5 Steps to a 5: AP Chemistry 2017 (2016)

STEP 4

Review the Knowledge You Need to Score High

CHAPTER 11

Bonding

IN THIS CHAPTER

Summary: The difference between elements and compounds was discussed in the Basics chapter, and chemical reactions were discussed in the Reactions and Periodicity chapter. 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. The following general guidelines apply:

metals + nonmetals → ionic bonds

nonmetal + nonmetal → covalent bonds

metal + metal → metallic bonding

We will discuss the first two types of bonding, ionic and covalent, in some depth. Metallic bonding is a topic that is very rarely encountered on the AP exam. 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.

The basic concept that drives bonding is related to the stability of the noble gas family (the group VIIIA or group 18 elements). Their extreme stability (lower energy state) is due to the fact that they have 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 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. There are numerous exceptions to the octet rule; for example, some atoms may have more 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 11.1 shows the Lewis symbol for several different 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.

Figure 11.1   Lewis electron-dot symbols for selected elements.

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 11.2 for the reaction of sodium and chlorine to form 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 in order 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.

Figure 11.2   Formation of sodium chloride.

If it lost that one, the valence shell, now energy level 2, would be full (a more common way of showing this is with zero electrons). Chlorine, having seven valence electrons, needs to gain one more in order to complete its octet. So, an electron is transferred from sodium to chlorine, completing the octet for both.

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 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—6. Two aluminum atoms would each lose 3 electrons (total of 6 electrons lost) to three oxygen atoms, which would each gain 2 electrons (total 6 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 11.3 illustrates the crisscross rule for the reaction between aluminum and oxygen.

Figure 11.3   Using the crisscross rule.

If magnesium reacts with oxygen, then automatic application of the crisscross rule would lead to the formula Mg2 O2 , which is incorrect because the subscripts are not in the lowest whole-number ratio. For the same reason, lead(IV) oxide would have the formula PbO2 and not Pb2 O4 . Make sure the formula has the lowest whole-number ratio of elements.

Ionic bonding may also involve polyatomic ions. The polyatomic ion(s) simply replace(s) one or both of the monoatomic ions.

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 H2 C═CH2 ; or triple bonds, three shared pairs of electrons, N≡N. 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 non-polar 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 (greater electronegativity) and so takes on a partial negative charge. Many times, instead of 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:

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 to be mostly ionic in nature. If the difference is slight (<0.4), it is mostly nonpolar covalent. Anything in between is polar covalent.

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, 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 CH4 O.

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:

N = sum of valence electrons needed for each atom. The two allowed values are two for hydrogen and eight for all other elements.

A = sum of all available valence electrons

S = # of electrons shared and S/2 = # bonds

For CH4 O, 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 11.4 shows the Lewis structural formula of CH4 O.

Figure 11.4   Lewis structure of CH4 O.

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

This process 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:

Formal Charge = (number of valence electrons) – (number of nonbonding electrons + 1/2 number of bonding electrons)

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 these rules: 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.

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 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.

Molecular Geometry—VSEPR

The shape of a molecule has quite a bit to do with its reactivity. 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 (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 are 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 11.5 , on the next page, shows the electron-group and molecular geometry for two to six electron pairs.

For example, let’s determine the electron-group and molecular geometry of carbon dioxide, CO2 , and water, H2 O. 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 11.6 shows the Lewis structures of these compounds.

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.

Figure 11.5   Electron-group and molecular geometry.

Figure 11.6   Lewis structures of carbon dioxide and water.

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 atomic orbitals formed as a result of mixing the 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 11.7 shows the hybrid orbitals resulting from the mixing of s, p, and d orbitals.

Figure 11.7   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.

sp 2 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.

sp 3 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.

sp 3 d hybridization results from the blending of an s orbital, three p orbitals, and one d orbital. The result is five sp3 d 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.

sp 3 d 2 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 gives a square planar shape.

Figure 11.8 shows the hybridization that occurs in ethylene, H2 C=CH2 . Each carbon has undergone 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 bond. 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 an 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 11.8   Hybridization in ethylene, H2 C =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 11.9 shows the molecular orbital diagram for H2 .

Figure 11.9   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 11.9 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 11.10 , on the next page, shows the MO diagram for O2 . For the sake of simplicity, the 1s orbitals of each oxygen and the MOs for these elections are not shown, just the valence-electron orbitals.

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.)

Figure 11.10   Molecular orbital diagram of valence-shell electrons of O2 .

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. The individual structures are called resonance structures (or forms) and are written with a two-headed arrow (↔) between them. Figure 11.11 shows the three resonance forms of the nitrate ion.

Figure 11.11   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 11.10 , the MO diagram for diatomic oxygen. Note that it does have two unpaired electrons in the  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.

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.
  2. The octet rule does not always work, but for the representative elements, it works the majority of the time.
  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 thevalence 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: 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 16 multiple-choice questions similar to what you will encounter in Section I of the AP Chemistry exam. Following those is a multipart 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 20 minutes. You may not use a calculator. 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) H2 O

(B) HNO3

(C) O2

(D) N2

4 . Which of the following is polar?

(A) SF4

(B) XeF4

(C) CF4

(D) SbF5

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

(A) H2 O

(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 . What is the expected hybridization of the central atom in a molecule of TiCl4 ? This molecule is tetrahedral.

(A) sp3

(B) sp3 d

(C) sp

(D) sp2

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

(A) Na2 CO3

(B) HClO2

(C) H2 O

(D) NaCl

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 molecule or ion in the following list has the greatest number of unshared electrons around the central atom?

(A) CF4

(B) ClF3

(C) BF3

(D) NH4 +

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

(A) PH3

(B) CH4

(C) H2 O

(D) NO2

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

(A) NH3

(B) N2

(C) CH3 I

(D) BF3

16 . Which of the following processes involves breaking an ionic bond?

(A) H2 (g) + Cl2 (g) → 2 HCl(g)

(B) 2 KBr(s) → 2 K(g) + Br2 (g)

(C) Na(s) → Na(g)

(D) 2 C2 H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2 O(g)

Answers and Explanations for the Multiple-Choice Questions

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 the bonds except 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 or ions 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 . A —The VSEPR model predicts all the other molecules to be nonpolar.

5 . C —All the 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 do not need to draw all of them. A and B have one unshared pair, while C does not have an unshared pair. D has two unshared pairs of electrons.

9 . A —Tetrahedral molecules are normally sp3 hybridized.

10 . A —Only A and D are ionic. The chloride ion has no internal bonds, so σ and π bonds are not possible.

11 . A —This hybridization requires a geometrical shape with three corners.

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

13 . B —A has 0. B has 2. C and D have 0. You may need to draw one or more Lewis structures.

14 . B —All the 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 —C is breaking metallic bonding. All the others involve covalently bonded molecules.

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 1

Answer the following questions about structure and bonding.

(a) Which of the following tetrafluoride compounds is nonpolar? 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 (C2 O4 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.

Answer and Explanation for the Free-Response Question

(a) Silicon tetrafluoride is the only one of the three compounds that is not polar.

You get 1 point if you correctly predict only SiF4 to be nonpolar. You get 1 additional point for each correct Lewis structure.

(b) The order is KrF2 < SeF2 < SnF2 .

The Lewis structure indicates that KrF2 is nonpolar. Thus, it only has 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. 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 C2 O4 2– and 1 point for showing or discussing resonance.

(d) PCl5 must ionize. 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 you get 1 point for either of the equations.

Total your points for the different parts. 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 mixed together.
  • There are a number of different types of hybrid orbital, such as sp, sp2, and sp3 .
  • In the valence bond theory, sigma bonds overlap on a line drawn between the two nuclei, while pi 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.
  • In the molecular orbital (MO) theory of covalent bonding, atomic orbitals form molecular orbitals that encompass the entire molecule.
  • The MO theory uses bonding and antibonding molecular orbitals.
  • The bond order is (# electrons in bonding MOs – # electrons in anti-bonding MOs)/2.
  • 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.