CHEM 481Assignment 5. Answers.
1.
Sketch the p block of the periodic table (you may use the blank table distributed in the course notes). Identify as many
elements as you can that form Lewis acids in one of their oxidation states and give the formula of a representative Lewis
base for each element.
Consult the lecture notes and your textbook to check your answer to this question. Be sure to learn some of the
representative Lewis bases and acids.
2.
For each of the following processes, identify the acids and bases involved and characterize the process as complex
formation or acid–base displacement. Identify the species that exhibit Brønsted acidity as well as Lewis acidity.
a) SO3 + H2O Æ HSO4– + H+
b) CH3[B12] + Hg 2+ Æ [B12] + + CH3Hg +
(B12 designates the cobalt macrocycle vitamin B12.
c) KCl + SnCl2 Æ K+ + [SnCl3] –
d) AsF3 (g) + SbF 5 (l) Æ [AsF2] +[SbF6] – (s)
e) Ethanol dissolves in pyridine to produce a non-conducting solution.
(a)
SO3 + H2O Æ HSO4– + H+
The acids in this reaction are the Lewis acids SO3 and H+ and the base is the Lewis base OH–. The complex (or adduct)
HSO4- is formed by the displacement of the proton from the hydroxide ion by the stronger acid SO3. In this way, the water
molecule is thought of as an adduct of H+ and OH–. Since the proton must be bound to a solvent molecule, even though
this fact is not explicitly shown in the reaction, the water molecule exhibits Bronsted acidity. Note that it is easy to tell that
this is a displacement reaction instead of just a complex formation reaction because, while there is only one base in the
reaction, there are two acids. A complex formation reaction only occurs with a single acid and a single base. A double
displacement, or metathesis, reaction only occurs with two acids and two bases.
(b)
CH3[B12] + Hg 2+ Æ [B12]+ + CH3Hg + ; [B12] designates the Co-macrocycle, vitamin B12.
(Note: [B12] designates the Co center of the macrocyclic complex called vitamin B12). This is a displacement reaction. The
Lewis acid Hg 2+ displaces the Lewis acid [B12] from the Lewis base CH–.
(c)
KCl + SnCl2 Æ K+ + [SnCl3]–
This is also a displacement reaction. The Lewis acid SnCl2 displaces the Lewis acid K+ from the Lewis base Cl-.
(d)
AsF3 (g) + SbF5 (l) Æ [AsF2]+ [SbF6]– (s)
Even though this reaction is the formation of an ionic substance, it is not simply a complex formation reaction. It is a
displacement reaction. The very strong Lewis acid SbF5 (one of the strongest known) displaces the Lewis acid [AsF2]+
from the Lewis base F–.
(e)
Ethanol dissolves in pyridine to produce a nonconducting solution.
A Lewis acid/base complex formation reaction between EtOH (the acid) and py (the base) produces the adduct EtOH-py,
which is held together by the kind of dative bond that you refer to as a hydrogen bond.
3.
Select the compound in each set (indicated by {}) with the named characteristic, and justify your choice:
{BeCl2 ; BCl3}
{B( nBu) 3 ; B( tBu) 3}
a) Strongest Lewis acid:
{BF3 ; BCl3 ; BBr3}
{(CH3) 3N ; (CH3CH2) 3N}
{2-methylpyridine ; 4-methylpyridine}
b) More basic towards B(CH3) 3:
(a) Strongest Lewis acid:
BF3
<
BCl 3
<
BBr3
1
The simple argument that more electronegative substituents lead to a stronger Lewis acid does not work in this case.
Boron tribromide is observed to be the strongest Lewis acid of these three compounds. The shorter boron-halogen bond
distances in BF3 and BCl3 than in BBr3 are believed to lead to stronger halogen-to-boron p-p n bonding (see Section 5.10).
According to this explanation, the acceptor orbital (empty p orbital) on boron is involved to a greater extent in n bonding in
BF3 and BCl3 than in BBr3, the Lewis acidity of BF3 and BCl3 are diminished relative to
BBr3.
BCl 3
BeCl 2 <
Boron trichloride is expected to be the stronger Lewis acid of the two for two
reasons. The first reason, which is more obvious, is that the oxidation number of
boron in BC13 is +3 while for the beryllium atom in BeCl2 it is only +2. The second
reason has to do with structure. The boron atom in BCl3 is only three-coordinate,
A piece of the infinite linear chain
leaving a vacant site to which a Lewis base can coordinate. Since BeCl2 is polymeric,
structure of BeCl2. Each Be atom
each beryllium atom is four-coordinate, and some Be-Cl bonds must be broken before
is four-coordinate, and each Cl
adduct formation can take place.
atom is two-coordinate. The
polymeric chains are formed by
extending this piece to the right
and to the left
>
B(t-Bu)3
B(n-Bu)3
The Lewis acid with the unbranched substituents, B(n-Bu)3, is the stronger of the two because, once the complex is
formed, steric repulsions between the substituents and the Lewis base will be less than with the bulky, branched
substituents in B(t-Bu)3.
(b) More basic toward B(CH3)3:
>
Et3N
Me 3N
These two bases have nearly equal basicities towards the proton in aqueous solution or in the gas phase. Steric
repulsions between the substituents on the bases and the proton are negligible, since the proton is very small. However,
steric repulsions between the substituents on the bases and molecular Lewis acids like BMe3 are an important factor in
complex stability, and so the smaller Lewis base NMe3 is the stronger in this case.
<
4- CH3C5H4N
2-CH3C5H4N
As above, steric factors favor complex formation with the smaller of two bases that have nearly equal Bronsted basicities.
Therefore, 4-Me-py is the stronger base toward BMe3, since the methyl substituent in this base cannot affect the strength
of the B-N bond by steric repulsions with the methyl substituents on the Lewis acid.
4.
Using HSAB concepts, which of the following reactions are predicted to have an equilibrium consent greater than 1?
Unless otherwise stated, assume gas phase or hydrocarbon solution and 25 °C.
(Note: I have to use ⇔ to indicate equilibrium.)
a) R3PBBr3 + R3NBF3 ⇔ R3PBF3 + R3NBBr3
b) SO2 + (C 6H5) 3PHOC(CH3) 3 ⇔ (C 6H5) 3P SO2 + HOC(CH3) 3
c) CH3HgI + HCl ⇔ CH3HgCl + HI
d) [AgCl2] – (aq) + 2 CN – (aq) ⇔ [AgCN2] – (aq) + 2 Cl – (aq)
(a) R3PBBr3 + R3NBF3 ⇔ R3PBF3 + R3NBBr3
From the discussion in Section 5.14, you know that phosphines are softer bases than amines. So, to determine the position
of this equilibrium, you must decide which Lewis acid is softer, since the softer acid will preferentially form a complex with a
soft base than with a hard base of equal strength. Boron tribromide is a softer Lewis acid than BF3, a consequence of the
relative hardness and softness of the respective halogen substituents. Therefore, the equilibrium position for this reaction
will lie to the left, the side with the soft-soft and hard-hard complexes, so the equilibrium constant is less than 1. In general,
it is found that soft substituents (or ligands) lead to a softer Lewis acid than for the same central element with harder
substituents.
(b) SO2 + (C6H5)3PHOC(CH3)3 ⇔ (C6H5)3P SO2 + HOC(CH3)3
In this reaction, the soft Lewis acid sulfur dioxide displaces the hard acid t-butyl alcohol from the soft base
triphenylphosphine. The soft-soft complex is favored, so the equilibrium constant is greater than 1.
2
(c) CH3HgI + HCl ⇔ CH3HgCl + HI
Iodide is a softer base than chloride, an example of the general trend that elements later in a group are softer than their
progenors. The soft acid CH3Hg + will form a stronger complex with iodide than with chloride, while the hard acid H+ will
prefer chloride, the harder base. Thus, the equilibrium constant is less than 1.
(d) [AgCl 2]– (aq) + 2 CN– (aq) ⇔ [Ag(CN)2]– (aq) + 2 Cl – (aq)
Cyanide is a softer and generally stronger base than chloride (see Table 5.4). Therefore, cyanide will displace the relatively
harder base from the soft Lewis acid Ag+ . The equilibrium constant is greater than 1.
5.
The molecule (CH3) 2N–PF2 has two basic atoms , P and N (draw the Lewis structure!) One is bound to B in a complex
with BH3, the other to B in a complex with BF3. Decide which is which and state your reasons.
The phosphorus atom in Me2NPF2 is the softer of the two basic sites, so it will bond more strongly with the softer Lewis
acid BH3. The hard nitrogen atom will bond more strongly to the hard Lewis acid BF3.
6.
The enthalpies of reaction of trimethylboron with NH3, CH3NH2, (CH3) 2NH, and (CH3) 3N are –58, –74, –81, and –74 kJ
mol–1, respectively. Why is trimethylamine out of line?
Since trimethylamine is the strongest Brønsted base in the gas phase, the reason that it does not form the most stable
complex with trimethylboron can only be steric repulsions between the methyl substituents on the acid and those on the
base.
7.
8.
Into which group of the qualitative analysis scheme will each of the following ions fall? Use acid-base classifications to
make your judgments. If the metal ion gives a precipitate, write the formula of the precipitate. Tl +; Cu+; Rb+; Cr3+; Sn2+;
Pr3+; Sb3+; Au+; Hg 22+
Tl+
Group I
Cu +
Group II
Rb +
Cr3+
Group V
Group III
Sn 2+
Group II
Pr3+
Group III
Sb 3+
Group II
Au +
Group I
Hg 22+
Group I
Answer the following questions about the following metal ions. Use all the acid-base principles discussed in the course
so far to guide your responses: (a) Zr4+; (b) Ag+
-
9.
Very soft acid cation precipitates with the borderline base chloride
Tl+ + Cl– Æ TlCl (s)
Forms complex CuCl2– ion in group I, so carries over to Group II as the sulfide
2 Cu + + S2– Æ Cu 2S (s)
Non-acidic cation (Brønsted theory); does not precipitate at any pH
Hard acid cation precipitates with the hard hydroxide base because of strong Brønsted acidity
Cr3+ + 3 OH– Æ Cr(OH)3 (s)
Borderline soft acid precipitates with very soft sulfide base
Sn 2+ + S2– Æ SnS (s)
Hard acid cation precipitates with the hard hydroxide base because of strong Brønsted acidity
Pr3+ + 3 OH– Æ Pr(OH)3 (s)
Borderline soft acid precipitates with very soft sulfide base
Sb 3+ + S2– Æ Sb 2S3 (s)
Forms complex AuCl2– ion in group I, so carries over to Group II as the sulfide
2 Au+ + S2– Æ Au 2S (s)
Very soft acid cation (this is the calomel ion) precipitates with the borderline base chloride
Hg 2+ + Cl– Æ Hg 2Cl2 (s)
is its bromide soluble or insoluble?
is its selenide soluble or insoluble?
is its oxide soluble or insoluble?
in which group of the qualitative analysis
scheme does this metal ion occur?
is this metal a lithophile?
which is the most likely mineral source:
a silicate, a sulfide, or sea water?
(a) Zr4+ soluble
(b) Ag+ insoluble
4+
(a) Zr soluble in acidic solution (b) Ag+ insoluble
4+
(b) Ag+ soluble
(a) Zr insoluble
(a) Zr4+ Group III
(a) Zr yes
(b) Ag+ Group I
(b) Ag no
(a) Zr4+ silicate
(b) Ag+ sulfide
4+
+
Classify each Lewis acid and each Lewis base in the following complexes as hard, soft or borderline:
a) Zn(OH) 42b) Ag(CN) 2c) TiF4
a) Zn(OH)42-
Zn borderline
OH– hard
3
b) Ag(CN)2c) TiF4
Ag + soft
Ti4+ hard
CN– soft
–
F hard
10. In which "group" of the qualitative analysis scheme do the following ions fall? If a precipitate forms, give a balanced
ionic equation for its formation. Justify your answer in detail using the theoretical principles developed in this course.
(b) Sr2+
a) Tl +
+
(a) Tl It is in group I of the qualitative analysis scheme. That is, it ppts as the Cl– from and HCl-containing solution
Tl+ + Cl– Æ TlCl (s)
It is a very soft Lewis acid. It reacts sufficiently strongly with the borderline Lewis base Cl– that the adduct
precipitates. (Another way to express this is in terms of the affinity of Tl+ for Cl– is very high)
2+
(b) Sr It is in group IV of the scheme. That is , it precipitates as the carbonate from a basic solution to which sodium carbonate
is added.
Sr2+ + CO32– Æ SrCO3 (s)
It is a hard Lewis acid, but an insufficiently strong Brønsted acid to precipitate the hydroxide from aqueous solution.
However, the carbonate will precipitate because of a favorable lattice energy for the large 2+ ion with the tetra-atomic
dianion.
11. Predict the product(s) of the following reactions. Use the Lewis acid/base approach, and any principles developed in the
course to guide your predictions. Identify the acids and bases involved and categorize each reaction as (i) acid
displacement, (ii) base displacement, (iii) complex formation or (iv) metathesis.
Me3P-BF3 + Et3N-BBr3 Æ Me3P-BBr3 + Et3N-BF3
B1 A1
B2 A2
B1 A2
B2 A1
A metathesis reaction, driven by HSAB factors, since BF3 is harder than BBr3, due to the greater inductive effect of F over
Br, such that BF3 prefers the harder base Et 3N over the softer base Me3P (both bases have a lone pair on N or P).
(a)
(b)
SO2 + AsMe3 Æ O2S←:AsMe 3
Complex (adduct) formation between the acid sulfur dioxide and the base trimethylarsine.
(c)
KF + PF5 Æ [K]+ [PF6]–
A Lewis acid displacement reaction where the acid PF5 displaces the acid K+ from the base F–.
12. In which group of the qualitative analysis scheme do the following ions fall? If a precipitate forms, give a balanced ionic
equation for its formation. Justify your answer in detail using the theoretical principles developed in this course.
(b) Ce 4+
(c) Sr2+
(a) Pb 2+
(a) Pb 2+ It is in group I of the qualitative analysis scheme. That is, it ppts as the Cl– from and HCl-containing solution
Pb 2+ + 2 Cl– Æ PbCl2 (s)
It is a very soft Lewis acid (χ = 2.33!). It reacts sufficiently strongly with the borderline Lewis base Cl– that the adduct
precipitates.
(b) Ce 4+ It is in group IV of the scheme. That is, it precipitates as the carbonate from a basic solution to which sodium carbonate
is added.
Ce4+ + 4 OH– Æ Ce(OH)4 (s)
It is a hard Lewis acid, and a very strong Brønsted acid, so it precipitates with the hard base to give an insoluble
hydroxide.
(c) Sr2+ It is in group IV of the scheme. That is, it precipitates as the carbonate from a basic solution to which sodium carbonate
is added.
Sr2+ + CO32– Æ SrCO3 (s)
It is a hard Lewis acid, but an insufficiently strong Brønsted acid to precipitate the hydroxide from aqueous solution.
However, the carbonate will precipitate because of a favorable lattice energy for the large 2+ ion with the tetra-atomic
dianion.
13. Predict the product(s) of the following reactions. Use the Lewis acid/base approach, and any principles developed in the
course to guide your predictions. Identify the components and categorize each reaction as (i) acid displacement, (ii)
base displacement, or (iii) complex formation.
Start by writing out the full Lewis structures of each component in these reactions! This will allow you to identify which are
acids, which bases.
(a)
KF + SnF2 Æ [K]+ [SnF3]–
A Lewis acid displacement reaction where the acid SnF2 displaces the acid K+ from the base F–.
(b)
(CH3)3B + S(CH3)2 Æ (CH3)3B←:S(CH3)2
4
Complex (adduct) formation between the acid trimethylboron and the base dimethylsulfide.
(c)
(CH3)2Se:→BF3 + NC5H5 Æ C5H5N:→BF3 + (CH3)2Se
A Lewis base displacement reaction where the base pyridine displaces the base dimethylselenide from the acid boron
trifluoride. The driving force here is again the HSAB principle, since BF3 is hard, as is pyridine, while dimethylselenide is
soft.
14. a) Predict the product(s) of the following reactions. Use the Lewis acid/base approach, and any principles developed in
the course to guide your predictions.
(i) SbCl5 + Me2S →
(ii) Me3Al + Me3P →
Me2S:→SbCl5 Complex formation as indicated by dative bond.
Me3P:→Al Me3
Complex formation as indicated by dative bond.
b) Identify the components and categorize each reaction as acid displacement, base displacement, complex formation or
metathesis.
(i) IBr + Et 2O: → Et2O:→BrI
LA
LB
A/B complex
A complex formation reaction. Note that here the hard base diethylether adds to the more electronegative of the two
halogen atoms in molecular IBr.
(ii) ClMe3Sn←:OC4H8 + CH3CN → ClMe3Sn←:NCCH3 + C4H8O:
LA
LB1
LB2
LA
LB2 LB1
This is a base displacement reaction, in which the softer Lewis base (N - borderline donor atom) displaces the harder Lewis
base (O- hard donor atom) from the soft Lewis acid (Sn - soft acceptor atom)
15. The f-block elements are found as M3+ lithophiles in silicate minerals. What does this indicate about their hardness?
Compare these elements with Hg(II) and Zn(II). Since the trivalent lanthanides and actinides are found as complexes with
hard oxygen bases (i.e. silicates) and not with soft bases such as sulfide, they must be hard. Since they are found
exclusively as silicates, they must be considered very hard, unlike the borderline behavior of Zn(II).
16. Consider the reaction forming metasilicates (chain silicates) from carbonates:
n CaCO3 (s) + n SiO2 (s) Æ [CaSiO3] n (s) + n CO2 (g)
Identify the stronger acid between SiO2 and CO2
Remember, we are asking about Lewis acidity: in this reaction, the Si atom accepts one additional oxygen atom, as shown in
the sketch below. Oxygen is a hard Lewis acid. Although carbon is normally categorized as a Lewis base, its role in the
linear O=C=O molecule is clearly that of an acid, with the oxygens as the bases. Carbon is soft, so it has less tendency to
coordinate the hard oxygen base than the much harder (more electropositive) silicon. Thus in this reaction, silica is a
stronger Lewis acid than is Carbon
O
C
O
O
+
O
O
Si
O
O
O
O
Si
O
+
O
C
O
O
17. The ores of titanium, tantalum, and niobium may be brought into solution near 800 °C using sodium disulfate. A
simplified version of the reaction is:
TiO2 + Na2S 2O7 Æ Na 2SO4 + TiO(SO4)
Identify the acids and bases.
The key to this, and all such reactions, is to identify the ions involved. We do so by recognizing those ions that we know,
such as Na + and SO42–. This means that in the starting materials we are dealing with the S2O72– anion, and in the products
the TiO2+ cation. You would probably need to look up the structure of the disulfate ion in the text or a reference book. Once
we do this, we recognize that, effectively, in this reaction the acid SO3 is transferred from SO42– to O2– from titanium. It is
therefore a base displacement reaction, with the stronger base O2– displacing the weaker base SO42–. The next ionic
equation becomes as follows (the transferred SO3 is shown in bold):
5
2-
O
O O
S
O S
O
O
O
O
+
2
O
O
S
O
O
(PS: why would oxide be a stronger Lewis base than sulfate?)
18. Sketch an outline of the s and p blocks of the periodic table and indicate on it the elements that form (a) strongly acidic
oxides and (b) strongly basic oxides, and (c) show the regions for which amphoterism is common.
See the diagram at right. If you cannot write out the s-, p- and d blocks from
memory, you should spend some time learning that part of the periodic
table. This knowledge will permit you to integrate many chemical facts into a
logical pattern of trends. The elements that form basic oxides in plain type,
those forming acidic oxides in outline type, and those forming amphoteric
oxides in boldface type. Note the diagonal region from upper left to lower
right that includes the elements forming amphoteric oxides. The elements Ge,
Sn, Pb, As, Sb, and Bi form amphoteric oxides only in their lower oxidation
state (II for Ge, Sn, and Pb, III for As, Sb, and Bi). They form acidic oxides in
their higher oxidation state (IV for Ge, Sn, and Pb, V for As, Sb, and Bi).
19. Identify the conjugate bases corresponding to the following acids:,
(a) [Co(NH3) 5(OH2)] +3: A conjugate base is a species with one fewer proton than the parent acid. Therefore, the conjugate
base in this case is [Co(NH3)5(OH)]2+, shown below right (L = NH3).
(b) HSO4–: The conjugate base is S042–.
(c) CH3OH,: The conjugate base is CH3O–.
(d) H2PO4–: The conjugate base is HPO42–.
(e) Si(OH) 4,: The conjugate base is SiO(OH)3–.
(f) HS –: The conjugate base is S2–.
(g) [Fe(OH2) 6] 2+: The conjugate base is [Fe(OH)(OH2)5]+
(similar to (a) with L = H2O and Fe in place of Co).
20. Identify the conjugate acids of the bases:
(a) Sb(OH) 3,: A conjugate acid is a species with one more proton than the parent
base. Therefore, the conjugate acid in this case is (H2O)Sb(OH)2+ .
(b) C5H5N (pyridine): The conjugate acid is the pyridinium ion, C5H6N+ , shown at
right.
(c) HPO42–: The conjugate acid is H2PO4–.
(d) O2–: The conjugate acid is OH–.
(e) CH3COOH (acetic acid): The conjugate acid is CH3C(OH)2+ , shown above next to the
pyridinium ion.
(f) [Co(CO) 4] –: The conjugate acid is HCo(CO)4, shown at right.
(g) CN–: The conjugate acid is HCN
21. List the bases HS –, F–, I–, and NH2– in order of increasing proton affinity.
You should make use of Table 5.2 (notes p.52) to answer this question. The species with the greatest proton affinity will be
the strongest base, and its conjugate acid will be the weakest acid. The weakest acid will have the smallest value of Ka (or
the most positive value of pK a ). Since Table 5.2 shows that HI is a stronger acid than HF which is a stronger acid than H2S,
a partial order of proton affinity is I– < F– < HS–. Since NH3 is a very weak acid, NH2– must be a very strong base. Therefore,
our final list, in order of increasing proton affinity, is I– < F– < HS– < NH2–.
22. Using Pauling's rules (consult the table of examples given in the notes, and consider also their conjugates), and taking
the concept of solvent leveling into account, identify which bases from the following lists are (a) too strong to be studied
experimentally; (b) too weak to be studied experimentally; or (c) of directly measurable base strength under the
conditions specified
(i) In water solution: CO32–, O2–, ClO4–, and NO3–. You can interpret the term "studied experimentally" to mean that the base
in question exists in water (i.e. it is not completely protonated to its conjugate acid) and that the base in question can be
partially protonated (i.e. it is not so weak that the strongest acid possible in water, H3O+ , will fail to produce a measurable
6
amount of the conjugate acid). The pKb of H2O is 15.7 (the weakest base in water), while pKb of OH– is –1.7 (the strongest
base in water). Thus bases with pKb between these two values can be studied in water. Using these criteria:
(a) CO32– has a pKa for its conjugate of 8 - 5 + 5 = 8, hence a pKb of 14 - 8 = 6. This fits the range of study in water.
(b) O2– is derived from water, for which Pauling's first rule doesn't work. But we know this species well. Since the base OH–
is the strongest base in water is its conjugate acid, it stands to reason that oxide must be a more powerful base. We can
estimate its pKa using Pauling's second rule by subtracting 5 from the value for hydroxide, which gives about -7. Oxide can
definitely not be studied in aqueous solution. It is instantly solvent leveled to hydroxide.
(c) ClO4– has a pKa for its conjugate of 8 - 15 = -7. Hence its pKb of 14 - (-7) = 21. It is a weaker base than water, so cannot
be studied in aqueous solution.
(d) NO3 has a pKa for its conjugate of 8 - 10 = -2. Hence its pKb of 14 - (-2) = 16. It is thus also a weaker base than water, so
cannot be studied in aqueous solution.
(ii) In liquid H2SO4 solution: HSO4–, NO3–, ClO4–
(a) HSO4–: The hydrogen sulfate ion, HSO4– is the strongest base possible in liquid sulfuric acid and it has a pKa for its
conjugate of 8 - 10 = -2. Hence its pKb of 14 - (-2) = 16. However, since acids can protonate it, it is not too strong to be
studied experimentally.
(b) NO3–: It also has an (estimated) pKb of 14 - (-2) = 16 by Pauling's rules. We expect it to be quite comparable to hydrogen
sulfate ion, and hence it should be possible to study it experimentally.
(c) ClO4– has a pKa for its conjugate of 8 - 15 = -7. Hence its pKb of 14 - (-7) = 21. It is a weaker base than hydrogen sulfate
ion, so cannot be studied in liquid sulfuric acid solution.
23. The aqueous solution pKa values of HOCN, H2NCN, and CH3CN are approximately 4, 10.5, and 20 (estimated),
respectively. Explain the trend in these compounds in which a –C≡N group has replaced a hydrogen in H2O, NH3 and
CH4 by comparison with the acidity of the parent hydrides. Is the cyano group electron donating or electron
withdrawing?
A comparison of the aqueous pKa values is necessary to answer this question. These are:
CH3CN, 20
HOCN, 4
H2NCN, 10.5
NH3, > 14
CH4, very large
H2O, 14
You know that the values of pKa are very large for ammonia and methane because these compounds are not normally
thought of as acids (this implies that they are extremely weak acids). Now, in all three cases, the cyanocontaining
compound has a lower pKa (a higher acidity) than the parent compound. In the case of H2O and HOCN, the latter
compound is 10 orders of magnitude more acidic than water. The deprotonation equilibrium involves the formation of an
anion, the conjugate base of the acid in question. For example:
Since a lower pKa means a larger Ka , this suggests that the anion OCN– is better stabilized than OH–. This occurs because
the -CN group is more electron withdrawing than the -H substituent.
24. The pKa value of HAsO42– is 11.6. Is this value consistent with the two Pauling rules?
Pauling's first rule for predicting the pKa of a mononuclear oxoacid is pKa = 8 - 5p (where p represents the number of oxo
groups attached to the central element).
Since p = 1, the predicted value of pKa (l) for H3AsO4 is 8 - 5 x 1 = 3.
Pauling's second rule for predicting the pKa of a mononuclear oxoacid is that successive pKa
values for polyprotic acids increase by five units for each successive proton transfer. Since
pKa (l) for H3AsO4 was predicted to be 3, the predicted value of pKa for HAs04–, which is pKa (3)
for H3AsO4, is 3 + (2)(5) = 13. The actual value, which differs by 1.5 pKa units, is 11.5. This
illustrates that Pauling's rules are only approximate.
25. Draw the structures (VSEPR) and indicate the charges of the tetraoxoanions of Si, P, S and Cl. What are their point
groups? Summarize and account for the trends in the pKa values of their conjugate acids.
The structures of these four anions, which can be determined to be tetrahedral using VSEPR, are shown below. They are all
tetrahedral, and thus belong to the point group Td. As can be seen, the charge on the anions decreases from -4 for the
silicon-containing species to -1 for the chlorine-containing species. The charge differences alone would make SiO44– the
most basic species. Hence HSiO43– is the least acidic conjugate acid. The acidity of the four conjugate acids increases in
the order HSiO43– < HPO42– < HSO4– < HClO4.
7
26. Which of the following pairs is the stronger Brønsted acid? Give reasons for your choice. Where possible, calculate the
pK a values to confirm your answer.
(a) [Fe(OH2)6]3+ or [Fe(OH2)6]2+ (using radii from p.55 of notes and χ = 1.83 from p.32)

 32
pKa = 1514
. − 0.8816
+ 9.60(183
. − 150
. ) = 2.2 The stronger acid.

 0.78
2

 2
For the Fe(II) complex: pKa = 1514
. − 0.8816
+ 9.60(183
. − 150
. ) = 8.5 The weaker acid.

 0.92
For the Fe(III) complex:
(b) [Al(OH2)6]3+ or [Ga(OH2)6]3+

 32
pKa = 1514
. − 0.8816
+ 9.60(161
. − 150
. )  = 2.4 The slightly stronger acid.

 0.67
2

 3
For the Ga compound: pKa = 1514
. − 0.8816
+ 9.60(181
. − 150
. ) = 2.1 The slightly weaker acid. Note

 0.76
For the Al compound:
that the difference is entirely due to the higher electronegativity of Ga, a consequence of the so-called Scandide
contraction, the fact that Ga is immediately proceeded by the 3d series, and that the filled 3d 10 shell is poor at shielding
the valence electrons from the extra nuclear charge gained along with the 10 extra electrons. Thus gallium chemistry is
significantly out of line with that expected for a heavier analog of aluminum, and this is just one example of this
occurrence. (In the table on p.55, you can see experimental pKa data for the 3+ ions Al, Ga, In and Tl. They oscillate in
acid strength as follows: 5.0, 2.6, 4.0, 0.6. Thus Tl is the strongest acid (it follows the first 4f14 series, known as the
Lanthanide contraction), followed by Ga, then comes In and Al is the weakest. Note that all of these values run
counter to the trnds in ionic radii, which normally are dominant for acid trends down a group in the periodic table.
(c) Si(OH)4 or Ge(OH)4: Our equation for the hydrolysis of aquated cations is not directly applicable to these two species.
The equilibria that they undergo in aqueous solution result in the loss of two protons, so that they are definitely still
"acidic" compounds, even if they have formally already produced 4 hydronium ions in reaction with water:
M(OH)4 + 2 H2O ⇔ MO2(OH)22– + 2 H3O+
How will the differences between the two work out? We see that χ for Si and Ge are 1.90 and 2.01, while the radii are
0.40 and 0.53 Å. Again the two work in opposite directions. It would be very difficult to predict which of these species
were the more acidic given the structural changes from normal aqua cations.
(d) HClO3 or HClO4 These two acids are shown at right. According to
Pauling's rule 1 for mononuclear oxoacids, the species with more oxo
groups has the lower pKa and is the stronger acid. Thus, HClO4 is a
stronger acid (pKa = -7) than HClO3 (pKa = -2). Note that the oxidation
state of the central chlorine atom in the stronger acid (+7) is higher than in
the weaker acid (+5).
(e) H2CrO4 or HMnO4 As in part (d), above, the oxidation states of these two
acids are different, +6 for the chromium atom in H2CrO4 and +7 for the manganese atom in HMnO4. The species with the
higher central atom oxidation state, HMnO4, is the stronger acid (pKa = -7). Note that this acid has more oxo groups,
three, than H2CrO4 (pKa = -2), which has two.
(f) H3PO4 or H2SO4. The oxidation state of sulfur in H2SO4 (pKa = -2) is +6 while the oxidation state of phosphorus in H3PO4
(pKa = +3) is only +5. Furthermore, sulfuric acid has two oxo groups attached to the central sulfur atom while
phosphoric acid has only one oxo group attached to the central phosphorus atom. Therefore, on both counts (which
by now you can see are really manifestations of the same underlying atomic properties) H2SO4 is a stronger acid than
H3PO4.
8
27. Arrange the oxides Al2O3, B2O3, BaO, CO2, Cl2O7, SO3 in order from the most acidic through amphoteric to the most basic.
First you pick out the intrinsically acidic oxides, since these will be the least basic. The compounds B2O3, CO2, Cl2O7, and
SO3 are acidic, since the central element for each of them is found in the acidic region of the periodic table (see the s and p
block diagram in the answer to question #1). The most acidic compound, Cl207, has the highest central-atom oxidation state,
+7, while the least acidic, B203, has the lowest, +3. Of the remaining compounds, A1203 is amphoteric, which puts it on the
borderline between acidic and basic oxides, and BaO is basic. Therefore, a list of these compounds in order of increasing
basicity is Cl2O7 < SO3 < CO2 < B2O3 < Al2O3 < BaO.
28. Arrange the acids HSO4–, H3O+, H4SiO4, CH3GeH3, NH3, HSO3F in order of increasing acid strength.
The weakest acids, CH3GeH3 and NH3, are easy to pick out of this group since they do not contain any -OH bonds.
Ammonia is the weaker acid of the two, since it has a lower central-atom oxidation state, +3, than that for the germanium
atom in CH3GeH3, which is +4. Of the remaining species, note that HSO3F is very similar to H2SO4 as far as structure and
sulfur oxidation state (+6) are concerned, so it is reasonable to suppose that HSO3F is a very strong acid, which it is. The
anion HSO4– is a considerably weaker acid than HSO3F, for the same reason that it is a considerably weaker acid than H2SO4,
namely Pauling's rule 2 for mononuclear oxoacids. Since HSO4– is not completely
deprotonated in water, it is a weaker acid than H3O+ , which is the strongest possible acidic
species in water. Finally, it is difficult to place exactly Si(OH)4 in this group, for reasons
described in #9. It is certainly more acidic than NH3 and CH3GeH3, and it turns out to be less
acidic than HSO4–, despite the negative charge of the latter species. Therefore, a list of
The structures of H2SO4
these species in order of increasing acidity is NH3 < CH3GeH3 < ??Si(OH)4 < HSO4– < H3O+ <
and HSO3F.
HSO3F.
29. The ions Na+ and Ag + have similar radii. Which aqua ion is the stronger acid? Why? Confirm your answer by
calculating the pK a values of the aqua ions [M(OH2) 6] +.
Even though these two ions have about the same ionic radius, Ag+ –OH2 bonds are much more covalent than Na + –OH2
bonds, because of the much higher electronegativity of silver. The greater covalence of the Ag+ –OH2 bonds has the effect
of delocalizing the positive charge of the cation over the whole aqua complex. As a consequence, the departing proton is
repelled more by the positive charge of Ag+ (aq) than by the positive charge of Na + (aq), and the former ion is the stronger
aqua cation acid. The reason that silver is more electronegative is due to the filled 3d 10 level that is poor at shielding the
valence electrons from the high nuclear charge.
12
= 14.4
116
.
 12

For Ag+ : pKa = 1514
. − 0.8816
+ 9.60( 193
. − 150
. )  = 10.8
.
129

For Na + :
pKa = 1514
. − 0.8816
30. Which of the elements Al, As, Cu, Mo, Si, B, Ti form oxide polyanions and which form oxide polycations?
The aqua ions of metals that have amphoteric oxides generally undergo polymerization to polycations. The elements Al, Cu,
and Ti fall into this category. On the other hand, polyoxoanions (oxide polyanions) are important for some of the early dblock metals, especially for V, Mo, and W. (See SAL sections 5.7 and 5.8 for more details.)
31. When a pair of aqua cations forms an M–O–M bridge with the elimination of water, what is the general rule for the
change in charge per M atom on the ion?
Consider the following example of the formation of a polymerized aqua cation:
The charge per aluminum atom is +3 for the mononuclear species on the left hand side
of the equation but only +2 for the dinuclear species on the right hand side. Thus,
polycation formation reduces the average positive charge per central M atom by +1
per M. (The molecular structure of the dimerized aluminum ion is shown at right.)
9
32. (a) Write a balanced equation for the formation of P2O74– from PO43–. (b) Write a balanced equation for the dimerization
of the complex [Fe(OH2) 6] 3+ to give [(H2O) 4Fe(OH) 2Fe(OH2) 4] 4+.
The two balanced equations are shown below. Note that the condensation reactions
involve a neutralization of charge, either by adding H+ to a highly charged anion or by
removing H+ from a highly charged cation. The structure of P4O124– , which is called
cyclotetrametaphosphate, is also shown at right. It is the dimer of the formula given in the
question, which is commonly used, but which species actually does not exist! (That is,
P2O74– is the empirical formula, and P4O124– is the correct molecular formula.)
The
structure of the
ion in the salt
[NH4+][P4O12]
P4O124–
33. Consider the aqua ions resulting from the hydration of the following cations: U3+, Ag+, Pa5+, C 4+, As3+, Tl +, Th 4+
(a) Calculate the pK a and classify the acidity of each aqua ion.
NB: you need the radii provided in the table from the section on ionic solids to answer this question! Assume CN 6.
(b) Write an equation for the reaction of their chlorides with water. Which of these would give cloudiness or
precipitation? The ones with a (s) in the product produce cloudiness or precipitate. CCl4 is immiscible with water.
UCl3 (s) + 6 H2O Æ U(OH)3 (s) + 3 H3O+ (aq) + 3 Cl– (aq)
AgCl (s) + H2O Æ no reaction
2 PaCl5 (s) + 15 H2O Æ Pa2O5 (s) + 10 H3O+ (aq) + 10 Cl– (aq)
CCl4 (l) + H2O Æ no reaction (but for kinetic reasons)
AsCl3 (s) + 6 H2O Æ As(OH)3 (s) + 3 H3O+ (aq) + 3 Cl– (aq)
TlCl (s) + (H2O ) Æ Tl+ (aq) + Cl– (aq)
ThCl4 (s) + 8 H2O Æ ThO2 (s) + 4 H3O+ (aq) + 4 Cl– (aq)
(c) If the solutions were adjusted to final pH's of 5.5 to 7, in what chemical form would each element be present? (See
page 57 in the lecture notes.)
U3+ (aq) (not UO22+ because of oxidation state); Ag+ (aq); Pa2O5 (s); CCl4 (l); As(OH)3 (not HAsO42– because of oxidation
state); Tl+ (aq); ThO2 (s)
34. Extend Pauling's rules to predict the approximate basicity of the following polyoxo anions: (O3P-O-PO3) 4–,
pyrophosphate ion; (O3P-O-PO2-O-PO3) 5–, tripolyphosphate ion. (Hint: consider the acidity of the conjugate acids).
For the conjugate H2O3P-O-PO3H2, we apply Pauling's rule per phosphorus atom: pKa = 8 – 5 = 3. Now we deprotonate on
average twice per phosphorus, so 3 +10 = 13. The pKb = 1.
For the congjugate H2O3P-O-P(OH)O-O-PO3H2, we apply Pauling's rule for the terminal atoms to get 3 as above, while for the
central atom we get pKa = 8 – 5 = 3. Deprotonate an average of 1.67 per phosphorus, so 3+ 8.35 = 11.35. The pKb = 2.7.
Hence pyrophosphate is more basic (smaller pKb) than tripolyphosphate.
35. Consider the chlorine oxoanions (Cl is the central atom in all): ClO–, ClO2–, ClO3– & ClO4–.
(a) For each of these ions: (i) determine the Lewis structure, including resonance if applicable; (ii) determine the shape
by VSEPR theory, and sketch the molecule; (iii) assign the point group. Final answers should be as follows:
O C
∞v
Cl
:
O
C2v
Cl
:
O C
3v
Cl
O
O
Td
Cl
O
O
(b) Rank these anions in order of increasing Brønsted BASICITY in water, clearly indicating which is least and which is
most basic. Explain the origin of this series using your knowledge of acid-base theory in conjunction with the
structures of these anions which you have determined in part (a).
Consider their conjugates, which by Pauling's rules should be in the order of increasing acid strength in the order:
ClOH < OClOH < O2ClOH < O3ClOH
Hence the basicity of their conjugates will increase in the opposite order, with ClO4– (weakest base) < ClO3– (2nd weakest
base) < ClO2– (2nd strongest base) < ClO– (strongest base)
:
:
O
:
O
:
O
36. Give formulas for the following common acids: nitric acid, sulfuric acid, hydrobromic acid, perchloric acid, carbonic
acid. What is the oxidation number of the central atom in each of these compounds? How are these related to the periodic
group number of the element?
N +5
Group 15 (Group number - 10)
HNO3
10
"
H2SO4
HBr
HClO4
H2CO3"
S +6
Br +1
Cl +7
C +4
Group 16 (Group number - 10)
Group 17 (18 -Group number). This example is not an oxoanion.)
Group 17 (Group number - 10)
Group 14 (Group number - 10) - actually exists as the HCO3– ion in solution.
37. Define the term ‘‘amphoteric.’’ Write chemical equations to illustrate the amphoteric character of Al(OH) 3.
An amphoteric hydroxide will be neither basic nor acidic, and in the case of aluminum is a very insoluble solid. However,
aluminum hydroxide will react with either strong acid or strong base, and in the course of the reaction it will be dissolved.
See the following two equations:
Al(OH)3 (s) + 3 H+ (aq) Æ Al3+ (aq) + 3 H2O (l)
Al(OH)3 (s) + OH– (aq) Æ Al(OH)4– (aq)
38. Give examples of two basic oxides. Write equations illustrating the formation of each oxide from its component elements.
Write another chemical equation that illustrates the basic character of each oxide.
Li2O (s) + H2O (l) Æ 2 LiOH (aq)
Group 1: 2 Li (s) + ½ O2 (g) Æ Li2O (s)
BaO (s) + H2O (l) Æ Ba(OH)2 (aq)
Group 2: Ba (s) + ½ O2 (g) Æ BaO (s)
39. Give examples of two acidic oxides. Write equations illustrating the formation of each oxide from its component elements.
Write another chemical equation that illustrates the acidic character of each oxide.
P4O10 (s) + 6 H2O (l) Æ 4 H3PO4 (aq)
Group 15: 4 P4 (s) + 5 O2 (g) Æ P4O10 (s)
SO2 (s) + H2O (l) Æ "H2SO3" (aq)
Group 16: S8 (s) + 8 O2 (g) Æ 8 SO2 (s)
11