22
Chemistry of the
Nonmetals
Visualizing Concepts
22.2
22.5
(a)
Acid-base (Brønsted)
(b)
Charges on species from left to right in the reaction: 0, 0, 1+, 1–
(c)
NH 3(aq) + H 2O(l)
⇌NH
+
4
–
(aq) + OH (aq)
Analyze. Given: space-filling models of molecules containing nitrogen and oxygen
atoms.
Find: molecular formulas and Lewis structures.
Plan. Nitrogen atoms are blue, and oxygen atoms are red. Count the number of spheres
of each color to determine the molecular formula. From each molecular formula, count
the valence electrons (N = 5, O = 6) and draw a correct Lewis structure. Resonance
structures are likely.
Solve.
(a)
N 2O 5
–
40 valence electrons, 20 e pairs
Many other resonance structures are possible. Those with double bonds to the
central oxygen (like the right-hand structure above) do not minimize formal
charge and are less significant in the net bonding model.
(b)
N 2O 4
–
–
34 e , 17 e pairs
Other equivalent resonance structures with different arrangement of the double
bonds are possible.
(c)
NO 2
–
–
17 e , 8.5 e pairs
We place the odd electron on N because of electronegativity arguments.
(d)
N 2O 3
–
–
28 e , 14 e pairs
635
22 Chemistry of the Nonmetals
(e)
NO
–
Solutions to Exercises
–
11 e , 5.5 e pairs
We place the odd electron on N because of electronegativity arguments.
(f)
N 2O
–
–
16 e , 8 e pairs
The right-most structure above does not minimize formal charge and makes
smaller contribution to the net bonding model.
22.8
Analyze/Plan. Evaluate the graph, describe the trend in data, recall the general trend for
each of the properties listed, and use details of the data to discriminate between
possibilities.
Solve. The general trend is an increase in value moving from left to right across the
period, with a small discontinuity at S. Considering just this overall feature, both (a)
first ionization energy and (c) electronegativity increase moving from left to right, so
these are possibilities. (b) Atomic radius decreases, and can be eliminated. Since Si is a
solid and Cl and Ar are gases at room temperature, melting points must decrease across
the row; (d) melting point can be eliminated. According to data in Tables 22.2, 22.5, 22.7,
and 22.8, (e) X-X single bond enthalpies show no consistent trend. Furthermore, there is
no known Ar-Ar single bond, so no value for this property can be known; (e) can be
eliminated.
Now let’s examine trends in (a) first ionization energy and (c) electronegativity more
closely. From electronegativity values in Chapter 8, we see a continuous increase with
no discontinuity at S, and no value for Ar. Values in (a) first ionization energy from
Chapter 7 do match the pattern in the figure. The slightly lower value of I 1 for S is due
to a decrease in repulsion by removing an electron from a fully occupied orbital. In
summary, only (a) first ionization energy fits the property depicted in the graph.
Periodic Trends and Chemical Reactions
22.11
Analyze/Plan. Use the color-coded periodic chart on the front-inside cover of the text to
classify the given elements.
Solve.
Metals: (b) Sr, (c) Mn, (e) Rh;
22.13
22.15
nonmetals: (a) P, (d) Se, (f) Kr;
Analyze/Plan. Follow the logic in Sample Exercise 22.1.
(a)
O
(b)
Br
(d)
O
(e)
Co
(c)
metalloids: none
Solve.
Ba
Analyze/Plan. Use the position of the specified elements on the periodic chart, periodic
trends, and the arguments in Sample Exercise 22.1 to explain the observations.
Solve.
636
22 Chemistry of the Nonmetals
22.17
Solutions to Exercises
(a)
Nitrogen is too small to accommodate five fluorine atoms about it. The P and As
atoms are larger. Furthermore, P and As have available 3d and 4d orbitals,
respectively, to form hybrid orbitals that can accommodate more than an octet of
electrons about the central atom.
(b)
Si does not readily form π bonds, which would be necessary to satisfy the octet
rule for both atoms in SiO.
(c)
A reducing agent is a substance that readily loses electrons. As has a lower
electronegativity than N; that is, it more readily gives up electrons to an acceptor
and is more easily oxidized.
Analyze/Plan. Follow the logic in Sample Exercise 22.2.
(a)
Solve.
Mg 3N 2(s) + 6H 2O(l) → 2NH 3(g) + 3Mg(OH) 2(s)
Because H 2O(l) is a reactant, the state of NH 3 in the products could be expressed
as NH 3(aq).
(b)
2C 3H 7OH(l) + 9O 2(g) → 6CO 2(g) + 8H 2O(l)
(c)
(d)
AlP(s) + 3H 2O(l) → PH 3(g) + Al(OH) 3(s)
(e)
Na 2S(s) + 2HCl(aq) → H 2S(g) + 2NaCl(aq)
Hydrogen, the Noble Gases, and the Halogens
22.19
Analyze/Plan. Use information on the isotopes of hydrogen in Section 22.2 to list their
symbols, names, and relative abundances.
Solve.
a)
b)
22.21
The order of abundance is proteum > deuterium > tritium.
Analyze/Plan. Consider the electron configuration of hydrogen and the Group 1A
elements. Solve.
Like other elements in group 1A, hydrogen has only one valence electron and its most
common oxidation number is +1.
22.23
Analyze/Plan. Use information on the descriptive chemistry of hydrogen in Section 22.2
to formulate the required equations. Steam is H 2O(g).
Solve.
(a)
+
2+
Mg(s) + 2H (aq) → Mg (aq) + H 2(g)
(b)
(c)
637
22 Chemistry of the Nonmetals
22.25
Solutions to Exercises
Analyze/Plan. Use information on the descriptive chemistry of hydrogen given in
Section 22.2 to complete and balance the equations.
Solve.
(a)
NaH(s) + H 2O(l) → NaOH(aq) + H 2(g)
(b)
Fe(s) + H 2SO 4(aq) → Fe (aq) + H 2(g) + SO 4 (aq)
(c)
H 2(g) + Br 2(g) → 2HBr(g)
(d)
2Na(l) + H 2(g) → 2NaH(s)
2+
2–
(e)
22.27
Analyze/Plan. If the element bound to H is a nonmetal, the hydride is molecular. If H is
bound to a metal with integer stoichiometry, the hydride is ionic; with noninteger
stoichiometry, the hydride is metallic.
Solve.
(a)
ionic (metal hydride)
(b)
molecular (nonmetal hydride)
(c)
metallic (nonstoichiometric transition metal hydride)
22.29
Vehicle fuels produce energy via combustion reactions. The reaction
H 2(g) + 1/2 O 2(g) → H 2O(g) is very exothermic, producing 242 kJ per mole of H 2
burned. The only product of combustion is H 2O, a nonpollutant (but like CO 2, a
greenhouse gas).
22.31
Analyze/Plan. Consider the periodic properties of Xe and Ar.
Solve.
Xenon is larger, and can more readily accommodate an expanded octet. More important
is the lower ionization energy of xenon; because the valence electrons are a greater
average distance from the nucleus, they are more readily promoted to a state in which
the Xe atom can form bonds with fluorine.
22.33
22.35
22.37
Analyze/Plan. Follow the rules for assigning oxidation numbers in Section 4.4 and the
logic in Sample Exercise 4.8.
Solve.
(a)
ClO 3 , +5
–
(b)
HI, –1
(c)
ICl 3; I, +3; Cl, –1
(d)
NaOCl, +1
(e)
HClO 4, +7
(f)
XeF 4, +4; F, –1
Analyze/Plan. Review the nomenclature rules and ion names in Section 2.8.
(a)
iron(III) chlorate
(b)
chlorous acid
(c)
xenon hexafluoride
(d)
bromine pentafluoride
(e)
xenon oxide tetrafluoride
(f)
iodic acid
Solve.
Analyze/Plan. Consider intermolecular forces and periodic properties, including
oxidizing power, of the listed substances.
Solve.
(a)
Van der Waals intermolecular attractive forces increase with increasing numbers
of electrons in the atoms.
(b)
F 2 reacts with water: F 2(g) + H 2O(l) → 2HF(aq) + 1/2 O 2(g). That is, fluorine is
too strong an oxidizing agent to exist in water.
638
22 Chemistry of the Nonmetals
22.39
Solutions to Exercises
(c)
HF has extensive hydrogen bonding.
(d)
Oxidizing power is related to electronegativity. Electronegativity decreases in the
order given.
–
If the lifetime of perchlorate anion, ClO 4 , in soils and water is decades, the ion must be
extremely stable (unreactive) in aqueous solutions and aerobic environments; it is not
–
easily oxidized by O 2. Although chlorine is in a very high oxidation state in ClO 4 , it is
not readily reduced, because the ion has a stable, symmetric structure that protects it
against reactions. The anion is a symmetrical tetrahedron with several plausible Lewis
structures when expanded octets about Cl are considered.
(minimum formal charges)
More structures with alternate locations of the single and double bonds can be drawn.
Resonance stabilization could certainly contribute to the ion’s high stability.
Oxygen and the Group 6A Elements
22.41
22.43
Analyze/Plan. Consider the industrial uses of oxygen and ozone given in Section 22.5.
Solve.
(a)
As an oxidizing agent in steel-making; to bleach pulp and paper; in oxyacetylene
torches; in medicine to assist in breathing
(b)
Synthesis of pharmaceuticals, lubricants, and other organic compounds where
bonds are cleaved; in water treatment
Analyze/Plan. Use information on the descriptive chemistry of oxygen given in Section
22.5 to complete and balance the equations.
Solve.
(a)
(b)
(c)
PbS(s) + 4O 3(g) → PbSO 4(s) + 4O 2(g)
(d)
2ZnS(s) + 3O 2(g)
(e)
2K 2O 2(s) + 2CO 2(g) → 2K 2CO 3(s) + O 2(g)
2ZnO(s) + 2SO 2(g)
639
22 Chemistry of the Nonmetals
22.45
22.47
Solutions to Exercises
Analyze/Plan. Oxides of metals are bases, oxides of nonmetals are acids, oxides that act
as both acids and bases are amphoteric and oxides that act as neither acids nor bases are
neutral.
Solve.
(a)
acidic (oxide of a nonmetal)
(b)
acidic (oxide of a nonmetal)
(c)
amphoteric
(d)
basic (oxide of a metal)
Analyze/Plan. Follow the rules for assigning oxidation numbers in Section 4.4 and the
logic in Sample Exercise 4.8. Solve.
(a)
H 2SeO 3, +4
(b)
KHSO 3, +4
(d)
CS 2, –2
(e)
CaSO 4, +6
(c)
H 2Te, –2
Oxygen (a group 6A element) is in the –2 oxidation state in compounds (a), (b),
and (e).
22.49
Analyze/Plan. The half-reaction for oxidation in all these cases is:
+
–
H 2S(aq) → S(s) + 2H + 2e (The product could be written as S 8(s), but this is not
necessary. In fact it is not necessarily the case that S 8 would be formed, rather than
some other allotropic form of the element.) Combine this half-reaction with the given
reductions to write complete equations. The reduction in (c) happens only in acid
solution. The reactants in (d) are acids, so the medium is acidic.
Solve.
3+
2+
+
(a)
2Fe (aq) + H 2S(aq) → 2Fe (aq) + S(s) + 2H (aq)
(b)
Br 2(l) + H 2S(aq) → 2Br (aq) + S(s) + 2H (aq)
(c)
2MnO 4 (aq) + 6H (aq) + 5H 2S(aq) → 2Mn (aq) + 5S(s) + 8H 2O(l)
(d)
2NO 3 (aq) + H 2S(aq) + 2H (aq) → 2NO 2(aq) + S(s) + 2H 2O(l)
–
–
–
+
+
2+
+
22.51
Analyze/Plan. For each substance, count valence electrons, draw the correct Lewis
structure, and apply the rules of VSEPR to decide electron domain geometry and
geometric structure. Solve.
22.53
Analyze/Plan. Use information on the descriptive chemistry of sulfur given in Section
22.6 to complete and balance the equations.
Solve.
⇌
+
(a)
SO 2(s) + H 2O(l) → H 2SO 3(aq)
(b)
ZnS(s) + 2HCl(aq) → ZnCl 2(aq) + H 2S(g)
(c)
8SO 3 (aq) + S 8(s) → 8S 2O 3 (aq)
2–
2–
640
–
H (aq) + HSO 3 (aq)
22 Chemistry of the Nonmetals
(d)
Solutions to Exercises
SO 3(aq) + H 2SO 4(l) → H 2S 2O 7(l)
Nitrogen and the Group 5A Elements
22.55
22.57
Analyze/Plan. Follow the rules for assigning oxidation numbers in Section 4.4 and the
logic in Sample Exercise 4.8.
Solve.
(a)
NaNO 2, +3
(b)
NH 3, –3
(c)
N 2O, +1
(d)
NaCN, –3
(e)
HNO 3, +5
(f)
NO 2, +4
Analyze/Plan. For each substance, count valence electrons, draw the correct Lewis
structure, and apply the rules of VSEPR to decide electron domain geometry and
geometric structure. Solve.
(a)
The molecule is bent around the central oxygen and nitrogen atoms; the four
atoms need not lie in a plane. The right-most form does not minimize formal
charges and is less important in the actual bonding model.
The geometry is tetrahedral
around the left nitrogen,
trigonal pyramidal around the right.
22.59
(three equivalent resonance forms)
The ion is trigonal planar.
Analyze/Plan. Use information on the descriptive chemistry of nitrogen given in Section
22.7 to complete and balance the equations.
Solve.
(a)
Mg 3N 2(s) + 6H 2O(l) → 2NH 3(g) + 3Mg(OH) 2(s)
Because H 2O(l) is a reactant, the state of NH 3 in the products could be expressed
as NH 3(aq).
22.61
(b)
2NO(g) + O 2(g) → 2NO 2(g)
(c)
N 2O 5(g) + H 2O(l) → 2H (aq) + 2NO 3 (aq)
(d)
NH 3(aq) + H (aq) → NH 4 (aq)
(e)
N 2H 4(l) + O 2(g) → N 2(g) + 2H 2O(g)
+
+
–
+
Analyze/Plan. Follow the method for writing balanced half-reactions given in Section
20.2 and Sample Exercises 20.2 and 20.3 Solve.
–
+
–
(a)
HNO 2(aq) + H 2O(l) → NO 3 (aq) + 3H (aq) + 2e ,
(b)
N 2(g) + H 2O(l) → N 2O(g) + 2H (aq) + 2e ,
+
641
–
22 Chemistry of the Nonmetals
22.63
22.65
Analyze/Plan. Follow the rules for assigning oxidation numbers in Section 4.4 and the
logic in Sample Exercise 4.8.
Solve.
(a)
H3PO 3, +3
(b)
H 4P 2O 7, +5
(d)
Mg 3As 2, –3
(e)
P 2O 5, +5
(c)
SbCl 3, +3
Analyze/Plan. Consider the structures of the compounds of interest when explaining the
observations.
Solve.
(a)
(b)
22.67
Solutions to Exercises
Phosphorus is a larger atom and can more easily accommodate five surrounding
atoms and an expanded octet of electrons than nitrogen can. Also, P has
energetically “available” 3d orbitals which participate in the bonding, but
nitrogen does not.
Only one of the three hydrogens in H 3PO 2 is bonded to oxygen. The other two
are bonded directly to phosphorus and are not easily ionized because the
bond is not very polar.
+
+
(c)
PH 3 is a weaker base than H 2O (PH 4 is a stronger acid than H 3O ). Any attempt
+
to add H to PH 3 in the presence of H 2O merely causes protonation of H 2O.
(d)
White phosphorus consists of P 4 molecules, with
bond angles of 60°.
Each P atom has four VSEPR pairs of electrons, so the predicted electron pair
geometry is tetrahedral and the preferred bond angle is 109°. Because of the
severely strained bond angles in P 4 molecules, white phosphorus is highly
reactive.
Analyze/Plan. Use information on the descriptive chemistry of phosphorus given in
Section 22.8 to complete and balance the equations.
Solve.
(a)
(b)
PBr 3(l) + 3H 2O(l) → H 3PO 3(aq) + 3HBr(aq)
(c)
4PBr 3(g) + 6H 2(g) → P 4(g) + 12HBr(g)
Carbon, the Other Group 4A Elements, and Boron
22.69
Analyze/Plan. Review the nomenclature rules and ion names in Section 2.8.
(a)
HCN
(b) Ni(CO) 4
(c) Ba(HCO 3) 2
(d)
Solve.
CaC 2
22.71
Analyze/Plan. Use information on the descriptive chemistry of carbon given in
Section 22.9 to complete and balance the equations.
Solve.
(a)
22.73
2+
–
(b)
BaC 2(s) + 2H 2O(l) → Ba (aq) + 2OH (aq) + C 2H 2(g)
(c)
2C 2H 2(g) + 5O 2(g) → 4CO 2(g) + 2H 2O(g)
(d)
CS 2(g) + 3O 2(g) → CO 2(g) + 2SO 2(g)
(e)
Ca(CN) 2(s) + 2HBr(aq) → CaBr 2(aq) + 2HCN(aq)
Analyze/Plan. Use information on the descriptive chemistry of carbon given in Section
22.9 to complete and balance the equations.
Solve.
642
22 Chemistry of the Nonmetals
Solutions to Exercises
(a)
22.75
22.77
22.79
22.81
+
+
(b)
NaHCO 3(s) + H (aq) → CO 2(g) + H 2O(l) + Na (aq)
(c)
2BaCO 3(s) + O 2(g) + 2SO 2(g) → 2BaSO 4(s) + 2CO 2(g)
Analyze/Plan. Follow the rules for assigning oxidation numbers in Section 4.4 and the
logic in Sample Exercise 4.8.
Solve.
(a)
H 3BO 3, +3
(b)
SiBr 4, +4
(d)
Na 2B 4O 7•10H 2O, +3
(e)
B 2O 3, +3
(c)
PbCl 2, +2
Analyze/Plan. Consider periodic trends within a family, particularly metallic character,
as well as descriptive chemistry in Sections 22.9 and 22.10.
Solve.
(a)
Sn; see Table 22.8. The filling of the 4f subshell at the beginning of the sixth row
of the periodic table increases Z and Zeff for later elements. This causes the
ionization energy of Pb to be greater than that of Sn.
(b)
Carbon, Si, and Ge; these are the nonmetal and metalloids in group 4A. They
form compounds ranging from XH 4 (–4) to XO 2 (+4). The metals Sn and Pb are
not found in negative oxidation states.
(c)
Silicon; silicates are the main component of sand.
Analyze/Plan. Consider the structural chemistry of silicates discussed in Section 22.10
and shown in Figures 22.45-22.46.
Solve.
(a)
Tetrahedral
(b)
Metasilicic acid will probably adopt the single-strand silicate chain structure
shown in Figure 22.46(a). The empirical formula shows 3 O and 2 H atoms per Si
atom. The chain has the same Si to O ratio as metasilicic acid. Furthermore, in the
chain structure, there are two terminal (not bridging) O atoms on each Si. These
can accommodate the 2 H atoms associated with each Si atom of the acid. The
sheet structure does not fulfill these requirements.
(a)
Diborane (Figure 22.48 and below) has bridging H atoms linking the two B
atoms. The structure of ethane shown below has the C atoms bound directly,
with no bridging atoms.
(b)
B 2H 6 is an electron deficient molecule. It has 12 valence electrons, while C 2H 6 has
14 valence electrons. The 6 valence electron pairs in B 2H 6 are all involved in
sigma bonding, so the only way to satisfy the octet rule at B is to have the
bridging H atoms shown in Figure 22.48.
(c)
A hydride ion, H , has two electrons while an H atom has one. The term hydridic
indicates that the H atoms in B 2H 6 have more than the usual amount of electron
density for a covalently bound H atom.
–
643
22 Chemistry of the Nonmetals
Solutions to Exercises
Additional Exercises
–
–
22.85
BrO 3 (aq) + XeF 2(aq) + H 2O(l) → Xe(g) + 2HF(aq) + BrO 4 (aq)
22.87
(a)
H 2SO 4 – H 2O → SO 3
(b)
2HClO 3 – H 2O → Cl 2O 5
(c)
2HNO 2 – H 2O → N 2O 3
(d)
H 2CO 3 – H 2O → CO 2
(e)
2H 3PO 4 – 3H 2O → P 2O 5
(a)
PO 4 , + 5; NO 3 , + 5
(a)
The Lewis structure for NO 4 would be:
22.90
3–
–
3–
The formal charge on N is +1 and on each O atom is –1. The four electronegative
oxygen atoms withdraw electron density, leaving the nitrogen deficient. Since N
can form a maximum of four bonds, it cannot form a π bond with one or more of
3–
the O atoms to regain electron density, as the P atom in PO 4 does. Also, the
short
distance would lead to a tight tetrahedron of O atoms subject to
steric repulsion.
22.93
Ge(l) + 2Cl 2(g) → GeCl 4(l)
GeCl 4(l) + 2H 2O(l) → GeO 2(s) + 4HCl(g)
GeO 2(s) + 2H 2(g) → Ge(s) + 2H 2O(l)
22.95
Assume that the reactions occur in acidic solution. The half-reaction for reduction of
+
–
H 2O 2 is in all cases H 2O 2(aq) + 2H (aq) + 2e → 2H 2O(aq).
(a)
N 2H 4(aq) + 2H 2O 2(aq) → N 2(g) + 4H 2O(l)
(b)
SO 2(g) + H 2O 2(aq) → SO 4 (aq) + 2H (aq)
(c)
NO 2 (aq) + H 2O 2(aq) → NO 3 (aq) + H 2O(l)
(d)
H 2S(g) + H 2O 2(aq) → S(s) + 2H 2O(l)
2–
–
+
–
(e)
644
22 Chemistry of the Nonmetals
Solutions to Exercises
Integrative Exercises
22.97
(a)
(b)
22.100
First calculate the molar solubility of Cl 2 in water.
Assuming that x is small compared with 0.1384:
3
–4
–5
x = (0.1384)(4.7 × 10 ) = 6.504 × 10 ; x = 0.0402 = 0.040 M
We can correct the denominator using this value, to get a better estimate of x:
One more round of approximation gives x = 0.0364 = 0.036 M. This is the equilibrium
concentration of HClO.
22.103
(a)
SO 2(g) + 2H 2S(aq) → 3S(s) + 2H 2O(g) or, if we assume S 8 is the product,
8SO 2(g) + 16H 2S(aq) → 3S 8(s) + 16H 2O(g).
(b)
Assume that all S in the coal becomes SO 2 upon combustion, so that
1 mol S (coal) = 1 mol SO 2.
3
3
= 1.98 × 10 = 2.0 × 10 mol H 2S
(c)
This is about 210 lb S per ton of coal combusted. (However, two-thirds of this
comes from the H 2S, which presumably at some point was also obtained from
coal.)
22.106
(a)
MnSi: more than one element, so not metallic; high melting, so not molecular;
insoluble in water, so not ionic; therefore covalent network.
(b)
MnSi(s) + HF(aq) → SiH 4(g) + MnF 4(s)
645
22 Chemistry of the Nonmetals
Solutions to Exercises
Reduction of Mn(IV) to Mn(II) is unlikely, because F
–
–
reducing agent.
for F 2(g) + 2 e → 2F (aq) = 2.87 V
22.109
–
is an extremely weak
(a)
(b)
(c)
about 1.23 Å;
1.34 Å or less. Since consecutive
bonds
require sp hybrid orbitals on C (as in allene, C 3H 4), we might expect the orbital
overlap requirements of this bonding arrangement to require smaller than usual
distances.
(d)
The product has the formula C 3H 4O 2.
–
–
28 valence e , 14 e pr
Three possibilities are shown above. The
group on the lower structure
is uncommon and less likely than the two symmetrical structures.
646