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CHAPTER TWENTY
THE REPRESENTATIVE ELEMENTS:
GROUPS 5A THROUGH 8A
For Review
1.
Group 5A: ns2np3; As with groups IIIA and IVA, metallic character increases going down a
group. Nitrogen is strictly a nonmetal in properties, while bismuth, the heaviest Group 5A
element, has mostly metallic physical properties. The trend of increasing metallic character
going down the group is due in part to the decrease in electronegativity. Nitrogen, with its
high electronegativity, forms covalent compounds as nonmetals do. Bismuth and antimony,
with much lower electronegativities, exhibit ionic character in most of their compounds.
Bismuth and antimony exist as +3 metal cations in these ionic compounds.
NH3, 5 + 3(1) = 8 e
AsCl5, 5 + 5(7) = 40 e
Cl
N
H
H
Cl
H
As
Cl
Cl
Cl
trigonal pyramid; sp3
trigonal bipyramid; dsp3
PF6, 5 + 6(7) + 1 = 48 e
F
F
P
F
F
F
F
Octahedral; d2sp3
Nitrogen does not have low energy d orbitals it can use to expand its octet. Both NF5 and
NCl6 would require nitrogen to have more than 8 valence electrons around it; this never
happens.
2.
N: 1s22s22p5; The extremes of the oxidation states for N can be rationalized by examining
the electron configuration of N. Nitrogen is three electrons short of the stable Ne electron
configuration of 1s22s22p6. Having an oxidation state of 3 makes sense. The +5 oxidation
state corresponds to N “losing” its 5 valence electrons. In compounds with oxygen, the NO
bonds are polar covalent, with N having the partial positive end of the bond dipole. In the
751
752
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
world of oxidation states, electrons in polar covalent bonds are assigned to the more
electronegative atom; this is oxygen in NO bonds. N can form enough bonds to oxygen to
give it a +5 oxidation state. This loosely corresponds to losing all of the valence electrons.
NH3: fertilizers, weak base properties, can form hydrogen bonds; N2H4: rocket propellant,
blowing agent in manufacture of plastics, can form hydrogen bonds; NH2OH: weak base
properties, can form hydrogen bonds; N2: makes up 78% of air, very stable compound with a
very strong triple bond, is inert chemically; N2O: laughing gas, propellant in aerosol cans,
effect on earth’s temperature being studied; NO: toxic when inhaled, may play a role in
regulating blood pressure and blood clotting, one of the few odd electron species that forms;
N2O3, least common of nitrogen oxides, a blue liquid that readily dissociates into NO(g) and
NO2(g); NO2: another odd electron species, dimerizes to form N2O4, plays a role in smog
production; HNO3: important industrial chemical, used to form nitrogen-based explosives,
strong acid and a very strong oxidizing agent.
3.
Hydrazine also can hydrogen bond because it has covalent NH bonds as well as having a
lone pair of electrons on each N. The high boiling point for hydrazine’s relatively small size
supports this.
N2(g) + 3 H2(g) ⇌ 2 NH3(g) + heat
a. This reaction is exothermic, so an increase in temperature will decrease the value of K
(see Table 13.3 of text.) This has the effect of lowering the amount of NH3(g) produced at
equilibrium. The temperature increase, therefore, must be for kinetics reasons. When the
temperature increases, the reaction reaches equilibrium much faster. At low temperatures,
this reaction is very slow, too slow to be of any use.
b. As NH3(g) is removed, the reaction shifts right to produce more NH3(g).
c. A catalyst has no effect on the equilibrium position. The purpose of a catalyst is to speed
up a reaction so it reaches equilibrium quicker.
d. When the pressure of reactants and products is high, the reaction shifts to the side that has
fewer gas molecules. Since the product side contains 2 molecules of gas compared to 4
molecules of gas on the reactant side, the reaction shifts right to products at high
pressures of reactants and products. Also, a high pressure indicates that reactants are
present in large quanties. The more reactants present, the further right the reaction shifts.
The pollution provides nitrogen and phosphorous nutrients so the algae can grow. The algae
consume oxygen, causing fish to die.
4.
White phosphorus consists of discrete tetrahedral P4 molecules. The bond angles in the P4
tetrahedrons are only 60°, which makes P4 very reactive, especially towards oxygen. Red and
black phosphorus are covalent network solids. In red phosphorus, the P4 tetrahedra are
bonded to each other in chains, making them less reactive than white phosphorus. They need
a source of energy to react with oxygen, such as when one strikes a match. Black phosphorus
is crystalline, with the P atoms tightly bonded to each other in the crystal, and is fairly
unreactive towards oxygen.
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
753
Even though phosphine and ammonia have identical Lewis structures, the bond angles of PH3
are only 94, well below the predicted tetrahedral bond angles of 109.5. PH3 is an unusual
exception to the VSEPR model.
The acidic hydrogens in the oxyacids of phosporus all are bonded to oxygen. The hydrogens
bonded directly to phosphorus are not acidic. H3PO4 has three oxygen bonded hydrogens, and
it is a triprotic acid. H3PO3 has only two of the hydrogens bonded to oxygen and it is a
diprotic acid. The third oxyacid of phosphorus, H3PO2, has only one of the hydrogens bonded
to an oxygen; it is a monoprotic acid.
5.
Group 6A: ns2np4; As expected from the trend in other groups, oxygen has properties which
are purely nonmetal. Polonium, on the other hand, has some metallic properties. The most
significant property differences are radioactivity and toxicity. Polonium is only composed of
radioactive isotopes, unlike oxygen, and polonium is highly toxic, unlike oxygen.
The two allotropic forms of oxygen are O2 and O3.
O2, 2(6) = 12 e
O
O3, 3(6) = 18 e
O
O
O
O
O
O
O
The MO electron configuration of O2 has two unpaired electrons in the degenerate pi
antibonding ( *2 p ) orbitals. A substance with unpaired electrons is paramagnetic (see Figure
9.40). Ozone has a V-shape molecular structure with bond angles of 117, slightly less than
the predicted 120 trigonal planar bond angle.
In the upper atmosphere, O3 acts as a filter for UV radiation:
O3
hv
O2 + O
O3 is also a powerful oxidizing agent. It irritates the lungs and eyes, and, at high concentration, it is toxic. The smell of a "fresh spring day" is O3 formed during lightning discharges.
Toxic materials don't necessarily smell bad. For example, HCN smells like almonds.
6.
Both rhombic and monoclinic sulfur exist in S8 rings. The difference between the two is that
the S8 rings are stacked together differently giving different solid structures.
Oxygen forms strong  bonds and, because of this, exists in nature as O2 molecules. Sulfur
forms much stronger sigma bonds than  bonds. Therefore, elemental sulfur is found in
nature singly bonded to other sulfur atoms. We assume SO doesn’t form because of the
difference in ability of oxygen and sulfur to form  bonds. Sulfur forms relatively weak 
bonds as compared to oxygen.
SO2(aq) + H2O(l) → H2SO3(aq); SO3(aq) + H2O(l) → H2SO4(aq); SO2 and SO3 dissolve in
water to form the acids H2SO3 and H2SO4, respectively. Figures 20.18 and 20.19 show the
Lewis structures for SO2 and SO3. The molecular structure of SO2 is bent with a 119 bond
angle (close to the predicted 120 trigonal planar geometry). The molecular structure of SO3
754
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
is trigonal planar with 120 bond angles. Both SO2 and SO3 exhibit resonance. Both sulfurs
in SO2 and SO3 are sp2 hybridized. To explain the equal bond lengths that occur in SO2 and
SO3, the molecular orbital model assumes that the  electrons are delocalized over the entire
surface of the molecule. The orbitals that form the delocalized  bonding system are
unhybridized p atomic orbitals from the sulfurs and oxygens in each molecule. When all of
the p atomic orbitals overlap together, there is a cloud of electron density above and below
the entire surface of the molecule. Because the  electrons are delocalized over the entire
surface of the molecule in SO2 and SO3, all of the SO bonds in each molecule are
equivalent.
A dehydrating agent is one that has a high affinity for water. Sulfuric acid grabs water
whenever it can. When it reacts with sugar (C12H22O11) it removes the hydrogen and oxygen
in a 2:1 ratio even though there are no H2O molecules in sugar. H2SO4 is indeed a powerful
dehydrating agent.
7.
Group 7A: ns2np5; The diatomic halogens (X2) are nonpolar, so they only exibit London
dispersion intermolecular forces. The strength of LD forces increases with size. The boiling
points and melting points steadily increase from F2 to I2 because the strength of the
intermolecular forces are increasing.
Fluorine is the most reactive of the halogens because it is the most electronegative atom and
the bond in the F2 molecule is very weak.
One reason is that the H ‒ F bond is stronger than the other hydrohalides, making it more
difficult to form H+ and F. The main reason HF is a weak acid is entropy. When F (aq)
forms from the dissociation of HF, there is a high degree of ordering that takes place as water
molecules hydrate this small ion. Entropy is considerably more unfavorable for the formation
of hydrated F than for the formation of the other hydrated halides. The result of the more
unfavorable ΔS° term is a positive ΔG° value, which leads to a Ka value less than one.
HF exhibits the relatively strong hydrogen bonding intermolecular forces, unlike the other
hydrogen halides. HF has a high boiling point due to its ability to form these hydrogen
bonding interactions.
The halide ion is the 1 charged ion that halogens form when in ionic compounds. As can be
seen from the positive standard reduction potentials in Table 20.6, the halogens energetically
favor the X form over the X2 form. Because the reduction potentials are so large, this give an
indication of the relative ease to which halogens will grab electrons to form the halide ion. In
general, the halogens are highly reactive; that is why halogens exist as cations in various
minerals and in seawater as opposed to free elements in nature.
Some compounds of chlorine exhibiting the 1 to +7 oxidation state are: HCl(1), HOCl
(+1), HClO2 (+3), HClO3 (+5), and HClO4 (+7). Note that these are all acids. HCl is a strong
acid, and of the oxyacids, only HClO4 is a strong acid. The oxyacid strength increases as the
number of oxygens in the formula increase. Therefore, the order of the oxyacids from
weakest to strongest acid is HOCl < HClO2 < HClO3 < HClO4.
8.
Most of the compounds in Table 20.11 have the following molecular structures, bond angles,
and hybridization. Examples found in Table 20.11 are listed for each.
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
755
trigonal planar: 120, sp2, e.g., BX3
V-shape: < 109.5, sp3, e.g., OF2, OCl2, OBr2, SF2, SCl2, SeCl2
trigonal pyramid: < 109.5, sp3, e.g., NX3, PX3, AsF3, SbF3
tetrahedral: 109.5, sp3, e.g., BF4, CX4, SiF4, SiCl4, GeF4, GeCl4
T-shape: 90, dsp3, e.g., ClF3, BrF3, ICl3, IF3
see-saw: 90 and ~120, dsp3, e.g., SF4, SCl4, SeF4, SeCl4, SeBr4, TeBr4, TeCl4, TeBr4, TeI4
trigonal bipyramid: 90 and 120, dsp3, e.g., PF5, PCl5, PBr5, AsF5, SbF5
square pyramid: 90, d2sp3, e.g., ClF5, BrF5, IF5
octahedral: 90, d2sp3, e.g., SiF62, GeF62, SF6, SeF6, TeF6
ICl, IBr, BrF, BrCl, and ClF have no molecular structure or bond angles. The predicted
hybridization for each halogen is sp3. N2F4 is trigonal pyramid about both nitrogens, with <
109.5 bond angles and sp3 hybridization. O2F2, S2Cl2, S2F2, and S2Cl2 is V-shape about both
central oxygens or sulfurs with < 109.5 bond angles and sp3 hybridization.
Some of the compounds in Table 20.11 are exceptions to the octet rule, like ICl 3. The row
three halogens (Cl) and heavier (Br and I) have low lying empty d-orbitals available to
expand their octet when they have to. Fluorine, with its valence electrons in the n = 2 level,
does not have low energy d-orbitals available to expand its octet. When F is the central atom,
its compounds always obeys the octet rule.
9.
The noble gases have filled s and p valence orbitals (ns2np6 = valence electron configuration). They don’t need to react like other representative elements in order to achieve the
stable ns2np6 configuration. Noble gases are unreactive because they do not want to lose their
stable valence electron configuration.
Noble gases exist as free atoms in nature. They only exhibit London dispersion forces in the
condensed phases. Because LD forces increase with size, as the noble gas gets bigger, the
strength of the intermolecular forces get stronger leading to higher melting and boiling points.
Helium is unreactive and doesn't combine with any other elements. It is a very light gas and
would easily escape the earth's gravitational pull as the planet was formed.
In Mendeleev's time, none of the noble gases were known. Since an entire family was
missing, no gaps seemed to appear in the periodic arrangement. Mendeleev had no evidence
to predict the existence of such a family. The heavier members of the noble gases are not
really inert. Xe and Kr have been shown to react and form compounds with other elements.
10.
XeF2: 180, dsp3; XeO2F2: ~90 and ~120, dsp3; XeO3: < 109.5, sp3; XeO4: 109.5,
sp3; XeF4: 90, d2sp3; XeO3F2: 90 and 120, dsp3; XeO2F4, 90, d2sp3
Questions
1.
This is due to nitrogen’s ability to form strong  bonds whereas heavier group 5A elements
do not form strong  bonds. Therefore, P2, As2, and Sb2 do not form since two  bonds are
required to form these diatomic substances.
756
2.
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
Nitrogen fixation is the process of transforming N2 to other nitrogen-containing compounds.
Some examples are:
N2(g) + 3 H2(g) → 2 NH3(g)
N2(g) + O2(g) → 2 NO(g)
N2(g) + 2 O2(g) → 2 NO2(g)
3.
There are medical studies that have shown an inverse relationship between the incidence of
cancer and the selenium levels in soil. The foods grown in these soils and eventually digested
are assumed to somehow furnish protection from cancer. Selenium is also involved in the
activity of vitamin E and certain enzymes in the human body. In addition, selenium deficiency has been shown to be connected to the occurrence of congestive heart failure.
4.
Chlorine is a good oxidizing agent. Similarly, ozone is a good oxidizing agent. After chlorine
reacts, residues of chloro compounds are left behind. Long term exposure to some chloro
compounds may cause cancer. Ozone would not break down and form harmful substances.
The major problem with ozone is that because virtually no ozone is left behind after initial
treatment, the water supply is not protected against recontamination. In contrast, for
chlorination, significant residual chlorine remains after treatment, thus reducing (eliminating)
the risk of recontamination.
5.
+6 oxidation state: SO42, SO3, SF6
+4 oxidation state: SO32, SO2, SF4
+2 oxidation state: SCl2
0 oxidation state: S8 and all other elemental forms of sulfur
2 oxidation state: H2S, Na2S
6.
sp3 hybridization: HBr, 1 + 7 = 8 e
H
IBr, 7 + 7 = 14 e
Br
I
Br
Diatomic molecules don’t have a molecular structure because they have no bond angles.
dsp3 hybridization: BrF3, 7 + 3(7) = 28 e
F
F
T-shaped
Br
F
d2sp3 hybridization: BrF5, 7 + 5(7) = 42 e
F
F
F
Br
F
square pyramid
F
CHAPTER 20
7.
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
757
a. H2(g) + Cl2(g) → 2 HCl(g); this reaction produces a lot of energy which can be used in
a cannon apparatus to send a stopper across the room. To initiate this extremely slow
reaction, light of specific wavelengths is needed. This is the purpose of lighting the
magnesium strip. When magnesium is oxidized to MgO, an intense white light is produced. Some of the wavelengths of this light can break ClCl bonds and get the reaction
started.
b. Br2 is brown. The disappearance of the brown color indicates that all of the Br2 has
reacted with the alkene (no free Br2 is remaining).
c. 2 Al(s) + 3 I2(s) → 2 AlI3(s); This is a highly exothermic reaction, hence the sparks that
accompany this reaction. The purple smoke is excess I2(s) being vaporized [the purple
smoke is I2(g)].
8.
One would expect RnF2 and RnF4 to form in fashion similar to XeF2 and XeF4. The chemistry
of radon is difficult to study because radon isotopes are all radioactive. The hazards of
dealing with radioactive materials are immense.
Exercises
Group 5A Elements
9.
NO43
3-
O
N
O
O
O
Both NO43 and PO43 have 32 valence electrons, so both
have similar Lewis structures. From the Lewis structure
for NO43, the central N atom has a tetrahedral
arrangement of electron pairs. N is small. There is
probably not enough room for all 4 oxygen atoms around
N. P is larger, thus, PO43 is stable.
PO3
O
P
O
10.
a. PF5;
O
PO3 and NO3 each have 24 valence electrons so both
have similar Lewis structures. From the Lewis structure
for PO3, PO3 has a trigonal planar arrangement of
electron pairs about the central P atom (two single bonds
and one double bond). P=O bonds are not particularly
stable, while N=O bonds are stable. Thus, NO3 is stable.
N is too small and doesn't have low energy d-orbitals to expand its octet to form
NF5.
b. AsF5; I is too large to fit 5 atoms of I around As.
c. NF3; N is too small for three large bromine atoms to fit around it.
758
11.
CHAPTER 20
a. NO: %N =
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
14.01 g N
× 100 = 46.68% N
30.01 g NO
b. NO2: %N =
14.01 g N
× 100 = 30.45% N
46.01 g NO2
c. N2O4: %N =
28.02 g N
× 100 = 30.45% N
92.02 g N 2 O 4
d. N2O: %N =
28.02 g N
× 100 = 63.65% N
44.02 g N 2 O 4
The order from lowest to highest mass percentage of nitrogen is: NO2 = N2O4 < NO < N2O.
12.
1.0 × 106 kg HNO3 ×
1000 g HNO3
1 mol HNO3
= 1.6 × 107 mol HNO3

kg HNO3
63.02 g HNO3
We need to get the relationship between moles of HNO3 and moles of NH3. We have to use
all 3 equations.
2 mol HNO3
2 mol NO2
4 mol NO 16 mol HNO3



3 mol NO2
2 mol NO
4 mol NH3
24 mol NH3
Thus, we can produce 16 mol HNO3 for every 24 mol NH3 we begin with:
1.6 × 107 mol HNO3 ×
24 mol NH3
17.03 g NH3
= 4.1 × 108 g or 4.1 × 105 kg

16 mol HNO3
mol NH3
This is an oversimplified answer. In practice, the NO produced in the third step is recycled
back continuously into the process in the second step. If this is taken into consideration, then
the conversion factor between mol NH3 and mol HNO3 turns out to be 1:1, i.e., 1 mol of NH3
produces 1 mol of HNO3. Taking into consideration that NO is recycled back gives an answer
of 2.7 × 105 kg NH3 reacted.
13.
a. NH4NO3(s)
heat
N2O(g) + 2 H2O(g)
b. 2 N2O5(g) → 4 NO2(g) + O2(g)
c. 2 K3P(s) + 6 H2O(l) → 2 PH3(g) + 6 KOH(aq)
d. PBr3(l) + 3 H2O(l) → H3PO3(aq) + 3 HBr(aq)
e. 2 NH3(aq) + NaOCl(aq) → N2H4(aq) + NaCl(aq) + H2O(l)
CHAPTER 20
14.
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
759
4 As(s) + 3 O2(g) → As4O6(s); 4 As(s) + 5 O2(g) → As4O10(s)
As4O6(s) + 6 H2O(l) → 4 H3AsO3(aq); As4O10(s) + 6 H2O(l) → 4 H3AsO4(aq)
15.
Unbalanced equation:
CaF23Ca3(PO4)2(s) + H2SO4(aq) → H3PO4(aq) + HF(aq) + CaSO42H2O(s)
Balancing Ca2+, F-, and PO43:
CaF23Ca3(PO4)2(s) + H2SO4(aq) → 6 H3PO4(aq) + 2 HF(aq) + 10 CaSO42H2O(s)
On the right hand side, there are 20 extra hydrogen atoms, 10 extra sulfates, and 20 extra
water molecules. We can balance the hydrogen and sulfate with 10 sulfuric acid molecules.
The extra waters came from the water in the sulfuric acid solution. The balanced equation is:
CaF23Ca3(PO4)2(s) + 10 H2SO4(aq) + 20 H2O(l) →
6
H3PO4(aq)
+
2
HF(aq)
+
10
CaSO42H2O(s)
16.
a. NO2, 5 + 2(6) = 17 e-
N2O4, 2(5) + 4(6) = 34 eO
N
O
N
O
N
O
plus other resonance structures
b. BH3, 3 + 3(1) = 6 e
O
plus other resonance structures
NH3, 5 + 3(1) = 8 e
H
N
H
B
H
O
H
H
H
BF3NH3, 6 + 8 = 14 e
H
H
H
B
N
H
H
H
In reaction a, NO2 has an odd number of electrons, so it is impossible to satisfy the octet rule.
By dimerizing to form N2O4, the odd electron on two NO2 molecules can pair up, giving a
species whose Lewis structure can satisfy the octet rule. In general, odd electron species are
very reactive. In reaction b, BH3 is electron deficient. Boron has only six electrons around
760
17.
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
it. By forming BH3NH3, the boron atom satisfies the octet rule by accepting a lone pair of
electrons from NH3 to form a fourth bond.
2 NaN3(s) → 2 Na(s) + 3 N2(g)
PV
1.00 atm  70.0 L
= 3.12 mol N2 needed to fill air bag.

RT 0.08206 L atm  273 K
mol K
2 mol NaN3
mol NaN3 reacted = 3.12 mol N2 ×
= 2.08 mol NaN3
3 mol N 2
n N2 
18.
For ammonia (in one minute):
PV
90. atm  500 . L
= 1.1 × 103 mol NH3
n NH3 

0
.
08206
L atm
RT
 496 K
mol K
NH3 flows into the reactor at a rate of 1.1 × 103 mol/min.
For CO2 (in one minute):
n CO 2 
PV
45 atm  600 . L
= 6.6 × 102 mol CO2

0
.
08206
L
atm
RT
 496 K
mol K
CO2 flows into the reactor at 6.6 × 102 mol/min.
To react completely with 1.1 × 103 mol NH3/min, we need:
1.1  10 3 NH3
1 mol CO 2

= 5.5 × 102 mol CO2/min
min
2 mol NH3
Because 660 mol CO2/min are present, ammonia is the limiting reagent.
1.1  10 3 NH3
1 mol urea
60.06 g urea


= 3.3 × 104 g urea/min
min
2 mol NH3
mol urea
19.
H
H
N
H
(l)
N
+
O
O (g)
N
N (g) + 2 H
O
H
Bonds broken:
Bonds formed:
1 N‒N (160. kJ/mol)
1 N≡ N (941 kJ/mol)
4 N‒H (391 kJ/mol)
1 O=O (495 kJ/mol)
2 × 2 O‒H (467 kJ/mol)
H (g)
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
761
ΔH = 160. + 4(391) + 495  [941 + 4(467)] = 2219 kJ  2809 kJ = 590. kJ
20.
5 N2O4(l) + 4 N2H3CH3(l) → 12 H2O(g) + 9 N2(g) + 4 CO2(g)

  242 kJ 
  393 .5 kJ   
  20. kJ 
 54 kJ  
ΔH° = 12 mol
  4 mol
   5 mol
  4 mol

 mol 
 mol   
 mol 
 mol  

= 4594 kJ
Using bond energies, ΔH = 5.0 × 103 kJ (from Sample Exercise 20.2). When using bond
energies to calculate ΔH, the enthalpy change is assumed to be due only to the difference in
bond strength between reactants and products. Bond energies generally give a very good
estimate for ΔH for gas phase reactions. However, when solids and liquids are present, ΔH
estimates from bond energy differences are not as good. This is because the difference in the
strength of the intermolecular forces between reactants and products is not considered when
using bond energies. Here, the reactants are in the liquid phase. The loss in strength of the
intermolecular forces as the liquid reactants are converted to the gaseous products was not
considered when using bond energies in Sample Exercise 20.2; hence, the large difference
between the two calculated ΔH values.
21.
1/2 N2(g) + 1/2 O2(g) → NO(g) ΔG° = ΔG of , NO = 87 kJ/mol; By definition, ΔG of for a
compound equals the free energy change that would accompany the formation of 1 mol of
that compound from its elements in their standard states. NO (and some other oxides of
nitrogen) have weaker bonds as compared to the triple bond of N2 and the double bond of O2.
Because of this, NO (and some other oxides of nitrogen) have higher (positive) standard free
energies of formation as compared to the relatively stable N2 and O2 molecules.
22.
ΔH° = 2(90. kJ)  [0 + 0] = 180. kJ; ΔS° = 2(211 J/K)  [192 + 205] = 25 J/K
ΔG° = 2(87 kJ)  [0] = 174 kJ
At the high temperatures in automobile engines, the reaction N2 + O2 → 2 NO becomes
spontaneous since the favorable ΔS° term will become dominate. In the atmosphere, even
though 2 NO → N2 + O2 is spontaneous at the cooler temperatures of the atmosphere, it
doesn't occur because the rate is slow. Therefore, higher concentrations of NO are present in
the atmosphere as compared to what is predicted by thermodynamics.
23.
MO model:
NO+: (σ2s)2(σ2s*)2(π2p)4(σ2p)2, Bond order = (8  2)/2 = 3, 0 unpaired e (diamagnetic)
NO: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)1, B.O. = 2.5, 1 unpaired e (paramagnetic)
NO-: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2, B.O. = 2, 2 unpaired e (paramagnetic)
762
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
N
Lewis structures: NO+:
NO:
N
NO:
O
+
N
O
N
O
N
O
O
The two models give the same results only for NO+ (a triple bond with no unpaired
electrons). Lewis structures are not adequate for NO and NO. The MO model gives a better
representation for all three species. For NO, Lewis structures are poor for odd electron
species. For NO, both models predict a double bond, but only the MO model correctly
predicts that NO is paramagnetic.
24.
For NCl3 → NCl2 + Cl, only the N‒Cl bond is broken. For O=N‒Cl → NO + Cl, the NO
bond gets stronger (bond order increases from 2.0 to 2.5) when the N‒Cl bond is broken. This
makes ΔH for the reaction smaller than just the energy necessary to break the N‒Cl bond.
25.
a. H3PO4 > H3PO3; The strongest acid has the most oxygen atoms.
b. H3PO4 > H2PO4 > HPO42; Acid strength decreases as protons are removed.
26.
TSP = Na3PO4; PO43 is the conjugate base of the weak acid HPO42 (Ka = 4.8 × 10 13 ). All
conjugate bases of weak acids are effective bases (Kb = Kw/Ka = 1.0 × 10 14 /4.8 × 10 13 =
2.1 × 10 2 ). The weak base reaction of PO43 with H2O is: PO43 + H2O ⇌ HPO42 + OH
Kb = 2.1 × 10 2 .
27.
The acidic protons are attached to oxygen.
H4P2O6 (50 valence e):
O
H
P
H4P2O5 (44 valence e):
O
O
O
P
O
O
H
H
O
H
H
O
P
H
O
O
P
H
O
H
CHAPTER 20
28.
a.
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
SbF5, 40 valence e
F
Sb
F
F
F
F
dsp3
O
O
S
S
O
O
sp3
F
Sb
F
F
H
F5SbOSO2FH, 72 valence e
F
H2SO3F+, 32 valence e
HSO3F, 32 valence e-
F
F
O
F
O
H
S
F
763
+
H
O
O
H
sp3
F5SbOSO2F-, 72 valence e
F
F
Sb
F
O
Sb: d2sp3; S: sp3
F
O
F
O
S
F
O
Sb: d2sp3; S: sp3
b. The active protonating species is H2SO3F+, the species with two OH bonds.
Group 6A Elements
29.
O=O‒O → O=O + O
Break O‒O bond: ΔH =
146 kJ
1 mol
= 2.42 × 10 22 kJ = 2.42 × 10 19 J

23
mol
6.022  10
A photon of light must contain at least 2.42 × 10 19 J to break one O‒O bond.
Ephoton =
hc
(6.626  10 34 J s) (2.998  10 8 m / s)
, λ 
= 8.21 × 10 7 m = 821 nm
λ
2.42  10 19 J
30.
From Figure 7.2 in the text, light from violet to green will work.
31.
a. 2 SO2(g) + O2(g) → 2 SO3(g)
b. SO3(g) + H2O(l) → H2SO4(aq)
c. 2 Na2S2O3(aq) + I2(aq) → Na2S4O6(aq) + 2 NaI(aq)
d. Cu(s) + 2 H2SO4(aq) → CuSO4(aq) + 2 H2O(l) + SO2(aq)
32.
H2SeO4(aq) + 3 SO2(g) → Se(s) + 3 SO3(g) + H2O(l)
764
33.
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
a. SO32, 6 + 3(6) + 2 = 26 e
b. O3, 3(6) = 18 e
2S
O
O
O
O
O
trigonal pyramid; ≈ 109.5°; sp3
c. SCl2, 6 + 2(7) = 20 e
O
O
O
V-shaped; ≈ 120°; sp2
d. SeBr4, 6 + 4(7) = 34 e
S
Cl
O
Br
a
Br
Cl
V-shaped; ≈ 109.5°; sp3
b Br
Se
b Br
see-saw; a ≈ 120°, b ≈ 90°; dsp3
e. TeF6, 6 + 6(7) = 48 e
F
F
F
Te
F
F
F
octahedral; 90°; d2sp3
34.
35.
S2N2 has 2(6) + 2(5) = 22 valence electrons.
S
N
S
N
S
N
S
N
N
S
N
S
N
S
N
S
1.50 g BaO2 ×
25.0 mL ×
1 mol BaO2
= 8.86 × 10 3 mol BaO2
169 .3 g BaO2
0.0272 g HCl 1 mol HCl
= 1.87 × 10 2 mol HCl

mL
36.46 HCl
The required mole ratio from the balanced reaction is 2 mol HCl to 1 mol BaO2. The actual
ratio is:
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
1.87  10 2 mol HCl
8.86  10 3 mol BaO2
765
= 2.11
Because the actual mole ratio is larger than the required mole ratio, the denominator (BaO2)
is the limiting reagent.
8.86 × 10 3 mol BaO2 ×
1 mol H 2O2 34.02 g H 2O 2
= 0.301 g H2O2

mol BaO2
mol H 2O2
The amount of HCl reacted is:
8.86 × 10 3 mol BaO2 ×
2 mol HCl
mol BaO2
= 1.77 × 10 2 mol HCl
excess mol HCl = 1.87 × 10 2 mol  1.77 × 10 2 mol = 1.0 × 10 3 mol HCl
mass of excess HCl = 1.0 × 10 3 mol HCl ×
36.
1.00 L 
36.46 g HCl
= 3.6 × 10 2 g HCl
mol HCl
0.200 mol Na 2S2O3
1 mol AgBr
187.8 mol AgBr
= 18.8 g AgBr


L
2 mol Na 2S2O3
mol AgBr
Group 7A Elements
37.
O2F2 has 2(6) + 2(7) = 26 valence e; From the following Lewis structure, each oxygen atom
has a tetrahedral arrangement of electron pairs. Therefore, bond angles ≈ 109.5° and each O
is sp3 hybridized.
F
O
O
F
Formal Charge
0
0
0
0
Oxid. Number
-1
+1
+1
-1
Oxidation numbers are more useful. We are forced to assign +1 as the oxidation number for
oxygen. Oxygen is very electronegative, and +1 is not a stable oxidation state for this
element.
38.
a. CCl2F2, 4 + 2(7) + 2(7) = 32 e
F
b. HClO4, 1 + 7 + 4(6) = 32 e
F
O
C
Cl
Cl
Cl
O
O
H
O
766
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
tetrahedral; 109.5°; sp3
c. ICl3, 7 + 3(7) = 28 e
About Cl: tetrahedral; 109.5°; sp3
About O: V-shaped; ≈ 109.5°; sp3
d. BrF5, 7 + 5(7) = 42 e
F
Cl
I
F
Cl
Cl
T-shaped; ≈ 90°; dsp3
39.
F
Br
F
F
square pyramid; ≈ 90°; d2sp3
a. BaCl2(s) + H2SO4(aq) → BaSO4(s) + 2 HCl(g)
b. BrF(s) + H2O(l) → HF(aq) + HOBr(aq)
c. SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l)
40.
a. F2 + H2O → HOF + HF; 2 HOF → 2 HF + O2; 3 HOF + H2O → 3 HF + H2O2 + O2;
HOF + H2O → HF + H2O2 (dilute acid)
In dilute base, HOF exists as OF and HF exists as F. The balanced reaction is:
(2e + H2O + OF → F + 2 OH) × 2
4 OH → O2 + 2 H2O + 4e
________________________________
2 OF → O2 + 2 F
b. HOF: Assign +1 to H and 1 to F. The oxidation state of oxygen is then zero. Oxygen is
very electronegative. A zero oxidation state is not very stable since oxygen is a very
good oxidizing agent.
41.
ClO + H2O + 2 e → 2 OH + Cl
E° = 0.90 V
2 NH3 + 2 OH → N2H4 + 2 H2O + 2 e
E° = 0.10 V
_________________________________________________________
ClO(aq) + 2 NH3(aq) → Cl(aq) + N2H4(aq) + H2O(l) E ocell = 1.00 V
Because E ocell is positive for this reaction, ClO , at standard conditions, can spontaneously
oxidize NH3 to the somewhat toxic N2H4.
42.
A disproportion reaction is an oxidation-reduction reaction in which one species will act as
both the oxidizing agent and reducing agent. The species reacts with itself , forming products
with higher and lower oxidation states. For example, 2 Cu+ → Cu + Cu2+ is a disproportion
reaction.
HClO2 will disproportionate at standard conditions because E ocell > 0:
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
767
HClO2 + 2 H+ + 2 e → HClO + H2O
E° = 1.65 V

+

HClO2 + H2O → ClO3 + 3 H + 2 e
E° = 1.21 V
_________________________________________________________________
2 HClO2(aq) → HClO(aq) + ClO3(aq) + H+(aq)
E ocell = 0.44 V
Group 8A Elements
43.
Xe has one more valence electron than I. Thus, the isoelectric species will have I plus one
extra electron substituted for Xe, giving a species with a net minus one charge.
a. IO4
44.
b. IO3
c. IF2
d. IF4
a. KrF2, 8 + 2(7) = 22 e
F
Kr
e. IF6
b. KrF4, 8 + 4(7) = 36 e
F
F
F
Kr
F
linear; 180°; dsp3
F
square planar; 90°; d2sp3
c. XeO2F2, 8 + 2(6) + 2(7) = 34 e
F
O
O
O
F
Xe
or
F
or
Xe
O
F
F
Xe
O
O
F
All are: see-saw; ≈ 90° and ≈ 120°; dsp3
d. XeO2F4, 8 + 2(6) + 4(7) = 48 e
O
F
F
F
F
F
or
Xe
F
O
O
Xe
F
O
F
All are: octahedral; 90°; d2sp3
45.
XeF2 can react with oxygen and water to produce explosive xenon oxides and oxyfluorides,
and react with water to form HF.
46.
10.0 m × 10.0 m × 10.0 m = 1.00 × 103 m3; From Table 20.12, volume % Ar = 0.9%.
768
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
3
1L
0.9 L Ar
 10 dm 
1.00 × 103 m3 × 
= 9 × 103 L of Ar in the room

 
3
m
100
L
air
dm


PV = nRT, n =
PV
(1.0 atm)(9  10 3 L)

= 4 × 102 mol Ar
RT
(0.08206 L atm / mol  K )(298 K )
4 × 102 mol Ar ×
39.95 g
= 2 × 104 g Ar in the room
mol
4 × 102 mol Ar ×
6.022  10 23 atoms
= 2 × 1026 atoms Ar in the room
mol
A 2 L breath contains: 2 L air ×
n=
0.9 L Ar
= 2 × 10 2 L Ar
100 L air
PV
(1.0 atm)(2  10 2 L)

= 8 × 10 4 mol Ar
RT
(0.08206 L atm / mol  K )(298 K )
8 × 10 4 mol Ar ×
6.022  10 23 atoms
= 5 × 1020 atoms of Ar in a 2 L breath
mol
Because Ar and Rn are both noble gases, both species will be relatively unreactive. However,
all nuclei of Rn are radioactive, unlike most nuclei of Ar. It is the radioactive decay products
of Rn that can cause biological damage when inhaled.
47.
Release of Sr is probably more harmful. Xe is chemically unreactive. Strontium is in the same
family as calcium and could be absorbed and concentrated in the body in a fashion similar to
Ca. This puts the radioactive Sr in the bones, and red blood cells are produced in bone
marrow. Xe would not be readily incorporated into the body.
The chemical properties determine where a radioactive material may concentrate in the body
or how easily it may be excreted. The length of time of exposure and what is exposed to
radiation significantly affects the health hazard.
48.
a.
238
92 U
4
0
→ 222
86 Rn + ? 2 He + ? 1 e ; To account for the mass number change, 4 alpha
particles are needed. To balance the number of protons, 2 beta particles are needed.
222
86 Rn
→ 42 He +
218
84 Po;
Polonium-218 is produced when 222Rn decays.
b. Alpha particles cause significant ionization damage when inside a living organism.
Because the half-life of 222Rn is relatively short, a significant number of alpha particles
will be produced when 222Rn is present (even for a short period of time) in the lungs.
Additional Exercises
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
769
49.
As the halogen atoms get larger, it becomes more difficult to fit three halogen atoms around
the small nitrogen atom, and the NX3 molecule becomes less stable.
50.
a. The Lewis structures for NNO and NON are:
N
N
O
N
N
O
N
N
O
N
O
N
N
O
N
N
O
N
The NNO structure is correct. From the Lewis structures, we would predict both NNO
and NON to be linear. However, we would predict NNO to be polar and NON to be
nonpolar. Since experiments show N2O to be polar, NNO is the correct structure.
b. Formal charge = number of valence electrons of atoms  [(number of lone pair electrons)
+ 1/2 (number of shared electrons)].
N
N
O
-1
+1
0
N
N
O
N
N
O
0
+1
-1
-2
+1
+1
The formal charges for the atoms in the various resonance structures are below each
atom. The central N is sp hybridized in all of the resonance structures. We can probably
ignore the third resonance structure on the basis of the relatively large formal charges
compared to the first two resonance structures.
c. The sp hybrid orbitals on the center N overlap with atomic orbitals (or hybrid orbitals) on
the other two atoms to form the two sigma bonds. The remaining two unhybridized p
orbitals on the center N overlap with two p orbitals on the peripheral N to form the two π
bonds.
2px
sp
sp
z axis
2py
51.
OCN has 6 + 4 + 5 + 1 = 16 valence electrons.
Formal
charge
O
C
N
O
C
N
O
C
N
0
0
-1
-1
0
0
+1
0
-2
Only the first two resonance structures should be important. The third places a positive
formal charge on the most electronegative atom in the ion and a -2 formal charge on N.
CNO:
770
CHAPTER 20
Formal
charge
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
C
N
O
C
N
O
C
N
O
-2
+1
0
-1
+1
-1
-3
+1
+1
All of the resonance structures for fulminate (CNO) involve greater formal charges than in
cyanate (OCN), making fulminate more reactive (less stable).
52.
a.
(2e + NaBiO3 → BiO33 + Na+) × 5
(4 H2O + Mn2+ → MnO4 + 8 H+ + 5e) × 2
____________________________________________________________________
8 H2O(l) +2Mn2+(aq) + 5 NaBiO3(s) → 2 MnO4(aq) + 16 H+(aq) + 5 BiO33(aq)
+ 5 Na+(aq)
b. Bismuthate exists as a covalent network solid: (BiO3)x.
53.
1.0 × 104 kg waste 


1 mol C5 H 7 O 2 N
3.0 kg NH4
1 mol NH4
1000 g





100 kg waste
kg
18.04 g NH4
55 mol NH4

113.12 g C5 H 7 O 2 N
= 3.4 × 104 g tissue if all NH4+ converted
mol C5 H 7 O 2 N
Since only 95% of the NH4+ ions react:
mass of tissue = (0.95) (3.4 × 104 g) = 3.2 × 104 g or 32 kg bacterial tissue
54.
This element is in the oxygen family as all oxygen family members have ns2np4 valence
electron configurations.
a. As with all elements of the oxygen family, this element has 6 valence electrons.
b. The nonmetals in the oxygen family are O, S, Se and Te, which are all possible identities
for the element.
c. Ions in the oxygen family are 2 charged in ionic compounds. Li2X would be the formula between Li+ and X2 ions.
d. In general, radii increase from right to left across the periodic table and increase going
down a family. From this trend, the radius of the unknown element must be smaller than
the Ba radius.
e. The ioniation energy trend is the opposite of the radii trend indicated in the previous
answer. From this trend, the unknown element will have a smaller ionization energy than
fluorine.
55.
TeF5 has 6 + 5(7) + 1 = 42 valence electrons.
-
F
F
F
Te
F
F
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
771
The lone pair of electrons around Te exerts a stronger repulsion than the bonding pairs,
pushing the four square planar F's away from the lone pair and thus reducing the bond angles
between the axial F atom and the square planar F atoms.
56.
hv
a. AgCl(s)
Ag(s) + Cl; The reactive chlorine atom is trapped in the crystal. When
light is removed, Cl reacts with silver atoms to reform AgCl, i.e., the reverse reaction
occurs. In pure AgCl, the Cl atoms escape, making the reverse reaction impossible.
b. Over time, chlorine is lost and the dark silver metal is permanent.
57.
As temperature increases, the value of K decreases. This is consistent with an exothermic
reaction. In an exothermic reaction, heat is a product and an increase in temperature shifts the
equilibrium to the reactant side (as well as lowering the value of K).
58.
7.15 = log (6.2 × 10 8 ) + log
2
[HPO4 ]

[H 2 PO4 ]
= 10 0.06 = 0.9,
2
[HPO4 ]

[H 2 PO4 ]
2
[HPO4 ]

[H 2 PO4 ]
2
, 7.15 = 7.21 + log

[HPO4 ]

[H 2 PO4 ]
1
= 1.1 ≈ 1
0.9
A best buffer has approximately equal concentrations of weak acid and conjugate base so that
pH ≈ pKa for a best buffer. The pKa value for a H3PO4/H2PO4 buffer is log (7.5 × 10 3 ) =
2.12. A pH of 7.1 is too high for a H3PO4/H2PO4 buffer to be effective. At this high a pH,
there would be so little H3PO4 present that we could hardly consider it a buffer. This solution
would not be effective in resisting pH changes, especially when a strong base is added.
59.
MgSO4(s) → Mg2+(aq) + SO42(aq); NH4NO3(s) → NH4+(aq) + NO3(aq)
Note that the dissolution of MgSO4 used in hot packs is an exothermic process, and the
dissolution of NH4NO3 in cold packs is an endothermic process.
60.
Strong acids have a Ka >> 1 and weak acids have Ka < 1. Table 14.2 in the text lists some Ka
values for weak acids. Ka values for strong acids are hard to determine so they are not listed
in the text. However, there are only a few common strong acids, so if you memorize the
strong acids, then all other acids will be weak acids. The strong acids to memorize are HCl,
HBr, HI, HNO3, HClO4 and H2SO4.
a.
b.
c.
d.
HClO4 is a strong acid.
HOCl is a weak acid (Ka = 3.5 × 10 8 ).
H2SO4 is a strong acid.
H2SO3 is a weak diprotic acid with Ka1 and Ka2 values less than one.
772
61.
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
EO3 is the formula of the ion. The Lewis structure has 26 valence electrons. Let x = number
of valence electrons of element E.
26 = x + 3(6) + 1, x = 7 valence electrons
Element E is a halogen because halogens have 7 valence electrons. Some possible identities
are F, Cl, Br and I. The EO3 ion has a trigonal pyramid molecular structure with bond angles
 109.5°.
62.
The formula is EF2O2- and the Lewis structure has 28 valence electrons.
28 = x + 2(7) + 6 + 2, x = 6 valence electrons for element E
63.
Element E must belong to the group 6A elements since E has 6 valence electrons. E must also
be a row 3 or heavier element since this ion has more than 8 electrons around the central E
atom (row 2 elements never have more than 8 electrons around them). Some possible
identities for E are S, Se and Te. The ion has a T-shaped molecular structure with bond angles
of  90°.
1 / 8 Xe
1/ 4 F
8 corners ×
+ 1 Xe inside cell = 2 Xe; 8 edges ×
+ 2 F inside cell = 4 F
edge
corner
Empirical formula is XeF2. This is also the molecular formula.
Challenge Problems
64.
In order to form a π bond, the d and p orbitals must overlap “side to side” instead of “head to
head” as in sigma bonds. A representation of the “side to side” overlap follows. For a
bonding orbital to form, the phases of the lobes must match (positive to positive and negative
to negative).
d
65.
+
p
For the reaction:
O
N
N
O
NO2 + NO
O
the activation energy must in some way involve breaking a nitrogen-nitrogen single bond.
For the reaction:
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
773
O
N
N
O2 + N 2O
O
O
at some point nitrogen-oxygen bonds must be broken. N‒N single bonds (160. kJ/mol) are
weaker than N‒O single bonds (201 kJ/mol). In addition, resonance structures indicate that
there is more double bond character in the N‒O bonds than in the N‒N bond. Thus, NO2 and
NO are preferred by kinetics because of the lower activation energy.
66.
Mg2+ + P3O105 ⇌ MgP3O103 K = 4.0 × 108
[Mg2+]o =
50.  10 3 g
1 mol

= 2.1 × 10 3 M
L
24.31 g
[P3O105]o =
40. g Na 5 P3O10
1 mol
= 0.11 M

L
367.86 g
Assume the reaction goes to completion because K is large. Then solve the back equilibrium
problem to determine the small amount of Mg2+ present.
Mg2+
Before
Change
After
Change
Equil.
P3O105
+
3
67.
MgP3O103
2.1 × 10 3 M
0.11 M
0
3
3
2.1 × 10
2.1 × 10 →
+2.1 × 10 3 React completely
0
0.11
2.1 × 10 3 New initial condition
x mol/L MgP3O103 dissociates to reach equilibrium
+x
+x
←
x
x
0.11 + x
2.1 × 10 3  x
K = 4.0 × 108 =
4.0 × 108 ≈
⇌
[MgP3O10 ]
5
[Mg 2 ][P3O10 ]

2.1  10 3  x
(assume x << 2.1 × 10 3 )
x(0.11  x)
2.1  10 3
, x = [Mg2+] = 4.8 × 10 11 M; Assumptions good.
x(0.11)
a. NO is the catalyst. NO is present in the first step of the mechanism on the reactant side,
but it is not a reactant because it is regenerated in the second step and does not appear in
the overall balanced equation.
b. NO2 is an intermediate. Intermediates also never appear in the overall balanced equation.
In a mechanism, intermediates always appear first on the product side while catalysts
always appear first on the reactant side.
774
CHAPTER 20
c. k = A exp (Ea/RT);
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
k cat
A exp E a (cat) / RT
 E (un )  E a (cat) 

 exp a

k un
A exp E a (cat) / RT
RT



 0.85
k cat
2100 J / mol
 = e = 2.3
 exp 
k un
 8.3145 J / K  mol  298 K 
The catalyzed reaction is approximately 2.3 times faster than the uncatalyzed reaction at
25°C.
d. The mechanism for the chlorine-catalyzed destruction of ozone is:
O3(g) + Cl(g) → O2(g) + ClO(g) slow
ClO(g)+ O(g) → O2(g) + Cl(g)
fast
__________________________________
O3(g) + O(g) → 2 O2(g)
e. Because the chlorine atom-catalyzed reaction has a lower activation energy, the Cl catalyzed rate is faster. Hence, Cl is a more effective catalyst. Using the activation energy,
we can estimate the efficiency that Cl atoms destroy ozone as compared to NO molecules.
At 25°C:
 (2100  11,900) J / mol 
k Cl
  E (Cl) E a ( NO) 

 exp a

  exp
k NO
RT 
 RT
 (8.3145  298) J / mol 
= e3.96 = 52
At 25°C, the Clcatalyzed reaction is roughly 52 times faster (more efficient) than the
NOcatalyzed reaction, assuming the frequency factor A is the same for each reaction
and assuming similar rate laws.
68.
3 O2(g)
ln K =
⇌
2 O3(g); ΔH° = 2(143 kJ) = 286 kJ; ΔG° = 2(163 kJ) = 326 kJ
 ΔG o
 326  10 3 J

= 131.573, K = e 131.573 = 7.22 × 10 58
RT
8.3145 J / K  mol  298 K
Note: We carried extra significant figures for the K calculation.
We need the value of K at 230. K. From Section 16.8 of the text: ln K =
For two sets of K and T:
ln K1 =
 ΔH o  1  ΔSo
 ΔH o  1  ΔSo
  
 
; ln K2 =
R  T1 
R
R  T2 
R
Subtracting the first expression from the second:
 ΔH o
ΔSo

RT
R
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
ln K2  ln K1 =
775
K
ΔH o  1
1 
ΔH o  1
1 

 or ln 2 




R  T1
T2 
K1
R  T1
T2 
Let K2 = 7.22 × 10 58 , T2 = 298; K1 = K230, T1 = 230. K; ΔH° = 286 × 103 J
ln
7.22  10 58
286  10 3  1
1 



 = 34.13 (Carrying extra sig. figs.)
K 230
8.3145  230 . 298 
7.22  10 58
= e34.13 = 6.6 × 1014, K230 = 1.1 × 10 72
K 230
K230 = 1.1 × 10 72 =
PO23
PO3 2

PO23
3 3
(1.0  10 )
,
PO3 = 3.3 × 10 41 atm
The volume occupied by one molecule of ozone is:
V=
nRT (1 / 6.022  10 23 mol )  0.08206 L atm / mol  K  230 . K

, V = 9.5 × 1017 L
 41
P
3.3  10 atm
Equilibrium is probably not maintained under these conditions. When only two ozone
molecules are in a volume of 9.5 × 1017 L, the reaction is not at equilibrium. Under these
conditions, Q > K and the reaction shifts left. But with only 2 ozone molecules in this huge
volume, it is extremely unlikely that they will collide with each other. In these conditions,
the concentration of ozone is not large enough to maintain equilibrium.
69.
NH3 + NH3
⇌
NH4+ + NH2
K = [NH4+] [NH2] = 1.8 × 10 12
NH3 is the solvent, so it is not included in the K expression. In a neutral solution of ammonia:
[NH4+] = [NH2]; 1.8 × 10 12 = [NH4+]2, [NH4+] = 1.3 × 10 6 M = [NH2]
We could abbreviate this autoionization as: NH3
⇌ H+
+ NH2, where [H+] = [NH4+].
This abbreviation is synonomous to the abbreviation of the autoionization of water (H2O ⇌
H+ + OH). So: pH = pNH4+ = log(1.3 × 10 6 ) = 5.89.
70.
Let’s consider a reaction between 3.00 x mol N2 and 3.00 x mol H2 (equimolar).
N2(g)
Before
Change
Equil.
3.00 x mol
1.00 x mol
2.00 x mol
+
3 H2(g)
3.00 x mol
3.00 x mol
0
→
2 NH3(g)
0
+2.00 x mol
2.00 x mol
When an equimolar mixture is reacted, the number of moles of gas present decreases from
6.00 x moles initially to 4.00 x moles after completion.
776
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
a. The total pressure in the piston apparatus is a constant 1.00 atm. After the reaction, we
have 2.00 x moesl N2 and 2.00 x moles NH3. One-half of the moles of gas present are
NH3 molecules, so one-half of the total pressure is due to the NH3 molecules. PNH3 =
0.500 atm.
b.
χ NH3 
mol NH3
2.00 mol x
= 0.500

total mol (2.00 x  2.00 x) mol
c. At constant P and T, volume is directly proportional to n. Because n decreased from 6.00
x moles to 4.00 x moles, the volume will decrease by the same factor.
Vfinal = 15.0 L (4/6) = 10.0 L
71.
Let n SO 2 = initial mol SO2 present. The reaction is summarized in the following table (O2 is
in excess).
2 SO2
Initial
n SO 2
Change
 n SO 2
Final
0
+
→
O2(g)
2.00 mol
2 SO3(g)
0
 n SO 2 /2
+ n SO 2
2.00 - n SO 2 /2
n SO 2
d = mass/volume; Let di = initial density of gas mixture and df = final density of gas mixture
after reaction. Because mass is conserved in a chemical reaction, massi = massf.
df
massf / Vf
V

 i
di
massi / Vi
Vf
At constant P and T, V  n, so:
df
V
n
 i  i ; Setting up an equation:
di
Vf
nf
n SO 2  2.00
n SO 2  2.00
df
n
0.8471 g / L
= 1.059, 1.059 = i =
=

(2.00  n SO 2 / 2)  n SO 2
2.00  n SO 2 / 2
di
0.8000 g / L
nf
Solving: n SO 2 = 0.25 mol; so, 0.25 moles of SO3 formed
0.25 mol SO3 ×
72.
80.07 g
= 20. g SO3
mol
Ca(IO3)2(s)
⇌
Ca2+(aq) + 2 IO3(aq)
Ksp = [Ca2+][ IO3]2
CHAPTER 20
Initial
Equil.
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
s = solubility (mol/L)
mol IO3 present = 0.0149 L ×
0
s
0.100 mol S2O3
mol
0
2s
Ksp = s(2s)2 = 4s3
2

777
1 mol I 2
2 mol S2O3
2

1 mol IO3
3 mol I 2

= 2.48 × 10 4 mol IO3
[IO3] =
2.48  10 4 mol
= 2.48 × 10 2 M = 2s, s = 1.24 × 10 2 M
0.0100 L
Ksp = 4s3 = 4(1.24× 10 2 )3 = 7.63 × 10 6
Integrative Problems
73.
a. 307 kJ = [1136 + x] – [(-254 kJ) + 3(96 kJ)], x = ΔHof , NI3 = 287 kJ/mol
b. IF2+, 7 + 2(7) – 1 = 20 e
BF4, 3 + 4(7) + 1 = 32 e
F
+
I
F
B
F
F
F
F
V-shaped; sp3
74.
tetrahedral; sp3
a. Because the hydroxide ion has a 1 charge, Te has a +6 oxidation state.
b. mol Te = (0.545 cm)3 
mol F2 = n =
6.240 g 1 mol Te
= 7.92 × 10 3 mol Te

3
127.6
cm
PV
1.06 atm  2.34 L
= 0.101 mol F2

RT 0.08206 L atm / K  mol  298 K
mol F2
0.101 mol
= 12.8
(actual) 
mol Te
7.92  10 3 mol
The balanced reaction requires a 3:1 mol ratio of F2 to Te. Because actual > theoretical,
the denominator (Te) is limiting. Assuming 115 mL of solution:
7.92  10 3 mol Te 
[TeF6]o =
0.115 L
1 mol TeF6
mol Te
= 6.89 × 10 2 M
778
CHAPTER 20
THE REPRESENTATIVE ELEMENTS: GROUPS 5A8A
Because K a1  K a 2 , the amount of protons produced by the K a 2 reaction will be
insignificant.
Te(OH)6
⇌
Te(OH)5O +
Initial 0.0689 M
Equil. 0.0689  x
K a1 = 2.1 × 10 8 =
0
x
H+
K a1 = 10 7.68 = 2.1 × 10 8
~0
x
x2
x2

, x = [H+] = 3.8 × 10 5 M
0.0689  x
0.0689
pH = log(3.8 × 10 5 ) = 4.42; Assumptions good.
Marathon Problem
75.
The answers to the clues are:
(1) HI has the second highest boiling point; (2) HF is the weak hydrogen halide acid; (3)
He was first discovered from the sun’s emission spectrum; (4) Both Bi and Sb form MOCl
precipitates. For a message that makes sense, Bi is the correct choice; (5) Te is a semiconductor; (6) S has both rhombic and monoclinic solid forms; (7) Cl2 is a yellow-green gas
and Cl- forms the indicated precipitates; (8) O is the most abundant element in and near the
earth’s crust; (9) Se appears to furnish some form of protection against cancer; (10) Kr
forms compounds. The symbol in reverse order is rk; (11) As forms As4 molecules; (12)
N2 is a major inert component of air and N is often found in fertilizers and explosives.
Filling in the blank spaces with the answers to the clues, the message is “If he bites, close
ranks.”