ChemTours

Chemistry involves the study of both the physical and chemical properties of substances, seeking to explain observed behavior in molecular terms.

H₂O
molecular
melts at 0°C

CO₂
molecular
sublimes at −78.5°C

0°C

−78.5°C

For example, why is H₂O a liquid at room temperature, while CO₂ is a gas, although both substances are molecular compounds?

We find that physical properties such as melting point and solubility are determined by the type and strength of the interactions that form and break during phase changes and dissolution.

What changes when a molecular substance changes phase? When we examine H₂O in its different phases, we see that individual H₂O molecules are present in all three physical forms.

We conclude that phase changes for this type of substance do NOT typically involve the breaking or formation of covalent bonds.

H₂O (g)

H₂O (s)

H₂O (l)

Covalent bonds do NOT break during a phase change.

In the solid and liquid phases of a molecular substance, individual molecules are held together by attractions called intermolecular forces, also called nonbonding interactions.

H₂O (s)

covalent bonds

intermolecular attractions

Frequently, these attractive forces are many orders of magnitude weaker than the covalent bonds that hold the molecule together internally.

125 - 1200 kJ/mol

0.05 - 40 kJ/mol

When the substance changes phase from solid to liquid to gas, it is these weaker attractions that are weakened or broken. Individual molecules remain intact because the covalent bonds are unaffected.

H₂O (l)

intermolecular attractions overcome by increased movement of particles

When a molecular substance such as H2O changes phase from solid to liquid to gas, it is the

that are weakened and/or broken.

remain intact

during the change. Thus, the melting and boiling point of a such a substance depends on its

The stronger these attractions, the

the melting and boiling

points of the substance will be.

There are three main types of intermolecular attractions that may occur between the individual molecules of a pure molecular substance.

These are listed here in order from weakest to strongest, although note that each type of attraction has a range of strengths. All of these types of interactions are typically many orders of magnitude weaker than either ionic bonds or covalent bonds.

Hydrogen Bonding

Dipole-Dipole Interactions

Dispersion Forces

strongest

weakest

stronger for larger, more polarizable particles

stronger for more polar molecules

10 - 40 kJ/mol

5 - 25 kJ/mol

0.05 - 40 kJ/mol

Dispersion forces, also called London forces, are experienced by all atoms, ions, and molecules. These attractive forces are the result of the distortion of the electron clouds of particles when they move close together.

nucleus

electron cloud

As the particles move together, their electron clouds will become temporarily distorted so that a region of high electron density in one particle will be adjacent to a region of low electron density on the other.

We say that a temporary dipole is induced in one particle by the other.

As the particles move away from each other, the distortion disappears.

The temporary electron-rich region of one particle, marked δ−, attracts the temporarily electron-poor or δ+ region of the other. This attraction is relatively weak because it is fleeting.

δ+

δ−

Although the weakest dispersion forces are weaker than any other type of intermolecular attraction, there will be many such interactions between individual atoms or molecules in a substance.

two atoms near each other showing the dispersion forces as dotted lines

Larger molecules are more polarizable, meaning that their electron clouds distort more easily. They therefore experience stronger dispersion forces as well as more such interactions between individual molecules.

δ+

δ−

Question 1:

A) Rank the following substances in order of increasing strength of the dispersion forces their atoms or molecules experience in the gas phase.

(Hint: What parameter could you use to approximate a size for the constituent atoms or molecules of each substance?)

weakest
dispersion forces

strongest
dispersion forces

CH₄

CO₂

Br₂

B) Which of these substances do you expect to have the highest boiling point?

A) CH₄

B) CO₂

C) He

D) Br₂

Dipole-dipole interactions are attractive forces that exist between polar molecules, such as HCl. The electron-rich δ− region of one molecule attracts the electron-poor δ+ region of another.

The greater the overall dipole moment (and therefore polarity) of the individual molecules, the stronger these attractions will be.

Polar molecules also experience dispersion forces, which enhance their overall polarity and increase the attraction between them

For molecules of similar size and shape, molecules that experience dipole-dipole attractions in addition to dispersion forces will have higher melting and boiling points.

image of C 4 H 10 and its electron cloud

image of C 3 H 6 O and its electron cloud

	methylpropane C₄H₁₀	acetone C₃H₆O
molar mass	58.12 g/mol	58.08 g/mol
dipole moment (μ)	0.132 D (essentially nonpolar)	2.88 D
intermolecular forces experienced	dispersion forces	dispersion forces dipole-dipole interactions
boiling point	−12°C (261 K)	56°C (329 K)

Question 2:

Which of the following compounds experience dipole-dipole attractions between their molecules?

A) CCl₄

B) Cl₂O (O is central)

C) CO₂

D) PCl₃

E) Cl₂

A special interaction called a hydrogen bond is possible for molecules that contain a H atom covalently bonded to N, O, or F.

3 images of lewis dot diagrams for H F, H O, and H N

These partner atoms are small and highly electronegative, causing the covalent bond to H to be very strongly polarized. The H atom is called a hydrogen-bond donor atom, or H-bond donor atom.

Diagram with H bond X, where is can be N, O, or F. With an arrow labeling the X as small, highly electronegative

electron cloud with delta plus on the left and delta minus on the right, labeled strongly polarized

This extremely electron-poor H is strongly attracted to a hydrogen-bond acceptor atom on an adjacent molecule.

Two H bond X structures shown connected with a dotted line between one X and one H. The X is labeled h bond acceptor, the H as H bond donor. The h atoms are laled as delta plus, the x's as delta minus.

The H-bond acceptor atom must be N, O, or F with an available lone pair. The hydrogen bond itself is the strong interaction between the δ+ H atom and the lone pair of the strongly δ− acceptor atom.

Unlike other types of intermolecular attractions, hydrogen bonds are directional and require the molecules to be oriented in a specific spatial relationship.

This is what gives rise to the highly ordered cyrstal structure in solid H₂O, or ice.

It is very important to remember that although a hydrogen bond has the word "bond" in its name, it is still a nonbonding interaction and is many orders of magnitude weaker than a covalent or ionic bond.

intermolecular attractions

covalent bonds

125 - 1200 kJ/mol

0.05 - 40 kJ/mol

H₂O (s)

Hydrogen bonding plays a very important role in determining the three-dimensional shapes of synthetic polymers as well as biological molecules. The famous double-helix structure of DNA is held together by hydrogen bonding. This type of interaction also allows the base-pair recognition that is so important for preserving and transmitting the genetic code during RNA transcription and translation to proteins.

Question 3:

Which of the following compounds can form hydrogen bond between their molecules?

(Hint: Draw a molecular skeleton or a Lewis structure for each molecule before trying to answer the question.)

A) CH₃F

B) CH₃OH

C) CH₃OCH₃

D) CH₃NH₂

Question 4:

Rank the following substances in order of increasing strength of the nonbonding interactions between constituent particles. (The term "nonbonding" includes all attractive forces that are not covalent bonds.)

weakest
nonbonding interactions

strongest
nonbonding interactions

C₂H₆

NF₃

CH₃NH₂

Question 5:

Identify ALL the types of intermolecular forces experienced by each of the following substances. Drag the relevant types of intermolecular forces to the boxes under each formula. If a box is not needed, leave it blank.

Br₂

NH₃

CH₂F₂

Select by dragging and dropping under the relevant substance.

hydrogen bonding

dipole-dipole attractions

dispersion forces

hydrogen bonding

dipole-dipole attractions

dispersion forces

hydrogen bonding

dipole-dipole attractions

dispersion forces

We often express the rule of thumb "like dissolves in like" as a rough guide to determining solubility. This "rule" expresses the general observation that substances will dissolve in solvents that experience similar types of noncovalent interactions.

Question 6:

Using this general principle of solubility, drag each solute to the solvent in which it is most likely to be soluble. (If a box is not needed, leave it blank.)

CH₃OH

C₆H₁₄

I₂
(iodine)

HOCH₂CH₂OH
(ethylene glycol)

C₁₀H₈
(naphthalene)

C₆H₁₂O₆
(glucose)

Congratulations! You have completed the ChemTour on noncovalent interactions. These types of interactions determine macroscopic properties such as melting point, boiling point, and solubility. You are now ready to examine the formation and behavior of solutions in more detail. You are also ready to understand the interactions responsible for determining the structure and function of important biological molecules like proteins and DNA.

The question of solubility is a complex one and involves more than just the ability of solute and solvent to interact in similar ways. Just as particles can induce dipoles in each other, polar molecules such as H₂O can induce dipoles
in nonpolar molecules such as O₂ and N₂. The ability of H₂O to dissolve these small, nonpolar molecules is very important for sustaining life.

Dispersion forces are stronger for larger, more polarizable molecules or atoms. Molar mass can give a first approximation of size, particularly for simple atoms and molecules like these. Br₂ has the largest molar mass (160 g/mol). With two Br atoms covalently bonded together, we can expect this to be the largest molecule in this series. It will have the strongest dispersion forces.

He is an atomic element, composed only of small He atoms. This substance is expected to exhibit the weakest dispersion forces.

CO₂ has a larger molar mass (44.01 g/mol) than CH₄ (16.04 g/mol). It also makes sense that one C bonded to two O atoms in a linear structure would give a larger molecule than one C atom bonded to four H atoms, since H is so much smaller than O. In addition, the two C=O double bonds in CO₂ make it more polarizable.

CH₄

CO₂

Br₂

Br₂ molecules are larger than the atoms or molecules of any of the other substances listed here. It will experience the strongest dispersion forces between its molecules and will therefore have the highest melting and boiling point. In fact, Br₂ is the only one of these substances that is a liquid at room temperature—all the other substances are gases. This behavior is directly related to the strength of the dispersion forces between their particles.

A) CCl₄ has a tetrahedral geometry. Altough all the C–Cl bonds are polar, the bond dipoles cancel due to the shape, and the molecule is nonpolar overall.

bond dipoles cancel

Lewis dot diagram and a geometry image for C C L 4

B) Cl₂O has a bent geometry. The two bond dipoles do not cancel, making this molecule polar overall. This substance therefore experiences dipole-dipole attractions between its molecules.

bond dipoles do not cancel

Lewis dot diagram and a geometry image for C L 2 O

C) CO₂ has a linear geometry and its bond dipoles cancel. As it is a nonpolar molecule, it cannot experience dipole-dipole interactions.

bond dipoles cancel

Lewis dot diagram and a geometry image for C O 2

D) PCl₃ has a trigonal pyramidal geometry. The P–Cl bonds are all polar, and bond dipoles do not cancel due to the presence of the lone pair. The molecule is polar overall, and this substance experiences dipole-dipole interactions between its molecules.

bond dipoles do not cancel

Lewis dot diagram and a geometry image for P C L 3

E) Cl₂ is a nonpolar molecule because it is composed of two identical atoms covalently bonded together.
It will not experience dipole-dipole interactions between its molecules.

Start by drawing a Lewis structure for each substance to determine if it contains both H-bond donor atoms (i.e., H covalently bonded to N, O, or F) and H-bond acceptor atoms. If not, the compound cannot form H-bonds between its own molecules.

A) CH₃F contains no H-bond donor
atoms and therefore cannot form H-bonds between its molecules. However, the F is an H-bond acceptor, so a molecule of CH₃F could accept a hydrogen bond
from an H-bond donor in a molecule such as H₂O. This increases its solubility in
H₂O and other solvents that donate hydrogen bonds.

B) CH₃OH contains an H covalently bonded to O. This H atom acts as an H-bond donor, while the O atom in an adjacent molecule can accept the H-bond. CH₃OH therefore experiences hydrogen bonding between its molecules.

2 different lewis dot diagrams for C H 3 O h, one showing the h bond donor, the other a hydrogen bond

C) CH₃OCH₃ has no H atoms covalently bonded to N, O, or F. This substance cannot form hydrogen bonds between its molecules. However, the O is an H-bond acceptor, so a molecule of CH₃OCH₃could accept a hydrogen bond from an H-bond donor in a molecule such as H₂O. This increases its solubility in H₂O and other solvents that donate hydrogen bonds.

D) CH₃NH₂ has two H atoms covalently bonded to N and therefore has two H-bond donor atoms per molecule. Either of these H atoms can form a hydrogen bond to the lone pair on a nitrogen atom of a neighboring molecule.

Lewis dot diagram for C H 3 N H 2 and labeling the h bond donors as the h's bonded to the N atom

C₂H₆

NF₃

CH₃NH₂

When ranking substances according to the strength of their noncovalent interactions, you should first identify the type of substance. For molecular substances, the types of intermolecular attractions experienced depend on polarity and the ability to form hydrogen bonds.

C₂H₆, CH₃NH₂, and NF₃ are molecular compounds. CH₃NH₂ is polar and can form hydrogen bonds, as it has two H-bond donor atoms per molecule. NF₃ is polar, but cannot form hydrogen bonds because it has no hydrogen bond donor atoms. We therefore expect the combination of dispersion forces, dipole-dipole attractions, and hydrogen bonds between CH₃NH₂ molecules to be stronger than the combination of dispersion forces and dipole-dipole attractions between NF₃ molecules.

C₂H₆ is a hydrocarbon and is therefore nonpolar. Like Ne, which is an atomic substance, this substance exhibits only dispersion forces. Because each C₂H₆ molecule is composed of multiple atoms with more electrons than a single Ne atom, it is more polarizable than Ne. C₂H₆ therefore experiences stronger dispersion forces than Ne.

Remember that ALL particles experience dispersion forces.

nonpolar

Br₂

dispersion forces

Br₂ is nonpolar and therefore
experiences dispersion forces only.

polar
has H-bond donors
has N with lone pairs

NH₃

dispersion forces

dipole-dipole
attractions

hydrogen bonding

NH₃ is polar overall and has H atoms bonded to N. The N atom also has a lone pair. NH3 molecules therefore experience dispersion forces, dipole-dipole interactions, and hydrogen bonding between them.

Lewis dot diagram and geometry image of N H 3

polar
no H-bond donors

CH₂F₂

dispersion forces

dipole-dipole
attractions

CH₂F₂ contains F but neither H atom is bonded to F. There are no H-bond donor atoms, and therefore hydrogen bonding is not possible. The molecule is polar overall, however, and therefore experiences dipole-dipole attractions in addition to dispersion forces.

Lewis dot diagram and geometry image of C H 2 F 2

polar

CH₃OH

HOCH₂CH₂OH
(ethylene glycol)

C₆H₁₂O₆
(glucose)

nonpolar

C₆H₁₄

I₂
(iodine)

C₁₀H₈
(naphthalene)

CH₃OH (methanol) is polar and is capable of hydrogen bonding. Polar molecular solutes such as ethylene gylcol and glucose are more likely to dissolve in methanol than in C₆H₁₄, which is nonpolar.

C₆H₁₄ (hexane) is a hydrocarbon and is therefore essentially nonpolar. Nonpolar solutes such as iodine and naphthalene (another hydrocarbon) are more likely to dissolve in hexane than in the polar solvent methanol.