Fundamentals of Chemistry

53 Pages·1980·1.47 MB·English

by  Frank Brescia

Fundamentals
of
Chemistry
FOURTH EDITION
FRANK BRESCIA
JOHN ARENTS
HERBERT MEISLICH
AMOS TURK
The City College of the
City University of New York
ACADEMIC PRESS
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Cover: Chromium compounds illustrating the colors
associated with different states of oxidation,
polymerization, and hydration. From front cover to back:
chromium (VI) oxide, 00 ; ammonium
3
dichromate, (NH)Cr0; sodium chromate, NaCrÖ;
42 2 7 2 4
chromium (III) sulfate/water (1/10), Cr2(SOA)3(H2O)l0; chromium (III) oxide, Cr203;
chromium (III) chloride, CrCl; chromium (III) chloride/
3
water (1/10), CrCl(HO) . The kind cooperation of
3 2 l0
Pfaltz and Bauer, Inc., Division of Aceto Chemical
Company, Inc., is gratefully acknowledged.
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ISBN: 0-12-132392-7
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Printed in the United States of America
8.4 sizes of atoms and ions 159
TABLE 8.6 Atomic (covalent)t and ionic radii in Angstrom unitst
1A 2 A 3A 4 A 5A 6A 7A
H 0.28 to
1 0.38
H" 2.08
Li 1.33 Be 0.90! B 0.80 C 0.77 |N 0.73; O 0.74 i F 0.71
2 0.16 !
Li+ 0.60 Be2+ 0.31 B3+ 0.23 C4_ 2.60 N3" 1.71 O2- 1.40 F- 1.36
Na 1.54 Mg 1.36 AI 1.25 Si 1.15 I P 1.1 ! S 1 .02 CI 0.99
3
I
Na+ 0.95 Mg2+ 0.65 Al3+ 0.50 Si4+ 1.84 ci- 1.81
0.42 ;P3~ 2.12;S2-
O K 1.96 Ca 1.74 Ga 1.26 Ge 1.16 Br 1.14
O 1.22 ; As 1.19 Se
œ 4
LCUL K+ 1.33 Ca2+ 0.99 Ga3+ 0.62 Ge4+ 0.53: As3" 2.221 Se2" 1.98 Br- 1.95
Rb 2.16 Sr Ûîa In 1.44 Sn 1.41 Sb 1.381 Te 1.35 I 1.33
5
Rb+ 1.48 Sr2+ 1.131 ln3+ 0.81 Sn4+ 0.71 Sb3" 2.45; Te2" 2.21 i- 2.16
Cs 2.35 Ba 1.9^' In 1.48 Pb 1.47 Bi 1.46 Po At
6
Cs+ 1.69 Ba2+ 1.3^ Bi3+ 0.96
Fr Ra !
7
M
3B 4B 5B 6B 7B 8B . 1B 2B
Se 1.44 Ti 1.32 V 1.22 Cr 1.18 Mn 1.17 Fe 1.17 Co 1.16 Ni 1.15 Cu 1.17 Zn 1.25
4 Ti2+ 0.901 \p+ 0.74 Cr3+ 0.65 Mn2+ 0.80 Fe2+ 0.75 Co2+ 0.82 Ni2+ 0.78 Cu+ 0.96
Sc3+ 0.81 Ti4+ 0.68 V5+ 0.59 Cr6* 0.52 Mn7+ 0.46 Fe3+ 0.67 Co3+ 0.29 Ni3+ 0.35 Cu2+ 0.72 Zn2+ 0.74
Y 1.62 Zr 1.45 Nb 1.34 Mo 1.30 Tc 1.27 Ru 1.25 Rh 1.25 Pd 1.28 Ag 1.44 Cd 1.48
Q 5
O γ3 + 0.93 Zr4+ 0.80 Ag+ 1.26 Cd2+ 0.97
LL
LU La 1.69 Hi 1.44| Ta 1.34 W 1.30 Re 1.281 Os 1.26 Ir 1.27 Pt 1.30 Au 1.44 Hg 1.49
CL 6 I I l
Au+ 1.37 Hg2+ 1.10
Ac
I \
tCovalent radii (taken mostly from Table ot Interatomic Distances, London, Chem. Soc, Special Publ., 1958 and from the publications of R. T. Sanderson) are
applicable only to single-bonded atoms in mainly covalent molecules (Chapter 9). Ionic radii are from the publications of Linus Pauling.
t Recall that 1 angstrom = 10 w m, or 0.1 nm. It is especially convenient to use angstroms for atomic radii and bond lengths because the numbers then range
around 1 or 2.
electrons more than offsets the attraction of the increased nuclear charge, and
thus the atomic radius of oxygen is slightly higher than that of nitrogen. This
effect constitutes a minor exception to the previous generalization that atomic
radius decreases with increasing atomic number within a given period.
Within a given group the atomic radius generally increases with atomic
number. In going from one period to the next, electrons are added to higher
shells. For example, the radius of Mg is larger than that of Be, as shown in Figure
8.5b. As we progress down Group 2A, each step corresponds to an increase in the
9.2 Lewis theory of bonding 1 79
FIGURE 9.3 (a) Structure of water, an individual covalent molecule, (b) Structure of silicon dioxide, a
network covalent substance, (c) Structure of diamond, a network covalent substance.
The bonding between Si and O goes on and on in three dimensions; there are
no individual molecules of Si02. The entire piece of solid can be considered a
giant molecule. Si02 is called a network covalent substance. The formula Si0 2
is really the empirical formula—not the molecular formula—of the network
Diamond is a
network covalent substance, substance. The unit species in the solid is the bonded atom. Therefore Si0 2 melts
shown in Figure 9.3c
and boils only when covalent bonds between atoms are broken. It takes much
(m.p. 3800°C, b.p.
4830°C). more energy to break covalent bonds than it does to separate molecules. There-
fore network covalent substances have much higher melting and boiling points
than molecular covalent substances. Table 9.1 summarizes the differences in
properties. In this chapter we will examine the formation of ionic and covalent
bonds. Metallic bonding will be discussed in Chapter 20.
9.2 Before discussing the way atoms bond, we will first consider the symbols used
LEWIS THEORY by G. N. Lewis (1916) to represent the atoms. A Lewis symbol is the symbol of
OF BONDING the element surrounded by dots to represent electrons in the highest shell. The
electrons in the highest shell are called valence electrons. These are the electrons
which affect the chemical behavior of atoms. For atoms of representative
elements, the number of valence electrons equals the periodic group number.
Lewis symbols for the first ten elements and some of their ions are shown.
H·
H+ He:
j Hi-
Li- Be: B: •c·. •N· :Ö- :F·
Li+ Be2+ :N:3- : Ö : 2- :F:- :Ne:
Note that · N :3 , : O ·2 , and '· F : have the same electronic arrangement as
' Ne : ; they are isoelectronic. : H", Li+, and Be2+ are isoelectronic with He :
Transition elements are because they have the Is2 electron arrangement. Electrons in filled 5 and p
discussed in Chapter 21. orbitals are shown as a pair of dots in the Lewis symbol. It makes no difference
a single dot or pairs of dots are placed around the symbol. This symbolism is
180 types of chemical bonds
TABLE 9.1 A comparison of distinguishable properties of the main types of substances
MOLECULAR NETWORK
IONIC COVALENT COVALENT METALLIC
Electrical
conductivity
Solid Not No No Yes (high)
Liquid Yes Not No Yes (high)
Approximate 700° to -253 to 2000° to 357° to
range of 3600( DC + 600°C 6000°C 6000°C
boiling points
|A few exceptions exist.
very useful for the representative elements, but is not generally used for the
transition elements. Bonding of the transition elements can involve electrons in
more than one shell.
EXAMPLE 1 Write the Lewis symbols for AI, Al3+, Sn, Sn2+, Se, Se2", P, and P3".
ANSWER These are all atoms or ions of representative elements. For the uncharged atoms the number
of valence electrons is the same as the periodic group number. Therefore we have · AI :
(Group 3A), : Sn ■ (Group 4A), : Se : (Group 6A), and :P· (Group 5A).
Positive ions result when the atom loses electrons; negative ions result when the atom
gains electrons. Thus : Sn · loses 2 electrons to become : Sn 2+and "AI· loses 3 electrons
to become Al3+ (no valence electrons remain so no dots are shown). Note that the number
of electrons lost is the charge on the positive ion. For the negative ions, :Se : becomes
: Se :2" on gaining 2 electrons, and : P · becomes : P : 3_ on gaining 3 electrons. Note that
the number of electrons gained is the charge on the negative ion.
An important clue to the understanding of chemical bonding was the discov-
ery, in the late nineteenth century, of the noble gases and their apparent resist-
ance to chemical change. The relationship between the bonding of atoms and
the electronic configuration of noble gases was proposed in 1916 independently
by Walther Kossel and G. N. Lewis, and extended in 1919 by Irving Langmuir.
We imagine that they reasoned as follows: Electrons in atoms are involved in
chemical bonding. Atoms bond with one another so as to acquire new and more
stable electron configurations. Since the atoms of the noble gases were not
known to form chemical bonds, they must already have a stable arrangement of
electrons. Therefore all other atoms undergo bonding by gaining, or losing, or
Compounds of the
noble gases were first sharing electrons so as to acquire the electronic arrangement of a noble gas. With
discovered in 1962. the exception of : He, which has a Is 2 electron arrangement, each noble gas has
8 electrons with an s2p6 distribution in its highest shell (Section 8.2):
:Ne: :Ar: :Kr: :Xe: :Rn:
neon argon krypton xenon radon
2s22p6 3523p* 4s 2V 5s25p6 6s26p6
The stable ions of N, O, and F, for example, also have an octet of electrons:
9.3 ionic bond 181
:N: :0: F:
nitride oxide fluoride
ion ion ion
The need for 8 electrons gives the name octet rule to this concept. The elements
hydrogen, lithium (Li), and beryllium (Be) are close to He in the periodic table
and therefore tend to acquire a duet (instead of an octet) of electrons. There
are, however, many exceptions to the octet rule—even compounds of some of the
noble gases have been synthesized.
9.3 Recall that salts such as NaCl are ionic compounds. Somehow the ions of this
IONIC BOND salt form from the atoms. We use potassium bromide, KBr, a typical ionic salt, to
examine how this is done. A potassium atom, K · (atomic number 19), loses 1
electron to become K+, an ion isoelectronic with argon (atomic number 18), a
noble gas. A bromine atom, : Br · (atomic number 35), gains 1 electron to
become : Br : ~, an ion isoelectronic with krypton (atomic number 36), another
noble gas. Such transfer of electrons results in the formation of the ionic bond.
The formation of the ionic bond in potassium bromide is shown using Lewis
symbols:
ïElectron is transferred2
A half-headed arrow {-^) K· [Ar]45! + :Br· [Ar]3dX04s24p K+[Ar] + :Br:-[Kr]
indicates movement of a
single electron. argon krypton
configuration configuration
or
K· + :Br· K+:Br:~ (KBr, an ionic salt)
Lewis symbols a Lewis formula
Formulas of compounds using Lewis symbols are called Lewis formulas.
When atoms react by electron transfer, the number of electrons gained and
lost must be equal because the resulting ionic salt is neutral. We illustrate this
idea by the formation of ionic aluminum fluoride, A1F , from aluminum (Group
3
3A) and fluorine (Group 7A) using Lewis symbols:
+ 3 : F : (AIF )
3
Al '· has to lose 3 electrons, but · F : can gain only 1 electron. Therefore 3 · F ·
atoms are needed to accept the 3 electrons lost by 1 Al : atom.
EXAMPLE 2 Use Lewis symbols for the atoms and Lewis formulas for the ionic compounds to depict the
formation of (a) calcium hydride; (b) lithium oxide; (c) magnesium nitride.
ANSWER (a) The Lewis symbols are H · and Ca : (Group 2A). Two H · atoms are needed to accept
the 2 electrons from one Ca: atom, Ca: + 2H· -> Ca2+ + 2H:-(CaH ).
2
(b) Two Li · (Group 1 A) atoms are needed to furnish the 2 electrons needed by the single
: O : (Group 6A) atom:
2Li· + :Q: 2Li+ + :0:: (LioO)
182 types of chemical bonds
(c) Mg : (Group 2A) must lose 2 electrons, but · N · (Group 5A) needs 3 electrons. The
numbers 2 and 3 are both common factors of 6, so that a total of 6 electrons must be
transferred as shown:
3Mg: + 2·Ν· > 3Mg2+ + 2 : N : 3~ (Mg N )
3 2
loses gains
6 electrons 6 electrons
Notice that the number of electrons lost or gained by an atom in forming an
ionic bond is equal to its valence. The charge on the ion formed from the atom is
the valence of the element. Atoms that lose electrons to form positive ions are
generally the metals. Atoms that gain electrons to form negative ions are gener-
ally the nonmetals. The loss of electrons is called oxidation; the gain of electrons
is called reduction.f By these definitions, the formation of an ionic bond from
elements must involve an oxidation-reduction (or redox for short) reaction. The
metallic element is oxidized, and the nonmetallic element is reduced.
9.4 The formation of a solid ionic compound such as KBr from its elements at 25°C
FORMATION OF is exothermic (see Color Plate III). Does this mean that the transfer of an
IONIC SOLIDS electron from K to Br is exothermic? To approach the question, the process is
broken down into steps. Electron transfer can be represented by two steps:
K · (g) > K+(g) + e~ AH = + 418 kJ
ionization energy
: Br · (g) + e~ > : Br : ~(g) ΔΗ = -341.4 kJ
electron affinity
K · (g) + : Br · (g) > K+(g) + : Br : ~(g) Δ// = +77.0 kJ (1)
The gain of an electron by a Br atom is exothermic. The loss of an electron by K
(or by any other element) is endothermic. The sum of the two steps is endother-
mic, not exothermic as we might have expected. But we have not told the
complete story. As previously stated (Section 6.3) the ΔΗοί a reaction depends in
part on the states of the reactants and products. In Equation (1) the reactants
and products are gaseous, but the actual reaction occurs between solid potas-
sium and liquid Br :
2
K(c) + ^Br(X) > KBr(c) àH = -392.0 kJ
2 f
heat of formation
Somewhere in this reaction the potassium and the liquid Br molecules must be
2
broken up and solid, crystalline KBr salt must be produced. To understand why
this reaction is exothermic we need not know in what sequence these events
occur because the enthalpy change, ΔΗ, is independent of the pathway. We just
assume reasonable steps with measurable AH values. Then, according to Hess's
law, the sum of these individual ΔΗ values must equal the heat of formation:
fOxidation originally referred to the addition of oxygen to elements. Reduction meant return to the
metallic state. The broader concept of oxidation as a loss of electrons and reduction as a gain was
first suggested by Wilhelm Ostvvald in 1903.
9.4 formation of ionic solids 183
ΔΗ in kJ
Vaporization of èBr2(X) JBr2(g) +15.0 (J heat of vaporization)
liquid Br
2
Dissociation of Br2(g) iBr2(g) : Br · (g) +96.6 (J bond energy)
Conversion of K(c) to K(c) • K(g) +89.9 (heat of sublimation)
individual atoms
lonization of K atoms ■K(g) K+(gO + e~ +418.4 (ionization energy)
Formation of Br : Br · (gf) + e- : Br : ~(s0 -341 ·4 (electron affinity)
Total: JBr (X) +}£*fâ) + · K(c) + - J^ + i3^CgT+^'
2
> i ß ^ )+ LB^tgy+ -J^rr-+ K+(c/) + :B'r:-(g) +^
Net reaction: · K(c) + ^Br2(X) > K+(g) + : Br : ~{g) ΔΗ = +274.5 kJ (2)
The sum of these ΔΗ values is even more positive than the ΔΗ for Equation (1),
and does not account for the exothermicity of the reaction. However, we have
thus far accounted only for the formation of the gaseous ionic state. Therefore
we must add one more equation, the one for the combination of the individual
gaseous K+ and '- Br '· ~ ions to form the solid crystalline KBr:
K+(#)+ :Br:-(£) KBr(c) Mi = - 668.4 kJ (3)
When this last addition is made, we get the net equation for the reaction and the
ΔΗ, calculated by Hess's law, is +278.4 — 668.4 = — 390 kJ, which is close to
the observed value of —392.0 kJ for the heat of formation (AH). This thermo-
f
chemical analysis makes it clear that the exothermicity of the net reaction comes
mostly from the formation of crystalline KBr from the individual gaseous K+
and Br~ ions.
When the electrical attraction binds the ions together in the solid (as in
Equation (3)) energy is released. This energy is called the crystal lattice energy.
In the crystal, each positive ion is surrounded by a number of negative ions.
Likewise, each negative ion is surrounded by a number of positive ions (see
Figure 10.26). In an ionic crystal no one positive ion is the mate of any one
negative ion. There are no molecules in an ionic solid. This is why the formulas
of ionic solids are actually empirical formulas.
Before we leave the ionic bond, there is one further question to consider:
What are the factors that favor the formation of an ionic bond between two
atoms? This question can be approached by using the kind of thermochemical
Called the
Born-Haber analysis. analysis just made for KBr. The enthalpy change for the reaction we are asking
about
element! + element - ionic compound(c)
2
metal nonmetal
is the enthalpy of formation of the compound, Mi. The more negative is ΔΗ,
{ (
the more likely is the ionic compound to be formed. If &H is positive, the
f
elements are likely to form a covalent bond or else not combine at all. There are
three factors that contribute significantly to &H: (1) the ionization energy, ΔΗ ,
f ΙΕ
of the metal; (2) the electron affinity, ΔΗ , of the nonmetal; and (3) the crystal
ΕΑ
lattice energy, ΔΗ , of the ionic solid.
ΙΕ
184 types of chemical bonds
In all cases, ionization to give a positive ion is endothermic; ΔΗ is always
ίΕ
positive:
elementj element,+ + e ΔΗΙΕ is positive
metal
This means that ΔΗ always represents an energy barrier and therefore it is
ΙΕ
advantageous for that barrier to be as small as possible. On the other hand the
ΔΗ is usually negative:
ΕΑ
element -f e~ element ΔΗ is usually negative for first electron
2 ΕΑ
nonmetal
Therefore to get a large negative heat of formation, &H, it helps to have a large
f
negative ΔΗ . Crystal lattice energies are always negative, so it also helps to
ΕΑ
have a large negative ΔΗ .
ΙΕ
On the basis of these general principles we can account for the following
trends.
FOR FORMATION OF POSITIVE IONS In going across a period of the periodic table
from left to right the positive charge on the ion becomes larger, ΔΗ becomes
ΙΕ
more positive, and the formation of positive ions becomes less likely. On going
down a group, ΔΗ becomes less positive and formation of a positive ion
ΙΕ
becomes more likely.
FOR FORMATION OF NEGATIVE IONS In going across a period from left to right,
the negative charge on an ion becomes smaller, ΔΗ becomes more negative,
ΕΑ
and formation of a negative ion becomes more likely. On going down a group,
ΔΗ becomes less negative and formation of a negative ion becomes less likely.
ΕΑ
FOR FORMATION OF THE IONIC COMPOUND The obstacles against forming
highly charged negative and positive ions can be overcome by a large crystal
lattice energy (large negative ΔΗ^).
These trends are illustrated in the following examples.
EXAMPLE 3 Explain why Li tends to form ionic bonds, whereas B (boron) forms covalent bonds.
ANSWER Li loses one electron to give Li+ and therefore has a smaller positive ΔΗ than B, which
ΙΕ
must lose 3 electrons to give B3+. (Note that the comparison is made between elements in
the same period.)
EXAMPLE 4 Explain why ionic fluorides outnumber ionic iodides.
ANSWER F and I are in the same group. F is smaller than I, meaning that the added electron is closer
to the nucleus. F therefore has a more negative ΔΗ , and is more likely to form an ion.
ΕΑ
EXAMPLE 5 Explain why aluminum oxide, Al 0 , is an ionic compound, whereas AICI is not.
2 3 3
ANSWER The O atom gains 2 electrons to give O2- and has a net positive ΔΗΕΑ. The CI atom, which
only gains 1 electron to give CI-, has a negative ΔΗ . On this basis we might expect the
ΕΑ
opposite result. The explanation is that the crystal lattice energy is greater between a +3 ion
■0:(g) + 2e- —> :Ö:2- (Al3+) and a —2 ion (02~) than it would be between a -f 3 ion and a —1 ion (Cl~). àH is
LE
+ 653.5 kJ the overriding factor.
198 types of chemical bonds
(b) CI is the central atom because it is less electronegative than the F atom
:Cl: + 3-F: : F : CI : F : (Cl has 2 unshared pairs of electrons)
' " : F : "
To form this compound more than 8 electrons must surround the CI atom.
(c) This is an ionic compound (a salt) and each ion is therefore shown independently.
+
H :F:
H:N:H :F:B:F:
H :F:
9.11 Before continuing the discussion of covalent bonding we consider how covalent
PROPERTIES OF bond length and bond energy are related. The following generalization can be
COVALENT BONDS made: The shorter the distance between a given pair of covalently bonded atoms,
the greater is the bond energy, and hence the stronger is the bond. This general-
ization is supported by the observed distances and energies of the bonds be-
tween the two carbon atoms in ethane, ethylene, and acetylene:
H3C—CH3 H2C—CH2 HCE=CH
bond length (in A) 1.54 1.33 1.20
bond energy (in kJ/mol) 333.9 589.9 811.7
These values also show that the distance between two given atoms decreases
when the number of bonds between the atoms increases. Multiple bonding results
in a greater electron density between the bonded nuclei. The nuclei are more
strongly attracted by this greater electron density and are brought closer to-
gether. Multiple-bond covalent radii are shown in Table 9.6 for C, N, O, and S.
Bond polarity may also cause contraction of bond length. For example, the
length of the polar Si—F bond, calculated from Table 8.6 (page 159) by adding r
Si
and r , is 1.86 Â. However, the measured bond length in SiF is only 1.54 Â. On
F 4
the other hand, when a comparison is made of the calculated and observed
lengths of a nonpolar bond, the correspondence is very good. Thus in bromine
chloride, Br—Cl, the calculated bond length is 2.13 A and the observed bond
length is 2.14 A. Along with contraction of a bond length goes an increase in
bond energy. Hence an increase in polarity of a covalent bond strengthens the
covalent bond.
TABLE 9.6 Multiple-bond covalent radii (in angstrom units, Â)
SINGLE-BOND DOUBLE-BOND TRIPLE-BOND
ELEMENT RADIUS RADIUS RADIUS
C 0.77 0.665 0.60
N 0.73 0.60 0.55
0 0.74 0.55
S 1.02 0.94