The formulas for chemical compounds are not accidents. NaCl is a compound, but NaCl2 is not. CaF2 is a compound, but CaF is not. On the other hand, there are elements that form two, or even more, different compounds; Cu2O and CuO exist and so do N2O, NO, and NO2. In the case of ionic compounds, the relative number of positive and negative ions in a formula is governed by establishing a combination that is electrically neutral (overall charge zero). In covalent compounds, structures are formed by establishing covalent bonds (electrons are shared). There is a range of covalent bonds from truly equal sharing of electrons to an extremely unequal sharing.
Ionic bonding is the term used to indicate that a compound is composed of oppositely charged particles called ions. We can determine the charges on the ions by thinking in terms of the electron configuration (electronic structure) of the atoms and comparing them to the noble gas (Group VIIIA) electron configuration. Generally speaking, those metals whose atomic numbers are within 2 or 3 of a noble gas tend to form ionic compounds with nonmetals within 2–3 atomic numbers of a noble gas. The vocabulary associated with ionic compounds identifies the positive ion as the cation (cat′i′-on) and the negative ion as the anion (an′i′-on).
EXAMPLE 1 What is the empirical formula for potassium chloride? Potassium loses one electron to achieve argon’s electron configuration (8e– in the outside orbit) and, because of this loss, forms the K+ cation. Chlorine gains one electron to achieve the argon electron configuration and forms the chloride anion, Cl–. Since compounds must be neutral, the ratio of one potassium ion to one chloride ion is the empirical formula, KCl (and also is the molecular formula). A note of interest: KCl is often used as a substitute for NaCl for those who cannot tolerate sodium, such as heart patients.
For transition metals, the lanthanides, and the actinides, no such simple rule exists. If we accept the ions’ charges as chemical facts, we can still write the empirical formulas for ionic compounds so that the net (overall) charge is zero. If we had Fe2+ and O2–, the compound would require a minimum of one of each of the elements for a neutral formula, FeO, whereas Fe3+ and O2– would have the formula Fe2O3.
The naming of the binary (two participants) compounds is set by using the name of the element from which the cation is derived, e.g., Li+ is the lithium ion. The names of the anions composed of a single element are formed by using the name of the element and changing the word ending to -ide, e.g., O2– is the oxide ion. When two cations exist for the same element, the accepted procedure is to write the charge per atom in Roman numerals within parentheses directly after the name of the metal, e.g., Fe2+ is the iron(II) ion. Note that there is still some use of the older system of differentiating the two charge states with the use of -ous to represent the lower charge and -ic for the higher charge. Applying the older system, Fe2+ is the ferrous ion and Fe3+ is the ferric ion.
EXAMPLE 2 Name the two oxides of lead, PbO and PbO2. Because oxide is –2, the lead must be +2 in the first case and +4 in the other. The names are lead(II) oxide and lead(IV) oxide, pronounced “lead two oxide” and “lead four oxide,” respectively. In the older system, the Latin names were used, providing us with the names plumbous oxide and plumbic oxide, respectively. Notice that the name in the newer system gives us the charge on the cation, but the older system just tells us which cation has the higher charge or lower charge without giving us the magnitude.
Table 9-1 lists a few of the more commonly encountered polyatomic ions. These ions, their names, and their charges should be memorized.
Table 9-1 Common Polyatomic Ions
Although hard-and-fast rules cannot be given, the following guidelines are helpful. Compounds of metals with nonmetal or polyatomic anions tend to be ionic. Compounds of nonmetals with nonmetals tend to be covalent. Physical properties can help classify compounds, also. Being hard, brittle, and having a high melting point indicates ionic compounds. Compounds liquid at room temperature and gases generally are covalent compounds. Your textbook provides you with more complete (and complex) rules for determining the character of a compound, as well as naming them.
The complex oxy-anions are named in a somewhat systematic matter by using -ate for the most common or most stable and -ite for the ion containing less oxygen. The prefix per- is added to an -ate name to indicate even more oxygen, and the prefix hypo- is added to an -ite name to indicate a lower oxygen content than in the -ite ion. The system takes a bit of “getting used to,” but does work well.
The covalent force between atoms sharing two or more electrons involved in forming a chemical bond is related to delocalization of those electrons. The meaning of “delocalization of electrons” is that they are not where they are predicted—in the case of sharing electrons, one electron is originally in orbit about an atom and the other electron is in orbit about a different atom. The delocalization of the two electrons results in their orbiting both nuclei, not just one. Think in terms of smearing the electron cloud so that it no longer is located around one nucleus, but surrounds two.
When a covalent bond is formed, the distance between the two nuclei involved becomes much less than if the bond were not formed. As an example, the distance between the hydrogens in H2 is 74 pm, whereas the sum of the distance of the van der Waals (nonbonding) radii of the two hydrogen atoms is 240 pm.
This bonding of two atoms at close approach is also reflected in energetic considerations. The energy of a bonded pair of atoms is less than the sum of the energies of the separated atoms. The bond energy is the magnitude of this energy lowering. From another viewpoint, the bond energy is the amount of energy, ΔE, required to break a chemical bond into two nonbonded fragments. Bond formation is exothermic and bond rupture is endothermic. Covalent bond energies range from about 150 to 550 kJ/mol for single electron-pair bonds formed among elements in the first three periods at normal temperatures.
A factor associated with the covalent bonding process is that the spins of the bond-forming electrons, which are unpaired in separate atoms, become paired during the bond formation.
Up to this point, many of the examples provided make use of the octet rule. The octet rule states that the stable atom possesses eight valence electrons (the electron configuration of the previous inert gas, most of which have eight valence electrons). Some of the lighter elements, such as hydrogen and lithium, will have helium’s electron configuration (two electrons in the outside shell). The electrons do not necessarily have to be in pairs and, further, electrons shared in bonding are both counted for each atom in the bond. Worthy of note is that a single covalent bond consists of a pair of shared electrons, a double bond has two shared pairs, and a triple bond has three shared pairs. Bond distances are usually shorter and bond energies are usually greater for multiple bonds than for single bonds.
Structural formulas, such as those in Fig. 9-1, are typical of the valence electron distributions in covalent molecules and ions. These structures are not meant to indicate actual bond angles, or lengths, in three-dimensional varieties like methyl chloride, ammonia, and the ammonium ion; they merely show the number of bonds connecting the participating atoms. In these Lewis formulas (structures), a line between two atoms represents a pair of shared electrons and a dot represents an unshared electron. Unshared electrons usually occur in pairs, lone pairs, on a single atom. Two lines constitute a double bond, and three lines constitute a triple bond. The total number of electrons shown in such a molecular structure is equal to the sum of the numbers of valence electrons in the free atoms: 1 for H, 4 for C, 5 for N, 6 for O, and 7 for Cl. For an ionic structure, one additional electron must be added to this sum for each unit of negative charge on the entire ion, as in OH–. Conversely, one electron must be subtracted from the sum for each unit of positive charge on the ion, as in the ammonium ion. The number of pairs of electrons shared by an atom is called its covalence.
Fig. 9-1
The covalence of hydrogen is always one because it cannot form more than one chemical bond. The covalence of oxygen is almost always two and occasionally one. The covalence of carbon is four in almost all of its stable compounds—there may be single, double, or triple bonds involved, but the total number of bonds is four. Although the octet rule is not a rigid rule for chemical bonding, it is obeyed for C, N, O, and F in almost all their compounds. The octet is exceeded commonly for elements in the third and higher periods.
Sometimes more than one Lewis structure can be written and there is no reason to select one over another—all of the structures must be used to represent the substance correctly. The true structure is said to be a resonance hybrid of those Lewis structures.
EXAMPLE 3 Experimentation has shown that the two terminal oxygens in ozone are equivalent, that is, they are equidistant from the central oxygen. If only one of the resonance diagrams in Fig. 9-2(a) were written, it would appear that one of the terminal oxygens is bonded more strongly to the central oxygen (double bond) than is the other (single bond) and that the more strongly bonded atom should be closer to the central atom. The hybrid of the two ozone structures gives equal weight to the extra bonding of the two terminal oxygen atoms. Similarly, the three resonance structures of carbonate, Fig. 9-2(b), are needed to account for the experimental fact that all three oxygens are equidistant from the central carbon.
Fig. 9-2
The total bond energy of a substance for which resonance structures are written is greater than would be expected if there were only one formal Lewis structure. This additional stabilization is called resonance energy. It arises from the same principle that is responsible for covalent bond energy, the delocalization of electrons around the atoms forming the bond. As a result of resonance in ozone, for example, the electrons constituting the second pair of the double bond are delocalized around the three oxygen atoms. The drawing of two or more resonance structures is a way to present a clear picture of the delocalization that may not be clearly presented or even possible with a single sketch.
Although a molecule is electrically neutral, there is a technique for identifying local charges associated with the various parts of the molecule. The algebraic sum of those charges in a molecule, of course, must equal zero. In the case of an ion, there is a net charge which is the algebraic sum of the local charges in the ion. In one method of apportioning charges to an atom in a molecule or polyatomic ion, the shared electrons in a covalent bond are divided equally between the two atoms forming the bond. (Recall that covalent bonds are formed by sharing electrons, one from one member of the bond and one from the other.) Unshared valence electrons within an atom are assigned exclusively to that atom. Each atom is then assigned a formal charge, FC, which is equal to the number of valence electrons possessed by that atom in the neutral free state minus the total number of valence electrons assigned to it (unshared e’s + 
each covalent bonding pair of e’s). These charges may be written into the diagram of the structure.
EXAMPLE 4 Figure 9-3 shows a single resonance structure for ozone. Each oxygen is assigned a 6 due to the number of valence electrons present (a Group VIA element—6 electrons).
Fig. 9-3
(1) The central oxygen is assigned just five electrons (two in the unshared pair + of the three pairs in the bonds); this atom, being one electron short of the complement of six in a free oxygen atom, is assigned +1 (FC = 6 – 5 = +1).
(2) The terminal oxygen connected by the single bond is assigned 7 (6 in unshared pairs and 1 from the bond); therefore, FC = 6 – 7 = –1.
(3) The doubly bound oxygen on the left is assigned 6 (4 in the unshared pairs plus half of the two shared pairs), providing us with the FC = 6 – 6 = 0.
A rough rule useful in predicting one allowable Lewis structure over another is that structures minimize the number of formal charges. Especially to be avoided are formal charges of magnitude greater than 1 and structures in which formal charges of the same sign are located on adjacent atoms.
There are some experimental procedures that give information about the actual distribution of charges within a molecule (as distinct from the arbitrary assignment of formal charges). One such procedure is the measurement of dipole moment. An electric dipole is a neutral object that has a positive charge of a magnitude q and a separately located equal, but opposite, negative charge (the charges cancel to neutrality). The extent to which a dipole is aligned in an electric field is dependent on the dipole moment (measurable), which is defined as the product of q and the distance, d, separating the (+) and (–) charges.
In a covalent diatomic molecule the dipole member would be expected to be zero if the bonding electrons were shared truly equally by the two atoms. This is the case in molecules which are of the type X2 (H2, N2, etc.), where two identical atoms are bonded. In the more general type, XY, two different kinds of atoms are bonded (C—H, N=O, etc.), and a dipole moment is usually observed. This is explained by imagining that one of the atoms, Y, has a greater attraction for the shared electrons in the bond than does X. Y is said to have a greater electronegativity than X. Electronegativity correlates with other atomic properties; in general, atoms with high ionization energies and/or high electron affinities tend to have high electronegativities. The most electronegative elements in order of decreasing electronegativity are F > O > N 
Cl. Metals are less electronegative than nonmetals. Of note is that carbon is only slightly more electronegative than hydrogen, especially when considering organic molecules.
The assignment of electrons to the various atomic orbitals (Chapter 8) concerned the electron distribution in the ground state (the state having the lowest energy) of an isolated free atom. From the following ground-state configurations of elements in the second period except fluorine (can you recall why?), one might predict that the maximum covalence of an element would be equal to the number of unpaired electrons. This would be assumed since an unpaired electron from each of the bonding atoms participates in the covalent bond formation, as described above.
Noticing where the unpaired electrons are located, the maximum covalences of B, C, N, and O should then be 1, 2, 3, and 2, respectively. These predictions account for most of the known compounds and complex ions of nitrogen (e.g., NH3, 
) and of oxygen (e.g., H2O, CH3OH, HOOH), but not for the commonly observed trivalence (+3) of B (e.g., BI3), or the tetravalence (±4) of carbon (e.g., CH4, CCl4, CH3OH).
Let us take a second look at the electron configurations above and attempt an explanation for the charge (oxidation number, valence) of +3 for boron, as in BI3, just mentioned. The electronegativities of these two elements are very close, implying that the bond between them is, in fact, covalent. Since there are to be three pairs of shared electrons, but the electron configuration provided, 
only allows one covalent bond to form, there must have been another electron configuration before all three bonds formed. Since the 2s and 2p subshells (orbitals) are in the same shell (orbit) and their energy levels are reasonably close, the application of sufficient energy to the 2s electrons will result in one of the electrons being boosted out of the next empty subshell, 2px, resulting in three unpaired electrons.
Your textbook tells you that the arrows represent the opposite characteristics of electrons, ↑↓ represents a pair of electrons, but ↑ is an unpaired electron. Notice that the 2s electron that is boosted to the 2py subshell reversed its characteristics, as indicated by the downward arrow in 2s becoming an upward arrow in the 2pt subshell. Some energy is required to form this hybrid orbital and, more loosely, bond hybridization, as the electron is boosted from a lower energy level to a higher one; also, some energy is used to reverse the electron’s characteristics.
The above explanation accounts for the number of covalent bonds formed by boron and carbon, but not for the equivalence of these bonds. The difference in spatial character of s and p orbitals (Fig. 8-3) and their energies might suggest differences in the bonds they formed, as measured by bond energy, bond distance, or bond angles. Experiment shows that all three bonds in BF3 are equivalent, that the angles between any two bonds are the same, and that the three fluorine atoms lie at the corners of an equilateral triangle with the boron atom in the center. Similarly, in CH4 all four bonds are equivalent, the angles between any two bonds are the same, and the four hydrogen atoms lie at the corners of a regular tetrahedron (Fig. 9-4) with the carbon atom at the center.
Fig. 9-4
The equivalence of the bonds in BF3, CH4, and similar compounds was first explained by Linus Pauling. To accommodate the surrounding atoms, s and p orbitals of a given atom may mix with each other, or hybridize. The mathematical formulations of the hybrid orbitals are simple linear combinations of the mathematical formulations of the separate s and p orbitals. The geometrical description of each hybrid orbital is a kind of superposition of the mappings of the separate s and p orbitals. Pauling showed that one s and three p orbitals are hybridized to give maximum concentration of the electron probability distribution along some particular direction (to best form a bond along this direction); the four resulting so-called sp3 hybrid orbitals indeed point to the corners of a regular tetrahedron. The angle of 109°289′ is made by any two bonds (lines) connecting the center of a regular tetrahedron with atoms at the corners (Problem 9.18). This is the observed situation in CH4, CCl4, SiF4, and many compounds of the Group IVA elements.
A similar hybridization of one s with two p orbitals leads to a set of sp2 hybrid orbitals which have their maximum concentrations along a set of lines in one plane forming angles of 120° with each other. This is the observed situation in BF3. The angles are close to 120 in C2H4, in which each of the two carbon atoms forms a set of sp2 hybrid orbitals, as in Fig. 9-5. The first of two other important types of hybrid orbitals are the sp type in which the two orbitals point in directions leading to a 180° bond angle (Fig. 9-6). The other hybridization results in the d2sp3 type in which the six orbitals point to the corners of a regular octahedron (Fig. 9-7), the situation in SF6 and many coordination compounds.
Fig. 9-5
Fig. 9-6
Fig. 9-7
The valence electrons in Lewis structures are described as being either unshared electrons localized on particular atoms or shared electrons assigned to bonds linking particular pairs of atoms. An alternative representation handles the electrons by assigning them to molecular orbitals (MOs) appropriate to the molecule as a whole. Just as an atomic orbital is the mathematical solution of a Schrödinger equation describing the probability distribution of the various positions that an electron having a given set of quantum number may occupy around the atomic nucleus, so a molecular orbital describes the distribution of positions in a molecule available to an electron having a given set of quantum number. Molecular orbitals may be approximated by writing mathematical combinations of the atomic orbitals of the constituent atoms or they may be pictured, at least in qualitative terms, as geometrical combinations of the contributing atomic orbitals. The rules governing the use of molecular orbitals are:
1. The total number of molecular orbitals (MOs) equals the sum of the number of atomic orbitals (AOs) of the constituent atoms.
2. Each orbital can hold 0, 1, or 2 electrons, corresponding to the possibility of two different directions of electron spin and the application of the Pauli principle.
3. When several MOs of equal or almost equal energy exist, electrons tend to fill those orbitals so as to maximize the number of unpaired electron spins. The more nearly equal the energy levels, the greater this tendency (Hund’s rule).
4. The directions in space describing the orientation of the orbitals, although arbitrary in the case of the free atom (Fig. 8-3), are related to the positions of neighboring atoms in the case of molecules or complex ions.
5. A molecular orbital is most likely to be composed of AOs of similar energy levels.
6. In diatomic molecules, or more generally in localized two-center bonds, orbitals of two atoms that may be combined to form an MO are those that have the same symmetry about the axis between the two atomic centers in the bond. This rule does not extend to constructing MOs that extend over three or more atoms; more complicated rules would be necessary.
7. A two-centered orbital with high electron probability in the region between the nuclei is a bonding orbital. A stable chemical bond exists when the number of electrons in bonding orbitals exceeds the number in antibonding orbitals.
The direction properties of molecular orbitals are governed by quantum numbers analogous to the atomic quantum number l and ml. Greek letters are used in molecular orbital notation to designate increasing values of the l-type quantum number, σ, π, δ, etc., analogous to the Latin letters s, p, d, etc., for atomic orbitals. Only σ and π orbitals will be considered in this book. There are two kinds of σ orbitals, bonding and antibonding (Fig. 9-8). An antibonding orbital is designated with a superscript asterisk applied to the Greek letter, as σ*, π*, etc. A bonding orbital has a region of electron overlap (high probability) between the bonded atoms and has a lower energy than either of the component AOs. An antibonding orbital has a nodal plane (region of zero probability) between the bonding atoms and is perpendicular to the bond axis; its energy is greater than that of either constituent AO.
Fig. 9-8
If the bond axis is designated by x, σ bonds can be formed by combining any two of the following atomic orbitals, each of which has a region of high electron probability lying along the x axis and is cylindrically symmetrical about it: s, px, dx2–y2, or hybrid orbitals pointing along the x axis. If the bond axis is z, σ orbitals may be formed from s, pz, dz2, or appropriate hybrids. If the bond axis is y, the component atomic orbitals for σ orbitals are s, py, dx2–y2, or appropriate hybrids.
A π orbital related to a bond in the x direction is characterized by having a zero value at all points along the x axis. πy orbitals, having their maximum probabilities in the +y and –y directions (that is, above and below the xz plane), may be formed from py atomic orbitals. Similarly, πz orbitals having their maximum probabilities above and below the xy plane may be formed from pz AOs, π orbitals may also be bonding or antibonding (Fig. 9-9). A dxy orbital may combine with a py orbital, or a dxz with a pz to form π orbitals.
Fig. 9-9
Nonbonding AOs are those which do not interact with orbitals of other atoms because:
1. The two atoms are too far apart for good orbital overlap (e.g., the case of nonadjacent atoms).
2. The energy of the nonbonding orbital is not close to that of any orbital on the adjacent atom (e.g., the 3s orbital of Cl is of much lower energy than the 1s orbital of H in HCl).
3. The nonbonding orbital is in an inner shell and would not overlap with an orbital even of the neighboring atom (e.g., the K electrons in F2).
4. The nonbonding orbital does not match its symmetry with any available orbital of the adjacent atom (e.g., the 3py of Cl does not have a symmetry match with the 1s of H in HCl, where x is the bonding direction and the 2py orbital of H is too high in energy to enter the picture).
A nonbonding orbital has the same energy as in the free atom. Electrons occupying nonbonding orbitals correspond to the unshared electrons in Lewis structures.
A buildup principle analogous to that for atoms exists for molecules. The order of filling MOs from the valence shells in the case of homonuclear diatomic molecules where x is the bond axis is
The ordering may change for heteronuclear diatomic molecules and for homonuclear molecules at the point where this complete set of orbitals is about half-filled.
The bond order in a diatomic molecule is defined as one-half the difference between the number of electrons in bonding orbitals and the number of antibonding orbitals. The factor one-half preserves the concept of the electron pair and makes the bond order correspond to the multiplicity in the valence-bond formulation: one for a single bond, two for a double bond, and three for a triple bond. Fractional bond orders are allowed, but are not within the scope of this discussion.
Ethylene, C2H4, has a basic framework established by combining five two-centered bonding σ orbitals, four of which are made from a 1s orbital on each hydrogen and an sp2 orbital on a carbon. The remaining σ orbital is made from one sp2 orbital on each of the carbons. The combinations of these two-centered σ orbitals are bonding MOs that extend over the whole molecular framework. These extended MOs, represented by the lightly shaded regions of Fig. 9-10, may be referred to as σ -type because their electron density is concentrated mainly along the axes connecting pairs of adjacent atoms.
If the plane of each H2C—C group is designated as the xy plane, the px and py AOs are used to form the sp2 hybrid orbitals. The pz orbitals on the two carbons (the darkly shaded regions of Fig. 9-10) are then available for forming πz orbitals, as indicated by the lines between the darkly shaded orbitals in Fig. 9-10 representing the overlap of orbitals. After filling the five σ-type bonding orbitals, the remaining electron pair (of the total of 12 valence electrons on the two carbons and four hydrogens) goes into the πz orbital. The two carbons are bonded partly by the electrons in the σ-type framework, constituting the equivalent of a single bond between the two carbons (plus the equivalent of a single bond connecting each of the hydrogens to its adjacent carbon), and partly by the pair of π electrons forming the second part of the double bond indicated in Fig. 9-5. The π bond, which is rigid, prevents rotation about the C=C axis and constrains all six atoms in C2H4 to the same plane.
Fig. 9-10
In acetylene, C2H2, the σ-type skeleton of three bonding orbitals (the lightly shaded regions in Fig. 9-11) is formed from the 1s orbitals on the hydrogen atoms and the hybrid sp orbitals on the carbon atoms. The px AOs are used to form the hybrid orbitals which point along the bonding x direction and the remaining p orbitals are free to form the πy and πz orbitals, represented by the lines joining the darkly shaded regions of Fig. 9-11. The ten valence electrons (one from each hydrogen and four from each carbon) fill the three σ-type orbitals and πy and πz orbitals. The carbons are held together by the equivalent of a triple bond (one σ bond and two π bonds), as indicated in Fig. 9-6.
Fig. 9-11
In ozone, O3, the atoms are held together in the first instance by a σ-type framework in the xy plane of the molecule represented by the lightly shaded regions of Fig. 9-12. The electron density is concentrated mainly along the axes connecting pairs of nearest-neighbor oxygen atoms. The pz orbitals of all three oxygen atoms have the symmetry with respect to the plane of the molecule and can combine to form π orbitals. The pz orbital of the central atom overlaps with the pz orbitals of both terminal oxygen atoms. As a result, a π bonding orbital and a π* antibonding orbital extend over all three atoms in the molecule. Further, a π nonbonding orbital involves the two terminal atoms. The electrons in ozone occupy the π bonding orbital, enveloping all three nuclei, with equal probability on the two sides of the molecule. This representation is an alternative to resonance. It was conceptualized in the attempt to conform to the octet rule. The π nonbonding orbital is also occupied in ozone. The occupied π bonding orbital is represented by the lines joining the darkly shaded regions of Fig. 9-12.
Fig. 9-12
Multicentered π orbitals are involved in the molecular-orbital representation of most structures for which resonance must be invoked in the valence-bond representation. In a long chain of atoms bonded within a planar configuration, such as the plant pigment carotene, C40H56, or in planar ring compounds, like naphthalene, C10H8, each π orbital extends over many carbon atoms, all those in the basic molecular plane, since the overlap of the pz orbital of any nonterminal carbon with those of its two or three neighbors allows for the buildup of long chains or rings of extensively overlapping electron probability distributions.
Bond lengths between a given pair of atoms are approximately constant from compound to compound if the nature of the bond (single, double, or triple) is the same. If it is assumed that the length of a single covalent bond is the sum of the covalent radii of the two bonding atoms, then quick and fairly reliable estimates of bond lengths can be obtained from readily available information, such as in Table 9-2.
Table 9-2 Single-Bond Covalent Radii
From the precisely measured lengths 154 pm, 133 pm and 120 pm for H3C—CH3, H2C=CH2, and HC≡CH, respectively, we can conclude that the bonds become shorter in the progression from a single bond to a double bond to a triple bond. A rule of thumb has been developed to allow a 21 pm shortening for any double bond and a 34 pm shortening for any triple bond, assuming the bonds are located between the same atoms. In the case of resonance or multicenter π orbitals, a bond length is intermediate between the values it would have in the separate resonance structures. This concept is also true if the bond is between the values it would have in the absence of π bonding and in the presence of a localized two-centered π bond. Some applications of this rule are shown in Table 9-3.
Starting with the Lewis structure, it is possible to predict fairly accurately the bond angles in a molecule. The VSEPR method (Valence Shell Electron Pair Repulsion) focuses on a central atom and counts the number of atoms bonded to it plus the number of unshared pairs. Multiple bonds count the same as single bonds. This characteristic VSEPR number is the number of orbitals (each occupied by an electron pair) that must emanate from the central atoms. The angles between them are determined by the principle that the electron pairs will repel each other. The VSEPR method is a simple technique that does not require explicit identification of the orbitals. Invoking the hybrid orbital technique results in the same angles, but in a more formal manner linked to the mathematical treatments of bonding. VSEPR numbers, corresponding to angles, and hybrid orbital sets are listed in Table 9-4.
Deviations from the angles arise from unshared pairs, which repel more strongly than shared pairs. For example, the VSEPR number is 4 for CH4 and :NH3. The H—C—H angle in CH4 is a perfect tetrahedral angle (109°28′), but the H—N—H bond angle in :NH3 is compressed to about 107° by the unshared pair. The VSEPR for water is also 4. Water’s oxygen has two unshared pairs compressing the HOH angle to 104.5°.
Double bonds also contribute extra repulsion compressing adjacent bond angles slightly, as shown in Fig. 9-5. Deviations arise if the various atoms around the central atom are very different in size.
The electrons of an electron-pair bond need not be contributed by both of the bonding atoms, as is demonstrated in the formation of the ammonium ion by the addition of a proton to ammonia. The Lewis structures for these two entities are shown in Fig. 9-13; recall that H+ has no electrons. Such a bond is often called a coordinate covalent bond, but is essentially no different from any other covalent bond. The special name indicates that one of the members of the bond brings to the process of bonding any electrons. In this particular case, once the bond has formed, it becomes indistinguishable from the other three N—H bonds in the molecule with the structure being a regular tetrahedron.
Fig. 9-13
Coordinate covalent bonding is a common type of bonding in coordination compounds, in which a central metal atom or ion is bonded to one or more neutral or ionic ligands. A typical ligand, such as NH3, Cl–, or CO, has an unshared electron pair that forms the coordinate covalent bond by interacting with the unfilled orbitals of the central metal. The overall charge of a complex ion is the algebraic sum of the charge of the central metal and the charges of the ligands.
A number of rules have been adopted internationally for the naming of coordination compounds:
1. If the compound itself is ionic, the cation is named first.
2. A complex ion or nonionic molecule carries the name of the central metal last, with its oxidation state (charge per atom) in Roman numerals (or 0) enclosed in parentheses. (A detailed discussion of oxidation state is provided in Chapter 11.)
3. Ligands that are anions are named with the suffix -o, as in chloro, oxalato, cyano.
4. The number of ligands of a given type is indicated by a Greek prefix, like mono- (often omitted), di-, tri-, tetra-, penta-, hexa-.
5. If the name of a ligand contains a Greek prefix, the number of ligands is indicated by such prefixes as bis-, tris-, tetrakis-, for 2, 3, or 4, respectively, and the name of the ligand is enclosed in parentheses.
6. When the complex ion is an anion, the Latin name of the metal is used with the suffix -ate.
7. Some neutral ligands are given special names—ammine for NH3, aqua for H2O, carbonyl for CO.
8. When several ligands occur in the same complex, they are named in alphabetical order (ignore any numerical prefixes).
EXAMPLE 5 Names of compounds containing complex ions:
In many coordination compounds, the ligands are arranged around the central metal in regular geometrical forms, such as the octahedron, the tetrahedron, or the square. In the formulas listed above, the brackets define the complex made up of the central metal and its ligands. These brackets are often omitted where there is no problem determining the nature of the complex. Many of the compounds of complexes are colored. Some are paramagnetic because of the presence of unpaired electrons, while others with the same metal atom are not paramagnetic. The molecular-orbital energy-level diagram, Fig. 9-14, is the key to the explanation the properties. The use of the derivation is important in the solving of problems. Here we shall consider only the six-coordinated octahedral complexes.
Fig. 9-14
The hybridization on a metal atom leading to octahedral bonding is d2sp3. If the bonding axes are x, y, and z, the two d orbitals that are used for the hybridization are those that point along one or more of these axes, dz2 and dx2–y2 (Fig. 8-3). Each of the resulting six hybrid orbitals mixes with an orbital directed along a bonding axis in the ligand, such as a p or a tetrahedral hybrid, to form σ or σ* molecular orbitals. Each bonding σ orbital is occupied by a pair of electrons.
The full set of molecular orbitals for the complex may be constructed from the following atomic orbitals: all five d orbitals of the next-to-outermost shell of the metal, the s and three p orbitals of the metal’s outer shell, and one orbital from each of the six ligands pointed along a bond axis (such as a p or a hybrid orbital). The total number of participating atomic orbitals is 15; the total number of orbitals remaining after molecular-orbital formation must still be 15. The electrons to be accommodated in these orbitals include the six previously unshared pairs provided by the six ligands for bond formation to the metal plus the valence electrons of the metal (the number depending on the oxidation state).
Figure 9-14 displays a molecular-orbital diagram with the metal’s atomic orbitals shown on the left and the ligand’s on the right. A more complicated diagram would be needed if the ligands also had π orbitals that participate in the bonding to the metal. The actual energy spacing will depend on the particular case, but the relative ordering of the (n – 1)d, ns, and np metal orbitals (where n is the principal quantum number of the outermost shell) in general. Also, it is common for the ligand’s orbitals to be at lower energy than the metal’s orbitals. Note that, of the nine metal orbitals, only six contribute to primary bonding in the complex. These six are the same ones that can be hybridized to form the octahedral hybrids pointing along the ±x, ±y, and ±z directions, namely, the s, the three p, and the dx2, and dx2–y2. The six metal orbitals contributing to bonding mix with the six ligand orbitals to form six σ-type bonding and six σ*-type antibonding molecular orbitals that are delocalized over the whole complex. The bonding molecular orbitals have lower energies than the ligand’s orbitals and the antibonding have higher energies, as is usual. Also as usual, the lower the energy of a bonding orbital, the higher is its corresponding antibonding orbital. The three remaining metal d orbitals, not having σ character with respect to bond directions, remain nonbonding and their energy is unchanged to a first approximation. Several consequences can be seen from this diagram:
1. The 12 electrons supplied by the ligands can be thought to occupy and fill the six bonding σ-type orbitals.
2. The next-lowest-lying orbitals, available for the valence electrons supplied by the metal, are the three nonbonding orbitals, dxy, dxz, and dyz orbitals, which do not point toward the ligands.
3. Of the σ*-type antibonding orbitals, the lowest-lying are the two whose metal components are dx2 and dx2–y2, designated as 
. The energy difference between this level and the nonbonding level is denoted as Δ. (The symbol for this energy difference is not standard; various textbooks use X, 10Dq, or Δ0.) The and the nonbonding orbitals are referred to, respectively, as eg and t2g.
4. The first three valence electrons of the metal will occupy the t2g orbitals.
5. The next two valence electrons of the metal could occupy the t2g or the eg orbitals, depending on whether Δ is respectively greater than or less than the energy increase associated with pairing two electrons in the same orbital, in violation of rule 1 under “Magnetic Properties” in Chapter 8. If the t2g orbitals are preferentially filled, the complexes tend to have a low spin and the d3 and d6 configurations are stabilizing, in accordance with the features of filled or half-filled equal-energy orbitals. If the eg orbitals are filled before the t2g orbitals are double occupied, the complexes tend to have high spin and the d5 and d10 are stabilizing configurations.
6. The weak electron transitions responsible for the color of coordination complexes are correlated with the t2g to eg transition of energy Δ. The origin of the strong transitions responsible for the deep colors of some complexes of metals in high oxidation states, like the permanganate and the chromate ions, has a different explanation not covered here.
With regard to the options in rule 5, CN–, CO, and 
are strong-field ligands that tend to increase Δ and promote low-spin complexes; OH– and Cl– are weak-field ligands that lead to smaller values of Δ and high-spin complexes. A more complete listing of ligands in order of their field strength is known as the spectrochemical series.
Quite a few molecules and ions have the same chemical formula (numbers and kinds of atoms), but have different three-dimensional shapes and/or the atoms are assembled in different manners. These substances are referred to as being isomers and may differ in their physical properties (melting point, boiling point, density, color, etc.) and their chemical properties. There are three categories of isomers described below.
One way of describing a chemical compound or ion is to list for each atom the numbers of each kind of atom connected to it by covalent bonds. Isomers which differ in these listings are structural isomers.
EXAMPLE 6 Figure 9-15 depicts the two structures that can be drawn for butane, C4H10, that conform to the octet rule for carbon and hydrogen. In n-butane, two of the carbons are bonded to one carbon each and two are bonded to two each. In iso-butane, three of the carbons are bonded to one carbon each and the fourth to three carbons. n-Butane melts at –135°C and boils at –0°C, while iso-butane melts at –145°C and boils at –10°C.
Fig. 9-15
In the case of n-butane, within any one molecule there is a continuous, practically unrestricted, rotation of the atoms about any C—C bond, so that all structures on paper differ only by the angular position of an atom or group of atoms linked by a single bond and are really the same, as shown in Fig. 9-16. These diagrams are all representations of the same substance, n-butane. In each, we can number the carbon atoms 1 to 4 as we move from one end of the molecule’s carbon skeleton to the other. We see that the kinds of neighbors of a given numbered carbon atom are the same in all three sketches.
Fig. 9-16
Structural isomers also exist for some coordination compounds. One example is a set of compounds in which a ligand in one isomer may occupy a position outside the coordination sphere in another isomer, such as [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br. These two compounds are structural isomers, each recognizable by its own color and complete set of distinctive properties.
In some isomers the listing of bonded atoms around each atom is the same, but the compounds differ because at least two atoms, bonded to the same or to adjacent atoms but not to each other, are at different distances in several forms. Such isomers are referred to as geometrical isomers.
An important set of geometrical isomers occurs in compounds containing a carbon-carbon double bond. The double bond does not allow for a rotation because it is rigid, which also keeps the atoms bonded to the doubly bonded carbons coplanar (in the same plane). As a result, there are different positions available across the double bond, above it, and below, as shown in Example 7.
EXAMPLE 7 In Fig. 9-17 molecules (a) and (b) are structural isomers of each other, as are (a) and (c). These relationships are noted because the chlorines are located on different carbons; however, (b) and (c) have the chlorines located on the same carbons.
Fig. 9-17
A second isomerism is shown in the case of the (b) and (c). Notice that the chlorines are on the same carbons in the two compounds, but differ in their locations—in (b) the chlorines are on the same side of the double bond and (c) shows the chlorines are on opposite sides of the double bond. These relationships exist because there is no rotation allowed about a double bond. These are called geometrical isomers.
EXAMPLE 8 Geometrical isomerism in compounds with square symmetry is illustrated in Fig. 9-18. Because of the rigidity of the square planar arrangement in the Pt, N, and Cl atoms there are two distinct forms. Notice that the chlorines in (a) are opposite each other, but the chlorines in (b) are adjacent. This is because the four locations around the central metal are different, but if the structure had been tetrahedral, the locations of the atoms would not have been different and there would be no isomerism.
Fig. 9-18
Identical adjacent atoms are said to be cis, as in Fig. 9-17(b) and Fig. 9-18(b). When opposite each other, identical atoms are called trans, as in Fig. 9-17(c) and Fig. 9-18(a).
In octahedral complexes, structures of type MX4Y2 may exist as geometrical isomers. The two Y atoms (or groups) can occupy either adjacent sites, as in Fig. 9-19(a), or may occupy opposites sites, as in Fig. 9-19(b). Only two isomers exist in this case because there only two different distances between corners of a regular octahedron. The cis and trans forms of Pt(NH3)2Cl4 are isomers of this type. Possibilities for geometrical isomerism occur for other octahedral formula types, like MX3Y3, or complexes with more than two different kinds of ligands.
Fig. 9-19
Optical isomerism is the existence of molecules that are mirror images of each other (not superimposable); this means that there will be pairs of isomers. Among the important category of optical isomers are compounds containing four different groups singly bonded to a given carbon atom (Fig. 9-20) and octahedral complexes with three different kinds of ligands or with several multivalent ligands (Fig. 9-21).
Fig. 9-20
Fig. 9-21
Molecular orbital (MO) theory has been used to explain the bonding in metallic crystals, such as pure sodium or pure aluminum. Each MO, instead of dealing with a few atoms in a typical molecule, must cover the entire crystal (might be 1020 or more atoms!). Following the rule that the number of MOs must equal the number of atomic orbitals (AOs) combined, this many MOs must be so close on an energy level diagram that they form a continuous band of energies. Because of this factor, the theory is known as band theory.
Let us take a look at a piece of sodium metal. Of the 11 electrons, the 10 forming the neon core are localized around each Na nucleus, leaving one electron per atom to fill the MOs that are throughout the crystal. If there were N atoms in the crystal, N MOs could be formed by use of one 3s orbital from each. Although these MOs have varying amounts of bonding and nonbonding character, their energies form a continuum within the 3s band.
In metals beyond Group IA, the picture becomes more complicated as both the s and p orbitals are used to form the band of MOs, which then contains many more orbitals than the number of electron pairs available. The criterion for electrical conductivity, a property associated with metals, is that the energy band be only partially filled.
A simple alternative model, consistent with band theory, is the electron sea concept illustrated in Fig. 9-22 for sodium. The circles represent the sodium ions which occupy regular lattice positions (the second and fourth lines of atoms are in a plane below the first and third). The eleventh electron from each atom is broadly delocalized so that the space between sodium ions is filled with an “electron sea” of sufficient density to keep the crystal electrically neutral. The massive ions vibrate about the nominal positions in the electron sea, which holds them in place something like cherries in a bowl of gelatin. This model successfully accounts for the unusual properties of metals, such as the electrical conductivity and mechanical toughness. In many metals, particularly the transition elements, the picture is more complicated, with some electrons participating in local bonding in addition to the delocalized electrons.
Fig. 9-22
9.1. Write the formulas for the following ionic compounds: (a) barium oxide; (b) aluminum chloride; (c) magnesium phosphate.
(a) Barium displays a charge of +2 as an ion and the charge of an oxide ion is –2. It takes one each of the ions to provide the zero charge of a compound; the formula is BaO.
(b) Since aluminum’s charge is +3, it will take three chloride ions at –1 each; therefore, AlCl3.·
(c) The charges on the ions are +2 for Mg and –3 for phosphate. We are to use the lowest number with 2 and 3 as common factors, which is 6. That means using 3 magnesium ions and 2 phosphate ions, which provides us with the formula Mg3(PO4)2.
9.2. Name the following compounds: (a) Mg3P2; (b) Hg2(NO3)2; (c) NH4TcO4.
(a) Since binary compounds are named on the basis of the ions involved, this compound is magnesium phosphide. Recall that a negative ion composed of a single element has the word ending -ide, explaining the naming of P3+.
(b) The naming of (Hg2) requires that we calculate the charge on each mercury ion. Since the total charge on (NO3)2 is –2 and there are two mercury ions, then each mercury ion must be +1. The name of the compound is mercury(I) nitrate (mercurous nitrate).
(c) Since the charge on the ammonium ion is +1, the other ion must be –1. A look at the periodic table tells us that Tc is in the same group as Mn; Mn forms 
(permanganate), implying that the ion 
should be handled in the same manner. The name of the compound is ammonium pertechnetate.
9.3. Determine the charges of the complex ions written in bold italics:
(a) Na2MnO4
(b) H4[Fe(CN)6]
(c) NaCd2P3O10
(d) Na2B4O7
(e) Ca3(CoF6)2
(f) Mg3(BO3)2
(g) UO2Cl2
(h) (SbO)2SO4
(a) Since two sodium ions have a +2 total charge, then MnO4 must be –2.
(b) The ion in brackets must balance the charge of four H+’s; it must be –4.
(c) The ion must balance the charge of one Na+ and two Cd2+ ions (total +5). Then, the ion is –5.
(d) Since there are two sodium ions (total +2), the charge on the complex ion must be –2.
(e) Three calcium ions provides +6, total. Each of the two complex ions must supply –3.
(f) The three Mg2+ ions (+6, total) tells us that each complex displays a –3.
(g) The charge of the complex ion must balance two Cl– ions and, therefore, is +2.
(h) A single sulfate has a charge of –2. Then, each of the two complex ions must be +1.
9.4. The formula of calcium pyrophosphate is Ca2P2O7. Determine the formulas of sodium pyrophosphate and iron(III) pyrophosphate (ferric pyrophosphate).
Because the pyrophosphate ion forms a compound with two calcium ions totaling +4, the charge on pyrophosphate must be –4. The compounds requested must be Na4P2O7 and Fe4(P2O7)3.
9.5. Write the octet structural formulas for (a) CH4O; (b) C2H3F; (c) azide ion, 
.
(a) Rather than using the valence electrons, let us consider the number of bonds typically formed. Each hydrogen can form only one chemical bond. This means that no hydrogen can be between two other atoms. Carbon can form four chemical bonds and oxygen can form two chemical bonds. The only structure that will work is Fig. 9-23.
Fig. 9-23
(b) Knowing that carbon will bond to carbon with one, two, or three bonds, look at Fig. 9-24 and verify that this is the only possible structure.
Fig. 9-24
(c) The total number of available valence electrons in the azide ion is 16 (5 in each of the 3 free N atoms plus 1 arising from the net ionic charge of –1). From this information, it can be seen that a linear structure without multiple bonds is not possible (Fig. 9-25), because there is no way to utilize all of the electrons. However, multiple bonds between the atoms are possible allowing for more than one possible layout to satisfy this set of circumstances, as shown in Fig. 9-26.
Fig. 9-25
Fig. 9-26
Two resonance structures for the azide ion are shown with the triple bond [Fig. 9-26(b) and (c)], because the two terminal nitrogen atoms are the same (look at it as a 180° rotation of the molecule in space). Interestingly, a different structure can be proposed that is based on a three-member ring (Fig. 9-27). The problem with this structure is that the 60° angles required to close the ring place too much stress on the bonds for the structure to be stable and, therefore, is not a possible choice.
Fig. 9-27
9.6. Experimentally, the azide ion is found to be linear with each adjacent nitrogen-nitrogen distance equal to 116 pm. (a) Evaluate the formal charge at each nitrogen in each of the three linear octet structures shown in Fig. 9-26. (b) Predict the relative importance of the three resonance structures.
(a) In structure (a) of Fig. 9-26 the central N is assigned of the four shared pairs, or four electrons. This number is one less than the number of valence electrons in a free N atom; this atom has a formal charge of +1. Each terminal N is assigned four unshared electrons plus of the two shared pairs, or a total of six electrons. The formal charge is –1. The net charge of the ion, –1, is the sum 2(–1)+1.
In structures (b) and (c) of Fig. 9-26 the central N again has four assigned electrons with a resulting formal charge of +1. The terminal N with the triple bond has 2 plus of 3 pairs (a total of 5) with a formal charge of 0.
(b) Structure (a) is the most important because it has no formal charge of magnitude greater than 1. For structure (a) the N—N bond distance is predicted to be (70 + 70) minus the double-bond shortening of 21 pm, or 119 pm. The observed bond length is 116 pm.
9.7. The sulfate ion is tetrahedral with four equal S—O distances of 149 pm. Draw a reasonable structural formula consistent with these facts.
Each of the five atoms involved belong to Group VIA, which tells us there are 30 electrons (6e– for each atom). Further, there are two additional electrons for the net ion charge of –2. It is possible to place the 32 valence electrons in an octet structure having only single bonds. There are two objections to this structure (Fig. 9-28). The first is that the predicted bond distance, rS + rO = 104 + 66 = 170 pm, is much too high. Secondly, the calculated formal charge on the sulfur, +2, is rather high. Regardless, this structure appears widely in textbooks. The reasoning is that the short bond length is the result of the strong attraction between the +2 sulfur and –1 oxygens (formal charges).
Fig. 9-28
A structure like that in Fig. 9-29 places zero formal charge on the sulfur and –1 on each of the singly bonded oxygens. The shrinkage in bond length due to double-bond formation helps to account for the low observed bond distance (149 instead of 170 pm, calculated). Other resonance structures (making a total of six) with alternate locations of the double bonds are, of course, implied. Structures like this, with an expanded valence level beyond the octet, are generally considered to involve d orbitals of the central atom. This is the reason that second-period elements (C, N, O, F) do not form compounds requiring more than eight valence electrons per atom (the 2d subshell simply does not exist).
Fig. 9-29
9.8. Draw all the octet resonance structures of (a) benzene, C6H6, and (b) naphthalene, C10H8. Benzene is known to have a hexagonal symmetry, and the carbon framework of naphthalene consists of two fused hexagons in the same plane.
(a) There is only one structure that will satisfy the requirement of electron pairing and the dictates of the octet rule. That structure is a ring composed of carbons bonded to carbons forming a hexagon. Each of the carbons is bonded to 1 hydrogen. The remaining bonds between carbons are placed so that alternate bonds are double bonds. Figure 9-30 shows the two resonance structures possible in this arrangement of atoms and bonds.
Fig. 9-30
Figure 9-31 displays a shorthand notation for the benzene ring (Fig. 9-30) where each of the points represents a carbon with associated hydrogen(s). Note that the alternating double bonds are indicated. This structure tells us that we are dealing with aromatic compounds. Aromatic compounds are characterized by hydrocarbons forming rings with single bonds between carbons in the ring alternating with double bonds between carbons.
Fig. 9-31
(b) As shown in Fig. 9-32, the two carbons at the fusion of the rings reach their covalence of four without bonding to hydrogen. The shorthand notation is provided in Fig. 9-33. As with other notations of this type, there is a C—H bond at each of the locations not indicating four bonds per carbon.
Fig. 9-32
Fig. 9-33
There is another commonly used form of shorthand notation used to show the structure benzene, naphthalene, and other aromatic compounds. That notation involves indicating that the electrons that form the second bond between carbons are not fixed between two specific carbons, but are free to move within the ring. The notation for these delocalized electrons is a circle within the ring structure, as shown in Fig. 9-34.
Fig. 9-34
9.9. From the data in Problem 7.41, express the H—H bond energy in kJ/mol.
The bond energy is the energy needed to dissociate gaseous H2 into separate atoms.
H2 → 2H
ΔH for this reaction is twice ΔHf for 1 mol of H.
ΔH = 2(218 kJ/mol) = 436 kJ/mol
19.10. Does Table 7-1 give sufficient data for the evaluation of Br—Br bond energy?
No; the energy of Br is given in the table with respect to the standard state for Br2, which is the liquid state, not the gaseous state. The bond energy is the energy required to dissociate individual Br2 molecules (gaseous) into individual Br atoms.
19.11. The enthalpies of hydrogenation of ethylene (C2H4) and benzene (C6H6) have been measured where all reactants and products are gases. Estimate the resonance energy of benzene.
If C6H6 had three isolated carbon to carbon bonds, ΔH of hydrogenation would be close to three times ΔH of hydrogenation of C2H4, with one double bond (–411 kJ/mol). The fact that benzene’s hydrogenation is less exothermic by the calculation, 411 – 206 = 205 kJ/mol, means that benzene has been stabilized by resonance to the extent of 205 kJ/mol.
19.12. Estimate ΔH for the reaction:
The following are the average bond energies in kJ/mol:
Since we are provided bond energies with which to work this problem, we must first identify all bonds involved and whether they are broken or established during the reaction.
Note that bond breaking is endothermic (+ΔH) and bond formation is exothermic (–ΔH) ΔH for the reaction is calculated by the algebraic summation (+ 414 + 242 – 327 – 431 = –102kJ):
There is a difference between the two calculations. Bond energies are calculated from the energies of the specific bond found in many different compounds (an average). The calculations for heats of formation are for the particular molecule under consideration (the entire molecule). In reality, the energy required to break a bond is dependent on the location of that bond in a specific molecule because the bond energy for it is determined by the environment in which the bond is located. In other words, the bond energy depends not only on the specific bond, but on the influences of the bonds and atoms in the surroundings provided by the molecule in which that bond is found.
9.13. The dipole moment (μ) of LiH is 1.964 × 10–29 C · m, and the interatomic distance between Li and H in this molecule is 159.6 pm. Approximately what is the percent ionic character in LiH?
Let us calculate the dipole moment of a hypothetical, 100% ionized Li+ H– ion pair with a separation of 159.6 pm, that is, assuming a point charge at each nucleus.
The approximate percent ionic character equals 100% times the fraction determined by the actual dipole moment divided by the hypothetical dipole moment.
9.14. The dipole moments of SO2 and CO2 are 5.37 × 10–30 C · m and zero, respectively. What can be said of the shapes of the two molecules?
Oxygen is considerably more electronegative than either sulfur or carbon. Each sulfur-oxygen and carbon-oxygen bond should be polar with a net negative charge resides on the oxygen.
Since CO2 has no net dipole moment, the two C—O bond moments must exactly cancel. This can occur only if the two bonds are in a straight line, as in Fig. 9-35(a). (The net moment of the molecule is the vector sum of the bond moments.) The existence of a dipole moment in SO2 must mean that the molecule is not linear, but bent, as in Fig. 9-35(b).
Fig. 9-35
9.15. Using molecular-orbital considerations, account for the fact that oxygen gas is paramagnetic. What is the bond order in O2?
The oxygen atom has the 1s22s22p4 configuration in the ground state. Aside from the first orbit electrons of the two atoms in O2, which are so deeply imbedded in their respective atoms as to not overlap with other electrons, the remaining 12 electrons (6 from each atom) will fill the lowest of the available molecular orbitals as shown in Fig. 9-36. The bond axis is the x-axis. The electron configuration of O2 is: [He]
.
Fig. 9-36 Electrons are indicated by arrows, orbitals by squares, first orbit electrons are omitted. Note: The above energy level diagram is not relevant to the molecules B2 and C2, in which the order of the σpx and πy,z levels is reversed.
The last two electrons went into the antibonding equi-energetic π* orbitals, one into 
and one into 
, so as to maximize electron spin in accordance with Hund’s rule. These two unpaired electrons tell us that the molecule is paramagnetic. (BO, bonding orbitals; ABO, antibonding orbitals.)
9.16. Explain the observations that the bond length in 
is 2 pm greater than in N2, while the bond length in NO+ is 9 pm less than in NO.
The electron configurations are written for the four molecules according to the buildup principle.
The computed bond orders are 3 for N2 and 2 for . N2, therefore, has a stronger bond and should have the shorter bond length. The computed bond orders are 2 for NO and 3 for NO+. The NO+ cation has the stronger bond and should have the shorter bond length. In contrast to the ionization of N2, which involves the loss of an electron in the bonding orbital, the ionization of NO involves the loss of an electron in an antibonding orbital.
9.17. Two substances having the same molecular formula, C4H8O, were examined in the gaseous state by electron diffraction. The carbon-oxygen distance was found to be 143 pm in compound A and 124 pm in compound B. What can you conclude about the structures of these two compounds?
In compound A the carbon-oxygen distance is the sum of the single-bond covalent radii of carbon and oxygen, indicating a single bond (77 pm + 66 pm = 143 pm).
Then, the oxygen cannot be terminal. One such structure is the heterocyclic compound tetrahydrofuran [Fig. 9-37(a)].
Fig. 9-37
In compound B, the carbon-oxygen distance is close to that predicted for a double bond, 122 pm. Then, the oxygen must be terminal. A structure that conforms to these data is 2-butanone [Fig. 9-37(b)].
9.18. Verify the value Θ = 109°28’ for the central angles in a regular tetrahedron.
A simple way of constructing a regular tetrahedron is to select alternating corners of a cube and to connect each of the selected corners with each of the other three, as in Fig. 9-38(a). Figure 9-38(b) shows triangle OAB, determined by the center of the cube, O, which is also the center of the tetrahedron, and two corners of the tetrahedron, A and B. If P is the midpoint of AB, we see from right triangle OPA that the mathematical relationship is as follows:
Fig. 9-38
9.19. The C—C single-bond distance is 154 pm. What is the distance between the terminal carbons in propane, C3H8? Assume that the four bonds of any carbon atom are pointed toward the corners of a regular tetrahedron.
With reference to Fig. 9-38(b), two terminal carbons can be thought of as lying at A and B with the central atom at O. Then,
9.20. Sulfur and chlorine combine in various proportions forming S2Cl2, SCl2, and SCl4. Draw the Lewis structures of these molecules and, using VSEPR, predict their shapes.
In S2O2, Fig. 9-39(a), each S has a VSEPR number of 4 so that each CISS bond angle is about 109.5° (probably somewhat less because of extra repulsion from the unshared pairs on each sulfur), and the molecule cannot be linear (straight). There is free rotation around the S — S bond so that there is no fixed conformation of the molecule.
Fig. 9-39
In SCl2, Fig. 9-39(b), the S atom has a VSEPR number of 4, also. The Cl—S —Cl angle is a little less than 109° and the molecule is angular (bent).
In SCl4, Fig. 9-39(c), the VSEPR number is 5, but one of the positions has an unshared electron pair. That must be one of the trigonal positions in the trigonal pyramid, Fig. 9-40(a), in which there are two shared-unshared angles of 90°, rather than an axial position (b), which has three shared-unshared angles of 90°.
Fig. 9-40
The stick figure corresponding to Fig. 9-40(a) has been called a “see-saw” or a “saw-horse” in which the axial ClSCl group represents a horizontal beam. Note: The unshared pair bends the ClSCl slightly.
9.21. What are the O—N—O bond angles in the nitrate, NO
–, and nitrite, NO–, ions?
Only one Lewis structure for each ion needs to be drawn to deduce the VSEPR number, which is 3 in both cases for the central N. (In No– there are three σ -bonded neighbors and no unshared pairs and in NO– there are two neighbors and one unshared pair, as shown in Fig. 9-41.) The nominal bond angle is 120°, exactly what is found in nitrate since resonance makes all the angles equal. The nitrite ion is a different story because the unshared pair repels the bonding pairs more than they repel each other, forcing them closer to each other. We would expect a slightly smaller angle than 120°; it turns out to be 115°.
Fig. 9-41
9.22. The POCl3 molecule has the shape of an irregular tetrahedron with the P atom located centrally. The Cl—P—Cl bond angle is found to be 103.5°. Give a qualitative explanation for the deviation of this structure from a regular tetrahedron.
The VSEPR number for the P atom is 4, so that the nominal bond angles are 109°28′. However, the Lewis structure for POCl3 shows a double bond between P and O. (P is allowed to exceed the octet because of the availability of 3d orbitals.) The increased electron density in the P═O bond would make the repulsion between the P═O bond and the P—Cl bond greater than between two P—Cl bonds. The Cl—P—Cl angle is lowered and the Cl—P═O angle increased.
9.23. PCl5 has the shape of a trigonal bipyramid (Fig. 9-42) and IF5 has the shape of a square pyramid (Fig. 9-43). Account for the difference.
Fig. 9-42
Fig. 9-43
The Lewis structures of the singly bonded compounds (Fig. 9-44) reveal a VSEPR number of 5 in PCl5, for which the trigonal bipyramid structure with 90°, 120°, and 180° angles is predicted (Table 9-4). However, the unshared pair on the iodine raises its VSEPR number to 6 so that the nominal bond angles are 90°. The square pyramid structure of IF5 may be thought of as an octahedron (Fig. 9-7) with the unshared pair pointing to the corner below the indicated horizontal plane. Because of the repulsion of the unshared pair, the iodine atom is expected to be slightly below the base plane of the pyramid.
Fig. 9-44
9.24. Which of the four C—C bond types in naphthalene (Fig. 9-33) is predicted to be the shortest?
The four different kinds of carbon-carbon bonds are represented by 1-2, 1-9, 2-3, and 9-10. (Every other carbon-carbon bond is equivalent to one of these four. For example, 6-7 is equivalent to 2-3, 7-8 is equivalent to 1-2, etc.) The bond with the greatest double-bond character should be the shortest. Among the three resonance structures shown in Fig. 9-33, the frequency of double bonds for the various bond types is as follows: 2 in 1–2 [in (a) and (c)], 1 in 1–9 [in (b)], 1 in 2–3 [in (b)], and 1 in 9–10 [in (c)]. Bond 1–2 is expected to have the greatest double-bond character and to be the shortest. This prediction is found to be true experimentally. The above four bonds are found to have lengths of 136.5, 142.5, 140.4, and 139.3 pm, respectively.
Note that the method of counting the number of resonance structures containing a double bond between a given pair of carbon atoms is very crude and cannot distinguish among the last three of the listed bond types, each of which shows a double bond in only one resonance structure. Even within the framework of the limited resonance theory, it would be necessary to know the relative weighting of each of the two equivalent structures (a) and (b) with the nonequivalent structure (c).
9.25. What is the shape of the triple iodide ion, 
The Lewis structure reveals a VSEPR number of 5 for the central iodine atom, two bonded neighbors and three unshared pairs. To determine which corners of the trigonal bipyramid are occupied by the terminal iodine atoms, find the arrangement which maximizes the angles between the unshared pairs. The preferred arrangement, Fig. 9-45(a), must be the one in which the unshared pairs are all at 120° because any other alternative [Fig. 9-45(b) and (c)] would have two sets of pairs at 90°. Therefore, the two terminal iodine atoms must occupy the axial positions (180° to each other), making the molecule linear.
Fig. 9-45
9.26. Soluble compounds of the complex ion [Co(NH3)3]3+ have a maximum absorption of visible light at 437 nm. (a) What is the value of Δ for this complex ion; express the answer in cm–1. (b) What is the color of this ion in solution? (c) How many unpaired electrons would you expect this ion to have if it is considered low-spin, and how many if it is considered high-spin?
(a) 
(b) The color depends not only on the wavelengths of the absorption maximum, but also on the shape of the whole absorption band and on the color sensitivity of the human eye. Because of these factors, the prediction of the color from the data is not absolute; however, a decent prediction can be made. The absorption, which peaks in the blue-violet region of the spectrum, would be expected to cover most of the blue region and part of the green. The prediction of the ion’s color in solution is complementary (the opposite) of the light absorbed and expected to be yellow.
(c) The electron configuration of Co3+ is [Ar]3d6. The spin is due to unpaired d electrons. For low spin, the six d electrons would all be paired in the three t2g orbitals, and the spin would be zero. For high spin, the two eg molecular orbitals would be available as well (see Fig. 9-14). Four of the available orbitals would be half filled (1 electron each) and one would be filled (1 pair of electrons) in order to maintain the maximum number of unpaired electrons, four in this case. The Δ value is large enough to rule out high spin and the ion is diamagnetic.
9.27. Predict the magnetic properties of (a) [Rh(NH3)6]3+ and (b) [CoF6]3–.
(a) This problem can be approached by comparison with Problem 9.26. For analogous complexes of two different members of the same group in the periodic table, Δ increases with increasing atomic number. Since Δ for [Co(NH3)3]3+ is already so high that the ion is low-spin, [Rh(NH3)6]3+ is certainly low-spin and diamagnetic (observed Δ = 34,000 cm–1; therefore, diamagnetic).
(b) F– is a weak-field ligand, tending to form complexes with a low Δ value, so that the ion is expected to be high-spin with four unpaired and parallel electron spins [compare with Problem 9-26(c)]. The measured Δ is 13,000 cm–1, a low figure, and the ion is paramagnetic.
9.28. Write all the structural isomeric formulas for C4H9Cl.
The molecular composition resembles butane, C4H9, except that one of the hydrogens is replaced by a chlorine. Figure 9-15 is a good starting place because butane’s two carbon skeletons are shown.
Note that the two terminal (end) carbons of n-butane [Fig. 9-46(a) and (b)] are identical and that the two interior carbons are also identical. If a chlorine were to be substituted on the left carbon, it would be the same as substituting on the right if we were to rotate the molecule 180° and we would still have (a). The same concept is true of the interior carbons; substituting as in (b) or substituting at the left, interior carbon yields the same molecule on rotation.
Fig. 9-46
In iso-butane [Fig. 9-46(c) and (d)] one isomer (c) has the Cl on the central carbon and there is only one other isomer (d) because the three terminal carbons are identical. A point worth noting is that all of the C to C bonds are single bonds; this means that, since there is free rotation about a single bond, all of the hydrogens bonded to a particular carbon are identical. Because of this factor, replacing one H on a carbon with a Cl is the same as replacing any other.
9.29. Write formulas for all the structural and geometrical isomers of C4H8.
If the four carbons are in a row, there must be one double bond in order to satisfy the requirement that each carbon have four chemical bonds. The double bond occurs either in the center of the molecule or toward an end. If in the center, two geometrical isomers occur with different positions of the terminal carbons relative to the double bond. In the latter case, two structural isomers occur differing in the extent of the branching within the carbon skeleton. Additional possibilities are ring structures.
9.30. Which of the isomeric C4H9Cl from Fig. 9-46 would you expect to be optically active?
Compound (b) is the only one which would exist in optically active isomeric forms because it is the only one which has a carbon atom bonded to four different groups (the C bonded to the Cl). All other carbons have either two hydrogens or two methyl groups (—CH3).
9.31. How many geometrical isomers could [Rh(py)3Cl3] have? The abbreviation py is for the ligand pyridine.
One of the two possibilities, Fig. 9-47(a), has the three chlorines occupying bonding positions cis (adjacent) to each other on one face of the octahedron and the three pyridines on the opposite face. The other possibility, Fig. 9-47(b), indicates that there are two of the chlorines trans (opposite) to each other and two of the pyridines are also trans to each other.
Fig. 9-47
9.32. Some ligands are multifunctional, that is, they have two or more atoms that can bind to the central metal atom or ion. Each binding site occupies a different corner on the coordination surface. Ethylenediamine (abbreviated en) is such a ligand. The two binding atoms are nitrogens and the two binding sites must be cis to each other because of the shape and size of the en. How many geometrical isomers of [Cr(en)2Cl2]+ should exist and which isomer(s) might display optical activity?
Two geometrical isomers exist, cis and trans (Fig. 9-48). Each en can be represented by an arc terminating at the two binding sites. By drawing other arrangements of arcs while preserving positions of the chlorines, one can see that (b) is a distinct mirror image of (a); however, the mirror image of (c) is the exact same structure as (c). In other words, only the cis isomers can be optically active.
Fig. 9-48
9.33. Explain why metals are usually lustrous (mirror-like).
In the band model there is a continuum of empty energy levels, rather than discrete energy levels. This situation allows light quanta of all energies within a wide range of wavelengths to be absorbed equally, then the energized electrons will re-emit the light when they fall back into their ground-state orbitals. This is the mechanism for reflection of light of all frequencies, which we call “luster.”
9.34. Predict how the Group II metals differ from Group I in density, melting point, and mechanical strength.
In any given period, the Group II ions are smaller and may then approach each other more closely. At the same time, twice as many electrons are present in the electron sea. The closer approach and the much greater electrostatic interactions between the 2+ ions and the sea of high negative charge density lead to greater density and much greater bonding energy, which in turn leads to a much higher melting point and greater hardness. Actually, going from Group I to Group II increases the density by a factor of about 2 and raises the melting point by hundreds of degrees.
9.35. Metals feel cool to the touch compared with other materials because they are very good conductors of heat. How can we explain this unusual thermal conductivity?
In most material, heat is conducted by atom-to-atom transfer of vibrational motion from the hot end to the cooler end. In metals, the thermal energy is transferred primarily by the motion of the electron sea’s free electrons, which are very mobile.
9.36. Determine the charges on the groups written in bold italics, as the chloride in NaCl.
(a) CaC2O4; (b) Ca(C7H5O3)2 · 2H2O; (c) Mg3(AsO3)2; (d) MoOCl3; (e) CrO2F2; (f) PuO2Br; (g)(PaO)2S3
Ans. (a)–2; (b)–1; (c) –3; (d) +3; (e) +2; (f) +1; (g) +3
9.37. Heavy metal compounds tend to be toxic and require careful handling; some of those are compounds of lead, thallium, mercury, and barium. Provide the formulas of these heavy metal bromides, sulfides, nitrides, and carbonates. The more common oxidation numbers of the metals having more than one are Pb2+, Tl+, Hg2+.
Ans. PbBr2, PbS, Pb3N2, PbCO3 TlBr, Tl2S, Tl3N, Tl2CO3
HgBr2, HgS, Hg3N2, HgCO3 BaBr2, BaS, Ba3N2, BaCO3
9.38. Write the formulas for the following ionic compounds: (a) lithium hydride; (b) calcium bromate; (c) chromium(II) oxide; (d) thorium(IV) perchlorate; (e) nickel phosphate; (f) zinc sulfate.
Ans. (a) LiH; (b) Ca(BrO3)2; (c) CrO; (d) Th(ClO4)4; (e)Ni3(PO4)2; (f) ZnSO4
9.39. Write the chemical formulas for (a) gold(III) nitrate, cobalt(II) nitrate, bismuth(V) nitrate, radium nitrate, tin(IV) nitrate, and arsenic(III) nitrate. (b) Write the sulfite compounds.
Ans. (a) Au(NO3)3, Co(NO3)2, Bi(NO3)5, Ra(NO3)2, Sn(NO3)4, As(NO3)3
(b)Au2(SO3)3, CoSO3, Bi2(SO3)5, RaSO3, Sn(SO3)2, As2(SO3)3
9.40. Name the following compounds: (a) Al(NO3)3; (b) Al(NO2)3; (c) AlN; (d) Al2(SO4)3; (e) Al2(SO3)3; (f)Al2S3; (g)Sb2S3; (h)Sb2S5
Ans. (a) aluminum nitrate; (b) aluminum nitrite; (c) aluminum nitride; (d) aluminum sulfate; (e) aluminum sulfite; (f) aluminum sulfide; (g) antimony(III) sulfide ; (h) antimony(IV) sulfide.
9.41. Write the chemical formulas for (a) cupric sulfide; (b) stannous fluoride; (c) plumbous chloride; (d) ferric iodide; (e) auric nitrate; (f) mercuric sulfide.
Ans. (a) CuS; (b) SnF2; (c) PbCl2; (d) FeI3; (e) Au(NO3)3; (f) HgS
9.42. Name the following compounds: (a) Mg(IO)2; (b) Fe2(SO4)3; (c) CaMnO4; (d) KReO4; (e) CaWO4; (f) CoCO3
Ans. (a) Magnesium hypoiodite; (b) iron(III) sulfate or ferric sulfate; (c) calcium manganate; (d) potassium perrhenate; (e) calcium tungstate; (f) cobalt(II) carbonate
9.43. The formula for potassium arsenate is K3AsO4. The formula for potassium ferrocyanide, systematically called potassium hexacyanoferrate(II), is K4Fe(CN)6. Write the formulas for (a) calcium arsenate; (b) iron(III) arsenate; (c) barium ferrocyanide; (d) aluminum ferrocyanide.
Ans. (a)Ca3(AsO4)2; (b) FeAsO4; (c)Ba2Fe(CN)6; (d)Al4[Fe(CN)6]3
9.44. Draw Lewis structures for each of the following: (a)C2HCl; (b)C2H6O; (c)C2H4O; (d)NH3O; (e)
(both O’s terminal); (f)N2O4 (all O’s terminal); (g)OF2
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
Ans. Note that not all have resonance structures.
9.45. Complete the following structures by adding unshared electron pairs where necessary. Then evaluate the formal charges.
(a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
(i) 
(j) 
(k) 
(l) 
Ans. (a) all zero; (b) +1 on one N (which does not have an octet), —1 on the other; (c) all zero; (d) +1 on Cl, —1 on N; (e) —1 on terminal N, +1 on central N; (f) +1 on central N, —1 on O; (g) all zero; (h) +1 on doubly bonded Cl, —1 on O; (i) +1 on N, —1 on singly bonded O; (j) +1 on Cl, —1 on singly bonded O; (k) +1 on each N, —1 on each B; (l) all zero
9.46. Given that the formula for formaldehyde is CH2O, one could draw three Lewis structures starting with the skeletons:
There are 12 valence electrons of which six are used in each of the above skeletons. Use the other six to complete the octets about C and O; determine the formal charges; and decide which structure is correct.
Ans.
9.47. Three molecules of formaldehyde can condense into one cyclic molecule. Draw the Lewis structure of the molecule. Hint: The C and O atoms alternate.
Ans.
9.48. Are these two structures isomers? If not, why not? If they are isomers, what is the classification of the isomerism?
Ans. No. The right structure is the left structure rotated 180°.
9.49. What are the isomers of this compound holding the carbon skeleton stable?
Ans. There are five sketches including the above structure repeated in the list below. Note that changing positions for the OH on each of the carbons does not always provide us with different isomers; (e) and (e) are the same structure!
9.50. Of the isomers of the compound in the previous question, which is/are optically active?
Ans. Only those carbons with four different groups attached allow for optical activity; there are no structures that qualify, except for (c).
9.51. The chlorine-oxygen bond distance in ClO–4 is 144 pm. What do you conclude about the valence-bond structures for this ion?
Ans. Since the Cl—O single-bond length estimated from Table 9-2 is 165 pm, there must be considerable double-bond character in the bonds.
9.52. The VSEPR number for phosphorus is 4 for PH3. However, the bond angles noted are not 109°28′, the tetrahedral angle. Explain.
Ans. The unshared pair of electrons on the phosphorus tends to compress the angle.
9.53. Considering only double bonds between adjacent carbons, how many resonance structures can be written for each of the following aromatic hydrocarbons?
Ans. (a)4; (b)5; (c)5
9.54. The structure of 1,3-butadiene is often written as H2C═CH═CH═CH2. The distance between the central carbon atoms is 146 pm. Comment on the adequacy of the assigned structure.
Ans. The expected bond length is 77 + 77 = 154 pm for a purely single bond. There must be non-octet resonance structures involving double bonding between the central carbons, such as
9.55. The average C—C bond energy is 347 kJ/mol. What do you predict for the Si— Si single-bond energy and why?
Ans. Because Si is a much bigger atom than C, there will be less orbital overlap (less electron sharing) and probably less than 300 kJ/mol bond energy.
9.56. The average C—Cl bond energy is 330 kJ/mol. What do you predict for the C—N single-bond energy and why?
Ans. The chlorine atom is significantly larger than the nitrogen and has a much larger nuclear charge, leading us to conclude that the C—N bond energy is most likely less than 330 kJ/mol. The reported average C — N bond energy is 300 kJ/mol.
9.57. (a) What are the bond orders for CN–, CN, and CN+? (b) Which of these species should have the shortest bond length?
Ans. (a)CN–, 3; CN, 2;CN+, 2; (b)CN–
9.58. Other than oxygen, O2, which homonuclear diatomic molecule(s) of second period elements should be paramagnetic?
Ans. Boron, B2
9.59. Assuming the elements in the second period can form homonuclear diatomic molecules, which diatomic molecules should have zero bond order?
Ans. Be2, Ne2
9.60. Dipole moments are sometimes expressed in debyes, where
1 debye = 10–18 (esu of charge) × cm
The electrostatic unit (esu) of charge is defined by 1 C = 2.998 × 109 esu. What is the value of 1 debye in SI units?
Ans. 3.336 × 10–30 C · m
9.61. Where in space are the six electrons in benzene, normally drawn as a ring inside a hexagon?
Ans. The six electrons occupy molecular orbitals, which are made by combining p-orbitals that are perpendicular to the plane of the molecule. The electron density can be shown by two identical circular clouds, one above and one below the plane of the molecule.
9.62. If a hydrogen were to be removed from benzene and were to be substituted by a chlorine (or any other entity with a single bond), where would the chlorine be located with respect to the plane of the molecule?
Ans. It would be either above or below the plane of the molecule, but definitely not in the plane of the molecule. The bond is formed using a p-orbital and that limits the choice to above or below the plane of the molecule.
9.63. The dipole moment of HBr is 2.6 × 10–30 C · m and the interatomic spacing is 141 pm. What is the percent ionic character of HBr?
Ans. 11.5%
9.64. The dipole moments of NH3, AsH3, and BF3 are (4.97, 0.60, and 0.00) × 10–30 C · m, respectively. What can be concluded about the shapes of these molecules?
Ans. NH3 and AsH3 are both pyramidal and BF3 is planar. From the dipole moments alone, nothing can be concluded about the relative flatness of the NH3 and AsH3 pyramids because the electronegativities of nitrogen and arsenic are different.
9.65. The As—Cl bond distance in AsCL3 is 217 pm. Estimate the single-bond radius of As.
Ans. 118 pm
9.66. Predict the bond length between carbon and fluorine in 1-chloro-1-fluoroetheylene (refer to Table 9-2) and predict if there is likely to be variation from your calculations.
Ans. The predicted bond length is 141 pm; however, the presence of the C═C and the chlorine will tend to affect this prediction. Fluorine and chlorine are quite electronegative (compared with carbon), forming negative centers that tend to repel.
9.67. The carbon-carbon double-bond energy in C2H4 is 615 kJ/mol and the carbon-carbon single-bond energy in C2H6 is 347 kJ/mol. Why is the double-bond energy appreciably less than twice the single-bond energy?
Ans. The σ orbital has greater electron overlap between the atoms because its component atomic orbitals are directed toward each other, whereas the component p orbitals making up the π orbital are directed perpendicularly to the internuclear axis and have only side-to-side overlap.
9.68. Estimate ΔH for the reaction:
C2H5Cl(g) → HCl(g) + C2H4(g)
given the following bond energies in kJ/mol:
Ans. –64 kJ
9.69. Estimate the bond energy of the F—F bond given that ΔH°f of HF(g)is –271 kJ/mol. The bond energies are H—F, 565 kJ/mol, and H—H, 435 kJ/mol.
Ans. 153 kJ/mol
9.70. Using bond energies, calculate the energy involved in the burning of 1 mol octane, one of the components of gasoline (balance reaction first).
The bond energies (kJ/mol) are: C —C (346), C—H (413), C═O (732), C —O (358), O═O (498), and H—O (463).
Ans. –3965 kJ/mol octane (exothermic reaction)
9.71. Ethanol, C2H5OH, is in the spirits that caramelize the sugar on the top of crème brûlée. Calculate the amount of energy involved in the combustion of one mole of ethanol using the bond energies in Problem 9.70.
Ans. –980 kJ/mol C2H5OH (exothermic reaction)
9.72. The platinum-chlorine distance has been found to be 232 pm in several crystalline compounds. If this value applies to both of the compounds in Fig. 9-18, what is the Cl—Cl distance in (a) structure (a); (b) structure (b)?
Ans. (a) 464 pm, (b) 328 pm
9.73. What is the length of a polymer molecule containing 1001 carbon atoms singly bonded in a line if the molecule could be stretched to its maximum length consistent with maintenance of the normal tetrahedral angle within any C —C—C group?
Ans. 126 nm or 1260 Å
9.74. A plant virus was examined by the electron microscope and was found to consist of uniform cylindrical particles 15.0 nm in diameter and 300 nm long. The virus has a specific volume of 0.73 cm3/g. If the virus particle is considered to be one molecule, what is its molar mass?
Ans. 4.4 × 107 g/mol
9.75. Assuming the covalent radii in the C—Cl bond are additive, what would be the Cl—Cl distance in each of the three dichlorobenzenes (Fig. 9-49)? Assume that the ring is a regular hexagon and that each C—Cl bond lies on a line through the center of the hexagon. The distance between adjacent carbons is 140 pm.
Fig. 9-49
Ans. (a) 316 pm; (b) 547 pm; (c) 632 pm
9.76. Estimate the length and width of the carbon skeleton of anthracene, Problem 9-53(a). Assume hexagonal rings with equal C—C distances of 140 pm.
Ans. 730 pm long and 280 pm wide
9.77. BBr3 is a symmetrical planar molecule with all B—Br bonds lying at 120° to each other. The distance between Br atoms is found to be 324 pm. From this fact and given that the covalent radius of Br is 114 pm, estimate the covalent radius of boron. Assume all bonds are single bonds.
Ans. 73 pm
9.78. What is the VSEPR number for the central atoms in each of the following species? (Hint: Draw the Lewis structures first.) (a)SO2; (b)SO3; (c)
(d) 
(e) SF6; (f) 
Ans. (a)3; (b)3; (c)4; (d)4; (e)6; (f)4
9.79. What are the bond angles in each of the species in Problem 9.78?
Ans. (a) slightly less than 120°; (b) 120°; (c) slightly less than 109°28′; (d) 109°28′; (e)90°; (f) slightly less than 109°28′
9.80. Provide the VSEPR number, the general shape, and the bond angles for each of the following species: (a) XeF4; (b) XeO3; (c) XeF2
Ans. (a) VSEPR number 6, square planar, 90°; (b) VSEPR number 4, trigonal pyramid with one unshared pair at the apex, slightly less than 109°28′; (c) VSEPR number 5, linear, 180°
9.81. Which of these molecules or ions is/are straight (i.e., all bond angles 180°)? (a)OF2; (b) HCN; (c)H2S; (d)CO2; (e)
Ans. (b), (d), and (e)
9.82. Which of the these molecules or ions is/are bent? (a) BeCl2; (b) HOCl; (c)
(d) 
(e)
Ans. (b), (c), and (d)
9.83. Which of these molecules or ions is/are flat (all atoms in the same plane)? (a) BF3; (b) XeO4; (c) 
(d)C2H2; (e)HN3
Ans. (a); (c); (d); (e); (d) is perfectly linear; (e) is bent at the H end
9.84. Which of these molecules or ions is expected to have a dipole moment? (a)CH2Cl2; (b) cis-C2H2Cl2; (c) trans-C2H2Cl2; (d)CH2CCl2; (e)SF4; (f) XeF4; (g)C2F4; (h)H2SO4; (i)
(f) N2H4; (k) NCl3
Ans. (a); (b); (d); (e); (h); (k)
9.85. Disregarding any hydrogen atoms, which of these molecules or ions is/are expected to be flat? (a) CH3CHCHCl; (b) HNO3; (c)
(d) SOCl2; (e) C6H5OH
Ans. (a), (b), and (e)
9.86. Name the following compounds (en = ethylenediamine, py = pyridine):
(a) [Co(NH3)5Br]SO4
(b) [Cr(en)2Cl2]Cl
(c) [Pt(py)4][PtCl4]
(d) K2[NiF6]
(e) K3[Fe(CN)5CO]
(f) CsTeF5
Ans. (a) pentaamminebromocobalt(III) sulfate
(b) dichlorobis(ethylenediamine)chromium(III) chloride
(c) tetrapyridineplatinum(II) tetrachloroplatinate(II)
(d) potassium hexafluoronickelate(IV)
(e) potassium carbonylpentacyanoferrate(II)
(f) cesium pentafluorotellurate(IV)
9.87. Write the formulas for the named compounds using square brackets to enclose the complex ion.
(a) triamminebromoplatinum(II) nitrate; (b) dichlorobis(ethylenediamine)cobalt(II) monohydrate; (c) pentamminesulfatocobalt(III) bromide; (d) potassium hexafluoroplantinate(IV); (e) tetraaquadibromochromium(III) chloride; (f) ammonium heptafluorozirconate(IV)
Ans. (a) [Pt(NH3)3Br]NO3; (b) [Co(en)2Cl2] · H2O; (c) [Co(NH3)5SO4]Br (d)K2[PtF6]; (e) [Cr(H2O)4Br2]Cl; (f) (NH4)3[ZrF7]
9.88. Considering that Δ for IrCl
is 27,600 cm–1, (a) What is the wavelength of maximum absorption? (b) What do you predict for the magnetic behavior of this ion?
Ans. (a) 362 nm; (b) diamagnetic
9.89. (a) What is the maximum number of unpaired electrons which a high-spin octahedral complex of the first transition series could possess in the ground state? (b) Which first transition elements could show this maximum? Provide the oxidation states.
Ans. (a)5; (b) Mn(II) and Fe(III)
9.90. If the metal of the first transition series has a dn configuration (in its relevant oxidation state) and forms an octahedral complex, for what values of i could magnetic properties alone distinguish between strong-field and weak-field ligands?
Ans. 4, 5, 6, 7
9.91. Both [Fe(CN)6]4– and [Fe(H2O)6]2+ appear colorless in dilute solutions. The former ion is low-spin and the latter is high-spin. (a) How many unpaired electrons are in each of these ions? (b) In view of the apparent significant difference in their Δ values, why should both ions be colorless?
Ans. (a) 0 in [Fe(CN)6]4–, 4 in [Fe(H2O)6]2+, (b) Δ for [Fe(CN)6]4– is so large that the absorption peak is in the ultraviolet; Δ for [Fe(H2O)6]2+ is so small that its absorption peak is in the infrared. Both of these ions produce solutions that absorb practically no visible light.
9.92. The hexaaquairon(III) ion is practically colorless. Its solutions become red when NCS– is added. Explain. (Compare with Problem 9.19.)
Ans. Water is not a strong-field ligand (Problem 9.91). NCS– has vacant π* orbitals that overlap the t2g orbitals of the metal and so can accept electron density from the metal. This “back-bonding” increases the strength of the metal-ligand bond and lowers the t2g energy level, making NCS– a strong-field ligand. These ligands cause an increase in A; this produces a lowering of the wavelength of maximum d-d absorption from the near-infrared well into the visible region (into the blue-green).
9.93. How many structural isomers can be drawn for each of these compounds? (a)C5H12; (b)C3H7Cl; (c)C3H6Cl2; (d)C4H8Cl2; (e)C5H11Cl; (f) C6H14; (g) C7H16
Ans. (a) 3 (b) 2; (c) 4; (d) 9; (e) 8; (f) 5; (g) 9
9.94. Among the paraffin hydrocarbons (CnH2n+2, where n is a whole number), what is the empirical formula of the compound of lowest molar mass which could demonstrate optical activity in at least one of its structural isomers?
Ans. C7H16
9.95. How many structural and geometrical isomers can be written for the following without counting ring compounds: (a)C3H5Cl; (b)C3H4Cl2; (c)C4H7Cl; (d)C5H10?
Ans. (a)4; (b)7; (c) 11; (d)6
9.96. For the square coplanar complex [Pt(NH3)(NH2OH)py(NO2]+, how many geometrical isomers are possible? Describe them.
Ans. There are three isomers. Any one ligand can be trans to any of the other three; the two ligands not trans to the first have their positions automatically fixed as trans to each other.
9.97. Predict whether [Ir(en)3] 3+ should exhibit optical isomerism. If so, prove by diagrams that the two optical isomers are not simple rotational aspects of the same compound.
Ans. There are two optical isomers, which are represented in Fig. 9-50.
Fig. 9-50
9.98. How many isomers are there with the formula [M(en)XY3], where M is the central metal, en is the bidentate ligand ethylenediamine, and X and Y are monodentate ligands? Explain how they differ.
Ans. There are two geometric isomers. One of the isomers has two of the Y’s trans to each other. In the second isomer, all the Y’s are cis.
9.99. In silicon there is a gap in energy between the band of bonding MOs and the band of an equal number of antibonding MOs, all derived from the 3s and 3p atomic orbitals. Is silicon a metal electrical conductor? Explain.
Ans. No. In a crystal of N atoms there will be 4N MOs, 2Nof them bonding. The 4N valence electrons will just fill these 2N bonding MOs, but metallic conductivity requires that the band be only partially filled.
9.100. How does a rise in temperature affect the electrical conductivity of a metal? Explain.
Ans. A rise in temperature lowers the conductivity because vigorous atomic motion disrupts the long-range order of the lattice which obstructs the MOs that are throughout the whole crystal.
9.101. Most metals are ductile and malleable, in contrast to the brittleness of most other solids. Account for these properties in terms of the electron-sea model.
Ans. Since the atoms are rather far apart and the electron sea offers little resistance to deformation, it does not require much energy to cause one layer of atoms to slide past another. Because of this factor the crystal is deformed rather than being shattered by an external stress.
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