Chapt09

Accounting

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 9 Chemical Bonding I: Basic Concepts INTRODUCTION WHY DO ATOMS OF DIFFERENT ELEMENTS REACT? WHAT ARE THE FORCES 9.1 LEWIS DOT SYMBOLS THAT HOLD ATOMS TOGETHER IN MOLECULES AND IONS IN IONIC COM- 9.2 THE IONIC BOND POUNDS? 9.3 LATTICE ENERGY OF IONIC COMPOUNDS WHAT SHAPES DO THEY ASSUME? THESE QUESTIONS ADDRESSED IN THIS CHAPTER AND IN ARE SOME OF THE CHAPTER 10. WE BE- 9.4 THE COVALENT BOND GIN BY LOOKING AT THE TWO TYPES OF BONDS — IONIC AND COVA- 9.5 ELECTRONEGATIVITY LENT — AND THE FORCES THAT STABILIZE THEM. 9.6 WRITING LEWIS STRUCTURES 9.7 FORMAL CHARGE AND LEWIS STRUCTURE 9.8 THE CONCEPT OF RESONANCE 9.9 EXCEPTIONS TO THE OCTET RULE 9.10 BOND DISSOCIATION ENERGY 329 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 330 CHEMICAL BONDING I: BASIC CONCEPTS 9.1 LEWIS DOT SYMBOLS The development of the periodic table and concept of electron configuration gave chemists a rationale for molecule and compound formation. This explanation, formulated by Gilbert Lewis,† is that atoms combine in order to achieve a more stable electron configuration. Maximum stability results when an atom is isoelectronic with a noble gas. When atoms interact to form a chemical bond, only their outer regions are in contact. For this reason, when we study chemical bonding, we are concerned primarily with the valence electrons of the atoms. To keep track of valence electrons in a chemical reaction, and to make sure that the total number of electrons does not change, chemists use a system of dots devised by Lewis and called Lewis dot symbols. A Lewis dot symbol consists of the symbol of an element and one dot for each valence electron in an atom of the element. Figure 9.1 shows the Lewis dot symbols for the representative elements and the noble gases. Note that, except for helium, the number of valence electrons each atom has is the same as the group number of the element. For example, Li is a Group 1A element and has one dot for one valence electron; Be, a Group 2A element, has two valence electrons (two dots); and so on. Elements in the same group have similar outer electron configurations and hence similar Lewis dot symbols. The transition metals, lanthanides, and actinides all have incompletely filled inner shells, and in general, we cannot write simple Lewis dot symbols for them. In this chapter we will learn to use electron configurations and the periodic table to predict the type of bond atoms will form, as well as the number of bonds an atom of a particular element can form and the stability of the product. 9.2 Lithium fluoride. Industrially, LiF (like most other ionic compounds) is obtained by purifying minerals containing the compound. THE IONIC BOND In Chapter 8 we saw that atoms of elements with low ionization energies tend to form cations, while those with high electron affinities tend to form anions. As a rule, the elements most likely to form cations in ionic compounds are the alkali metals and alkaline earth metals, and the elements most likely to form anions are the halogens and oxygen. Consequently, a wide variety of ionic compounds combine a Group 1A or Group 2A metal with a halogen or oxygen. An ionic bond is the electrostatic force that holds ions together in an ionic compound. Consider, for example, the reaction between lithium and fluorine to form lithium fluoride, a poisonous white powder used in lowering the melting point of solders and in manufacturing ceramics. The electron configuration of lithium is 1s22s1, and that of fluorine is 1s22s22p5. When lithium and fluorine atoms come in contact with each other, the outer 2s1 valence electron of lithium is transferred to the fluorine atom. Using Lewis dot symbols, we represent the reaction like this: T Li 2 1 s 2s 1 SOT F Q 1s22s22p5 SOS F Q Li (or LiF) (9.1) 1s 1s22s22p6 2 For convenience, imagine that this reaction occurs in separate steps — first the ionization of Li: † Gilbert Newton Lewis (1875 – 1946). American chemist. One of the foremost physical chemists of this century, Lewis made many significant contributions in the areas of chemical bonding, thermodynamics, acids and bases, and spectroscopy. Despite the significance of Lewis’s work, he was never awarded a Nobel Prize. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9.2 331 THE IONIC BOND 1 1A 18 8A H 2 2A 13 3A 14 4A 15 5A 16 6A 17 7A He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Ga Ge As Se Br Kr Rb Sr In Sn Sb Te I Xe Cs Ba Tl Pb Bi Po At Rn Fr Ra 3 3B 4 4B 5 5B FIGURE 9.1 Lewis dot symbols for the representative elements and the noble gases. The number of unpaired dots corresponds to the number of bonds an atom of the element can form in a compound. 6 6B 7 7B 8 9 8B 10 11 1B 12 2B T Li Li e and then the acceptance of an electron by F: SOT F Q e SOS F Q Next, imagine the two separate ions joining to form a LiF unit: SOS F Q Li Li SOS F Q Note that the sum of these three reactions is We normally write the empirical formulas of ionic compounds without showing the charges. The and are shown here to emphasize the transfer of electrons. SOT F Q T Li O Li SQS F which is the same as Equation (9.1). The ionic bond in LiF is the electrostatic attraction between the positively charged lithium ion and the negatively charged fluoride ion. The compound itself is electrically neutral. Many other common reactions lead to the formation of ionic bonds. For instance, calcium burns in oxygen to form calcium oxide: 2Ca(s) O2(g) 88n 2CaO(s) Assuming that the diatomic O2 molecule first splits into separate oxygen atoms (we will look at the energetics of this step later), we can represent the reaction with Lewis symbols: TCa T [Ar]4s2 O TOT Q 1s22s22p4 Ca2 [Ar] O SOS2 Q [Ne] There is a transfer of two electrons from the calcium atom to the oxygen atom. Note that the resulting calcium ion (Ca2 ) has the argon electron configuration, the oxide ion (O2 ) is isoelectronic with neon, and the compound (CaO) is electrically neutral. In many cases the cation and the anion in a compound do not carry the same charges. For instance, when lithium burns in air to form lithium oxide (Li2O), the balanced equation is 4Li(s) Back Forward Main Menu TOC O2(g) 88n 2Li2O(s) Study Guide TOC Textbook Website MHHE Website 332 CHEMICAL BONDING I: BASIC CONCEPTS Using Lewis dot symbols, we write 2 T Li 2 1 s 2s 1 TOT O Q 224 1s 2s 2p O SOS2 Q 2Li [He] (or Li2O) [Ne] In this process the oxygen atom receives two electrons (one from each of the two lithium atoms) to form the oxide ion. The Li ion is isoelectronic with helium. When magnesium reacts with nitrogen at elevated temperatures, a white solid compound, magnesium nitride (Mg3N2), forms: 3Mg(s) N2(g) 88n Mg3N2(s) or 3 TMg T [Ne]3s 2 2 TOT N R 2 2 O 2 SNS3 Q 3Mg2 3 1s 2s 2p [Ne] (or Mg3N2) [Ne] The reaction involves the transfer of six electrons (two from each Mg atom) to two nitrogen atoms. The resulting magnesium ion (Mg2 ) and the nitride ion (N3 ) are both isoelectronic with neon. Because there are three 2 ions and two 3 ions, the charges balance and the compound is electrically neutral. In the following example, we apply the Lewis dot symbols to study the formation of an ionic compound. EXAMPLE 9.1 Use Lewis dot symbols to show the formation of aluminum oxide (Al2O3). We use the electroneutrality as our guide in writing formulas for ionic compounds. According to Figure 9.1, the Lewis dot symbols of Al and O are Answer TAl T R O TOT Q Since aluminum tends to form the cation (Al3 ) and oxygen the anion (O2 ) in ionic compounds, the transfer of electrons is from Al to O. There are three valence electrons in each Al atom; each O atom needs two electrons to form the O2 ion, which is isoelectronic with neon. Thus the simplest neutralizing ratio of Al3 to O2 is 2:3; two Al3 ions have a total charge of 6, and three O2 ions have a total charge of 6. Thus the empirical formula of aluminum oxide is Al2O3, and the reaction is 3 TOT O Q 2 TAl T R 2 [Ne]3s 3p 1 2 2 2Al3 4 1s 2s 2p [Ne] O 3 SQS2 O (or Al2O3) [Ne] Check to see that the number of valence electrons (number of dots) is the same on both sides of the equation (24). Comment Similar problems: 9.17, 9.18. PRACTICE EXERCISE Use Lewis dot symbols to represent the formation of barium hydride. 9.3 LATTICE ENERGY OF IONIC COMPOUNDS We can predict which elements are likely to form ionic compounds based on ionization energy and electron affinity, but how do we evaluate the stability of an ionic compound? Ionization energy and electron affinity are defined for processes occurring in Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9.3 LATTICE ENERGY OF IONIC COMPOUNDS 333 the gas phase, but at 1 atm and 25°C all ionic compounds are solids. The solid state is a very different environment because each cation in a solid is surrounded by a specific number of anions, and vice versa. Thus the overall stability of a solid ionic compound depends on the interactions of all these ions and not merely on the interaction of a single cation with a single anion. A quantitative measure of the stability of any ionic solid is its lattice energy, defined as the energy required to completely separate one mole of a solid ionic compound into gaseous ions. THE BORN-HABER CYCLE FOR DETERMINING LATTICE ENERGIES Lattice energy cannot be measured directly. However, if we know the structure and composition of an ionic compound, we can calculate the compound’s lattice energy by using Coulomb’s† law, which states that the potential energy (E) between two ions is directly proportional to the product of their charges and inversely proportional to the distance of separation between them. For a single Li ion and a single F ion separated by distance r, the potential energy of the system is given by QLi QF r E k QLi QF r (9.2) where QLi and QF are the charges on the Li and F ions and k is the proportionality constant. Since QLi is positive and QF is negative, E is a negative quantity, and the formation of an ionic bond from Li and F is an exothermic process. Consequently, energy must be supplied to reverse the process (in other words, the lattice energy of LiF is positive), and so a bonded pair of Li and F ions is more stable than separate Li and F ions. We can also determine lattice energy indirectly, by assuming that the formation of an ionic compound takes place in a series of steps. This procedure, known as the Born-Haber cycle, relates lattice energies of ionic compounds to ionization energies, electron affinities, and other atomic and molecular properties. It is based on Hess’s law (see Section 6.5). Developed by Max Born‡ and Fritz Haber,§ the Born-Haber cycle defines the various steps that precede the formation of an ionic solid. We will illustrate its use to find the lattice energy of lithium fluoride. Consider the reaction between lithium and fluorine: Li(s) 1 2 F2(g) 88n LiF(s) The standard enthalpy change for this reaction is 594.1 kJ. (Because the reactants and product are in their standard states, that is, at 1 atm, the enthalpy change is also the standard enthalpy of formation for LiF.) Keeping in mind that the sum of enthalpy † Charles Augustin de Coulomb (1736–1806). French physicist. Coulomb did research in electricity and magnetism and applied Newton’s inverse square law to electricity. He also invented a torsion balance. ‡ Max Born (1882–1970). German physicist. Born was one of the founders of modern physics. His work covered a wide range of topics. He received the Nobel Prize in Physics in 1954 for his interpretation of the wave function for particles. § Fritz Haber (1868–1934). German chemist. Haber ’s process for synthesizing ammonia from atmospheric nitrogen kept Germany supplied with nitrates for explosives during World War I. He also did work on gas warfare. In 1918 Haber received the Nobel Prize in Chemistry. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 334 CHEMICAL BONDING I: BASIC CONCEPTS changes for the steps is equal to the enthalpy change for the overall reaction ( 594.1 kJ), we can trace the formation of LiF from its elements through five separate steps. The process may not occur exactly this way, but this pathway allows us to analyze the energy changes of ionic compound formation, with the help of Hess’s law. 1. Convert solid lithium to lithium vapor (the direct conversion of a solid to a gas is called sublimation): Ho 1 Li(s) 88n Li(g) The F atoms in a F2 molecule are held together by a covalent bond. The energy required to break this bond is called the bond dissociation energy (Section 9.10). 2. 155.2 kJ The energy of sublimation for lithium is 155.2 kJ/mol. Dissociate 1 mole of F2 gas into separate gaseous F Atoms: 2 1 2 Ho 2 F2(g) 88n F(g) 75.3 kJ The energy needed to break the bonds in 1 mole of F2 molecules is 150.6 kJ. Here we are breaking the bonds in half a mole of F2, so the enthalpy change is 150.6/2, or 75.3, kJ. 3. Ionize 1 mole of gaseous Li atoms (see Table 8.3): Li(g) 88n Li (g) 4. 520 kJ This process corresponds to the first ionization of lithium. Add 1 mole of electrons to 1 mole of gaseous F atoms. As discussed on page 306, the energy change for this process is just the opposite of electron affinity (see Table 8.4): F(g) 5. Ho 3 e Ho 4 e 88n F (g) 328 kJ Combine 1 mole of gaseous Li and 1 mole of F to form 1 mole of solid LiF: Li (g) Ho 5 F (g) 88n LiF(s) ? The reverse of step 5, energy LiF(s) 88n Li (g) F (g) defines the lattice energy of LiF. Thus the lattice energy must have the same magnitude as Ho but an opposite sign. Although we cannot determine Ho directly, we can 5 5 calculate its value by the following procedure. 1. Li(s) 88n Li(g) 1 2. 2 F2(g) 88n F(g) 3. Li(g) 88n Li (g) 4. F(g) e 88n F (g) 5. Li (g) F (g) 88n LiF(s) Li(s) 1 2 Ho 1 Ho 2 Ho 3 Ho 4 Ho 5 e 155.2 kJ 75.3 kJ 520 kJ 328 kJ ? Ho overall F2(g) 88n LiF(s) 594.1 kJ According to Hess’s law, we can write Ho overall Ho 1 Ho 2 Ho 3 Ho 4 Ho 5 or 594.1 kJ 155.2 kJ 75.3 kJ 520 kJ 328 kJ Ho 5 Hence Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9.3 FIGURE 9.2 The Born-Haber cycle for the formation of solid LiF. LATTICE ENERGY OF IONIC COMPOUNDS 335 Li+(g) + F –(g) ∆ H 3 = 520 kJ ° ∆ H 4 = –328 kJ ° ∆ H 5 = –1017 kJ ° Li(g) + F(g) ∆ H 1 = 155.2 kJ ° Li(s) + ∆ H 2 = 75.3 kJ ° 1 F (g) 22 ∆ H overall = –594.1 kJ ° LiF(s) Ho 5 1017 kJ and the lattice energy of LiF is 1017 kJ/mol. Figure 9.2 summarizes the Born-Haber cycle for LiF. Steps 1, 2, and 3 all require the input of energy. On the other hand, steps 4 and 5 release energy. Because Ho is 5 a large negative quantity, the lattice energy of LiF is a large positive quantity, which accounts for the stability of solid LiF. The greater the lattice energy, the more stable the ionic compound. Keep in mind that lattice energy is always a positive quantity because the separation of ions in a solid into ions in the gas phase is, by Coulomb’s law, an endothermic process. Table 9.1 lists the lattice energies and the melting points of several common ionic compounds. There is a rough correlation between lattice energy and melting point. The larger the lattice energy, the more stable the solid and the more tightly held the ions. It takes more energy to melt such a solid, and so the solid has a higher melting point TABLE 9.1 Lattice Energies and Melting Points of Some Alkali Metal and Alkaline Earth Metal Halides and Oxides COMPOUND LATTICE ENERGY (kJ/mol) LiF LiCl LiBr LiI NaCl NaBr NaI KCl KBr KI MgCl2 Na2O MgO 1017 828 787 732 788 736 686 699 689 632 2527 2570 3890 MELTING POINT (°C) 845 610 550 450 801 750 662 772 735 680 714 Sub* 2800 *Na2O sublimes at 1275°C. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 336 CHEMICAL BONDING I: BASIC CONCEPTS than one with a smaller lattice energy. Note that MgCl2, Na2O, and MgO have unusually high lattice energies. The first of these ionic compounds has a doubly charged cation (Mg2 ) and the second a doubly charged anion (O2 ); in the third compound there is an interaction between two doubly charged species (Mg2 and O2 ). The coulombic attractions between two doubly charged species, or between a doubly charged ion and a singly charged ion, are much stronger than those between singly charged anions and cations. LATTICE ENERGY AND THE FORMULAS OF IONIC COMPOUNDS Because lattice energy is a measure of the stability of ionic compounds, its value can help us explain the formulas of these compounds. Consider magnesium chloride as an example. We have seen that the ionization energy of an element increases rapidly as successive electrons are removed from its atom. For example, the first ionization energy of magnesium is 738 kJ/mol, and the second ionization energy is 1450 kJ/mol, almost twice the first. We might ask why, from the standpoint of energy, magnesium does not prefer to form unipositive ions in its compounds. Why doesn’t magnesium chloride have the formula MgCl (containing the Mg ion) rather than MgCl2 (containing the Mg2 ion)? Admittedly, the Mg2 ion has the noble gas configuration [Ne], which represents stability because of its completely filled shells. But the stability gained through the filled shells does not, in fact, outweigh the energy input needed to remove an electron from the Mg ion. The reason the formula is MgCl2 lies in the extra stability gained by the formation of solid magnesium chloride. The lattice energy of MgCl2 is 2527 kJ/mol, which is more than enough to compensate for the energy needed to remove the first two electrons from a Mg atom (738 kJ/mol 1450 kJ/mol 2188 kJ/mol). What about sodium chloride? Why is the formula for sodium chloride NaCl and not NaCl2 (containing the Na2 ion)? Although Na2 does not have the noble gas electron configuration, we might expect the compound to be NaCl2 because Na2 has a higher charge and therefore the hypothetical NaCl2 should have a greater lattice energy. Again the answer lies in the balance between energy input (that is, ionization energies) and the stability gained from the formation of the solid. The sum of the first two ionization energies of sodium is 496 kJ/mol 4560 kJ/mol 5056 kJ/mol The compound NaCl2 does not exist, but if we assume a value of 2527 kJ/mol as its lattice energy (same as that for MgCl2), we see that the energy yield would be far too small to compensate for the energy required to produce the Na2 ion. What has been said about the cations applies also to the anions. In Section 8.5 we observed that the electron affinity of oxygen is 141 kJ/mol, meaning that the following process releases energy (and is therefore favorable): O(g) e 88n O (g) As we would expect, adding another electron to the O ion O (g) e 88n O2 (g) would be unfavorable because of the increase in electrostatic repulsion. Indeed, the electron affinity of O is negative. Yet compounds containing the oxide ion (O2 ) do exist and are very stable, whereas compounds containing the O ion are not known. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9.4 THE COVALENT BOND 337 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Sodium Chloride — A Common and Important Ionic Compound We are all familiar with sodium chloride as table salt. It is a typical ionic compound, a brittle solid with a high melting point (801°C) that conducts electricity in the molten state and in aqueous solution. The structure of solid NaCl is shown in Figure 2.12. One source of sodium chloride is rock salt, which is found in subterranean deposits often hundreds of meters thick. It is also obtained from seawater or brine (a concentrated NaCl solution) by solar evaporation. Sodium chloride also occurs in nature as the mineral halite. Sodium chloride is used more often than any other material in the manufacture of inorganic chemicals. World consumption of this substance is about 150 million tons per year. The major use of sodium chloride is in the production of other essential inorganic chemicals such as chlorine gas, sodium hydroxide, sodium metal, hydrogen gas, and sodium carbonate. It is also used to melt ice and snow on highways and roads. However, since sodium chloride is harmful to plant life and promotes corrosion of cars, its use for this purpose is of considerable environmental concern. Underground rock salt mining. Meat processing, food canning, water softening, paper pulp, textiles and dyeing, rubber and oil industry Chlor-alkali process (Cl2, NaOH, Na, H2) 50% Na2CO3 10% 4% 12% Melting ice on roads 17% 3% Domestic table salt Other chemical manufacture Solar evaporation process for obtaining sodium chloride. 4% Animal feed Uses of sodium chloride. Again, the high lattice energy resulting from the presence of O2 ions in compounds such as Na2O or MgO far outweighs the energy needed to produce the O2 ion. 9.4 THE COVALENT BOND Although the concept of molecules goes back to the seventeenth century, it was not until early in this century that chemists began to understand how and why molecules form. The first major breakthrough was Gilbert Lewis’s suggestion that a chemical Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 338 CHEMICAL BONDING I: BASIC CONCEPTS bond involves electron sharing by atoms. He depicted the formation of a chemical bond in H2 as HT This discussion applies only to representative elements. TH 88n HSH This type of electron pairing is an example of a covalent bond, a bond in which two electrons are shared by two atoms. Covalent compounds are compounds that contain only covalent bonds. For the sake of simplicity, the shared pair of electrons is often represented by a single line. Thus the covalent bond in the hydrogen molecule can be written as HOH. In a covalent bond, each electron in a shared pair is attracted to the nuclei of both atoms. This attraction holds the two atoms in H2 together and is responsible for the formation of covalent bonds in other molecules. Covalent bonding between many-electron atoms involves only the valence electrons. Consider the fluorine molecule, F2. The electron configuration of F is 1s22s22p5. The 1s electrons are low in energy and stay near the nucleus most of the time. For this reason they do not participate in bond formation. Thus each F atom has seven valence electrons (the 2s and 2p electrons). According to Figure 9.1, there is only one unpaired electron on F, so the formation of the F2 molecule can be represented as follows: SOT F Q TOS 88n SOSOS F FF Q QQ OO SQOQS FF or Note that only two valence electrons participate in the formation of F2. The other, nonbonding electrons, are called lone pairs — pairs of valence electrons that are not involved in covalent bond formation. Thus each F in F2 has three lone pairs of electrons: SOOOS FF QQ lone pairs lone pairs The structures we use to represent covalent compounds, such as H2 and F2, are called Lewis structures. A Lewis structure is a representation of covalent bonding in which shared electron pairs are shown either as lines or as pairs of dots between two atoms, and lone pairs are shown as pairs of dots on individual atoms. Only valence electrons are shown in a Lewis structure. Let us consider the Lewis structure of the water molecule. Figure 9.1 shows the Lewis dot symbol for oxygen with two unpaired dots or two unpaired electrons, as we expect that O might form two covalent bonds. Since hydrogen has only one electron, it can form only one covalent bond. Thus the Lewis structure for water is O HSOSH Q or O HOO OH Q In this case, the O atom has two lone pairs. The hydrogen atom has no lone pairs because its only electron is used to form a covalent bond. In the F2 and H2O molecules, the F and O atoms achieve the stable noble gas configuration by sharing electrons: OF S Q S OS FQ O H SOS H Q 8e 2e 8e 2e 8e The formation of these molecules illustrates the octet rule, formulated by Lewis: An atom other than hydrogen tends to form bonds until it is surrounded by eight valence electrons. In other words, a covalent bond forms when there are not enough electrons for each individual atom to have a complete octet. By sharing electrons in a covalent bond, the individual atoms can complete their octets. The requirement for hydrogen is that it attain the electron configuration of helium, or a total of two electrons. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9.4 HI FIGURE 9.3 Bond length (in pm) in H2 and HI. BOND TYPE COH COO CPO COC CPC CqC CON CPN CqN NOO NPO OOH BOND LENGTH (pm) 107 143 121 154 133 120 143 138 116 136 122 96 8e 8e or O O OPC PO Q Q H S TABLE 9.2 Average Bond Lengths of Some Common Single, Double, and Triple Bonds H OSSC SSO O O Q Q S Shortly you will be introduced to the rules for writing proper Lewis structures. Here we simply want to become familiar with the language associated with them. The octet rule works mainly for elements in the second period of the periodic table. These elements have only 2s and 2p subshells, which can hold a total of eight electrons. When an atom of one of these elements forms a covalent compound, it can attain the noble gas electron configuration [Ne] by sharing electrons with other atoms in the same compound. Later, we will discuss a number of important exceptions to the octet rule that give us further insight into the nature of chemical bonding. Atoms can form different types of covalent bonds. In a single bond, two atoms are held together by one electron pair. Many compounds are held together by multiple bonds, that is, bonds formed when two atoms share two or more pairs of electrons. If two atoms share two pairs of electrons, the covalent bond is called a double bond. Double bonds are found in molecules of carbon dioxide (CO2) and ethylene (C2H4): S H2 161 pm S 74 pm 339 THE COVALENT BOND CS C S H 8e H or H 8e H D G C PC G D H H 8e A triple bond arises when two atoms share three pairs of electrons, as in the nitrogen molecule (N2): SN O NS O O 8e 8e or SNqNS The acetylene molecule (C2H2) also contains a triple bond, in this case between two carbon atoms: H SC O C S H O O 8e or HO CqC OH 8e Note that in ethylene and acetylene all the valence electrons are used in bonding; there are no lone pairs on the carbon atoms. In fact, with the exception of carbon monoxide, stable molecules containing carbon do not have lone pairs on the carbon atoms. Multiple bonds are shorter than single covalent bonds. Bond length is defined as the distance between the nuclei of two covalently bonded atoms in a molecule (Figure 9.3) Table 9.2 shows some experimentally determined bond lengths. For a given pair of atoms, such as carbon and nitrogen, triple bonds are shorter than double bonds, which, in turn, are shorter than single bonds. The shorter multiple bonds are also more stable than single bonds, as we will see later. COMPARISON OF THE PROPERTIES OF COVALENT AND IONIC COMPOUNDS If intermolecular forces are weak, it is relatively easy to break up aggregates of molecules to form liquids (from solids) and gases (from liquids). Back Forward Main Menu Ionic and covalent compounds differ markedly in their general physical properties because of differences in the nature of their bonds. There are two types of attractive forces in covalent compounds. The first type is the force that holds the atoms together in a molecule. A quantitative measure of this attraction is given by bond energy, to be discussed in Section 9.10. The second type of attractive force operates between molecules and is called an intermolecular force. Because intermolecular forces are usually quite weak compared with the forces holding atoms together within a molecule, molecules of a covalent compound are not held together tightly. Consequently covalent compounds are usually gases, liquids, or low-melting solids. On the other hand, the electrostatic forces holding ions together in an ionic compound are usually very strong, so TOC Study Guide TOC Textbook Website MHHE Website 340 CHEMICAL BONDING I: BASIC CONCEPTS TABLE 9.3 Comparison of Some General Properties of an Ionic Compound and a Covalent Compound PROPERTY NaCl CCl4 Appearance Melting point (°C) Molar heat of fusion* (kJ/mol) Boiling point (°C) Molar heat of vaporization* (kJ/mol) Density (g/cm3) Solubility in water Electrical conductivity Solid Liquid White solid 801.0 30.2 1413.0 600.0 2.17 High Colorless liquid 23 2.5 76.5 30 1.59 Very low Poor Good Poor Poor *Molar heat of fusion and molar heat of vaporization are the amounts of heat needed to melt 1 mole of the solid and to vaporize 1 mole of the liquid, respectively. ionic compounds are solids at room temperature and have high melting points. Many ionic compounds are soluble in water, and the resulting aqueous solutions conduct electricity, because the compounds are strong electrolytes. Most covalent compounds are insoluble in water, or if they do dissolve, their aqueous solutions generally do not conduct electricity, because the compounds are nonelectrolytes. Molten ionic compounds conduct electricity because they contain mobile cations and anions; liquid or molten covalent compounds do not conduct electricity because no ions are present. Table 9.3 compares some of the general properties of a typical ionic compound, sodium chloride, with those of a covalent compound, carbon tetrachloride (CCl4). 9.5 Hydrogen fluoride is a clear, fuming liquid that boils at 19.8°C. It is used to make refrigerants and to prepare hydrofluoric acid. H F FIGURE 9.4 Electron density distribution in the HF molecule. The dots represent the positions of the nuclei. Back Forward ELECTRONEGATIVITY A covalent bond, as we have said, is the sharing of an electron pair by two atoms. In a molecule like H2, in which the atoms are identical, we expect the electrons to be equally shared — that is, the electrons spend the same amount of time in the vicinity of each atom. However, in the covalently bonded HF molecule, the H and F atoms do not share the bonding electrons equally because H and F are different atoms: F HOOS Q The bond in HF is called a polar covalent bond, or simply a polar bond, because the electrons spend more time in the vicinity of one atom than the other. Experimental evidence indicates that in the HF molecule the electrons spend more time near the F atom. We can think of this unequal sharing of electrons as a partial electron transfer or a shift in electron density, as it is more commonly described, from H to F (Figure 9.4). This “unequal sharing” of the bonding electron pair results in a relatively greater electron density near the fluorine atom and a correspondingly lower electron density near hydrogen. The HF bond and other polar bonds can be thought of as being intermediate between a (nonpolar) covalent bond, in which the sharing of electrons is exactly equal, and an ionic bond, in which the transfer of the electron(s) is nearly complete. A property that helps us distinguish a nonpolar covalent bond from a polar covalent bond is electronegativity, the ability of an atom to attract toward itself the elec- Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9.5 341 ELECTRONEGATIVITY Increasing electronegativity 1A 8A H 2A 3A 4A 5A 6A Li Increasing electronegativity 2.1 Be B C N O F 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Na Mg 0.9 1.2 K Ca Sc 0.8 1.0 1.3 7A Al 3B 4B 5B 6B 7B Ti V Cr Mn Fe Co 1.5 1.6 1.6 1.5 1.8 1.9 Si P S Cl 1B 2B 1.5 1.8 2.1 2.5 3.0 Ni Cu Zn Ga Ge As Se Br 1.9 1.9 1.6 1.6 1.8 2.0 2.4 2.8 8B Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I 0.8 1.0 1.2 1.4 1.6 1.8 1.9 2.2 2.2 2.2 1.9 1.7 1.7 1.8 1.9 2.1 2.5 Cs Ba La-Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At 0.7 0.9 1.0-1.2 1.3 1.5 1.7 1.9 2.2 2.2 2.2 2.4 1.9 1.8 1.9 1.9 2.0 2.2 Fr Ra 0.7 0.9 FIGURE 9.5 The electronegativities of common elements. Electronegativity values have no units. trons in a chemical bond. Elements with high electronegativity have a greater tendency to attract electrons than do elements with low electronegativity. As we might expect, electronegativity is related to electron affinity and ionization energy. Thus an atom such as fluorine, which has a high electron affinity (tends to pick up electrons easily) and a high ionization energy (does not lose electrons easily), has a high electronegativity. On the other hand, sodium has a low electron affinity, a low ionization energy, and a low electronegativity. Electronegativity is a relative concept, meaning that an element’s electronegativity can be measured only in relation to the electronegativity of other elements. Linus Pauling† devised a method for calculating relative electronegativities of most elements. These values are shown in Figure 9.5. A careful examination of this chart reveals trends and relationships among electronegativity values of different elements. In general, electronegativity increases from left to right across a period in the periodic table, as the metallic character of the elements decreases. Within each group, electronegativity decreases with increasing atomic number, and increasing metallic character. Note that the transition metals do not follow these trends. The most electronegative elements — the halogens, oxygen, nitrogen, and sulfur — are found in the upper right-hand corner of the periodic table, and the least electronegative elements (the alkali and alkaline earth metals) are clustered near the lower left-hand corner. These trends are readily apparent on a graph, as shown in Figure 9.6. Atoms of elements with widely different electronegativities tend to form ionic bonds (such as those that exist in NaCl and CaO compounds) with each other since the atom of the less electronegative element gives up its electron(s) to the atom of the more electronegative element. An ionic bond generally joins an atom of a metallic element and an atom of a nonmetallic element. Atoms of elements with comparable electronegativities tend to form polar covalent bonds with each other because the shift in † Linus Carl Pauling (1901–1994). American chemist. Regarded by many as the most influential chemist of the twentieth century, Pauling did research in a remarkably broad range of subjects, from chemical physics to molecular biology. Pauling received the Nobel Prize in Chemistry in 1954 for his work on protein structure, and the Nobel Peace Prize in 1962. He is the only person to be the sole recipient of two Nobel Prizes. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 342 CHEMICAL BONDING I: BASIC CONCEPTS F Electronegativity 4 Cl 3 Br I Ru H 2 Mn 1 Li Na FIGURE 9.6 Variation of electronegativity with atomic number. The halogens have the highest electronegativities, and the alkali metals the lowest. Rb K 10 0 Zn 20 30 Atomic number 40 50 electron density is usually small. Most covalent bonds involve atoms of nonmetallic elements. Only atoms of the same element, which have the same electronegativity, can be joined by a pure covalent bond. These trends and characteristics are what we would expect, given our knowledge of ionization energies and electron affinities. There is no sharp distinction between a polar bond and an ionic bond, but the following rule is helpful in distinguishing between them. An ionic bond forms when the electronegativity difference between the two bonding atoms is 2.0 or more. This rule applies to most but not all ionic compounds. Sometimes chemists use the quantity percent ionic character to describe the nature of a bond. A purely ionic bond would have 100 percent ionic character, although no such bond is known, whereas a nonpolar or purely covalent bond has 0 percent ionic character. Electronegativity and electron affinity are related but different concepts. Both indicate the tendency of an atom to attract electrons. However, electron affinity refers to an isolated atom’s attraction for an additional electron, whereas electronegativity signifies the ability of an atom in a chemical bond (with another atom) to attract the shared electrons. Furthermore, electron affinity is an experimentally measurable quantity, whereas electronegativity is an estimated number that cannot be measured. The following example shows how a knowledge of electronegativity can help us determine whether a chemical bond is covalent or ionic. EXAMPLE 9.2 Classify the following bonds as ionic, polar covalent, or covalent: (a) the bond in HCl, (b) the bond in KF, and (c) the CC bond in H3CCH3. Similar problems: 9.35, 9.36. Back Forward Answer (a) In Figure 9.5 we see that the electronegativity difference between H and Cl is 0.9, which is appreciable but not large enough (by the 2.0 rule) to qualify HCl as an ionic compound. Therefore, the bond between H and Cl is polar covalent. (b) The electronegativity difference between K and F is 3.2, well above the 2.0 mark; therefore, the bond between K and F is ionic. (c) The two C atoms are identical in every respect — they are bonded to each other and each is bonded to three H atoms. Therefore, the bond between them is purely covalent. Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9.6 WRITING LEWIS STRUCTURES 343 PRACTICE EXERCISE Which of the following bonds is covalent, which is polar covalent, and which is ionic? (a) the bond in CsCl, (b) the bond in H2S, (c) the NN bond in H2NNH2. ELECTRONEGATIVITY AND OXIDATION NUMBER In Chapter 4 we introduced the rules for assigning oxidation numbers of elements in their compounds. The concept of electronegativity is the basis for these rules. In essence, oxidation number refers to the number of charges an atom would have if electrons were transferred completely to the more electronegative of the bonded atoms in a molecule. Consider the NH3 molecule, in which the N atom forms three single bonds with the H atoms. Because N is more electronegative than H, electron density will be shifted from H to N. If the transfer were complete, each H would donate an electron to N, which would have a total charge of 3 while each H would have a charge of 1. Thus we assign an oxidation number of 3 to N and an oxidation number of 1 to H in NH3. Oxygen usually has an oxidation number of 2 in its compounds, except in hydrogen peroxide (H2O2), whose Lewis structure is OO HOO OOOH QQ A bond between identical atoms makes no contribution to the oxidation number of those atoms because the electron pair of that bond is equally shared. Since H has an oxidation number of 1, each O atom has an oxidation number of 1. Can you see now why fluorine always has an oxidation number of 1? It is the most electronegative element known, and it always forms a single bond in its compounds. Therefore, it would bear a 1 charge if electron transfer were complete. 9.6 WRITING LEWIS STRUCTURES Although the octet rule and Lewis structures do not present a complete picture of covalent bonding, they do help to explain the bonding scheme in many compounds and account for the properties and reactions of molecules. For this reason, you should practice writing Lewis structures of compounds. The basic steps are as follows: Write the skeletal structure of the compound, using chemical symbols and placing bonded atoms next to one another. For simple compounds, this task is fairly easy. For more complex compounds, we must either be given the information or make an intelligent guess about it. In general, the least electronegative atom occupies the central position. Hydrogen and fluorine usually occupy the terminal (end) positions in the Lewis structure. 2. Count the total number of valence electrons present, referring, if necessary, to Figure 9.1. For polyatomic anions, add the number of negative charges to that total. (For example, for the CO2 ion we add two electrons because the 2 charge indicates 3 that there are two more electrons than are provided by the neutral atoms.) For polyatomic cations, we subtract the number of positive charges from this total. (Thus, for NH4 we subtract one electron because the 1 charge indicates a loss of one electron from the group of neutral atoms.) 3. Draw a single covalent bond between the central atom and each of the surround1. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 344 CHEMICAL BONDING I: BASIC CONCEPTS ing atoms. Complete the octets of the atoms bonded to the central atom. (Remember that the valence shell of a hydrogen atom is complete with only two electrons.) Electrons belonging to the central or surrounding atoms must be shown as lone pairs if they are not involved in bonding. The total number of electrons to be used is that determined in step 2. 4. If the octet rule is not satisfied for the central atom, try adding double or triple bonds between the surrounding atoms and the central atom, using the lone pairs from the surrounding atoms. The following examples illustrate the four-step procedure for writing Lewis structures of compounds and an ion. EXAMPLE 9.3 Write the Lewis structure for nitrogen trifluoride (NF3) in which all three F atoms are bonded to the N atom. Answer Step 1: The N atom is less electronegative than F, so the skeletal structure of NF3 is NF3 FNF F Step 2: The outer-shell electron configurations of N and F are 2s22p3 and 2s22p5, re- spectively. Thus there are 5 (3 7), or 26, valence electrons to account for in NF3. Step 3: We draw a single covalent bond between N and each F, and complete the octets for the F atoms. We place the remaining two electrons on N: SOO O OOS F F QNQ A F SQS Similar problem: 9.43. Since this structure satisfies the octet rule for all the atoms, step 4 is not required. To check, we count the valence electrons in NF3 (in chemical bonds and in lone pairs). The result is 26, the same as the number of valence electrons on three F atoms and one N atom. PRACTICE EXERCISE Write the Lewis structure for carbon disulfide (CS2). EXAMPLE 9.4 Write the Lewis structure for nitric acid (HNO3) in which the three O atoms are bonded to the central N atom and the ionizable H atom is bonded to one of the O atoms. Answer Step 1: The skeletal structure of HNO3 is HNO3 ONOH O Step 2: The outer-shell electron configurations of N, O, and H are 2s22p3, 2s22p4, and 1s1, respectively. Thus there are 5 trons to account for in HNO3. Back Forward Main Menu TOC Study Guide TOC (3 6) 1, or 24, valence elec- Textbook Website MHHE Website 9.6 WRITING LEWIS STRUCTURES 345 Step 3: We draw a single covalent bond between N and each of the three O atoms and between one O atom and the H atom. Then we fill in electrons to comply with the octet rule for the O atoms: O O SOONOOOH Q Q A O SQS When we have done this, all 24 of the available electrons have been used. Step 4: We see that this structure satisfies the octet rule for all the O atoms but not for the N atom. Therefore we move a lone pair from one of the end O atoms to form another bond with N. Now the octet rule is also satisfied for the N atom: OPNOOOH O O Q Q A O SQS Similar problem: 9.43. PRACTICE EXERCISE Write the Lewis structure for formic acid (HCOOH). EXAMPLE 9.5 Write the Lewis structure for the carbonate ion (CO2 ). 3 Answer Step 1: We can deduce the skeletal structure of the carbonate ion by recognizing that C is less electronegative than O. Therefore, it is most likely to occupy a central position as follows: O CO2 3 O C O Step 2: The outer-shell electron configurations of C and O are 2s22p2 and 2s22p4, respectively, and the ion itself has two negative charges. Thus the total number of electrons is 4 (3 6) 2, or 24. Step 3: We draw a single covalent bond between C and each O and comply with the octet rule for the O atoms: O SOS A O O S O O C O QS O Q This structure shows all 24 electrons. Step 4: Although the octet rule is satisfied for the O atoms, it is not for the C atom. Therefore we move a lone pair from one of the O atoms to form another bond with C. Now the octet rule is also satisfied for the C atom: We use the brackets to indicate that the 2 charge is on the whole molecule. Similar problem: 9.44. SOS B O SOOCO OS O Q Q 2 As a final check, we verify that there are 24 valence electrons in the Lewis structure of the carbonate ion. PRACTICE EXERCISE Write the Lewis structure for the nitrite ion ( Back Forward Main Menu TOC Study Guide TOC 2 ). Textbook Website MHHE Website 346 CHEMICAL BONDING I: BASIC CONCEPTS 9.7 FORMAL CHARGE AND LEWIS STRUCTURE By comparing the number of electrons in an isolated atom to the number of electrons associated with the same atom in a Lewis structure, we can determine the distribution of electrons in the molecule and draw the most plausible Lewis structure. The bookkeeping procedure is as follows: In an isolated atom, the number of electrons associated with the atom is simply the number of valence electrons. (As usual, we need not be concerned with the inner electrons.) In a molecule, electrons associated with the atom are the lone pairs on the atom plus the electrons in the bonding pair(s) between the atom and other atom(s). However, because electrons are shared in a bond, we must divide the electrons in a bonding pair equally between the atoms forming the bond. An atom’s formal charge is the difference between the valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure. We can calculate the formal charge on an atom in a molecule using the equation formal charge on an atom in a Lewis structure total number of valence electrons in the free atom total number of nonbonding electrons 1 2 total number of bonding electrons (9.3) Let us illustrate the concept of formal charge using the ozone molecule (O3). Proceeding by steps, as we did in Examples 9.3 and 9.4, we draw the skeletal structure of O3 and then add bonds and electrons to satisfy the octet rule for the two end atoms: Liquid ozone below its boiling point ( 111.3°C). Ozone is a toxic, light-blue gas with a pungent odor. SOOOOOS O O QOQ You can see that although all available electrons are used, the octet rule is not satisfied for the central atom. To remedy this, we convert a lone pair on one of the end atoms to a second bond between that end atom and the central atom, as follows: OPOOO S OOO Q Q We can now use Equation (9.3) to calculate the formal charges on the O atoms as follows. • The central O atom. In the preceding Lewis structure the central atom has six valence electrons, one lone pair (or two nonbonding electrons), and three bonds (or six bonding electrons). Substituting in Equation (9.3), we write formal charge • 1 2 2 (6) 1 The end O atom in OPO. This atom has six valence electrons, two lone pairs (or four nonbonding electrons), and two bonds (or four bonding electrons). Thus we write formal charge • 6 6 1 2 4 (4) 0 The end O atom in OOO. This atom has six valence electrons, three lone pairs (or six nonbonding electrons), and one bond (or two bonding electrons). Thus we write formal charge 6 6 1 2 (2) 1 We can now write the Lewis structure for ozone and show the formal charges: OPOOOS O O QOQ For single positive and negative charges, we normally omit the numeral 1. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9.7 FORMAL CHARGE AND LEWIS STRUCTURE 347 When you write formal charges, the following rules are helpful: For neutral molecules, the sum of the formal charges must add up to zero. (This rule applies, for example, to the O3 molecule.) • For cations, the sum of the formal charges must equal the positive charge. • For anions, the sum of the formal charges must equal the negative charge. • Keep in mind that formal charges do not represent actual charge separation within the molecule. In the O3 molecule, for example, there is no evidence that the central atom bears a net 1 charge or that one of the end atoms bears a 1 charge. Writing these charges on the atoms in the Lewis structure merely helps us keep track of the valence electrons in the molecule. EXAMPLE 9.6 Write formal charges for the carbonate ion. Answer The Lewis structure for the carbonate ion was developed in Example 9.5: 2 SOS B O SO OCOOS O Q Q The formal charges on the atoms can be calculated as follows: The C atom: The O atom in CPO: The O atom in COO: formal charge formal charge formal charge 4 6 6 0 4 6 1 2 1 2 1 2 (8) (4) (2) 0 0 1 Thus the Lewis formula for CO2 with formal charges is 3 SO S B O OCOOS SQ O O Q Note that the sum of the formal charges is bonate ion. Similar problem: 9.44. 2, the same as the charge on the car- PRACTICE EXERCISE Write formal charges for the nitrite ion (NO2 ). Sometimes there is more than one acceptable Lewis structure for a given species. In such cases, we can often select the most plausible Lewis structure by using formal charges and the following guidelines: For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present. • Lewis structures with large formal charges ( 2, 3, and/or 2, 3, and so on) are less plausible than those with small formal charges. • Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms. • The following example shows how formal charges facilitate the choice of the correct Lewis structure for a molecule. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 348 CHEMICAL BONDING I: BASIC CONCEPTS EXAMPLE 9.7 Formaldehyde (CH2O), a liquid with a disagreeable odor, traditionally has been used to preserve laboratory species. Draw the most likely Lewis structure for the compound. Answer The two possible skeletal structures are H H CH2O C O H C O H (a) (b) We can draw a Lewis structure for each of these possibilities: H CO H OO P O OH H (a) G O C PQ O D (b) Although both structures satisfy the octet rule, (b) is the more likely structure because it carries no formal charges. Similar problem: 9.45. Comment Can you suggest two other reasons why (a) is less plausible? PRACTICE EXERCISE Draw the most reasonable Lewis structure of a molecule that contains a N atom, a C atom, and a H atom. 9.8 THE CONCEPT OF RESONANCE Our drawing of the Lewis structure for ozone (O3) satisfied the octet rule for the central atom because we placed a double bond between it and one of the two end O atoms. In fact, we can put the double bond at either end of the molecule, as shown by these two equivalent Lewis structures: O PO O OS O O QOQ OOO SQ OO P O O Q However, neither one of these two Lewis structures accounts for the known bond lengths in O3. We would expect the OOO bond in O3 to be longer than the OPO bond because double bonds are known to be shorter than single bonds. Yet experimental evidence shows that both oxygen-to-oxygen bonds are equal in length (128 pm). We resolve this discrepancy by using both Lewis structures to represent the ozone molecule: OPO OOS m8 OOO n Q Q O SO OOP O O QOQ Each of these structures is called a resonance structure. A resonance structure, then, is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure. The double-headed arrow indicates that the structures shown are resonance structures. The term resonance itself means the use of two or more Lewis structures to represent a particular molecule. Like the medieval European traveler to Africa who de- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9.8 THE CONCEPT OF RESONANCE 349 scribed a rhinoceros as a cross between a griffin and a unicorn, two familiar but imaginary animals, we describe ozone, a real molecule, in terms of two familiar but nonexistent structures. A common misconception about resonance is the notion that a molecule such as ozone somehow shifts quickly back and forth from one resonance structure to the other. Keep in mind that neither resonance structure adequately represents the actual molecule, which has its own unique, stable structure. “Resonance” is a human invention, designed to address the limitations in these simple bonding models. To extend the animal analogy, a rhinoceros is a distinct creature, not some oscillation between mythical griffin and unicorn! The carbonate ion provides another example of resonance: O S OS S OS B A O SO OCOOS m8 O PCOOS m8 O O nQ O n Q Q Q O SOS A SO O CPO O O Q Q According to experimental evidence, all carbon-to-oxygen bonds in CO2 are equiv3 alent. Therefore, the properties of the carbonate ion are best explained by considering its resonance structures together. The concept of resonance applies equally well to organic systems. A good example is the benzene molecule (C6H6): H H C C The hexagonal structure of benzene was first proposed by the German chemist August Kekulé (1829–1896). H H C C H H C C C C C H H H H C C C H H If one of these resonance structures corresponded to the actual structure of benzene, there would be two different bond lengths between adjacent C atoms, one characteristic of the single bond and the other of the double bond. In fact, the distance between all adjacent C atoms in benzene is 140 pm, which is shorter than a CO C bond (154 pm) and longer than a CP C bond (133 pm). A simpler way of drawing the structure of the benzene molecule and other compounds containing the “benzene ring” is to show only the skeleton and not the carbon and hydrogen atoms. By this convention the resonance structures are represented by Note that the C atoms at the corners of the hexagon and the H atoms are all omitted, although they are understood to exist. Only the bonds between the C atoms are shown. Remember this important rule for drawing resonance structures: The positions of electrons, but not those of atoms, can be rearranged in different resonance structures. In other words, the same atoms must be bonded to one another in all the resonance structures for a given species. The following example illustrates the procedure for drawing resonance structures of a molecule. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 350 CHEMICAL BONDING I: BASIC CONCEPTS EXAMPLE 9.8 Draw resonance structures (including formal charges) for the nitrate ion, NO3 , which has the following skeletal arrangement: O ONO Since nitrogen has five valence electrons and each oxygen has six valence electrons and there is a net negative charge, the total number of valence electrons is 5 (3 6) 1 24. The following resonance structures are all equivalent: Answer NO3 O SOS A O O OP N OOS m8 n Q Q Similar problems: 9.49, 9.54. SOS B O ON OOS m8 O SQ O n Q O SOS A O O SO O N P O Q Q PRACTICE EXERCISE Draw resonance structures for the nitrite ion (NO2 ). Finally, note that although it is more accurate to show all the resonance structures of an ion or a compound, we often use only one Lewis structure for simplicity. 9.9 EXCEPTIONS TO THE OCTET RULE As mentioned earlier, the octet rule applies mainly to the second-period elements. Exceptions to the octet rule fall into three categories characterized by an incomplete octet, an odd number of electrons, or more than eight valence electrons around the central atom. THE INCOMPLETE OCTET Beryllium, unlike the other Group 2A elements, forms mostly covalent compounds. In some compounds the number of electrons surrounding the central atom in a stable molecule is fewer than eight. Consider, for example, beryllium, which is a Group 2A (and a second-period) element. The electron configuration of beryllium is 1s22s2; it has two valence electrons in the 2s orbital. In the gas phase, beryllium hydride (BeH2) exists as discrete molecules. The Lewis structure of BeH2 is HOBeOH As you can see, only four electrons surround the Be atom, and there is no way to satisfy the octet rule for beryllium in this molecule. Elements in Group 3A, particularly boron and aluminum, also tend to form compounds in which they are surrounded by fewer than eight electrons. Take boron as an example. Since its electron configuration is 1s22s22p1, it has a total of three valence electrons. Boron reacts with the halogens to form a class of compounds having the general formula BX3, where X is a halogen atom. Thus, in boron trifluoride there are only six electrons around the boron atom: F SOS A O SQO B F A SQS F Boron trifluoride is a colorless gas (boiling point: 100°C) with a pungent odor. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9.9 EXCEPTIONS TO THE OCTET RULE 351 The following resonance structures all contain a double bond between B and F and satisfy the octet rule for boron: S OS F SFS SOS F B A A O PB m8 SOO B m8 SOO B F nQ F nQ F Q A B A S QS SQS SFS F F The fact that the BOF bond length in BF3 (130.9 pm) is shorter than a single bond (137.3 pm) lends support to the resonance structures even though in each case the negative formal charge is placed on the B atom and the positive formal charge on the F atom. Although boron trifluoride is stable, it readily reacts with ammonia. This reaction is better represented by using the Lewis structure in which boron has only six valence electrons around it: F SOS A O SQOB F A SQS F F H SOS H A A A O SNO H 88n SQO B O N O H F A A A SQS H F H It seems that the properties of BF3 are best explained by all four resonance structures. The BON bond in the above compound is different from the covalent bonds discussed so far in the sense that both electrons are contributed by the N atom. This type of bond is called a coordinate covalent bond (also referred to as a dative bond ), defined as a covalent bond in which one of the atoms donates both electrons. Although the properties of a coordinate covalent bond do not differ from those of a normal covalent bond (because all electrons are alike no matter what their source), the distinction is useful for keeping track of valence electrons and assigning formal charges. ODD-ELECTRON MOLECULES Some molecules contain an odd number of electrons. Among them are nitric oxide (NO) and nitrogen dioxide (NO2): O PO NO RQ O P N O OS O OP Q Q Since we need an even number of electrons for complete pairing (to reach eight), the octet rule clearly cannot be satisfied for all the atoms in any of these molecules. THE EXPANDED OCTET Atoms of the second-period elements cannot have more than eight valence electrons around the central atom, but atoms of elements in and beyond the third period of the periodic table form some compounds in which more than eight electrons surround the central atom. In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbitals that can be used in bonding. These orbitals allow an atom to form an expanded octet. One compound in which there is an expanded octet is sulfur hexafluoride, a very stable compound. The electron configuration of sulfur is [Ne]3s23p4. In SF6, each of sulfur ’s six valence electrons forms a covalent bond with a fluorine atom, so there are twelve electrons around the central sulfur atom: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 352 CHEMICAL BONDING I: BASIC CONCEPTS Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Just Say NO Colorless nitric oxide gas is produced by the action of Fe2 on an acidic sodium nitrite solution. The gas is bubbled through water and immediately reacts with oxygen to form the brown NO2 gas when exposed to air. Nitric oxide (NO), the simplest nitrogen oxide, is an odd-electron molecule, and therefore it is paramagnetic. A colorless gas (boiling point: 152°C), NO can be prepared in the laboratory by reacting sodium nitrite (NaNO2) with a reducing agent such as Fe2 in an acidic medium. NO2 (aq) Fe2 (aq) 2H (aq) 88n NO(g) Fe3 (aq) H2O(l ) Environmental sources of nitric oxide include the burning of fossil fuels containing nitrogen compounds and the reaction between nitrogen and oxygen inside the automobile engine at high temperatures N2(g) O2(g) 88n 2NO(g) Lightning also contributes to the atmospheric concentration of NO. Exposed to air, nitric oxide quickly forms brown nitrogen dioxide gas: 2NO( g) O2( g) 88n 2NO2(g) Nitrogen dioxide is a major component of smog. About ten years ago scientists studying muscle relaxation discovered that our bodies produce nitric oxide for use as a neurotransmitter. (A neurotransmitter is a small molecule that serves to facilitate cell-to-cell communications.) Since then, it has been detected in at least a dozen cell types in various parts of the body. Cells in the brain, the liver, the pancreas, the gastrointestinal tract, and the blood vessels can synthesize nitric oxide. This molecule also functions as a cellular toxin to kill harmful bacteria. And that’s not all: In 1996 it was reported that NO binds to hemoglobin, the oxygen-carrying protein in the blood. No doubt it helps to regulate blood pressure. The discovery of the biological role of nitric oxide has shed light on how nitroglycerin (C3H5N3O9) works as a drug. Nitroglycerin tablets are commonly prescribed for heart patients to relieve the pain (angina pectoris) caused by a brief interference in the flow of blood to the heart. It is now believed that nitroglycerin produces nitric oxide, which causes the muscles to relax and allows the arteries to dilate. That NO evolved as a messenger molecule is entirely appropriate. Nitric oxide is small and so can diffuse quickly from cell to cell. It is a stable molecule but under certain circumstances it is highly reactive, which accounts for its protective function. The enzyme that brings about muscle relaxation contains iron for which nitric oxide has a high affinity. It is the binding of NO to the iron that activates the enzyme. Nevertheless, in the cell, where biological effectors are typically very large molecules, the pervasive effects of one of the smallest known molecules are unprecedented. F SOS O SO A QS F QH E F S EH F F SO A OS Q SFS Q Q In the next chapter we will see that these twelve electrons, or six bonding pairs, are accommodated in six orbitals that originate from the one 3s, the three 3p, and two of Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9.9 EXCEPTIONS TO THE OCTET RULE 353 the five 3d orbitals. Sulfur also forms many compounds in which it obeys the octet rule. In sulfur dichloride, for instance, S is surrounded by only eight electrons: Sulfur dichloride is a toxic, foulsmelling cherry-red liquid (boiling point: 59°C). OSO SClOOO ClS QQQ The following examples concern compounds that do not obey the octet rule. EXAMPLE 9.9 Draw the Lewis structure for aluminum triiodide (AlI3). The outer-shell electron configuration of Al is 3s23p1. The Al atom forms three covalent bonds with the I atoms as follows: Answer O SIS A O SI OAl Q A SIS Q Although the octet rule is satisfied for the I atoms, there are only six valence electrons around the Al atom. This molecule is an example of the incomplete octet. Comment Similar problem: 9.60. PRACTICE EXERCISE Draw the Lewis structure for BeF2. EXAMPLE 9.10 Draw the Lewis structure for phosphorus pentafluoride (PF5), in which all five F atoms are bonded directly to the P atom. The outer-shell electron configurations for P and F are 3s23p3 and 2s22p5, respectively, and so the total number of valence electrons is 5 (5 7), or 40. Phosphorus, like sulfur, is a third-period element, and therefore it can have an expanded octet. The Lewis structure of PF5 is Answer F S OS O F A EQS F SOO P Q A HOS F SQS Q F PF5 Although the octet rule is satisfied for the F atoms, there are ten valence electrons around the P atom, giving it an expanded octet. Comment Similar problem: 9.62. PRACTICE EXERCISE Draw the Lewis structure for arsenic pentafluoride (AsF5). EXAMPLE 9.11 Draw a Lewis structure for the sulfate ion (SO2 ) in which all four O atoms are 4 bonded to the central S atom. Answer Step 1: The skeletal structure of SO2 4 Back Forward Main Menu TOC Study Guide TOC is Textbook Website MHHE Website 354 CHEMICAL BONDING I: BASIC CONCEPTS O O S O O Step 2: Both O and S are Group 6A elements and so have six valence electrons each. Including the two negative charges, we must therefore account for a total of 6 (4 6) 2, or 32, valence electrons in SO2 . 4 Step 3: We draw a single covalent bond between all the bonding atoms: O SOS A SO O S O OS O O Q Q A SQS O Next we show formal charges on the S and O atoms: S OS O A2 O S O O S O OS O Q Q A SO S Q Comment One of six other equivalent structures for SO2 is as follows: 4 SOS B SO O S O OS O O Q Q B SOS Similar problem: 9.83. This structure involves an expanded octet on S but may be considered more plausible because it bears fewer formal charges. However, detailed theoretical calculation shows that the most likely structure is the one that satisfies the octet rule, even though it has greater formal charge separations. The general rule for elements in the third period and beyond is that a resonance structure that obeys the octet rule is preferred over one that involves an expanded octet but bears fewer formal charges. PRACTICE EXERCISE Draw the Lewis structure of sulfuric acid (H2SO4). 9.10 BOND DISSOCIATION ENERGY A measure of the stability of a molecule is its bond dissociation energy (or bond energy), which is the enthalpy change required to break a particular bond in one mole of gaseous molecules. (Bond dissociation energies in solids and liquids are affected by neighboring molecules.) The experimentally determined bond dissociation energy of the diatomic hydrogen molecule, for example, is H2(g) 88n H(g) H(g) H° 436.4 kJ This equation tells us that breaking the covalent bonds in 1 mole of gaseous H2 molecules requires 436.4 kJ of energy. For the less stable chlorine molecule, Cl2(g) 88n Cl(g) Back Forward Main Menu TOC Study Guide TOC Cl(g) Textbook Website H° 242.7 kJ MHHE Website 9.10 BOND DISSOCIATION ENERGY 355 Bond energies can also be directly measured for diatomic molecules containing unlike elements, such as HCl, HCl(g) 88n H(g) Cl(g) H° 431.9 kJ as well as for molecules containing double and triple bonds: O2(g) 88n O(g) O(g) H° 498.7 kJ N2(g) 88n N(g) The Lewis structure of O2 MM is MPM. OO N(g) H° 941.4 kJ Measuring the strength of covalent bonds in polyatomic molecules is more complicated. For example, measurements show that the energy needed to break the first OOH bond in H2O is different from that needed to break the second OOH bond: H2O(g) 88n H(g) OH(g) 88n H(g) OH(g) H° 502 kJ O(g) H° 427 kJ In each case, an OOH bond is broken, but the first step is more endothermic than the second. The difference between the two H° values suggests that the second OOH bond itself undergoes change, because of the changes in chemical environment. Now we can understand why the bond energy of the same OOH bond in two different molecules such as methanol (CH3OH) and water (H2O) will not be the same: their environments are different. Thus for polyatomic molecules we speak of the average bond energy of a particular bond. For example, we can measure the energy of the OOH bond in ten different polyatomic molecules and obtain the average OOH bond energy by dividing the sum of the bond energies by 10. Table 9.4 lists the average bond energies of a number of diatomic and polyatomic molecules. As stated earlier, triple bonds are stronger than double bonds, which, in turn, are stronger than single bonds. USE OF BOND ENERGIES IN THERMOCHEMISTRY A comparison of the thermochemical changes that take place during a number of reactions (Chapter 6) reveals a strikingly wide variation in the enthalpies of different reactions. For example, the combustion of hydrogen gas in oxygen gas is fairly exothermic: H2(g) 1 2 O2(g) 88n H2O(l ) H° 285.8 kJ On the other hand, the formation of glucose (C6H12O6) from water and carbon dioxide, best achieved by photosynthesis, is highly endothermic: 6CO2(g) 6H2O(l ) 88n C6H12O6(s) 6O2(g) H° 2801 kJ We can account for such variations by looking at the stability of individual reactant and product molecules. After all, most chemical reactions involve the making and breaking of bonds. Therefore, knowing the bond energies and hence the stability of molecules tells us something about the thermochemical nature of reactions that molecules undergo. In many cases it is possible to predict the approximate enthalpy of reaction by using the average bond energies. Because energy is always required to break chemical bonds and chemical bond formation is always accompanied by a release of energy, we can estimate the enthalpy of a reaction by counting the total number of bonds broken and formed in the reaction and recording all the corresponding energy changes. The enthalpy of reaction in the gas phase is given by Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 356 CHEMICAL BONDING I: BASIC CONCEPTS TABLE 9.4 Some Bond Dissociation Energies of Diatomic Molecules* and Average Bond Energies for Bonds in Polyatomic Molecules BOND BOND ENERGY (kJ/mol) BOND BOND ENERGY (kJ/mol) HOH HON HOO HOS HOP HOF HOCl HOBr HOI COH COC CPC CqC CON CPN CqN COO CPO† COP 436.4 393.0 460.0 368.0 326.0 568.2 431.9 366.1 298.3 414.0 347.0 620.0 812.0 276.0 615.0 891.0 351.0 745.0 263.0 COS CPS NON NPN NqN NOO NOP OOO OPO OOP OPS POP PPP SOS SPS FOF ClOCl BrOBr IOI 255.0 477.0 193.0 418.0 941.4 176.0 209.0 142.0 498.7 502.0 469.0 197.0 489.0 268.0 352.0 156.9 242.7 192.5 151.0 *Bond dissociation energies for diatomic molecules (in color) have more significant figures than bond energies for bonds in polyatomic molecules because the bond dissociation energies of diatomic molecules are directly measurable quantities and not averaged over many compounds. † The CPO bond energy in CO2 is 799 kJ/mol. H° BE(reactants) total energy input BE(products) total energy released (9.4) where BE stands for average bond energy and is the summation sign. As written, Equation (9.4) takes care of the sign convention for H°. Thus, if the total energy input is greater than the total energy released, H° is positive and the reaction is endothermic. On the other hand, if more energy is released than absorbed, H° is negative and the reaction is exothermic (Figure 9.7). If reactants and products are all diatomic molecules, then Equation (9.4) will yield accurate results because the bond dissociation energies of diatomic molecules are accurately known. If some or all of the reactants and products are polyatomic molecules, Equation (9.4) will yield only approximate results because the bond energies used will be averages. For diatomic molecules, Equation (9.4) is equivalent to Equation (6.8), so the results obtained from these two equations should correspond, as Example 9.12 illustrates. EXAMPLE 9.12 Use Equation (9.4) to calculate the enthalpy of reaction for the process H2(g) Cl2(g) 88n 2HCl(g) Compare your result with that obtained using Equation (6.8). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9.10 FIGURE 9.7 Bond energy changes in (a) an endothermic reaction and (b) an exothermic reaction. Atoms Atoms ∑ BE Enthalpy Enthalpy – ∑ BE (products) Product molecules ∑ BE 357 BOND DISSOCIATION ENERGY (reactants) Reactant molecules – ∑ BE (products) (reactants) Reactant molecules Product molecules (a) (b) The first step is to count the number of bonds broken and the number of bonds formed. This is best done by creating a table: Answer Type of bonds broken Bond energy (kJ/mol) Energy change (kJ) HOH (H2) ClOCl (Cl2) 1 1 436.4 242.7 436.4 242.7 Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) HOCl (HCl) Refer to Table 9.4 for bond dissociation energies of these diatomic molecules. Number of bonds broken 2 431.9 863.8 Next, we obtain the total energy input and total energy released: total energy input 436.4 kJ total energy released 863.8 kJ 242.7 kJ 679.1 kJ Using Equation (9.4) we write H° 679.1 kJ 863.8 kJ 184.7 kJ Alternatively, we can use Equation (6.8) and the data in Appendix 3 to calculate the enthalpy of reaction: H° 2 Ho(HCl) f [ Ho(H2) f (2 mol)( 92.3 kJ/mol) Ho(Cl2)] f 0 0 184.6 kJ Since the reactants and products are diatomic molecules, we expect the results of Equations (9.4) and (6.8) to be nearly the same. Comment Similar problem: 9.102. PRACTICE EXERCISE Calculate the enthalpy of the reaction H2(g) F2(g) 88n 2HF(g) using (a) Equation (9.4) and (b) Equation (6.8). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 358 CHEMICAL BONDING I: BASIC CONCEPTS The following example uses Equation (9.4) to estimate the enthalpy of a reaction involving a polyatomic molecule. EXAMPLE 9.13 Estimate the enthalpy change for the combustion of hydrogen gas: 2H2(g) O2(g) 88n 2H2O(g) As in Example 9.12, we construct a table: Answer Type of bonds broken Number of bonds broken Bond energy (kJ/mol) Energy change (kJ) HOH (H2) OPO (O2) 2 1 436.4 498.7 872.8 498.7 Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) OOH (H2O) 4 460 1840 Next, we obtain the total energy input and total energy released: total energy input 872.8 kJ total energy released 498.7 kJ 1372 kJ 1840 kJ Using Equation (9.4) we write H° 1372 kJ 1840 kJ 468 kJ This result is only an estimate because the bond energy of OOH is an average quantity. Alternatively, we can use Equation (6.8) and the data in Appendix 3 to calculate the enthalpy of reaction: H° 2 Ho(H2O) f [2 Ho(H2) f (2 mol)( 241.8 kJ/mol) Ho(O2)] f 0 0 483.6 kJ The estimated value based on average bond energies is quite close to the true value calculated using Ho data. In general, Equation (9.4) works best for f reactions that are either quite endothermic or quite exothermic, that is, for H° 100 kJ or for H° 100 kJ. Comment Similar problem: 9.70. PRACTICE EXERCISE For the reaction H2(g) C2H4(g) 88n C2H6(g) (a) Estimate the enthalpy of reaction, using the bond energy values in Table 9.4. (b) Calculate the enthalpy of reaction, using standard enthalpies of formation. ( Ho f for H2, C2H4, and C2H6 are 0, 52.3 kJ/mol, and 84.7 kJ/mol, respectively.) SUMMARY OF KEY EQUATIONS formal charge on • an atom in a Lewis structure • Back Forward Main Menu H° total number of valence electrons in the free atom BE(reactants) TOC total number of nonbonding electrons BE(products) (9.4) Study Guide TOC 1 2 total number of bonding electrons (9.3) Calculating enthalpy change of a reaction from bond energies. Textbook Website MHHE Website QUESTIONS AND PROBLEMS SUMMARY OF FACTS AND CONCEPTS 359 1. A Lewis dot symbol shows the number of valence electrons possessed by an atom of a given element. Lewis dot symbols are useful mainly for the representative elements. 2. The elements most likely to form ionic compounds have low ionization energies (such as the alkali metals and the alkaline earth metals, which form cations) or high electron affinities (such as the halogens and oxygen, which form anions). 3. An ionic bond is the product of the electrostatic forces of attraction between positive and negative ions. An ionic compound consists of a large network of ions in which positive and negative charges are balanced. The structure of a solid ionic compound maximizes the net attractive forces among the ions. 4. Lattice energy is a measure of the stability of an ionic solid. It can be calculated by means of the Born-Haber cycle, which is based on Hess’s law. 5. In a covalent bond, two electrons (one pair) are shared by two atoms. In multiple covalent bonds, two or three pairs of electrons are shared by two atoms. Some covalently bonded atoms also have lone pairs, that is, pairs of valence electrons that are not involved in bonding. The arrangement of bonding electrons and lone pairs around atoms in a molecule is represented by a Lewis structure. 6. The octet rule predicts that atoms form enough covalent bonds to surround themselves with eight electrons each. When one atom in a covalently bonded pair donates two electrons to the bond, the Lewis structure can include the formal charge on each atom as a means of keeping track of the valence electrons. There are exceptions to the octet rule, particularly for covalent beryllium compounds, elements in Group 3A, and elements in the third period and beyond in the periodic table. 7. Electronegativity is a measure of an atom’s ability to attract electrons in a chemical bond. 8. For some molecules or polyatomic ions, two or more Lewis structures based on the same skeletal structure satisfy the octet rule and appear chemically reasonable. Taken together, such resonance structures represent the molecule or ion more accurately than any single Lewis structure does. 9. The strength of a covalent bond is measured in terms of its bond dissociation energy. Bond energies can be used to estimate the enthalpy of reactions. KEY WORDS Bond dissociation energy, p. 354 Bond length, p. 339 Born-Haber cycle, p. 333 Coordinate covalent bond, p. 351 Coulomb’s law, p. 333 Covalent bond, p. 338 Covalent compound, p. 338 Double bond, p. 339 Electronegativity, p. 340 Formal charge, p. 346 Ionic bond, p. 330 Lattice energy, p. 333 Lewis dot symbol, p. 330 Lewis structure, p. 338 Lone pair, p. 338 Multiple bond, p. 339 Octet rule, p. 338 Polar covalent bond, p. 340 Resonance, p. 348 Resonance structure, p. 348 Single bond, p. 339 Triple bond, p. 339 QUESTIONS AND PROBLEMS LEWIS DOT SYMBOLS Review Questions 9.1 What is a Lewis dot symbol? To what elements does the symbol mainly apply? 9.2 Use the second member of each group from Group 1A to Group 7A to show that the number of valence electrons on an atom of the element is the same as its group number. 9.3 Without referring to Figure 9.1, write Lewis dot sym- Back Forward Main Menu TOC bols for atoms of the following elements: (a) Be, (b) K, (c) Ca, (d) Ga, (e) O, (f) Br, (g) N, (h) I, (i) As, ( j) F. 9.4 Write Lewis dot symbols for the following ions: (a) Li , (b) Cl , (c) S2 , (d) Sr2 , (e) N3 . 9.5 Write Lewis dot symbols for the following atoms and ions: (a) I, (b) I , (c) S, (d) S2 , (e) P, (f) P3 , (g) Na, (h) Na , (i) Mg, ( j) Mg2 , (k) Al, (l) Al3 , (m) Pb, (n) Pb2 . Study Guide TOC Textbook Website MHHE Website 360 CHEMICAL BONDING I: BASIC CONCEPTS THE IONIC BOND Review Questions 9.6 Explain what an ionic bond is. 9.7 Explain how ionization energy and electron affinity determine whether elements will combine to form ionic compounds. 9.8 Name five metals and five nonmetals that are very likely to form ionic compounds. Write formulas for compounds that might result from the combination of these metals and nonmetals. Name these compounds. 9.9 Name one ionic compound that contains only nonmetallic elements. 9.10 Name one ionic compound that contains a polyatomic cation and a polyatomic anion (see Table 2.3). 9.11 Explain why ions with charges greater than 3 are seldom found in ionic compounds. 9.12 The term “molar mass” was introduced in Chapter 2. What is the advantage of using the term “molar mass” when we discuss ionic compounds? 9.13 In which of the following states would NaCl be electrically conducting? (a) solid, (b) molten (that is, melted), (c) dissolved in water. Explain your answers. 9.14 Beryllium forms a compound with chlorine that has the empirical formula BeCl2. How would you determine whether it is an ionic compound? (The compound is not soluble in water.) Problems 9.15 An ionic bond is formed between a cation A and an anion B . How would the energy of the ionic bond [see Equation (9.2)] be affected by the following changes? (a) doubling the radius of A , (b) tripling the charge on A , (c) doubling the charges on A and B , (d) decreasing the radii of A and B to half their original values. 9.16 Give the empirical formulas and names of the compounds formed from the following pairs of ions: (a) Rb and I , (b) Cs and SO2 , (c) Sr2 and N3 , 4 (d) Al3 and S2 . 9.17 Use Lewis dot symbols to show the transfer of electrons between the following atoms to form cations and anions: (a) Na and F, (b) K and S, (c) Ba and O, (d) Al and N. 9.18 Write the Lewis dot symbols of the reactants and products in the following reactions. (First balance the equations.) (a) Sr Se 88n SrSe (b) Ca H2 88n CaH2 (c) Li N2 88n Li3N (d) Al S 88n Al2S3 9.19 For each of the following pairs of elements, state whether the binary compound they form is likely to Back Forward Main Menu TOC be ionic or molecular. Write the empirical formula and name of the compound: (a) I and Cl, (b) Mg and F. 9.20 For each of the following pairs of elements, state whether the binary compound they form is likely to be ionic or covalent. Write the empirical formula and name of the compound: (a) B and F, (b) K and Br. LATTICE ENERGY OF IONIC COMPOUNDS Review Questions 9.21 What is lattice energy and what role does it play in the stability of ionic compounds? 9.22 Explain how the lattice energy of an ionic compound such as KCl can be determined using the Born-Haber cycle. On what law is this procedure based? 9.23 Specify which compound in the following pairs of ionic compounds has the higher lattice energy: (a) KCl or MgO, (b) LiF or LiBr, (c) Mg3N2 or NaCl. Explain your choice. 9.24 Compare the stability (in the solid state) of the following pairs of compounds: (a) LiF and LiF2 (containing the Li2 ion), (b) Cs2O and CsO (containing the O ion), (c) CaBr2 and CaBr3 (containing the Ca3 ion). Problems 9.25 Use the Born-Haber cycle outlined in Section 9.3 for LiF to calculate the lattice energy of NaCl. [The heat of sublimation of Na is 108 kJ/mol and Ho(NaCl) f 1 411 kJ/mol. Energy needed to dissociate 2 mole of Cl2 into Cl atoms 121.4 kJ.] 9.26 Calculate the lattice energy of calcium chloride given that the heat of sublimation of Ca is 121 kJ/mol and 795 kJ/mol. (See Tables 8.3 and 8.4 Ho(CaCl2) f for other data.) THE COVALENT BOND Review Questions 9.27 What is Lewis’s contribution to our understanding of the covalent bond? 9.28 Use an example to illustrate each of the following terms: lone pairs, Lewis structure, the octet rule, bond length. 9.29 What is the difference between a Lewis dot symbol and a Lewis structure? 9.30 How many lone pairs are on the underlined atoms in these compounds? HBr, H2S, CH4. 9.31 Compare single, double, and triple bonds in a molecule, and give an example of each. For the same bonding atoms, how does the bond length change from single bond to triple bond? 9.32 Compare the properties of ionic compounds and covalent compounds. Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS ELECTRONEGATIVITY AND BOND TYPE Review Questions 9.33 Define electronegativity, and explain the difference between electronegativity and electron affinity. Describe in general how the electronegativities of the elements change according to position in the periodic table. 9.34 What is a polar covalent bond? Name two compounds that contain one or more polar covalent bonds. 9.35 List the following bonds in order of increasing ionic character: the lithium-to-fluorine bond in LiF, the potassium-to-oxygen bond in K2O, the nitrogen-tonitrogen bond in N2, the sulfur-to-oxygen bond in SO2, the chlorine-to-fluorine bond in ClF3. 9.36 Arrange the following bonds in order of increasing ionic character: carbon to hydrogen, fluorine to hydrogen, bromine to hydrogen, sodium to iodine, potassium to fluorine, lithium to chlorine. 9.37 Four atoms are arbitrarily labeled D, E, F, and G. Their electronegativities are as follows: D 3.8, E 3.3, F 2.8, and G 1.3. If the atoms of these elements form the molecules DE, DG, EG, and DF, how would you arrange these molecules in order of increasing covalent bond character? 9.38 List the following bonds in order of increasing ionic character: cesium to fluorine, chlorine to chlorine, bromine to chlorine, silicon to carbon. 9.39 Classify the following bonds as ionic, polar covalent, or covalent, and give your reasons: (a) the CC bond in H3CCH3, (b) the KI bond in KI, (c) the NB bond in H3NBCl3, (d) the CF bond in CF4. 9.40 Classify the following bonds as ionic, polar covalent, or covalent, and give your reasons: (a) the SiSi bond in Cl3SiSiCl3, (b) the SiCl bond in Cl3SiSiCl3, (c) the CaF bond in CaF2, (d) the NH bond in NH3. Review Questions 9.41 Summarize the essential features of the Lewis octet rule. The octet rule applies mainly to the secondperiod elements. Explain. 9.42 Explain the concept of formal charge. Do formal charges represent actual separation of charges? Problems 9.43 Write Lewis structures for the following molecules: (a) ICl, (b) PH3, (c) P4 (each P is bonded to three other P atoms), (d) H2S, (e) N2H4, (f) HClO3, (g) COBr2 (C is bonded to O and Br atoms). Forward (a) H OOPO CQ N (f) H (b) HPCPCPH O Main Menu TOC QO G OOOS CQ F D (d) SO F Q OS F G DQ O B A S QS F (e) H OO POS OQ F F (g) SO F QG DOS Q N A SQS F 9.46 The skeletal structure of acetic acid shown below is correct, but some of the bonds are wrong. (a) Identify the incorrect bonds and explain what is wrong with them. (b) Write the correct Lewis structure for acetic acid. H SOS AA H PCOQOQOH CO A H THE CONCEPT OF RESONANCE Review Questions 9.47 Define bond length, resonance, and resonance structure. What are the rules for writing resonance structures? 9.48 Is it possible to “trap” a resonance structure of a compound for study? Explain. Problems LEWIS STRUCTURE AND THE OCTET RULE Back 9.44 Write Lewis structures for the following ions: (a) O2 , (b) C2 , (c) NO , (d) NH4 . Show formal 2 2 charges. 9.45 The following Lewis structures are incorrect. Explain what is wrong with each one and give a correct Lewis structure for the molecule. (Relative positions of atoms are shown correctly.) O (c) OOSn OO O Q Q Problems 361 9.49 Write Lewis structures for the following species, including all resonance forms, and show formal charges: (a) HCO2 , (b) CH2NO2 . Relative positions of the atoms are as follows: O HCO O H HC H O NO O 9.50 Draw three resonance structures for the chlorate ion, ClO3 . Show formal charges. 9.51 Write three resonance structures for hydrazoic acid, HN3. The atomic arrangement is HNNN. Show formal charges. Study Guide TOC Textbook Website MHHE Website 362 CHEMICAL BONDING I: BASIC CONCEPTS 9.52 Draw two resonance structures for diazomethane, CH2N2. Show formal charges. The skeletal structure of the molecule is HC C H N N N N 9.53 Draw three reasonable resonance structures for the OCN ion. Show formal charges. 9.54 Draw three resonance structures for the molecule N2O in which the atoms are arranged in the order NNO. Indicate formal charges. Review Questions 9.55 Why does the octet rule not hold for many compounds containing elements in the third period of the periodic table and beyond? 9.56 Give three examples of compounds that do not satisfy the octet rule. Write a Lewis structure for each. 9.57 Because fluorine has seven valence electrons (2s22p5), seven covalent bonds in principle could form around the atom. Such a compound might be FH7 or FCl7. These compounds have never been prepared. Why? 9.58 What is a coordinate covalent bond? Is it different from a normal covalent bond? Problems 9.59 The AlI3 molecule has an incomplete octet around Al. Draw three resonance structures of the molecule in which the octet rule is satisfied for both the Al and the I atoms. Show formal charges. 9.60 In the vapor phase, beryllium chloride consists of discrete BeCl2 molecules. Is the octet rule satisfied for Be in this compound? If not, can you form an octet around Be by drawing another resonance structure? How plausible is this structure? 9.61 Of the noble gases, only Kr, Xe, and Rn are known to form a few compounds with O and/or F. Write Lewis structures for the following molecules: (a) XeF2, (b) XeF4, (c) XeF6, (d) XeOF4, (e) XeO2F2. In each case Xe is the central atom. 9.62 Write a Lewis structure for SbCl5. Does this molecule obey the octet rule? 9.63 Write Lewis structures for SeF4 and SeF6. Is the octet rule satisfied for Se? 9.64 Write Lewis structures for the reaction Cl 88n AlCl4 What kind of bond joins Al and Cl in the product? Back Forward Main Menu Review Questions 9.65 What is bond dissociation energy? Bond energies of polyatomic molecules are average values, whereas those of diatomic molecules can be accurately determined. Why? 9.66 Explain why the bond energy of a molecule is usually defined in terms of a gas-phase reaction. Why are bond-breaking processes always endothermic and bond-forming processes always exothermic? Problems 9.67 From the following data, calculate the average bond energy for the NOH bond: EXCEPTIONS TO THE OCTET RULE AlCl3 BOND DISSOCIATION ENERGY TOC NH3(g) 88n NH2(g) NH2(g) 88n NH(g) NH(g) 88n N(g) H(g) H° 381 kJ H° H(g) 435 kJ H° H(g) 360 kJ 9.68 For the reaction O(g) O2(g) 88n O3(g) H° 107.2 kJ Calculate the average bond energy in O3. 9.69 The bond energy of F2(g) is 156.9 kJ/mol. Calculate Ho for F(g). f 9.70 For the reaction 2C2H6(g) 7O2(g) 88n 4CO2(g) 6H2O(g) (a) Predict the enthalpy of reaction from the average bond energies in Table 9.4. (b) Calculate the enthalpy of reaction from the standard enthalpies of formation (see Appendix 3) of the reactant and product molecules, and compare the result with your answer for part (a). ADDITIONAL PROBLEMS 9.71 Classify the following substances as ionic compounds or covalent compounds containing discrete molecules: CH4, KF, CO, SiCl4, BaCl2. 9.72 Which of the following are ionic compounds? Which are covalent compounds? RbCl, PF5, BrF3, KO2, CI4 9.73 Match each of the following energy changes with one of the processes given: ionization energy, electron affinity, bond dissociation energy, and standard enthalpy of formation. (a) F(g) e 88n F (g) (b) F2(g) 88n 2F(g) (c) Na(g) 88n Na (g) e (d) Na(s) 1 F2(g) 88n NaF(s) 2 9.74 The formulas for the fluorides of the third-period elements are NaF, MgF2, AlF3, SiF4, PF5, SF6, and ClF3. Classify these compounds as covalent or ionic. Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 9.75 Use ionization energy (see Table 8.3) and electron affinity (see Table 8.4) values to calculate the energy change (in kJ) for the following reactions: (a) Li(g) I(g) 88n Li (g) I (g) (b) Na(g) F(g) 88n Na (g) F (g) (c) K(g) Cl(g) 88n K (g) Cl (g) 9.76 Describe some characteristics of an ionic compound such as KF that would distinguish it from a covalent compound such as benzene (C6H6). 9.77 Write Lewis structures for BrF3, ClF5, and IF7. Identify those in which the octet rule is not obeyed. 9.78 Write three reasonable resonance structures for the azide ion N3 in which the atoms are arranged as NNN. Show formal charges. 9.79 The amide group plays an important role in determining the structure of proteins: S OS B OO CO ON A H 9.80 9.81 9.82 9.83 9.84 9.85 9.86 Draw another resonance structure for this group. Show formal charges. Give an example of an ion or molecule containing Al that (a) obeys the octet rule, (b) has an expanded octet, and (c) has an incomplete octet. Draw four reasonable resonance structures for the PO3F2 ion. The central P atom is bonded to the three O atoms and to the F atom. Show formal charges. Attempts to prepare the compounds listed below as stable species under atmospheric conditions have failed. Suggest possible reasons for the failure. CF2, LiO2, CsCl2, PI5 Draw reasonable resonance structures for the following sulfur-containing ions: (a) HSO4 , (b) PO3 , 4 (c) HSO3 , (d) SO2 . (Hint: See comment on p. 354.) 3 Are the following statements true or false? (a) Formal charges represent actual separation of charges. (b) H o can be estimated from the bond energies rxn of reactants and products. (c) All second-period elements obey the octet rule in their compounds. (d) The resonance structures of a molecule can be separated from one another. A rule for drawing plausible Lewis structures is that the central atom is invariably less electronegative than the surrounding atoms. Explain why this is so. Using the following information and the fact that the average COH bond energy is 414 kJ/mol, estimate the standard enthalpy of formation of methane (CH4). C(s) 88n C(g) Ho rxn 716 kJ 2H2(g) 88n 4H(g) Ho rxn 872.8 kJ 9.87 Based on energy considerations, which of the fol- Back Forward Main Menu TOC 363 lowing reactions will occur more readily? (a) Cl(g) CH4(g) 88n CH3Cl(g) H(g) (b) Cl(g) CH4(g) 88n CH3(g) HCl(g) (Hint: Refer to Table 9.4, and assume that the average bond energy of the COCl bond is 338 kJ/mol.) 9.88 Which of the following molecules has the shortest nitrogen-to-nitrogen bond? Explain. N2H4, N2O, N2, N2O4 9.89 Most organic acids can be represented as RCOOH, where COOH is the carboxyl group and R is the rest of the molecule. (For example, R is CH3 in acetic acid, CH3COOH). (a) Draw a Lewis structure for the carboxyl group. (b) Upon ionization, the carboxyl group is converted to the carboxylate group, COO . Draw resonance structures for the carboxylate group. 9.90 Which of the following species are isoelectronic? NH4 , C6H6, CO, CH4, N2, B3N3H6 9.91 The following species have been detected in interstellar space: (a) CH, (b) OH, (c) C2, (d) HNC, (e) HCO. Draw Lewis structures for these species and indicate whether they are diamagnetic or paramagnetic. 9.92 The amide ion, NH2 , is a Brønsted base. Represent the reaction between the amide ion and water. 9.93 Draw Lewis structures for the following organic molecules: (a) tetrafluoroethylene (C2F4), (b) propane (C3H8), (c) butadiene (CH2CHCHCH2), (d) propyne (CH3CCH), (e) benzoic acid (C6H5COOH). (To draw C6H5COOH, replace a H atom in benzene with a COOH group.) 9.94 The triiodide ion (I3 ) in which the I atoms are arranged in a straight line is stable, but the corresponding F3 ion does not exist. Explain. 9.95 Compare the bond dissociation energy of F2 with the energy change for the following process: F2(g) 88n F (g) F (g) Which is the preferred dissociation for F2, energetically speaking? 9.96 Methyl isocyanate (CH3NCO) is used to make certain pesticides. In December 1984, water leaked into a tank containing this substance at a chemical plant, producing a toxic cloud that killed thousands of people in Bhopal, India. Draw Lewis structures for CH3NCO, showing formal charges. 9.97 The chlorine nitrate molecule (ClONO2) is believed to be involved in the destruction of ozone in the Antarctic stratosphere. Draw a plausible Lewis structure for this molecule. 9.98 Several resonance structures for the molecule CO2 are shown on page 364. Explain why some of them are likely to be of little importance in describing the bonding in this molecule. Study Guide TOC Textbook Website MHHE Website 364 CHEMICAL BONDING I: BASIC CONCEPTS (a) O PCPO O O Q Q (c) SOqO OS CQ O 2 O (d) SO OCOOS O Q Q For each of the following organic molecules draw a Lewis structure in which the carbon atoms are bonded to each other by single bonds: (a) C2H6, (b) C4H10, (c) C5H12. For (b) and (c), show only structures in which each C atom is bonded to no more than two other C atoms. Draw Lewis structures for the following chlorofluorocarbons (CFCs), which are partly responsible for the depletion of ozone in the stratosphere: (a) CFCl3, (b) CF2Cl2, (c) CHF2Cl, (d) CF3CHF2. Draw Lewis structures for the following organic molecules. In each there is one CPC bond, and the rest of the carbon atoms are joined by COC bonds. C2H3F, C3H6, C4H8 Calculate H° for the reaction O (b) S OqCOOS Q 9.99 9.100 9.101 9.102 H2(g) 9.103 9.104 9.105 9.106 9.107 9.108 I2(g) 88n 2HI(g) using (a) Equation (9.4) and (b) Equation (6.8), given that Ho for I2(g) is 61.0 kJ/mol. f Draw Lewis structures for the following organic molecules: (a) methanol (CH3OH); (b) ethanol (CH3CH2OH); (c) tetraethyllead [Pb(CH2CH3)4], which is used in “leaded gasoline”; (d) methylamine (CH3NH2), which is used in tanning; (e) mustard gas (ClCH2CH2SCH2CH2Cl), a poisonous gas used in World War I; (f) urea [(NH2)2CO], a fertilizer; and (g) glycine (NH2CH2COOH), an amino acid. Write Lewis structures for the following four isoelectronic species: (a) CO, (b) NO , (c) CN , (d) N2. Show formal charges. Oxygen forms three types of ionic compounds in which the anions are oxide (O2 ), peroxide (O2 ), 2 and superoxide (O2 ). Draw Lewis structures of these ions. Comment on the correctness of the statement, “All compounds containing a noble gas atom violate the octet rule.” Write three resonance structures for (a) the cyanate ion (NCO ) and (b) the isocyanate ion (CNO ). In each case, rank the resonance structures in order of increasing importance. (a) From the following data calculate the bond energy of the F2 ion. F2(g) 88n 2F(g) Ho rxn 156.9 kJ F (g) 88n F(g) Ho rxn Ho rxn 290 kJ F2 (g) 88n F2(g) e e 333 kJ (b) Explain the difference between the bond energies of F2 and F2 . Back Forward Main Menu TOC 9.109 The resonance concept is sometimes described by analogy to a mule, which is a cross between a horse and a donkey. Compare this analogy with the one used in this chapter, that is, the description of a rhinoceros as a cross between a griffin and a unicorn. Which description is more appropriate? Why? 9.110 What are the other two reasons for choosing (b) in Example 9.7? 9.111 In the Chemistry in Action essay on p. 352, nitric oxide is said to be one of about ten of the smallest stable molecules known. Based on what you have learned in the course so far, write all the diatomic molecules you know, give their names, and show their Lewis structures. 9.112 The NOO bond distance in nitric oxide is 115 pm, which is intermediate between a triple bond (106 pm) and a double bond (120 pm). (a) Draw two resonance structures for NO and comment on their relative importance. (b) Is it possible to draw a resonance structure having a triple bond between the atoms? 9.113 Although nitrogen dioxide (NO2) is a stable compound, there is a tendency for two such molecules to combine to form dinitrogen tetroxide (N2O4). Why? Draw four resonance structures of N2O4, showing formal charges. 9.114 Another possible skeletal structure for the CO2 (car3 bonate) ion besides the one presented in Example 9.5 is O C O O. Why would we not use this structure to represent CO2 ? 3 9.115 Draw a Lewis structure for nitrogen pentoxide (N2O5) in which each N is bonded to three O atoms. 9.116 In the gas phase, aluminum chloride exists as a dimer (a unit of two) with the formula Al2Cl6. Its skeletal structure is given by Cl Cl GD Al DG Cl Cl GD Al DG Cl Cl Complete the Lewis structure and indicate the coordinate covalent bonds in the molecule. 9.117 The hydroxyl radical (OH) plays an important role in atmospheric chemistry. It is highly reactive and has a tendency to combine with a H atom from other compounds, causing them to break up. Thus OH is sometimes called a “detergent” radical because it helps to clean up the atmosphere. (a) Write the Lewis structure for the radical. (b) Refer to Table 9.4 and explain why the radical has a high affinity for H atoms. (c) Estimate the enthalpy change for the following reaction: OH(g) CH4(g) 88n CH3(g) H2O(g) (d) The radical is generated when sunlight hits water Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS vapor. Calculate the maximum wavelength (in nanometers) required to break an OOH bond in H2O. 9.118 Experiments show that it takes 1656 kJ/mol to break all the bonds in methane (CH4) and 4006 kJ/mol to break all the bonds in propane (C3H8). Based on these data, calculate the average bond energy of the COC bond. 9.119 Draw three resonance structures of sulfur dioxide (SO2). Indicate the most plausible structure(s). (Hint: See Example 9.11.) 9.120 Vinyl chloride (C2H3Cl) differs from ethylene (C2H4) in that one of the H atoms is replaced with a Cl atom. Vinyl chloride is used to prepare poly(vinyl chloride), which is an important polymer used in pipes. (a) Draw the Lewis structure of vinyl chloride. (b) The repeating unit in poly(vinyl chloride) is OCH2OCHClO. Draw a portion of the molecule showing three such repeating units. (c) Calculate the enthalpy change when 1.0 103 kg of vinyl chloride forms poly(vinyl chloride). Answers to Practice Exercises: 9.1 Ba 2 H 88n Ba2 2H: (or BaH2) 2 [Xe]6s 1s1 [Xe] [He] 9.2 (a) Ionic, (b) polar covalent, (c) covalent. S S 9.3 OP CPO Q Q SOS B 9.4 HOCOOOH O Q ONQ 9.5 OPOOOS O Q Forward Main Menu TOC O O 9.6 OPOOOS QNQ 9.7 HO CqNS 9.8 OPOOOS O O QNQ OO Be OOS F F 9.9 SQ Q ONO SOOOPQ O Q SOS F O A F QS OO AsE SQ F H A OS F SQS Q F 9.10 S OS B OO Se OOOH O 9.11 HOO Q Q Q 9.12 (a) 9.13 (a) Back 365 Study Guide TOC 543.1 kJ, (b) 543.2 kJ. 119 kJ, (b) 137.0 kJ. Textbook Website MHHE Website ...
View Full Document

This document was uploaded on 07/27/2009.

Ask a homework question - tutors are online