattic energy” and
then make sure you know how to “translate” that into a chemical equation representing a
process.
29. NT12. 9.46.
Arrange these compounds in order of increasing magnitude of lattice energy: KCl, SrO, RbBr, CaO.
Answer:
smallest magnitude
(easiest to
separate)
RbBr < KCl < SrO < CaO
largest magnitude:
(hardest to
separate)
Strategy / Reasoning:
PS10-8
Answer Key, Problem Set 10
1) By Coulomb’s Law, the strongest forces of attraction will act between the ions that are a) of
highest charge and b) closest to one another (center-to-center distance).
2) Recognize that the distance between the (centers of the) cation an anion is the sum of the two
ions’ radii.
3) So, determine the charges of the ions and the relative radii of each ion (as best you can) using
periodic trend / shell model ideas. Recall that:
a) more positively charged ions are smaller than less positively charged ones (all other things
being equal),
b) more negatively charged ions are larger than less negatively charged ones (other things
being equal), and
c) ions in the same family are larger the lower they are in the column (because outer electrons
are in a greater “n” level).
Execution of Strategy
largest radius
Rb+ > K+ ~ Sr2+ > Ca2+
smallest radius
(It’s hard to predict if K+ is larger or smaller than
Sr2+)
largest radius
Br- > Cl- ~ O2-
(It’s hard to predict if O2- is larger or smaller than Cl-)
O2- is likely smaller than Br- despite its greater charge because its valence shell is n = 2 (vs. n = 4 for Br-)
but it is hard to say for sure that it would be smaller than Cl-.
So,
1) RbBr is clearly the easiest to separate (largest radii and both charges are “1” [smaller])
2) Between CaO and SrO, the only difference is the cation, and Ca2+ is smaller than Sr2+ with the
same charge, so clearly SrO should be easier to separate than CaO.
3) With K+ and Sr2+ and Cl- and O2- having somewhat comparable radii, the charges are what
dominate the last two compounds: SrO has both ions with charges of “2” and the ions in KCl
both have charges of “1”, and so it clearly SrO should be harder to separate.
Thus: RbBr < KCl < SrO < CaO is the ordering.
Covalent Bonding and Lewis Dot Structures (I)
30. NT13. (a) Why don’t atoms of two nonmetal elements combine
chemically to form an ionic compound? (b) What is a covalent
bond, and how does this “concept” fit in with your answer to part
(a) [that is, why do atoms of two nonmetals combine to form
molecular compounds but they don’t form ionic compounds?]
O
Cl
K Ca
Rb Sr
(a) The one sentence answer: It won’t occur because it
would take too much energy to remove an electron from
either nonmetal to form the cation.
More detailed answer. In order for an ionic compound to form,
atoms from one element have to give up (transfer) at least one
electron “completely” to atoms of the other element. That is, one element’s atoms become cations,
and the other element’s atoms become anions. This will only occur if the energy “price” isn’t too high
(since energy will be “recovered” (lowered) when the ions come together after the transfer). It does
not require that much energy to remove electrons from atoms of metals because low Zeff’s lead to
the outer electrons in metals not being held that tightly, but it takes a LOT of energy to remove
PS10-9
Br
Answer Key, Problem Set 10
electrons from atoms of nonmetals (much higher Zeff). So ionic compounds form when nonmetals
meet with metals, but not when nonmetals meet with one another.
(b) A covalent bond is a lowered state of energy that results from the sharing of electrons between two
atoms. This relates to the answer in (a) as follows: when atoms of two nonmetals come close,
sharing electrons is a way to lower energy without having to completely remove an electron from
either atom. Both atoms get to “keep” their electron part of the time and actually get to “gain” the
other atom’s electron part of the time. Since the electrons from BOTH atoms each get to see two
nuclei instead of one (i.e., more positive charge), this represents a lower energy state than when the
atoms are apart. This is why it takes energy to pull the atoms apart if they are “bonded”—breaking a
bond means pulling each electron in the bond away from one of the nuclei to which it is attracted.
Note that the (potential) energy of the system is only lowered if the outer electrons of each atom
get to “see” positive charge from the other atom’s nucleus. That means that covalent bonds only
form when there is a “spot left” for an electron in the other atom’s outer energy level.
31. NT14. 9.52(a-c).
(a) NF3
Write a Lewis structure for each molecule.
v = 5 + 3(7) = 26 e-
F
N
(b) HBr
F
v = 1 + 7 = 8 e-
H—Br
F
(c) SBr2
v = 6 + 2(7) = 20 e-
Br
S
Br
These are all examples of ones in which after “Step 4”, you end up with 8 electrons around the
central atom and you need not do anything further.
32. NT15. 9.63(a-c) Just find ONE reasonable LDS and do not worry about resonance structures or formal charge)
(a) SeO2
v = 6 + 2(6) = 18 e-
O
Se
O
(b) CO32-
O
v = 4 + 3(6) + 2 = 24 e-
C
O
(The double bond could be on the other side as well)
2-
O
(c) ClOv = 7 + 6 + 1 = 14 e-
Cl—O
(The double bond could
be between either of
the other two C-O
b d )

33. NT16. 9.74(ab).
Write Lewis structures for each molecule or ion. (Use expanded octets as necessary). ← This
should not need to be stated!
(a) ClF5
v = 7 + 5(7) = 42 e-
F
F
(b) AsF6-
v = 5 + 6(7) +1 = 48 e-
F
F
F
F
A
Cl
F
F
F
F
F
PS10-10

Answer Key, Problem Set 10
34. NT17. 9.94.
Use Lewis structures to explain why Br3- and I3- are stable, while F3- is not.
Brief Answer: F is in the second row, and thus its valence shell can only hold 8 electrons. Br and I,
being in the 4th and 5th rows, have room for more than 8 electrons in their valence shell since they have
d orbitals available. The X3- LDS requires 10 electrons around the central X atom (see structure below;
this is not “obvious”!), so F3- isn’t predicted to be stable even though Br3- and I3- can be.
Full Answer: As can be seen below, these isoelectronic structures (in terms of valence electrons) each have
three atoms and 3(7) + 1 = 20 valence electrons. As such, the center in each ion ends up having 10
electrons around it (see below). This is “acceptable” for a Br or I atom, since these atoms have unfilled
d orbitals in their valence levels that can be used for bonding (Br’s valence electron configuration is
4s24p6, and the 4d orbitals are unfilled; I’s valence electron configuration is 5s25p6, and the 5d orbitals
are unfilled). However, it is considered “unacceptable” (energetically unfavorable) for F since it does not
have any more orbitals in its valence level (F’s valence electron configuration is 2s22p6, and there are no
d orbitals in the 2nd energy level!). Thus, one would not predict F3- to be stable, which is consistent with
the fact that it has never been observed.
v = 3(7) +1 = 22
e-
F
F
F
The F in the center of this structure
would have to have more than 8
electrons in its valence shell, which is not
“allowed” since the maximum is 8 for the
second energy level (no d orbitals).
========================A couple of extra problems for practice (left over from past)=============
8.64. Arrange these elements in order of decreasing atomic
radius: Cs, Sb, S, Pb, Se.
Answer:
Largest radius: Cs > Pb > Sb > Se > S
S
Se
Sb
Pb
Reasoning: Radii generally decrease as you go to the right Cs
and up in the periodic table. See video in Item 1
in Mastering to see how to “trace” the path from
one element to the other (sometimes in two steps) to
apply these “trends” to make your prediction. Please
note that (unlike what Tro says) using a trend to
make a prediction is not an “explanation” for “why” one atom is larger than another. (e.g.,
Do not say “Cs is bigger than Pb because it is farther to the left but in the same row on
the Periodic Table”. That is not an explanation of “why” it is bigger, it is just how you can
make the prediction that it is bigger (without explaining “why” it is so). To explain why Cs
is bigger, you must use “shell model” ideas along with Coulomb’s Law (see NT2 and
NT3 above).
8.74.
Choose the element with the higher first ionization energy from each pair.
(a) P or I
Cannot tell just from position in Periodic Table (I is farther right, but also lower)
(b) Si or Cl
Cl (farther to the right in the same row)
(c) P or Sb
P (same column, but higher in table)
(d) Ga or Ge
Ge (farther to the right in the same row)
(e) O or Ne
Ne (farther to the right in the same row)
PS10-11
Answer Key, Problem Set 10
Reasoning: Ionization energy generally increases across a row and up a column. [Although this
question does not ask you to explain the trends, I will do so briefly in this key for completeness. The
former is due to the increasing effective nuclear charge (and thus stronger force of attraction) across
a row, and the latter is due to the decreased distance of the valence shell to the nucleus (and thus
stronger force of attraction) up a column.
PS10-12