Chapter 12:  Crystal Types and Intermolecular Forces

 

 

Section 12-1: Crystal Types

Section 12-2: Intermolecular Forces

Section 12-3: "Like Dissolves Like" and Solubility

Chapter 12 Practice Exercises and Review Quizzes

 

 

 

 

 

Section 12-1:  Crystal Types

 

Many different substances in the solid state can be categorized into one of the following crystal types:

 

I. Network Covalent

 

In a network covalent crystal, the atoms are held together in a continuous three-dimensional array of covalent bonds.  For example, the carbon atoms in the network covalent C(diamond) are each bonded tetrahedrally to four neighboring carbon atoms as follows:

 

 

Other network covalent substances include SiC and SiO2.  Network covalent crystals generally have the following properties:

 

1.  High melting point.

2.  Physically hard. 

3.  Poor conductor of electricity.  For a substance to conduct electricity, charged particles must have the ability to move freely throughout the sample.  In the case of C(diamond), SiC, and SiO2, the negatively-charge electrons in the covalent bonds are localized between two particular atoms and, therefore, do not have the ability to move freely throughout the crystal.  As such, C(diamond), SiC, and SiO2 are poor conductors of electricity.      

 

C(graphite) is also considered network covalent because the carbon atoms are covalently bonded in a continuous trigonal planar pattern to form layers:

 

 

In contrast to the physically hard C(diamond), SiC, and SiO2, layers in a sample of C(graphite) will more readily flake off.  Each carbon atom in C(graphite) will use 3 of its 4 available valence electrons to form covalent bonds with 3 neighboring carbon atoms.  However, all the extra valence electrons will be delocalized between the layers of carbon and, thus, able to move freely throughout the crystal.  As such, C(graphite) will be a good conductor of electricity.

 

II. Ionic

 

Ionic crystals contain a continuous three-dimensional array of positively-charged metal cations or polyatomic cations such as ammonium (NH4+) and negatively-charged nonmetal anions or polyatomic anions.  Solid ionic crystals generally have the following properties:

 

1. High melting point.

2. Physically brittle.

3. Poor conductor of electricity because the charged cations and anions are held firmly in place and not able to move freely throughout the crystal (cations and anions are immobile).

 

Despite still being in close contact, the charged cations and anions in a liquid (molten) ionic compound are able to move freely throughout the sample (cations and anions are mobile).  Therefore, a liquid ionic compound is a good conductor of electricity.  In an aqueous ionic compound, the individual cations and anions are surrounded by water molecules and completely separated from each other.  Because these charged ions are able to move freely throughout the solution, an aqueous ionic compound is also a good conductor of electricity.

 

One aspect of Coulomb's Law essentially suggests that the attraction between two oppositely-charged particles will be greater when the magnitudes of the particles' charges are greater.  Greater Coulombic attraction between the cations and anions in an ionic solid will result in a higher melting point.  When comparing the melting points of ionic solids, a compound with a larger sum of one cation's charge magnitude + one anion's charge magnitude will often have the higher melting point (although ionic radii can also significantly affect melting point), as demonstrated in the following problem:    

 

Sample Exercise 12A:

 

Rank the compounds CsBr, MgO, and SrCl2 from lowest to highest melting point and justify your answer.

 

Solution:

 

First, we determine the sum of one cation's charge magnitude + one anion's charge magnitude for each compound:

 

CsBr = Cs+ and Br-, sum of charge magnitudes = 1 + 1 = 2

MgO = Mg2+ and O2-, sum of charge magnitudes = 2 + 2 = 4

SrCl2 = Sr2+ and Cl-, sum of charge magnitudes = 2 + 1 = 3

 

CsBr has the lowest sum and, therefore, the lowest melting point.  MgO has the highest sum and, therefore, the highest melting point.  As such, the ranking from lowest to highest melting point is CsBr < SrCl2 < MgO.

 

III. Metallic

 

Metallic crystals are often described as essentially being a three-dimensional array of metal cations in a sea of delocalized electrons.  Metallic crystals generally have the following properties:

 

1. Broad range of melting points, but all metals except mercury (Hg) are solids at room temperature.

2. Physically shiny, malleable, and ductile.

3. Good conductor of electricity because some electrons from each atom are delocalized and can move freely throughout the crystal. 

 

Liquid metals also have delocalized electrons and, therefore, are good conductors of electricity.

 

IV. Molecular

 

Many substances that are not network covalent, ionic, or metallic form molecular crystals in the solid state.  Molecular crystals generally have the following properties:

 

1. Low melting points because the intermolecular forces between molecules are relatively weak.  Generally lower melting points than network covalent, ionic, or metallic crystals.

2. Physically not hard.

3. Poor conductor of electricity because electrons within the molecules are localized and are not able to move freely throughout the crystal.

 

Molecular substances in the liquid state are poor conductors of electricity because electrons within the molecules are localized and are not able to move freely throughout the sample.  Molecular substances in the liquid state also generally have lower boiling points than ionic or metallic substances.

 

Sample Exercise 12B:

 

Categorize each substance below as being network covalent, ionic, metallic, or molecular:

 

(a) (NH4)2CO3

(b) Ni

(c) SiC

(d) (NH2)2CO

 

Solution:

 

(a) NH4+ and CO32- = ionic

(b) nickel = metallic

(c) SiC = network covalent

(d) molecular, Lewis structure:

 

 

Sample Exercise 12C:

 

Which of the following has the lowest melting point?

 

(a) CCl4               (b) Cu               (c) NaBr               (d) SiO2

 

Solution:

 

CCl4 = molecular = low melting point

Cu = metallic = generally higher melting point than molecular

NaBr = Na+ and Br- = ionic = high melting point

SiO2 = network covalent = high melting point

 

Therefore, (a) CCl4 has the lowest melting point.

 

Sample Exercise 12D:

 

State whether each of the following is a good or poor conductor of electricity in the solid state:

 

(a) BaCl2

(b) C(graphite)

(c) Kr

(d) Mn

 

Solution:

 

(a) Ba2+ and Cl- = ionic = poor conductor in solid state because cations and anions are immobile (but good conductor in liquid or aqueous state because cations and anions are mobile)

(b) network covalent but, unlike other network covalent crystals, has delocalized electrons between the planar layers = good conductor in solid state

(c) nonmetal = molecular = poor conductor in solid state because electrons are localized (and also poor conductor in liquid state because electrons are localized)

(d) manganese = metallic = good conductor in solid state because electrons are delocalized (and also good conductor in liquid state because electrons are delocalized)

 

 

 

 

Section 12-2:  Intermolecular Forces

 

Molecular substances with stronger intermolecular forces (IMFs) will have higher boiling points because the molecules will be more strongly held together.  The three different types of IMFs are described below:

 

I. Hydrogen Bonding

 

Hydrogen bonding occurs only between molecules containing hydrogen atoms that are bonded DIRECTLY to nitrogen, oxygen, or fluorine atoms.  For example, HF is capable of hydrogen bonding, whereas difluoromethane, CH2F2, is not capable of hydrogen bonding because the H atoms are not bonded directly to the F atoms, as shown in the following Lewis structure:

 

 

Sample Exercise 12E:

 

Which of the following is not capable of hydrogen bonding?

 

(a) C2H5OH               (b) CH3CN               (c) H2O2               (d) N2H4

 

Solution:

(b) CH3CN (H not bonded directly to N)

 

 

In a sample of HF, the covalent bond in each molecule is highly polar because of the large electronegativity difference between H (EN = 2.1) and F (EN = 4.0), leading to a partial negative charge (δ-) on the F end of the molecule and a partial positive charge (δ+) on the H end of the molecule.  The strong attraction between the F end of one molecule and the H end of a neighboring molecule is known as a hydrogen bond and is represented by the a dashed line below:

 

 

Molecular substances that are capable of hydrogen bonding generally have higher boiling points than molecular substances that are not capable of hydrogen bonding.

 

For the majority of substances, the density of the solid is higher than the density of the liquid because the molecules in the solid are packed closely together and then move slightly farther apart in the liquid.  However, the unique combination of small size, bent shape, and ability to hydrogen bond for H2O molecules allows ice to obtain a crystal structure wherein the water molecules are separated by a significant distance by hydrogen bonds, as shown below (red  = oxygen atoms, white = hydrogen atoms):

 

 

The significant empty space between molecules in ice results in a lower density for ice than liquid water when the two are compared at the freezing point of 0°C. 

 

II. London (Dispersion) Forces

 

Whereas only a limited number of molecular substances are capable of hydrogen bonding, all molecular substances are held together by London (dispersion) forces.  When more electrons happen to be on one side of a molecule than the other at a particular moment in time, an instantaneous dipole is created with a partial negative charge on the side with more electrons and a partial positive charge on the side with less electrons.  If a second molecule comes in close proximity to the negative end of the instantaneous dipole, the electron cloud in the second molecule will be repelled away from the negative end of the instantaneous dipole.  This creates an induced dipole in the second molecule with a partial positive charge closest to the first molecule and a partial negative charge furthest from the first molecule.  A London force is the attraction between the instantaneous dipole and the induced dipole and is represented by the double-headed arrow below:

 

 

Polarizability is essentially the ease with which the electron cloud can be shifted in a molecule to create the dipoles necessary for a London force.  Molecules with significantly more total electrons will generally be more polarizable and, therefore, have stronger London forces.  Molecular substances with stronger London forces will generally have higher boiling points.  Rather than counting total electrons to compare London forces in different molecular substances, we can usually obtain the same results by instead comparing molar masses.  Therefore, molecular substances with significantly larger molar masses will generally have stronger London forces and higher boiling points.

 

Sample Exercise 12F:

 

Explain why Br2 is a liquid at room temperature, whereas Cl2 is a gas.

 

Solution:

 

Both Br2 and Cl2 are molecular substances held together only by London forces.  Since Br2 has a significantly higher total number of electrons (2 x 35 = 70) than Cl2 (2 x 17 = 34), Br2 has London forces strong enough to hold the molecules together as a liquid, whereas the London forces in Cl2 are not as strong, so the Cl2 molecules separate into a gas.

 

III. Dipole-Dipole Forces

 

Molecular substances containing polar molecules will have dipole-dipole forces where the partial negative end of one polar molecule will be attracted to the partial positive end of another polar molecule.  This is similar to hydrogen bonding, but weaker in strength.  Two molecular substances that are not capable of hydrogen bonding and that have roughly equal London forces (as suggested by their similar total number of electrons) can have significantly different boiling points if one is polar and, therefore, is also held together by dipole-dipole forces while the other is nonpolar and, therefore, does not have the extra attraction due to dipole-dipole forces.

 

Sample Exercise 12G:

 

Which will have the higher boiling point, CO or N2?

Solution:

 

Both CO and N2 are molecular substances with the same total number of electrons (CO = 6 + 8 = 14, N2 = 2 x 7 = 14).  Therefore, CO and N2 are expected to have roughly equal London forces.  However, CO is polar and, therefore, has dipole-dipole forces to raise the boiling point above that of the nonpolar N2, which lacks the extra attraction of dipole-dipole forces.

 

 

 

 

 

Section 12-3:  "Like Dissolves Like" and Solubility

 

A common rule of thumb used to predict whether or not a solute will dissolve in a solvent is "like dissolves like":

 

a. Nonpolar solutes tend to dissolve in nonpolar solvents.  Note that the electronegativity difference between hydrogen and carbon is so small that we will consider bonds between hydrogen and carbon to be nonpolar.  Therefore, we will consider all compounds containing only hydrogen and carbon (hydrocarbons) to be nonpolar.  Oils, fats, and gasoline have in common significant hydrocarbon portions that make them effectively nonpolar.  As such, oils, fats, and gasoline will not dissolve in or mix with water, which is polar, to any significant extent.

 

b. Polar solutes tend to dissolve in polar solvents.

 

c. Solutes that are capable of hydrogen bonding tend to dissolve in solvents that are capable of hydrogen bonding. 

 

d. Ionic solutes tend to dissolve in polar solvents.  Although significant energy is required to separate the cations and anions in the solute during the dissolving process, significant energy is regained via ion-dipole interactions wherein the partial negative ends of the polar solvent molecules are attracted to the cations and the partial positive ends of the polar solvent molecules are attracted to the anions:

 

 

Sample Exercise 12H:

 

Predict whether each solute below will dissolve to a greater extent in carbon disulfide or water:

 

(a) HOCl

(b) I2

(c) KBr

(d) NH2OH

 

Solution:

 

H2O is polar and is capable of hydrogen bonding.  The Lewis structure of CS2 shows no lone pair on carbon, which indicates a linear and nonpolar molecule:

 

 

(a) The polar HOCl will dissolve to a greater extent in the polar water.

 

(b) The nonpolar I2 will dissolve to a greater extent in the nonpolar carbon disulfide.

 

(c) KBr = K+ and Br- = ionic, so will dissolve to a greater extent in the polar water (due to ion-dipole attraction).

 

(d) NH2OH is capable of hydrogen bonding, so will dissolve to a greater extent in water, which can hydrogen bond as well.

 

 

 

Chapter 12 Practice Exercises and Review Quizzes:

 

12-1) Rank the compounds CaO, K2S, RbI, and SCl2 from lowest to highest melting point and explain.

Click for Solution

 

12-1) SCl2 = molecular = lowest melting point

 

Other three compounds are ionic, so larger sum of one cation's charge magnitude + one anion's charge magnitude = higher melting point:

 

CaO = Ca2+ and O2-, sum of charge magnitudes = 2 + 2 = 4

K2S = K+ and S2-, sum of charge magnitudes = 1 + 2 = 3

RbI = Rb+ and I-, sum of charge magnitudes = 1 + 1 = 2

 

Therefore, the order of melting points = SCl2 < RbI < K2S < CaO.

 

 

 

12-2) State whether each of the following is a good or poor conductor of electricity in the solid state and explain:

 

(a) C(diamond)

(b) K

(c) MgBr2

(d) N2

Click for Solution

 

12-2)

(a) network covalent with localized electrons = poor conductor in solid state

(b) potassium = metallic = good conductor in solid state because electrons are delocalized (and also good conductor in liquid state because electrons are delocalized)

(c) Mg2+ and Br- = ionic = poor conductor in solid state because cations and anions are immobile (but good conductor in liquid state or aqueous state because cations and anions are mobile) 

(d) nonmetal = molecular = poor conductor in solid state because electrons are localized (and also poor conductor in liquid state because electrons are localized)

 

 

 

12-3) Rank each of the following groups from highest to lowest boiling point and explain:

 

(a) CO2, CH3OH, CH3OCH3, LiF

(b) Al, Br2, O2, ICl

Click for Solution

 

12-3)

(a) LiF = Li+ and F- = ionic = higher boiling point than the other three, which are all molecular substances.

 

CH3OH will have the second highest boiling point because it is capable of hydrogen bonding.  Note that CH3OCH3 is not capable of hydrogen bonding because H is not bonded directly to O.

 

The total number of electrons in CH3OCH3 (2 x 6 + 6 x 1 + 8 = 26) and CO2 (6 + 2 x 8 = 22) are similar, so these two are expected to have roughly equal London forces.  However, CH3OCH3 is polar and, therefore, is also held together by dipole-dipole forces that give it the third highest boiling point.  We have determined that CH3OCH3 is polar from the Lewis structure, which shows lone pairs on the oxygen atom that lead to a bent shape in that region:

 

 

On the other hand, CO2 is nonpolar and, therefore, lacks the extra dipole-dipole attraction, so CO2 will have the lowest boiling point in the group.  We have determined that CO2 is nonpolar from the Lewis structure, which shows no lone pairs on center carbon atom, indicating a linear shape for the molecule:

 

 

 

Therefore, the order of boiling points = LiF > CH3OH > CH3OCH3 > CO2.

 

(b) Al = metallic = higher boiling point than the other three, which are all molecular substances.

 

The total number of electrons in Br2 (2 x 35 = 70) and ICl (53 + 17 = 70) are the same, so these two are expected to have roughly equal London forces.  However, ICl is polar and, therefore, has dipole-dipole forces to raise the boiling point higher than that of the nonpolar Br2, which lacks the extra dipole-dipole attractions.  The nonpolar O2 has significantly fewer total electrons (2 x 8 = 16) than Br2, so the London forces in O2 will be weaker than those in Br2 and, therefore, O2 will have the lowest boiling point in the group.

 

Therefore, the order of boiling points = Al > ICl > Br2 > O2.   

 

 

 

12-4) Predict whether each solute below will dissolve to a greater extent in water or benzene, C6H6, and explain:

 

(a) C10H8

(b) H2CO

(c) NH3

(d) SrI2

Click for Solution

 

12-4) Water is polar and is capable of hydrogen bonding. Benzene is a hydrocarbon and, therefore, is considered nonpolar. 

 

(a) The nonpolar hydrocarbon C10H8 will dissolve to a greater extent in the nonpolar benzene.

 

(b) The polar H2CO will dissolve to a greater extent in the polar water.

 

 (c) NH3 is capable of hydrogen bonding, so will dissolve to a greater extent in water, which can hydrogen bond as well.

 

(d) SrI2 = Sr2+ and I- = ionic, so will dissolve to a greater extent in the polar water (due to ion-dipole attraction).

 

 

 

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