5.2: Application of Intermolecular Forces: Melting and Boiling Points

Learning Objectives

• Rank melting points and boiling points for molecules of approximately the same size
• Rank melting points and boiling points for atoms and molecules of different sizes

Introduction

The strength of intermolecular forces determines melting points (mp) and boiling points (bp) for substances. The stronger the IMF, the higher the melting point or boiling point. 

This is because IMFs are an attractive force holding the particles together and energy (in the form of heat) must be added to weaken or break the forces/attractions. Melting occurs when IMFs are weakened and boiling occurs when the IMFs are broken and the particles are allowed to completely separate from one another. 

Strength of Intermolecular Forces

The strength of IMFs increases as molecules become more polar (as measured by the dipole moment of the molecule), as shown in Table \(\PageIndex{1}\). Since the size of a molecule can also play a role, the table compares molecules of similar size as measured by the total number of electrons in the molecule. 

Table \(\PageIndex{1}\): Relationships between the Dipole Moment and the Boiling Point for Organic Compounds of Similar Size

Compound	Total Number of Electrons	Dipole Moment (D)	Boiling Point (K)
C3H6 (cyclopropane)	24	0	240
CH3OCH3 (dimethyl ether)	26	1.30	248
CH3CN (acetonitrile)	22	3.9	355

Example \(\PageIndex{1}\) Dipole-Dipole Forces and Their Effects

Predict which will have the higher boiling point: N₂ or CO. Explain your reasoning.

Solution

CO and N₂ are both small molecules with 14 electrons, so the polarity of the molecules will be the primary factor determining their boiling points. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because N₂ is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar N₂ molecules, so CO is expected to have the higher boiling point.

Exercise \(\PageIndex{1}\)

Predict which will have the higher boiling point: \(\ce{ICl}\) or \(\ce{Br_2}\). Explain your reasoning.

Answer

ICl. ICl and Br₂ have the same number of electrons, 70, and therefore their polarity will be the primary factor determining their boiling points. ICl is polar and thus participates in dipole-dipole attractions; Br₂ is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point 

Size Determines the Strength of Dispersion Forces

Larger atoms and molecules exhibit stronger dispersion forces than do smaller atoms and molecules because they have more electrons that may temporarily shift one way or another. F₂ and Cl₂ are gases at room temperature (reflecting weaker attractive forces); Br₂ is a liquid, and I₂ is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table \(\PageIndex{1}\).

Table \(\PageIndex{2}\): Melting and Boiling Points of the Halogens

Halogen	Number of Electrons	Atomic Radius	Melting Point	Boiling Point
fluorine, F2	18	72 pm	53 K	85 K
chlorine, Cl2	34	99 pm	172 K	238 K
bromine, Br2	70	114 pm	266 K	332 K
iodine, I2	106	133 pm	387 K	457 K
astatine, At2	170	150 pm	575 K	610 K

The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule’s charge distribution (its electron cloud) is known as polarizability. A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces.

Shape Also Affects Dispersion Forces

The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(\PageIndex{5}\) shows 2,2-dimethylpropane (neopentane) and n-pentane, both of which have the formula C₅H₁₂. Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n-pentane has an extended conformation that enables it to come into close contact with other n-pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n-pentane (36.1°C). The red lines show that there is more contact between two n-pentane molecules than there is between two neopentane molecules, even though they are both C₅H₁₂. 

Figure \(\PageIndex{5}\) Mass and Surface Area Affect the Strength of London Dispersion Forces. (a) In this series of four simple alkanes, larger molecules have stronger London forces between them than smaller molecules and consequently higher boiling points. (b) Linear n-pentane molecules have a larger surface area and stronger intermolecular forces than spherical neopentane molecules. As a result, neopentane is a gas at room temperature, whereas n-pentane is a volatile liquid.

London dispersion forces

All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present.

Example \(\PageIndex{2}\) London Forces and Their Effects

Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: CH₄, SiH₄, GeH₄, and SnH₄. Explain your reasoning.

Given: Molecular formulas

Asked for: Order of increasing boiling points

Strategy:

Determine the intermolecular forces in the compound. If the compounds can participate in the same type of IMFs, compare the size and surface area of compounds with similar types of intermolecular force. The substance with the weakest forces and the lowest lowest number of electrons and least surface area have the lowest boiling point.

Solution

Applying the skills acquired in the chapter on polarity all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces. The smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molecules of CH₄, SiH₄, GeH₄, and SnH₄ contain 10, 18, 36, and 54 electrons, respectively. Therefore, CH₄ is expected to have the lowest boiling point and SnH₄ the highest boiling point. The ordering from lowest to highest boiling point is expected to be

CH₄ < SiH₄ < GeH₄ < SnH₄

A graph of the actual boiling points of these compounds versus the period of the group 14 elements shows this prediction to be correct:

 

Comparing H-Bonding to Dispersion

The graph below is similar to the previous one. The Group 14 line (purple, bottom) shows the same boiling point data as the graph above. In Groups 14, 15, and 16 the same trend can be seen for hydrides from periods 3, 4, and 5. However, the boiling points for NH₃, HF, and H₂O are unusually high. This is because those molecules can form hydrogen bonds with their neighbors. Because hydrogen bonds are stronger interactions than dipole-dipole or dispersion forces, the boiling points are higher. 

Figure \(\PageIndex{9}\) The Effects of Hydrogen Bonding on Boiling Points. These plots of the boiling points of the covalent hydrides of the elements of groups 14–17 show that the boiling points of the lightest members of each series for which hydrogen bonding is possible (HF, NH₃, and H₂O) are anomalously high for compounds with such low molecular masses.

Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH₃. Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms four hydrogen bonds at a time.

Exercise \(\PageIndex{2}\)

Order the following hydrocarbons from lowest to highest boiling point: C₂H₆, C₃H₈, and C₄H₁₀.

Answer

All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore

C₂H₆ < C₃H₈ < C₄H₁₀.

Example \(\PageIndex{3}\): Buckyballs

Arrange C₆₀ (buckminsterfullerene, which has a soccer ball structure), NaCl, He, Ar, and N₂O in order of increasing boiling points.

Given: compounds

Asked for: order of increasing boiling points

Strategy:

Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.

Solution:

Interactions are strongest for an ionic compound which has full positive and negative charges, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their sizes (for London dispersion forces). Helium is nonpolar and by far the smallest, so it should have the lowest boiling point. Argon and N₂O have very similar numbers of electrons (18 and 22, respectively), but N₂O is polar while Ar is not. Consequently, N₂O should have a higher boiling point. A C₆₀ molecule is nonpolar, but it has 360 electrons, much greater than that of Ar or N₂O. Because the boiling points of nonpolar substances increase rapidly with size, C₆₀ should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses:

He (−269°C) < Ar (−185.7°C) < N₂O (−88.5°C) < C₆₀ (>280°C) < NaCl (1465°C).

Exercise \(\PageIndex{3}\)

Arrange C₉H₂₀, Ne, CS₂, Cl₂, and KBr in order of decreasing boiling points.

Answer

KBr (1435°C) > C₉H₂₀ (132.9°C for 2,4-dimethylheptane) > CS₂ (46.6°C) > Cl₂ (−34.6°C) > Ne (−246°C)

Summary

• If molecules are approximately the same size (same total number of electrons), then the type of intermolecular force will determine the relative melting and boiling points.
  • Hydrogen bonding forces are the strongest and result in higher mp and bp
  • Dispersion forces are the weakest and result in low mp and bp
• If atoms and molecules have significantly different numbers of electrons, then: 
  • Dispersion forces are highest in the atoms/molecules with the most electrons. This leads to higher mp and bp. 
  • Dispersion forces are lowest in atoms/molecules with the fewest electrons. This leads to low mp and bp.