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7: Temperature Conversions: An Approach to Problem Solving

7: Temperature Conversions: An Approach to Problem Solving

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384 Chapter 12 Chemical Bonding

Because each of the electron pairs is shared with a fluorine atom, the mo-lecular structure is

F

120Њ



F



F

120Њ



B

120Њ



or



F



F



B



F



This is a planar (flat) molecule with a triangular arrangement of F atoms,

commonly described as a trigonal planar structure. Whenever three pairs of

electrons are present around an atom, they should always be placed at the corners

of a triangle (in a plane at angles of 120° to each other).

Next let’s consider the methane molecule, which has the Lewis

structure

H

H



C



H



or



H



H

H C H

H



There are four pairs of electrons around the central carbon atom. What

arrangement of these electron pairs best minimizes the repulsions? First we

try a square planar arrangement:

90Њ



C

The carbon atom and the electron pairs are all in a plane represented by the

surface of the paper, and the angles between the pairs are all 90°.

Is there another arrangement with angles greater than 90° that would

put the electron pairs even farther away from each other? The answer is yes.

We can get larger angles than 90° by using the following three-dimensional

structure, which has angles of approximately 109.5°.

~109.5Њ



C



A tetrahedron has four

equal triangular faces.



In this drawing the wedge indicates a position above the surface of the paper and the dashed lines indicate positions behind that surface. The solid

line indicates a position on the surface of the page. The figure formed by

connecting the lines is called a tetrahedron, so we call this arrangement of

electron pairs the tetrahedral arrangement.



C



This is the maximum possible separation of four pairs around a given atom.

Whenever four pairs of electrons are present around an atom, they should always

be placed at the corners of a tetrahedron (the tetrahedral arrangement).

Now that we have the arrangement of electron pairs that gives the least

repulsion, we can determine the positions of the atoms and thus the molecular structure of CH4. In methane each of the four electron pairs is shared



12.9 Molecular Structure: The VSEPR Model



H



C

H

H

H



Figure 12.12

The molecular structure of

methane. The tetrahedral

arrangement of electron pairs

produces a tetrahedral arrangement of hydrogen atoms.



385



between the carbon atom and a hydrogen atom. Thus the hydrogen atoms

are placed as shown in Figure 12.12, and the molecule has a tetrahedral

structure with the carbon atom at the center.

Recall that the main idea of the VSEPR model is to find the arrangement of electron pairs around the central atom that minimizes the repulsions. Then we can determine the molecular structure by knowing how the

electron pairs are shared with the peripheral atoms. A systematic procedure

for using the VSEPR model to predict the structure of a molecule is outlined

below.



Steps for Predicting Molecular Structure Using

the VSEPR Model

Step 1 Draw the Lewis structure for the molecule.

Step 2 Count the electron pairs and arrange them in the way that minimizes

repulsion (that is, put the pairs as far apart as possible).

Step 3 Determine the positions of the atoms from the way the electron pairs

are shared.

Step 4 Determine the name of the molecular structure from the positions of

the atoms.



EXAMPLE 12.5



Predicting Molecular Structure Using the VSEPR Model, I

Ammonia, NH3, is used as a fertilizer (injected into the soil) and as a household cleaner (in aqueous solution). Predict the structure of ammonia using

the VSEPR model.

SOLUTION

Step 1 Draw the Lewis structure.

H



N



H



H

Step 2 Count the pairs of electrons and arrange them to minimize repulsions. The NH3 molecule has four pairs of electrons around the N atom: three

bonding pairs and one nonbonding pair. From the discussion of the methane molecule, we know that the best arrangement of four electron pairs is

the tetrahedral structure shown in Figure 12.13a.

Step 3 Determine the positions of the atoms. The three H atoms share

electron pairs as shown in Figure 12.13b.

Step 4 Name the molecular structure. It is very important to recognize that

the name of the molecular structure is always based on the positions of the

atoms. The placement of the electron pairs determines the structure, but the name

is based on the positions of the atoms. Thus it is incorrect to say that the NH3

molecule is tetrahedral. It has a tetrahedral arrangement of electron pairs but

not a tetrahedral arrangement of atoms. The molecular structure of ammonia is a trigonal pyramid (one side is different from the other three) rather

than a tetrahedron. ■



386 Chapter 12 Chemical Bonding



Lone

pair

Bonding

pair



N



N

H



H

H



Figure 12.13



a



Bonding

pair



c



b



The tetrahedral arrangement of electron

pairs around the

nitrogen atom in the

ammonia molecule.



Three of the electron pairs

around nitrogen are shared

with hydrogen atoms as

shown, and one is a lone

pair. Although the arrangement of electron pairs is

tetrahedral, as in the

methane molecule, the

hydrogen atoms in the

ammonia molecule occupy

only three corners of the

tetrahedron. A lone pair

occupies the fourth corner.



The NH3 molecule has

the trigonal pyramid

structure (a pyramid with

a triangle as a base).



Predicting Molecular Structure Using the VSEPR Model, II



EXAMPLE 12.6



Describe the molecular structure of the water molecule.

SOLUTION

Step 1 The Lewis structure for water is

H



O



H



Step 2 There are four pairs of electrons: two bonding pairs and two nonbonding pairs. To minimize repulsions, these are best arranged in a tetrahedral structure as shown in Figure 12.14a.



Lone

pair

Bonding

pair



O



O



Bonding

pair



H

H

Lone pair



Figure 12.14



a



The tetrahedral arrangement of the four

electron pairs around

oxygen in the water

molecule.



b



Two of the electron pairs are

shared between oxygen and the

hydrogen atoms, and two are

lone pairs.



c



The V-shaped molecular

structure of the water

molecule.



12.10 Molecular Structure: Molecules with Double Bonds



387



Step 3 Although H2O has a tetrahedral arrangement of electron pairs, it is

not a tetrahedral molecule. The atoms in the H2O molecule form a V shape, as

shown in Figure 12.14b and c.

Step 4 The molecular structure is called V-shaped or bent.



Self-Check EXERCISE 12.5 Predict the arrangement of electron pairs around the central atom. Then

sketch and name the molecular structure for each of the following molecules

or ions.

a. NH4ϩ



d. H2S



b. SO42Ϫ



e. ClO3Ϫ



c. NF3



f. BeF2

See Problems 12.81 through 12.84. ■



The various molecules we have considered are summarized in Table 12.4

on the following page. Note the following general rules.



Rules for Predicting Molecular Structure Using the VSEPR Model

1. Two pairs of electrons on a central atom in a molecule are always placed

180° apart. This is a linear arrangement of pairs.

2. Three pairs of electrons on a central atom in a molecule are always placed

120° apart in the same plane as the central atom. This is a trigonal planar

(triangular) arrangement of pairs.

3. Four pairs of electrons on a central atom in a molecule are always placed

109.5° apart. This is a tetrahedral arrangement of electron pairs.

4. When every pair of electrons on the central atom is shared with another

atom, the molecular structure has the same name as the arrangement of

electron pairs.

Number of Pairs

2

3

4



Name of Arrangement

linear

trigonal planar

tetrahedral



5. When one or more of the electron pairs around a central atom are unshared

(lone pairs), the name for the molecular structure is different from that for

the arrangement of electron pairs (see rows 4 and 5 in Table 12.4).



12.10

OBJECTIVE:



Molecular Structure: Molecules

with Double Bonds

To learn to apply the VSEPR model to molecules with double bonds.

Up to this point we have applied the VSEPR model only to molecules (and

ions) that contain single bonds. In this section we will show that this model

applies equally well to species with one or more double bonds. We will develop the procedures for dealing with molecules with double bonds by considering examples whose structures are known.



388 Chapter 12 Chemical Bonding

Table 12.4 Arrangements of Electron Pairs and the Resulting Molecular Structures for Two, Three, and Four Electron Pairs

Number of

Electron Pairs



Bonds



Electron Pair

Arrangement



2



2



Linear



3



3



Trigonal planar

(triangular)



Ball-and-Stick

Model



Molecular

Structure



Partial Lewis

Structure



Linear



180˚



120˚



AOBOA



Trigonal planar

(triangular)



Cl



A



A

4



Tetrahedral



Be Cl



F



B

4



Ball-and-Stick

Model



F



B



F



A



Tetrahedral



A



H



109.5˚



A



B



A



H



C H

H



H



N H

H



A

4



3



Tetrahedral



Trigonal pyramid

109.5˚



A



B



A



A



4



2



Tetrahedral



Bent or V-shaped

109.5˚



A



B



A



O

H



H



First we will examine the structure of carbon dioxide, a substance that

may be contributing to the warming of the earth. The carbon dioxide molecule has the Lewis structure

O C O

as discussed in Section 12.7. Carbon dioxide is known by experiment to be a

linear molecule. That is, it has a 180° bond angle.

C



O



O



180˚



Recall from Section 12.9 that two electron pairs around a central atom

can minimize their mutual repulsions by taking positions on opposite sides

of the atom (at 180° from each other). This causes a molecule like BeCl2,

which has the Lewis structure

Cl



Be



Cl



to have a linear structure. Now recall that CO2 has two double bonds and is

known to be linear, so the double bonds must be at 180° from each other.

Therefore, we conclude that each double bond in this molecule acts effectively as one repulsive unit. This conclusion makes sense if we think of a

bond in terms of an electron density “cloud” between two atoms. For example, we can picture the single bonds in BeCl2 as follows:

Cl



Be



Cl



C H E M I S T R Y I N F OCUS

Minimotor Molecule



O



ur modern society is characterized by a continual quest for miniaturization. Our computers,

cell phones, portable music players, calculators,

and many other devices have been greatly downsized over the last several years. The ultimate in

miniaturization—machines made of single molecules. Although this idea sounds like an impossible dream, recent advances place us on the

doorstep of such devices. For example, Hermann

E. Gaub and his coworkers at the Center for

Nanoscience at Ludwig-Maximilians University in

Munich have just reported a single molecule that

can do simple work.

Gaub and his associates constructed a polymer about 75 nm long by hooking together

many light-sensitive molecules called azobenzenes:

H

C

H



C

C



H

C



C



C



H



H



H



N



N



C

C



H



C



C



C

H



Azobenzene is ideal for this application because

its bonds are sensitive to specific wavelengths of

light. When azobenzene absorbs light of 420 nm,

it becomes extended; light at 365 nm causes the

molecule to contract.

To make their tiny machine, the German scientists attached one end of the azobenzene

polymer to a tiny, bendable lever similar to the

tip of an atomic-force microscope. The other end

of the polymer was attached to a glass surface.

Flashes of 365-nm light caused the molecule to

contract, bending the lever down and storing

mechanical energy. Pulses of 420-nm radiation

then extended the molecule, causing the lever to

rise and releasing the stored energy. Eventually,

one can imagine having the lever operate some

part of a nanoscale machine. It seems we are getting close to the ultimate in miniature machines.



H



C

H



The minimum repulsion between these two electron density clouds occurs

when they are on opposite sides of the Be atom (180° angle between them).

Each double bond in CO2 involves the sharing of four electrons between the carbon atom and an oxygen atom. Thus we might expect the

bonding cloud to be “fatter” than for a single bond:

O



C



O



However, the repulsive effects of these two clouds produce the same result as

for single bonds; the bonding clouds have minimum repulsions when they

are positioned on opposite sides of the carbon. The bond angle is 180°, and

so the molecule is linear:



In summary, examination of CO2 leads us to the conclusion that in using the VSEPR model for molecules with double bonds, each double bond

should be treated the same as a single bond. In other words, although a double bond involves four electrons, these electrons are restricted to the space



389



390 Chapter 12 Chemical Bonding

between a given pair of atoms. Therefore, these four electrons do not function as two independent pairs but are “tied together” to form one effective

repulsive unit.

We reach this same conclusion by considering the known structures of

other molecules that contain double bonds. For example, consider the ozone

molecule, which has eighteen valence electrons and exhibits two resonance

structures:

O



O



O ←⎯→ O



O



O



The ozone molecule is known to have a bond angle close to 120°. Recall that

120° angles represent the minimum repulsion for three pairs of electrons.



X



This indicates that the double bond in the ozone molecule is behaving as

one effective repulsive unit:

Lone pair



O



Double bond



O



O

Single bond



These and other examples lead us to the following rule: When using the

VSEPR model to predict the molecular geometry of a molecule, a double bond is

counted the same as a single electron pair.

Thus CO2 has two “effective pairs” that lead to its linear structure,

whereas O3 has three “effective pairs” that lead to its bent structure with a

120° bond angle. Therefore, to use the VSEPR model for molecules (or ions)

that have double bonds, we use the same steps as those given in Section 12.9,

but we count any double bond the same as a single electron pair. Although

we have not shown it here, triple bonds also count as one repulsive unit in

applying the VSEPR model.



EXAMPLE 12.7



Predicting Molecular Structure Using the VSEPR Model, III

Predict the structure of the nitrate ion.

SOLUTION

Step 1 The Lewis structures for NO3Ϫ are

Ϫ



O







N

O



O



Ϫ



O







N

O



O



Ϫ



O

N

O



O



Step 2 In each resonance structure there are effectively three pairs of electrons: the two single bonds and the double bond (which counts as one pair).



Chapter Review



391



These three “effective pairs” will require a trigonal planar arrangement (120°

angles).

Step 3 The atoms are all in a plane, with the nitrogen at the center and the

three oxygens at the corners of a triangle (trigonal planar arrangement).

Step 4 The NO3Ϫ ion has a trigonal planar structure. ■



C H A P T E R



12



REVIEW



F



Key Terms

bond (12.1)

bond energy (12.1)

ionic bonding (12.1)

ionic compound (12.1)

covalent bonding (12.1)

polar covalent

bond (12.1)

electronegativity (12.2)

dipole moment (12.3)

Lewis structure (12.6)

duet rule (12.6)

octet rule (12.6)

bonding pair (12.6)

lone (unshared)

pair (12.6)

single bond (12.7)

double bond (12.7)



triple bond (12.7)

resonance (12.7)

resonance structure (12.7)

molecular (geometric)

structure (12.8)

bond angle (12.8)

linear structure (12.8)

trigonal planar

structure (12.8)

tetrahedral

structure (12.8)

valence shell electron

pair repulsion (VSEPR)

model (12.9)

tetrahedral

arrangement (12.9)

trigonal pyramid (12.9)



Summary

1. Chemical bonds hold groups of atoms together. They

can be classified into several types. An ionic bond is

formed when a transfer of electrons occurs to form

ions; in a purely covalent bond, electrons are shared

equally between identical atoms. Between these extremes lies the polar covalent bond, in which electrons are shared unequally between atoms with different electronegativities.

2. Electronegativity is defined as the relative ability of

an atom in a molecule to attract the electrons shared

in a bond. The difference in electronegativity values

between the atoms involved in a bond determines

the polarity of that bond.

3. In stable chemical compounds, the atoms tend to

achieve a noble gas electron configuration. In the formation of a binary ionic compound involving representative elements, the valence-electron configuration of the nonmetal is completed: it achieves the

configuration of the next noble gas. The valence or-



VP



directs you to the Chemistry in Focus feature in the chapter

indicates visual problems

interactive versions of these problems are assignable in OWL



bitals of the metal are emptied to give the electron

configuration of the previous noble gas. Two nonmetals share the valence electrons so that both atoms

have completed valence-electron configurations (noble gas configurations).

4. Lewis structures are drawn to represent the arrangement of the valence electrons in a molecule. The rules

for drawing Lewis structures are based on the observation that nonmetal atoms tend to achieve noble

gas electron configurations by sharing electrons. This

leads to a duet rule for hydrogen and to an octet rule

for many other atoms.

5. Some molecules have more than one valid Lewis structure, a property called resonance. Although Lewis

structures in which the atoms have noble gas electron

configurations correctly describe most molecules,

there are some notable exceptions, including O2, NO,

NO2, and the molecules that contain Be and B.

6. The molecular structure of a molecule describes how

the atoms are arranged in space.

7. The molecular structure of a molecule can be predicted by using the valence shell electron pair repulsion (VSEPR) model. This model bases its prediction

on minimizing the repulsions among the electron

pairs around an atom, which means arranging the

electron pairs as far apart as possible.



Active Learning Questions

These questions are designed to be considered by groups of

students in class. Often these questions work well for introducing a particular topic in class.

1. Using only the periodic table, predict the most stable

ion for Na, Mg, Al, S, Cl, K, Ca, and Ga. Arrange these

elements from largest to smallest radius and explain

why the radius varies as it does.



392 Chapter 12 Chemical Bonding

19. How do we deal with multiple bonds in VSEPR theory?

2. Write the proper charges so that an alkali metal, a noble gas, and a halogen have the same electron con20. In Section 12.10 of your text, the term “effective

figurations. What is the number of protons in each?

pairs” is used. What does this mean?

The number of electrons in each? Arrange them from

3+

+

2+

2smallest to largest radii and explain your ordering ra- VP 21. Consider the ions Sc , Cl , K , Ca , and S . Match

these

ions

to

the

following

pictures

that

represent

the

tionale.

relative sizes of the ions.

3. What is meant by a chemical bond?

4. Why do atoms form bonds with one another? What

can make a molecule favored compared with the lone

atoms?

5. How does a bond between Na and Cl differ from a

bond between C and O? What about a bond between VP 22. Write the name of each of the following shapes of

N and N?

molecules.

6. In your own words, what is meant by the term electronegativity? What are the trends across and down the

periodic table for electronegativity? Explain them,

and describe how they are consistent with trends of

ionization energy and atomic radii.

7. Explain the difference between ionic bonding and

covalent bonding. How can we use the periodic table

to help us determine the type of bonding between

atoms?

8. True or false? In general, a larger atom has a smaller

electronegativity. Explain.

9. Why is there an octet rule (and what does octet mean)

in writing Lewis structures?

10. Does a Lewis structure tell which electrons came from

which atoms? Explain.

11. If lithium and fluorine react, which has more attraction for an electron? Why?

12. In a bond between fluorine and iodine, which has

more attraction for an electron? Why?

13. We use differences in electronegativity to account for

certain properties of bond.

What if all atoms had the same electronegativity values? How would bonding between atoms be affected?

What are some differences we would notice?

14. Explain how you can use the periodic table to predict

the formula of compounds.

15. Why do we only consider the valence electrons in

drawing Lewis structures?



a



b



c



Questions and Problems

12.1 Types of Chemical Bonds

QUESTIONS

1. In general terms, what is a chemical bond?

2. What does the bond energy of a chemical bond represent?

3. A What sorts of elements react to form ionic compounds?

4. In general terms, what is a covalent bond?

5. Describe the type of bonding that exists in the Cl2(g)

molecule. How does this type of bonding differ from

that found in the HCl(g) molecule? How is it similar?

6. Compare and contrast the bonding found in the

H2(g) and HF( g) molecules with that found in NaF(s).



12.2 Electronegativity

QUESTIONS



16. How do we determine the total number of valence

electrons for an ion? Provide an example of an anion

and a cation, and explain your answer.



7. The relative ability of an atom in a molecule to attract

electrons to itself is called the atom’s

.



17. What is the main idea in the valence shell electron

pair repulsion (VSEPR) theory?



8. What does it mean to say that a bond is polar? Give

two examples of molecules with polar bonds. Indicate

in your examples the direction of the polarity.



18. The molecules NH3 and BF3 have the same general

formula (AB3) but different shapes.

a. Find the shape of each of the above molecules.

b. Provide more examples of real molecules that have

the same general formulas but different shapes.



9. A bond between atoms having a (small/large) difference in electronegativity will be ionic.

10. What factor determines the relative level of polarity

of a polar covalent bond?



All even-numbered Questions and Problems have answers in the back of this book and solutions in the Solutions Guide.



Chapter Review

PROBLEMS

11. In each of the following groups, which element is

the most electronegative? Which is the least electronegative?



12. In each of the following groups, which element is

the most electronegative? Which is the least electronegative?

a. Rb, Sr, I

b. Ca, Mg, Sr

c. Br, Ca, K



a. OOO

b. AlOO

c. BOO



c. LiOCl or CsOCl

d. MgON or MgOP



a. NaOCl or CaOCl

b. CsOCl or BaOCl



c. FeOI or FeOF

d. BeOF or BaOF



12.3 Bond Polarity and Dipole Moments

21. What is a dipole moment? Give four examples of molecules that possess dipole moments, and draw the direction of the dipole as shown in Section 12.3.

22. Why is the presence of a dipole moment in the water

molecule so important? What are some properties of

water that are determined by its polarity?

PROBLEMS



14. On the basis of the electronegativity values given in

Figure 12.3, indicate whether each of the following

bonds would be expected to be covalent, polar covalent, or ionic.

a. KOCl

b. BrOCl

c. ClOCl



water, H2O

carbon monoxide, CO

fluorine, F2

nitrogen, N2



16. Which of the following molecules contain polar

covalent bonds?

sulfur, S8

fluorine, F2

iodine monochloride, ICl

hydrogen bromide, HBr



17. On the basis of the electronegativity values given in

Figure 12.3, indicate which is the more polar bond in

each of the following pairs.

HOF or HOCl

HOCl or HOI

HOBr or HOCl

HOI or HOBr



a. hydrogen chloride, HCl

b. carbon monoxide, CO

c. bromine monofluoride, BrF



a. hydrogen fluoride, HF

b. chlorine monofluoride, ClF

c. iodine monochloride, ICl

25. For each of the following bonds, draw a figure indicating the direction of the bond dipole, including which

end of the bond is positive and which is negative.

a. COF

b. SiOC



c. POS or POO

d. HOO or HON



c. COO

d. BOC



26. For each of the following bonds, draw a figure indicating the direction of the bond dipole, including which

end of the bond is positive and which is negative.

a. SOO

b. SON



c. SOF

d. SOCl



27. For each of the following bonds, draw a figure indicating the direction of the bond dipole, including which

end of the bond is positive and which is negative.

a. SiOH

b. POH



18. On the basis of the electronegativity values given in

Figure 12.3, indicate which is the more polar bond in

each of the following pairs.

a. OOCl or OOBr

b. NOO or NOF



23. In each of the following diatomic molecules, which

end of the molecule is negative relative to the other

end?



24. In each of the following diatomic molecules, which

end of the molecule is positive relative to the other

end?



15. Which of the following molecules contain polar

covalent bonds?



a.

b.

c.

d.



a. NaOF or NaOI

b. CaOS or CaOO



QUESTIONS



13. On the basis of the electronegativity values given in

Figure 12.3, indicate whether each of the following

bonds would be expected to be ionic, covalent, or polar covalent.



a.

b.

c.

d.



19. Which bond in each of the following pairs has the

greater ionic character?



20. Which bond in each of the following pairs has less

ionic character?



a. K, Na, H

b. F, Br, Na

c. B, N, F



a.

b.

c.

d.



393



c. SOH

d. ClOH



28. For each of the following bonds, draw a figure indicating the direction of the bond dipole, including which

end of the bond is positive and which is negative.

a. HOC

b. NOO



c. NOS

d. NOC



All even-numbered Questions and Problems have answers in the back of this book and solutions in the Solutions Guide.



394 Chapter 12 Chemical Bonding

12.4 Stable Electron Configurations

and Charges on Ions

QUESTIONS

29. What does it mean when we say that in forming

bonds, atoms try to achieve an electron configuration analogous to a noble gas?

30. The metallic elements lose electrons when reacting,

and the resulting positive ions have an electron configuration analogous to the

noble gas

element.

31. Nonmetals form negative ions by (losing/gaining)

enough electrons to achieve the electron configuration of the next noble gas.

32. Explain how the atoms in covalent molecules achieve

electron configurations similar to those of the noble

gases. How does this differ from the situation in ionic

compounds?

PROBLEMS

33. Which simple ion would each of the following elements be expected to form? What noble gas has an

analogous electron configuration to each of the ions?

a.

b.

c.

d.



chlorine, Z ϭ 17

strontium, Z ϭ 38

oxygen, Z ϭ 8

rubidium, Z ϭ 37



34. Which simple ion would each of the following elements be expected to form? Which noble gas has an

analogous electron configuration to each of the ions?

a.

b.

c.

d.



bromine, Z ϭ 35

cesium, Z ϭ 55

phosphorus, Z ϭ 15

sulfur, Z ϭ 16



35. For each of the following numbers of electrons, give

the formula of a positive ion that would have that

number of electrons, and write the complete electron

configuration for each ion.

a. 10 electrons

b. 2 electrons



c. 18 electrons

d. 36 electrons



36. Give the formula of a negative ion that would have

the same number of electrons as each of the following positive ions.

a. Naϩ

b. Ca2ϩ



c. Al3ϩ

d. Rbϩ



37. On the basis of their electron configurations, predict

the formula of the simple binary ionic compounds

likely to form when the following pairs of elements

react with each other.

a.

b.

c.

d.

e.



aluminum, Al, and sulfur, S

radium, Ra, and oxygen, O

calcium, Ca, and fluorine, F

cesium, Cs, and nitrogen, N

rubidium, Rb, and phosphorus, P



38. On the basis of their electron configurations, predict

the formula of the simple binary ionic compound

likely to form when the following pairs of elements

react with each other.

a.

b.

c.

d.



aluminum and bromine

aluminum and oxygen

aluminum and phosphorus

aluminum and hydrogen



39. Name the noble gas atom that has the same electron

configuration as each of the ions in the following

compounds.

a.

b.

c.

d.



barium sulfide, BaS

strontium fluoride, SrF2

magnesium oxide, MgO

aluminum sulfide, Al2S3



40. Atoms form ions so as to achieve electron configurations similar to those of the noble gases. For the following pairs of noble gas configurations, give the formulas of two simple ionic compounds that would

have comparable electron configurations.

a. [He] and [Ne]

b. [Ne] and [Ne]



c. [He] and [Ar]

d. [Ne] and [Ar]



12.5 Ionic Bonding and Structures

of Ionic Compounds

QUESTIONS

41. Is the formula we write for an ionic compound the

molecular formula or the empirical formula? Why?

42. Describe in general terms the structure of ionic solids

such as NaCl. How are the ions packed in the crystal?

43. Why are cations always smaller than the atoms from

which they are formed?

44. Why are anions always larger than the atoms from

which they are formed?

PROBLEMS

45. For each of the following pairs, indicate which

species is smaller. Explain your reasoning in terms of

the electron structure of each species.

a. H or HϪ

b. N or N3Ϫ



c. Al or Al3ϩ

d. F or Cl



46. For each of the following pairs, indicate which

species is larger. Explain your reasoning in terms of

the electron structure of each species.

a. Liϩ or FϪ

b. Naϩ or ClϪ



c. Ca2ϩ or Ca

d. Csϩ or IϪ



47. For each of the following pairs, indicate which is

smaller.

a. Fe or Fe3ϩ



b. Cl or ClϪ



c. Al3ϩ or Naϩ



48. For each of the following pairs, indicate which is larger.

a. I or F



b. F or FϪ



c. Naϩ or FϪ



All even-numbered Questions and Problems have answers in the back of this book and solutions in the Solutions Guide.



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