5 Steps to a 5: AP Chemistry 2024 - Moore J.T., Langley R.H. 2023

STEP 4 Review the Knowledge You Need to Score High
8 Bonding

IN THIS CHAPTER

In this chapter, the following AP topics are covered:

2.1 Types of Chemical Bonds

2.4 Structure of Metals and Alloys

2.5 Lewis Diagrams

2.6 Resonance and Formal Charge

2.7 VSEPR and Bond Hybridization

Summary: The difference between elements and compounds was discussed in Chapter 5, Basics. But what are the forces holding together a compound? What is the difference in bonding between table salt and sugar? What do these compounds look like in three-dimensional space?

Compounds have a certain fixed proportion of elements. The periodic table often can be used to predict the type of bonding that might exist between elements.

While there are exceptions, the following general guidelines normally apply:

metal + nonmetal → ionic bonds

nonmetal + nonmetal → covalent bonds

metal + metal → metallic bonding

Compounds exist in each category. In the ionic bonding category are substances like NaCl. A covalent example is H₂O. Examples in the metallic category are less well known, but materials such as MgNi₂ qualify. Metallic bonding in alloys are better-known examples of combinations with two or more metals.

We will discuss two types of bonding, ionic and covalent, in some depth. Suffice it to say that metallic bonding is a bonding situation between metals in which the valence electrons are donated to a vast electron pool (sometimes called a “sea of electrons”) so that the valence electrons are free to move throughout the entire metallic solid. One consequence of this free movement of electrons is that metals conduct electricity.

The basic concept that drives bonding is related to the stability of the noble gas family (the group 8A or group 18 elements). Their extreme stability (lower energy state) is a consequence of these elements having a filled valence shell, a full complement of eight valence electrons. (Helium is an exception. Its valence shell, the 1s, is filled with two electrons.) This is called the octet rule. During chemical reactions, atoms tend to lose, gain, or share electrons in order to achieve a filled valence shell, to complete their octet. By completing their valence shell in this fashion, they become isoelectronic, having the same number and arrangement of electrons, as the closest noble gas. Note: the nearest noble gas is the one with the closest atomic number. For example, both sodium and fluorine are “next” to neon.

There are numerous exceptions to the octet rule; for example, some atoms may have more than an octet, and their compounds may have one or more atoms with less than an octet.

Keywords and Equations

There are no keywords or equations on the AP Exam specific to this chapter.

Lewis Electron-Dot Structures

The Lewis electron-dot symbol is a way of representing the element and its valence electrons. The chemical symbol is written, which represents the atom’s nucleus and all inner-shell electrons. The valence, or outer-shell, electrons are represented as dots surrounding the atom’s symbol. Take the valence electrons, distribute them as dots one at a time around the four sides of the symbol, and then pair them up until all the valence electrons are distributed. Figure 8.1 shows the Lewis symbol for several different elements. In general, Lewis electron-dot structures do not work for the transition or inner transition metals.

Figure 8.1 Lewis electron-dot symbols for selected elements.

The Lewis symbols will be used in the discussion of bonding, especially covalent bonding, and will form the basis of the discussion of molecular geometry.

Ionic and Covalent Bonding

Ionic Bonding

Ionic bonding results from the transfer of electrons from a metal to a nonmetal with the formation of cations (positively charged ions) and anions (negatively charged ions). The attraction of the opposite charges forms the ionic bond. The metal loses electrons to form a cation (the positive charge results from having more protons than electrons), and the nonmetal becomes an anion by gaining electrons (it now has more electrons than protons). This is shown in Figure 8.2 for the reaction of sodium and chlorine to form sodium chloride.

Figure 8.2 Formation of sodium chloride.

The number of electrons to be lost by the metal and gained by the nonmetal is determined by the number of electrons lost or gained by the atom to achieve a full octet. There is a rule of thumb that an atom can gain or lose one or two and, on rare occasions, three electrons, but not more than that. Sodium has one valence electron in energy level 3, and chlorine has seven valence electrons in energy level 3. As shown in Figure 8.2, the lone sodium valence electron goes to the chlorine to produce ions.

If sodium loses the one electron from the valence shell, there are no electrons in that shell, and the highest energy electrons are now in energy level 2. Level 2 for the sodium ion now has an octet of electrons. In general, this is indicated with a Lewis-dot structure that has no electrons. Chlorine, having seven valence electrons, needs to gain one more to complete its octet. So, when an electron is transferred from sodium to chlorine, the outer shell of chlorine goes from seven to eight electrons. Thus, the transfer of one electron completes an octet for both elements.

If magnesium, with two valence electrons to be lost, reacts with chlorine (which needs one additional electron), then magnesium will donate one valence electron to each of two chlorine atoms, forming the ionic compound magnesium chloride, MgCl₂. Make sure the formula has the lowest whole-number ratio of elements.

If aluminum, with three valence electrons to be lost, reacts with oxygen, which needs two additional electrons to complete its octet, then the lowest common factor between 3 and 2 must be found; this value is 6. Two aluminum atoms would each lose three electrons (total of six electrons lost) to three oxygen atoms, which would each gain two electrons (total six electrons gained). The total number of electrons lost must equal the total number of electrons gained.

Another way of deriving the formula of the ionic compound is the crisscross rule. In this technique the cation and anion are written side by side. The numerical value of the superscript charge on the cation (without the sign) becomes the subscript on the nonmetal in the compound, and the superscript charge on the anion becomes the subscript on the metal in the compound. Figure 8.3 illustrates the crisscross rule for the reaction between aluminum and oxygen.

Figure 8.3 Using the crisscross rule.

If magnesium reacts with oxygen, then automatic application of the crisscross rule would lead to the formula Mg₂O₂, which is incorrect because the subscripts are not in the lowest whole-number ratio. This is a shortcoming of the crisscross rule that one must always consider. This means that Mg₂O₂ must be simplified to MgO. For the same reason, lead(IV) oxide (containing Pb⁴⁺ and O^2—) would have the formula PbO₂ and not Pb₂O₄. Make sure the formula has the lowest whole-number ratio of elements.

Ionic bonding may also involve polyatomic ions. The polyatomic ion(s) replace(s) one or both monatomic ions, and while the formula may look more complicated, the polyatomic ion behaves like the smaller monatomic ion.

Covalent Bonding

Consider two hydrogen atoms approaching each other. Both have only one electron, and each requires an additional electron to become isoelectronic with the nearest noble gas, He. One hydrogen atom could lose an electron; the other could gain that electron. One atom would have achieved its noble gas arrangement; but the other, the atom that lost its electron, has moved farther away from stability. The formation of the very stable H₂ cannot be explained by the loss and gain of electrons. In this situation, like that between any two nonmetals, electrons are shared, not lost, and gained. No ions are formed. It is a covalent bond that holds the atoms together. Covalent bonding is the sharing of one or more pairs of electrons. The covalent bonds in a molecule often are represented by a dash, which represents a shared pair of electrons. These covalent bonds may be single bonds, one pair of shared electrons as in H—H; double bonds, two shared pairs of electrons, H₂C==CH₂; or triple bonds, three shared pairs of electrons, N≡≡N. For the representative elements, three shared pairs are the maximum number of pairs shared between two atoms. The same driving force forms a covalent bond as an ionic bond—establishing a stable (lower energy) electron arrangement. In the case of the covalent bond, it is accomplished through sharing electrons.

In the hydrogen molecule, the electrons are shared equally. Each hydrogen nucleus has one proton equally attracting the bonding pair of electrons. A bond like this is called a nonpolar covalent bond. In cases where the two atoms involved in the covalent bond are not the same, the attraction is not equal, and the bonding electrons are pulled toward the atom with the greater attraction. The bond becomes a polar covalent bond, with the atom that has the greater attraction taking on a partial negative charge and the other atom a partial positive charge. Consider for example, HF(g). The fluorine has a greater attraction for the bonding pair of electrons and so takes on a partial negative charge at the expense of the hydrogen, which takes on a partially positive charge. Many times, in addition to using a single line to indicate the covalent bond, an arrow is used with the arrowhead pointing toward the atom that has the greater attraction for the electron pair and a cross at the opposite end:

The electronegativity (EN) is a measure of the attractive force that an atom exerts on a bonding pair of electrons. Electronegativity values are tabulated. In general, electronegativities increase from left to right on the periodic table, except for the noble gases, and decrease going from top to bottom. This means that fluorine has the highest electronegativity of any element. If the difference in the electronegativities of the two elements involved in the bond is great (>1.7), the bond is considered mostly ionic in nature. If the difference is slight (<0.4), it is mostly nonpolar covalent. Anything in between these two extremes is polar covalent. Many times the electronegativities of the transition and inner transition metals are exceptions to the general trend trends observed for the representative elements.

Many times the Lewis structure will be used to indicate the bonding pattern in a covalent compound. In Lewis formulas, the valence electrons that are not involved in bonding are shown as dots surrounding the element symbols (never between two atoms), while a bonding pair of electrons is represented as a dash. There are several ways of deriving the Lewis structure, but here is one that works well for those compounds that obey the octet rule.

Draw the Lewis structural formula for CH₄O.

First, write a general framework for the molecule. In this case, the carbon must be bonded to the oxygen, because hydrogen can form only one bond. Hydrogen is never central. Remember: carbon forms four bonds.

To determine where all the electrons are to be placed, apply the N — A = S rule, where:

For CH₄O, we would have:

Place the electron pairs, as dashes, between the adjacent atoms in the framework and then distribute the remaining available electrons so that each atom has its full octet, eight electrons—bonding or nonbonding, shared or not—for every atom except hydrogen, which gets two. Figure 8.4 shows the Lewis structural formula of CH₄O. (You should check to make sure you can locate the A = 14 electrons.)

Figure 8.4 Lewis structure of CH₄O.

Note: CH₄O is an organic compound, and while organic chemistry is not an AP topic, organic compounds make excellent examples and may appear on the AP Chemistry Exam in the form seen here.

Lewis structures may also be written for polyatomic anions or cations. The N — A = S rule can be used, but if the ion is an anion, extra electrons equal to the magnitude of the negative charge must be added to the electrons available. If the ion is a cation, electrons must be subtracted.

As we have mentioned previously, there are many exceptions to the octet rule. In these cases, the N — A = S rule does not apply, as illustrated by the following example.

Draw the Lewis structure for XeF₄.

Answer:

Each of the fluorine atoms will have an additional three pairs of electrons. Only the four fluorine atoms have their octets.

The process just outlined will usually result in the correct Lewis structure. However, there will be cases when more than one structure may seem to be reasonable. One way to eliminate inappropriate structures is by using the formal charge.

There is a formal charge associated with each atom in a Lewis structure. To determine the formal charge for an atom, enter the number of electrons for each atom into the following relationship:

A formal charge of zero for each atom in a molecule is a very common result for a favorable Lewis structure. In other cases, a favorable Lewis structure will follow the rules below. The formal charges are:

1. Small numbers, preferably 0.

2. No like charges are adjacent to each other, but unlike charges are close together.

3. The more electronegative element(s), the lower the formal charge(s) will be.

4. The total of the formal charges equals the charge on the ion or 0 for a compound.

Now we will apply this formal-charge concept to the cyanate ion OCN^—. We chose this example because many students incorrectly write the formula as CNO^— and then try to use this as the atomic arrangement in the Lewis structure. Based on the number of electrons needed, the carbon needing the most electrons should be the central atom. We will work this example using both the incorrect atom arrangement and the correct atom arrangement. Notice that in both structures all atoms have a complete octet.

The formal charges make the OCN arrangement the better choice.

Do a quick check of a Lewis structure to make sure the total number of electrons matches what the total should be. For example, based on the formula, the Lewis structure of must show 24 valence electrons (C has 4, each O has 6, and the charge gives 2 more).

Molecular Geometry—VSEPR

The shape of a molecule has quite a bit to do with its reactivity and physical properties. This is especially true in biochemical processes, where slight changes in shape in three-dimensional space might make a certain molecule inactive or cause an adverse side effect. One way to predict the shape of molecules is the VSEPR theory (valence-shell electron-pair repulsion theory). The basic idea behind this theory is that the valence electron pairs surrounding a central atom, whether involved in bonding or not, will try to move as far away from each other as possible to minimize the repulsion between the like charges. Two geometries can be determined; the electron-group geometry, in which all electron pairs surrounding a nucleus are considered, and molecular geometry, in which the nonbonding electrons become “invisible” and only the geometry of the atomic nuclei is considered. For the purposes of geometry, double and triple bonds count the same as single bonds. To determine the geometry:

1. Write the Lewis electron-dot formula of the compound.

2. Determine the number of electron-pair groups surrounding the central atom(s). Remember that double and triple bonds are treated as a single group.

3. Determine the geometric shape that maximizes the distance between the electron groups. This is the geometry of the electron groups.

4. Mentally allow the nonbonding electrons to become invisible. They are still there and are still repelling the other electron pairs, but we don’t “see” them. The molecular geometry is determined by the remaining arrangement of atoms (as determined by the bonding electron groups) around the central atom.

Figure 8.5, on the next page, shows the electron-group and molecular geometry for two to six electron pairs.

Figure 8.5 Electron-group and molecular geometry.

For example, let’s determine the electron-group and molecular geometry of carbon dioxide, CO₂, and water, H₂O. At first glance, one might imagine that the geometry of these two compounds would be similar, since both have a central atom with two groups attached. Let’s see if that is true.

First, write the Lewis structure of each. Figure 8.6 shows the Lewis structures of these compounds. Don’t forget a correct Lewis-dot structure may be drawn with more than one “geometry”; these shapes may or may not be related to the true shape of the molecule.

Figure 8.6 Lewis structures of carbon dioxide and water.

Next, determine the electron-group geometry of each. For carbon dioxide, there are two electron groups around the carbon, so it would be linear. For water, there are four electron pairs around the oxygen—two bonding and two nonbonding electron pairs—so the electron-group geometry would be tetrahedral.

Finally, mentally allow the nonbonding electron pairs to become invisible and describe what is left in terms of the molecular geometry. For carbon dioxide, all groups are involved in bonding, so the molecular geometry is also linear. However, water has two nonbonding pairs of electrons, so the remaining bonding electron pairs (and hydrogen nuclei) are in a bent arrangement. If you had drawn the Lewis-dot structure as H—O—H (linear), it would not alter the fact that the water molecule is bent.

This determination of the molecular geometry of carbon dioxide and water also accounts for the fact that carbon dioxide does not possess a dipole and water has one, even though both are composed of polar covalent bonds. Carbon dioxide, because of its linear shape, has partial negative charges at both ends and a partial charge in the middle. To possess a dipole, one end of the molecule must have a positive charge and the other a negative end. Water, because of its bent shape, satisfies this requirement. Carbon dioxide does not.

Valence Bond Theory

The VSEPR theory is only one way in which the molecular geometry of molecules may be determined. Another way involves the valence bond theory. The valence bond theory describes covalent bonding as the mixing of atomic orbitals to form a new kind of orbital, a hybrid orbital. Hybrid orbitals are new atomic orbitals formed as a result of mixing the original atomic orbitals of the atoms involved in the covalent bond. The number of hybrid orbitals formed is the same as the number of atomic orbitals mixed, and the type of hybrid orbital formed depends on the types of atomic orbital mixed. Figure 8.7 shows the hybrid orbitals resulting from the mixing of s, p, and d orbitals.

Figure 8.7 Hybridization of s, p, and d orbitals.

sp hybridization results from the overlap of an s orbital with one p orbital. Two sp hybrid orbitals are formed with a bond angle of 180°. This is a linear orientation.

sp² hybridization results from the overlap of an s orbital with two p orbitals. Three sp² hybrid orbitals are formed with a trigonal planar orientation and a bond angle of 120°.

One place this type of bonding occurs is in the formation of the carbon-to-carbon double bond, as will be discussed later.

sp³ hybridization results from the mixing of one s orbital and three p orbitals, giving four sp³ hybrid orbitals with a tetrahedral geometric orientation. This sp³ hybridization is found in carbon when it forms four single bonds.

sp³d hybridization results from the blending of an s orbital, three p orbitals, and one d orbital. The result is five sp³d orbitals with a trigonal bipyramidal orientation. This type of bonding occurs in compounds like PCl₅. Note that this hybridization is an exception to the octet rule.

sp³d² hybridization occurs when one s, three p, and two d orbitals are mixed, giving an octahedral arrangement. SF₆ is an example. Again, this hybridization is an exception to the octet rule. If one starts with this structure and one of the bonding pairs becomes a lone pair, then a square pyramidal shape results, while two lone pairs give a square planar shape.

Figure 8.8 shows the hybridization that occurs in ethylene, H₂C=CH₂. Each carbon atom has sp² hybridization. On each carbon, two of the hybrid orbitals have overlapped with an s orbital on a hydrogen atom to form a carbon-to-hydrogen covalent bond. The third sp² hybrid orbital has overlapped with the sp² hybrid on the other carbon to form a carbon-to-carbon covalent bond. Note that the remaining p orbital on each carbon that has not undergone hybridization is also overlapping above and below a line joining the carbons. In ethylene, there are two types of bonds. In sigma (σ) bonds, the overlap of the orbitals occurs on a line between the two atoms involved in the covalent bond. In ethylene, the C—H bonds and one of the C—C bonds are sigma bonds. In pi (π) bonds, the overlap of orbitals occurs above and below a line through the two nuclei of the atoms involved in the bond. A double bond always is composed of one sigma and one pi bond. A carbon-to-carbon triple bond results from the overlap of a sp hybrid orbital and two p orbitals on one carbon, with the same on the other carbon. In this situation, there will be one sigma bond (overlap of the sp hybrid orbitals) and two pi bonds (overlap of two sets of p orbitals).

Figure 8.8 Hybridization in ethylene, H₂C=CH₂.

Molecular Orbital Theory

Still another model to represent the bonding that takes place in covalent compounds is the molecular orbital theory. In the molecular orbital (MO) theory of covalent bonding, atomic orbitals (AOs) on the individual atoms combine to form orbitals that encompass the entire molecule. These are called molecular orbitals (MOs). These molecular orbitals have definite shapes and energies associated with them. When two atomic orbitals are added, two molecular orbitals are formed, one bonding and one antibonding.

The bonding MO is of lower energy than the antibonding MO. In the molecular orbital model, the atomic orbitals are added together to form the molecular orbitals. Then the electrons are added to the molecular orbitals, following the rules used previously when filling orbitals: lowest-energy orbitals get filled first, maximum of two electrons per orbital, and half-fill orbitals of equal energy before pairing electrons (see Chapter 5). When s atomic orbitals are added, one sigma bonding (σ) and one sigma antibonding (σ^*) molecular orbital are formed. Figure 8.9 shows the molecular orbital diagram for H₂.

Figure 8.9 Molecular orbital diagram of H₂.

Note that the two electrons (one from each hydrogen) have both gone into the sigma bonding MO. The bonding situation can be calculated in the molecular orbital theory by calculating the MO bond order. The MO bond order is the number of electrons in bonding MOs minus the number of electrons in antibonding MOs, divided by 2. For H₂ in Figure 8.9 the bond order would be (2 — 0)/2 = 1. A stable bonding situation exists between two atoms when the bond order is greater than zero. The larger the bond order, the stronger the bond.

When two sets of p orbitals combine, one sigma bonding and one sigma antibonding MO are formed, along with two bonding pi MOs and two pi antibonding (π) MOs. Figure 8.10 shows the MO diagram for O₂. For the sake of simplicity, the 1s orbitals of each oxygen and the MOs for these electrons are not shown, just the valence-electron orbitals.

Figure 8.10 MO diagram of O₂.

The bond order for O₂ would be (10 — 6)/2 = 2. (Don’t forget to count the bonding and antibonding electrons at energy level 1.) Unlike VSEPR or valence bond theory, MO theory predicts that the oxygen molecule has two unpaired electrons (the two highest energy electrons). Experimental observations demonstrate the oxygen molecules have two unpaired electrons each. MO theory is the only one of the three theories to give the correct answer.

Resonance

Sometimes when writing the Lewis structure of a compound, more than one possible structure is generated for a given molecule. The nitrate ion, NO₃^—, is a good example. Three possible Lewis structures can be written for this polyatomic anion, differing in which oxygen is double bonded to the nitrogen. None truly represents the actual structure of the nitrate ion; that would require an average of all three Lewis structures. Resonance theory is used to describe this situation. Resonance occurs when more than one Lewis structure can be written for a molecule (without rearranging the atoms). The individual structures are called resonance structures (or forms) and are written with a two-headed arrow (↔) between them. Figure 8.11 shows the three resonance forms of the nitrate ion.

Figure 8.11 Resonance structures of the nitrate ion, NO₃^—.

Again, let us emphasize that the actual structure of the nitrate is not any of the three shown. Neither is it flipping back and forth among the three. It is an average of all three. All the bonds are the same, intermediate between single bonds and double bonds in strength and length.

Bond Length, Strength, and Magnetic Properties

The length and strength of a covalent bond is related to its bond order. The greater the bond order, the shorter and stronger the bond. Diatomic nitrogen, for example, has a short, extremely strong bond due to its nitrogen-to-nitrogen triple bond.

One of the advantages of the molecular orbital model is that it can predict some of the magnetic properties of molecules. If molecules are placed in a strong magnetic field, they exhibit one of two magnetic behaviors—attraction or repulsion. Paramagnetism, the attraction to a magnetic field, is due to the presence of unpaired electrons; diamagnetism, the slight repulsion from a magnetic field, is due to the presence of only paired electrons. Look at Figure 8.10, the MO diagram for diatomic oxygen. Note that it does have two unpaired electrons in the π_2p* antibonding orbitals. Thus, one would predict, based on the MO model, that oxygen should be paramagnetic, and that is exactly what is observed in the laboratory.

Structure of Metals and Alloys

Approximately 80 percent of the elements are metals. There are three basic structures adopted by most metals. One of these basic structures is body-centered cubic. One of the metals adopting this structure is lithium. The structure of lithium is pictured in Figure 8.12. In the structure of lithium, there are eight lithium atoms at the corners of a cube with an additional lithium atom in the center of the cube. The structure is more intricate than this simple figure portrays. A piece of lithium metal consists of these cubes stacked together. The arrangement of these cubes is such that each corner of each cube is a corner of seven other cubes. The result is that corner lithium atom is surrounded by eight lithium atoms at the corners of a cube. In addition, if one examines the atoms at the center of the cubes, they also form a cube with a lithium atom at the center of a cube.

Figure 8.12 The body-centered cubic structure of lithium metal.

All the lithium atoms have the same electronegativity, and if they were not metal atoms, one would assume covalent bonding. However, covalent will not work. If we look at the lithium atom in the center of the cube, it is bonded to the eight corners. If these were covalent bonds, 16 electrons would be necessary. However, lithium atoms only have one valence electron, which gives a total of nine for the cube. Nine electrons are far less than the 16 required by covalent bonds. Compared to the bonding in covalent compounds, metals are electron deficient. Metallic bonding does not have localized bonds like covalent bonding has. One view of metallic bonding is to consider the metal nuclei and core electrons to be in fixed positions with the valence electrons freely moving. The freely moving electrons serve as a “glue” to keep the atoms together. The freely moving electrons are often referred to as a “sea of electrons.”

The other metal structures are similar except that each atom is surrounded by 12 neighbors instead of 8.

When more than one type of metal atom is present, there are two possible results. (A few nonmetallic atoms may also work.) An intermetallic compound may form. An intermetallic compound, like any compound, has a fixed composition. The more common situation is for an alloy to form. An alloy is a solution, and as such behaves like all other solutions. Solutions may be unsaturated, saturated, or supersaturated. The solvent and solute may be miscible (able to mix in any proportion). Alloys exhibit all these characteristics.

There are two broad categories of alloys. One category is the interstitial alloys. The other category is the substitutional alloys. Some alloys may be a mixture of both types. The different types of metal alloys are illustrated in Figure 8.13.

Figure 8.13 An example of a metal and the various types of alloys that metal may form. The metal could be nickel, the interstitial atoms could be hydrogen, and the substitutional atom could be cobalt.

Interstitial alloys form when a metal “dissolves” small atoms. These small atoms, such as boron, carbon, and hydrogen, fit into the spaces between the metal atoms. These spaces are “interstitial holes.” The metal atoms generally remain in or near their original positions. Many of these compounds are commercially important. For example, many metal hydrides are useful for the storage of hydrogen. Some of these metal hydrides can store more hydrogen per unit volume than an equal volume of liquid hydrogen. Most steel contains some interstitial carbon for added strength. Many interstitial hydrides are nonstoichiometric (no definite formula). For example, the maximum amount of hydrogen that can dissolve in palladium metal gives the formula PdH_0.7.

Substitutional alloys occur when one metal atom substitutes for another. Normally the atoms need to be close to the same size (the radii should not differ by more than 15%). If the radii differ by more than 15%, there may be limited dissolution of one metal into another. The radii used are metallic radii, not atomic radii, or ionic radii. For example, manganese will dissolve in iron as observed in most steels.

Experiments

There have been no experimental questions concerning this material on recent AP Chemistry Exams.

Common Mistakes to Avoid

1. Remember that metals + nonmetals form ionic bonds, while the reaction of two nonmetals forms a covalent bond. And finally, the reaction between two metals forms a metallic bond. While there are exceptions to these predictions, do not expect them to appear on the AP Chemistry Exam.

2. The octet rule does not always work, but for the representative elements, it works the majority of the time. Hydrogen will always be an exception because it aims to get two electrons.

3. Atoms that lose electrons form cations; atoms that gain electrons form anions.

4. In writing the formulas of ionic compounds, make sure the subscripts are in the lowest ratio of whole numbers.

5. When using the crisscross rule, be sure the subscripts are reduced to the lowest whole-number ratio.

6. When using the N — A = S rule in writing Lewis structures, be sure you add electrons to the A term for a polyatomic anion and subtract electrons for a polyatomic cation.

7. In the N — A = S rule, only the valence electrons are counted.

8. In using the VSEPR theory, when going from the electron-group geometry to the molecular geometry, start with the electron-group geometry; make the nonbonding electrons mentally invisible; and then describe what is left.

9. When adding electrons to the molecular orbitals, remember to begin with the lowest energy first. On orbitals with equal energies, half-fill and then pair up.

10. When writing Lewis structures of polyatomic ions, don’t forget to show the charge.

11. When you draw resonance structures, you can move only electrons (bonds). Never move the atoms.

12. When answering questions, the stability of the noble-gas configurations is a result, not an explanation. Your answers will require an explanation, i.e., lower-energy state.

Review Questions

Use these questions to review the content of this chapter and practice for the AP Chemistry Exam. First are 20 multiple-choice questions similar to what you will encounter in Section I of the AP Chemistry Exam. The multiple-choice portion also includes questions designed to help you review prior knowledge. Following those is a long free-response question like the ones in Section II of the exam. To make these questions an even more authentic practice for the actual exam, time yourself following the instructions provided.

Multiple-Choice Questions

Answer the following questions in 30 minutes. You may use the periodic table and the equation sheet at the back of this book.

1. VSEPR predicts an SbF₅ molecule will be which of the following shapes?

(A) tetrahedral

(B) trigonal bipyramidal

(C) square pyramid

(D) trigonal planar

2. The shortest bond would be present in which of the following substances?

(A) I₂

(B) CO

(C) CCl₄

(D) O₂^2—

3. Which of the following does NOT have one or more π bonds?

(A) H₂O

(B) HNO₃

(C) O₂

(D) N₂

4. Which of the following is nonpolar?

(A) IF₅

(B) IF₃

(C) SiF₄

(D) SeF₄

5. Resonance structures are necessary to describe the bonding in which of the following?

(A) H₂O

(B) ClF₃

(C) HNO₃

(D) CH₄

For questions 6 and 7, pick the best choice from the following:

(A) ionic bonds

(B) hybrid orbitals

(C) resonance structures

(D) van der Waals attractions

6. An explanation of the equivalent bond lengths of the nitrite ion is:

7. Most organic substances have low melting points. This may be because, in most cases, the intermolecular forces are:

8. Which of the following has more than one unshared pair of valence electrons on the central atom?

(A) BrF₅

(B) NF₃

(C) IF₇

(D) ClF₃

9. Which of the following groups contains only elements that can exceed an octet in at least some of their compounds?

(A) Sb, Br, N

(B) As, I, Si

(C) C, S, Bi

(D) O, Te, Ge

10. The only substance listed below that contains ionic, σ, and π bonds is:

(A) Ca₃P₂

(B) CO₂

(C) CaCO₃

(D) NH₃

11. The electron pairs point toward the corners of which geometrical shape for a molecule with sp² hybrid orbitals?

(A) trigonal planar

(B) octahedron

(C) trigonal bipyramid

(D) trigonal pyramid

12. Regular tetrahedral molecules or ions include which of the following?

(A) SF₄

(B) NH₄⁺

(C) XeF₄

(D) ICl₄^—

13. Which of the following molecules or ions has the greatest number of unshared electrons around the central atom?

(A) SO₂

(B) NO₃^—

(C) KrF₂

(D) SF₄

14. Which of the following molecules is the least polar?

(A) PH₃

(B) CH₄

(C) H₂O

(D) NO₂

15. Which of the following molecules is the most polar?

(A) NH₃

(B) N₂

(C) CH₃I

(D) BF₃

16. Each of the following processes involves breaking bonds, and most involve forming new bonds. Which of the following processes involves breaking an ionic bond?

(A) H₂(g) + I₂(g) → 2 HI(g)

(B) 2 KCl(s) → 2 K(g) + Cl₂(g)

(C) Ca(s) → Ca(g)

(D) 2 C₈H₁₈(g) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g)

17. Which of the following sets of bonds is listed in order of increasing covalent character?

(A) Na—Cl < Al—Cl < P—Cl < Cl—Cl

(B) Na—Cl < P—Cl < Cl—Cl < Al—Cl

(C) P—Cl < Cl—Cl < Al—Cl < Na—Cl

(D) Al—Cl < Na—Cl < P—Cl < Cl—Cl

18. Which of the following molecules does NOT contain one or more π bonds?

(A) CO₂

(B) SF₆

(C) CO

(D) SO₃

19. The above diagram illustrates a portion of a molecule of buckminsterfullerene (a “Bucky ball”). The corners are carbon atoms. Each carbon atom has four covalent bonds, allowing it to follow the octet rule. The structure also shows many single and double bonds. Carbon—carbon double bonds are shorter than carbon—carbon single bonds. In an analysis of the structure of buckminsterfullerene, all carbon—carbon bonds are found to be the same length and not a combination of short double bonds and longer single bonds. Which of the following best explains this observation?

(A) VSEPR

(B) resonance

(C) hybridization

(D) experimental error

20. There are three nitrogen-oxygen species known with a 1:2 nitrogen-to-oxygen ratio. The Lewis electron-dot diagrams for these three nitrogen oxygen species are shown in the above diagram. Which of the three has the largest bond angle?

(A) NO₂^—

(B) NO₂

(C) NO₂⁺

(D) All have a 180° angle.

Answers and Explanations

1. B—The Lewis (electron-dot) structure has five bonding pairs around the central Sb and no lone pairs. VSEPR predicts this number of pairs to give a trigonal bipyramidal structure.

2. B—All bonds except those in CO are single bonds. The CO bond is a triple bond. Triple bonds are shorter than double bonds, which are shorter than single bonds. Drawing Lewis structures might help you answer this question.

3. A—Answers B through D contain molecules with double or triple bonds. Double and triple bonds contain π bonds. Water has only single (σ) bonds. If any of these are not obvious to you, draw a Lewis structure.

4. C—Use VSEPR; only the tetrahedral SiF₄ is nonpolar. The other materials form square pyramidal (IF₅), T-shaped (IF₃), and irregular tetrahedral (SeF₄) shapes and are therefore polar.

5. C—All other answers involve species containing only single bonds. Substances without double or triple bonds seldom need resonance structures.

6. C—Resonance causes bonds to have the same average length.

7. D—Many organic molecules are nonpolar. Nonpolar substances are held together by weak van der Waals attractions.

8. D—Lewis structures are required. You may not need to draw all of them. A and B have one unshared pair, whereas C does not have an unshared pair. D has two unshared pairs of electrons.

9. B—For an atom to exceed an octet of electrons, it must be in the third period or lower on the periodic table. The elements N, C, and O are in the second period; therefore, the answers containing these elements contain an element that cannot exceed an octet. Only answer B is limited to elements that can exceed an octet.

10. C—The ionic bonds are present in the calcium compounds (eliminating all but A and C). The phosphide ion, P^3—, has no internal bonding (eliminating A); however, the carbonate ion, CO₃^2—, has both σ and π bonds.

11. A—This hybridization requires a geometrical shape with three corners. B is for six pairs. C is for five pairs. D has four pairs.

12. B—One or more Lewis structures may help you. A is an irregular tetrahedron (see-saw); C and D are square planar.

13. C—Draw the Lewis structures. The number of unshared pairs are as follows: (A) 1; (B) 0; (C) 3; (D) 1.

14. B—All molecules are polar except B.

15. A—Drawing one or more Lewis structures may help you. Only A and C are polar. Only the ammonia has hydrogen bonding, which is very, very polar.

16. B—Calcium is a metal; therefore, the process shown is breaking metallic bonds. All the species in A and D are covalently bonded molecules. Potassium chloride, KCl, is an ionic compound, and separating the ions involves breaking ionic bonds.

17. A—Increasing covalent character means decreasing electronegativity difference. Chlorine is a constant, so the electronegativity differences of concern are all relative to chlorine. The most covalent would be a bond with an electronegativity difference of 0, which occurs when chlorine is bonded to chlorine (Cl—Cl). In general, the covalent character will decrease (ionic character will increase) when moving from chlorine to the left on the periodic table. The Na—Cl bond would be the most ionic (least covalent). Only answer A has the bonds in the correct order.

18. B—The SF₆ molecule has six σ bonds and no π bonds. The other three molecules have double or triple bonds. All double bonds are a combination of an σ bond and a π bond, and all triple bonds are a combination of one σ bond and two π bonds.

19. B—The process of resonance tends to equalize the lengths of the bonds involved.

20. C—Just looking at the Lewis structures can be misleading. You should use VSEPR to predict the bond angles. The ion NO₂⁺ is a linear species (bond angle 180°) because there are no lone electron pairs on the nitrogen atom.

The other two molecules are bent species (bond angle < 180°) due to the lone electron (≈134°) or pair of electrons (≈118°) on the nitrogen. The presence of resonance does not alter the result.

Free-Response Question

You have 15 minutes to answer the following question. You may use a calculator and the tables in the back of the book.

Question

Answer the following questions about structure and bonding.

(a) Which of the following tetrafluoride compounds is polar? Use Lewis electron-dot structures to explain your conclusions.

(b) Rank the following compounds in order of increasing melting point. Explain your answer. Lewis electron-dot structures may aid you.

(c) Use Lewis electron-dot structures to show why the carbon—oxygen bonds in the oxalate ion (C₂O₄^2—) are all equal.

(d) When PCl₅ is dissolved in a polar solvent, the solution conducts electricity. Explain why. Use an appropriate chemical equation to illustrate your answer. Hint: there are no P⁵⁺ ions present.

Answer and Explanation

(a) Sulfur tetrafluoride is the only one of the three compounds that is polar.

You get 1 point if you correctly predict only SF₄ to be polar, and 1 point for having a correct Lewis structure for SF₄. You get 1 point each for a correct Lewis structure for SiF₄ and XeF₄.

(b) The order is KrF₂ < SeF₂ < SnF₂.

The Lewis structure indicates that KrF₂ is nonpolar. Thus, it has only very weak London dispersion forces between the molecules. SeF₂ is polar, and the molecules are attracted by dipole—dipole attractions, which are stronger than London (higher melting point). SnF₂ has the highest melting point because of the presence of strong ionic bonds.

You get 1 point for the order and 1 point for the discussion.

(c) It is possible to draw the following resonance structures for the oxalate ion. The presence of resonance equalizes the bonds.

You get 1 point for any correct Lewis structure for C₂O₄^2— and 1 point for showing or discussing resonance.

(d) PCl₅ must ionize (be an electrolyte). There are several acceptable equations, all of which must indicate the formation of ions. Here are two choices:

You get 1 point for the explanation and 1 point for any balanced equation showing the formation of ions unless the equation contains P⁵⁺ ions.

Total your points. There are 10 points possible.

Rapid Review

• Compounds are pure substances that have a fixed proportion of elements.

• Metals react with nonmetals to form ionic bonds, and nonmetals react with other nonmetals to form covalent bonds.

• The Lewis electron-dot structure is a way of representing an element and its valence electrons.

• Atoms tend to lose, gain, or share electrons to achieve the same electronic configuration as (become isoelectronic to) the nearest noble gas.

• Atoms are generally most stable when they have a complete octet (eight electrons).

• Ionic bonds result when a metal loses electrons to form cations and a nonmetal gains those electrons to form an anion.

• Ionic bonds can also result from the interaction of polyatomic ions.

• The attraction of the opposite charges (anions and cations) forms the ionic bond.

• The crisscross rule can help determine the formula of an ionic compound.

• In covalent bonding two atoms share one or more electron pairs.

• If the electrons are shared equally, the bond is a nonpolar covalent bond, but unequal sharing results in a polar covalent bond.

• The element that will have the greatest attraction for a bonding pair of electrons is related to its electronegativity.

• Electronegativity values increase from left to right on the periodic table and decrease from top to bottom.

• The N — A = S rule can be used to help draw the Lewis structure of a molecule.

• Molecular geometry, the arrangement of atoms in three-dimensional space, can be predicted using the VSEPR theory. This theory says the electron pairs around a central atom will try to get as far as possible from each other to minimize the repulsive forces.

• In using the VSEPR theory, first determine the electron-group geometry, then the molecular geometry.

• The valence bond theory describes covalent bonding as the overlap of atomic orbitals to form a new kind of orbital, a hybrid orbital.

• The number of hybrid orbitals is the same as the number of atomic orbitals that were combined.

• There are several types of hybrid orbital, such as sp, sp², and sp³.

• In the valence bond theory, sigma (σ) bonds overlap on a line drawn between the two nuclei, while pi (π) bonds result from the overlap of atomic orbitals above and below a line connecting the two atomic nuclei.

• A double or triple bond is always composed of one sigma bond and the rest pi bonds.

• Resonance occurs when more than one Lewis structure can be written for a molecule.

• The actual structure of the molecule is an average of the Lewis resonance structures.

• The higher the bond order, the shorter and stronger the bond.

• Paramagnetism, the attraction of a molecule to a magnetic field, is due to the presence of unpaired electrons. Diamagnetism, the repulsion of a molecule from a magnetic field, is due to the presence of paired electrons.