How Are Molecules Held Together

4.4: What makes molecules stick together? -- Intermolecular Forces

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Learning Objectives

To describe the intermolecular forces in liquids.

The properties of liquids are intermediate between those of gases and solids, but are more than similar to solids. In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in ane mol of water, only it takes only nigh 41 kJ to overcome the intermolecular attractions and catechumen 1 mol of liquid water to h2o vapor at 100°C. (Despite this seemingly depression value, the intermolecular forces in liquid h2o are amongst the strongest such forces known!) Given the big difference in the strengths of intra- and intermolecular forces, changes betwixt the solid, liquid, and gaseous states nearly invariably occur for molecular substances without breaking covalent bonds.

The backdrop of liquids are intermediate between those of gases and solids but are more similar to solids.

Intermolecular forces decide majority properties such every bit the melting points of solids and the humid points of liquids. Liquids boil when the molecules have plenty thermal energy to overcome the intermolecular attractive forces that concord them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire plenty thermal free energy to overcome the intermolecular forces that lock them into place in the solid.

Intermolecular forces are electrostatic in nature; that is, they ascend from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance betwixt molecules, intermolecular interactions are almost important for solids and liquids, where the molecules are close together.

In this section, we explicitly consider three intermolecular forces of attraction between neutral molecules: London dispersion, dipole-dipole and hydrogen bonding.

London Dispersion Forces

Thus far we have considered only interactions between polar molecules, merely other factors must exist considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature, and others, such every bit iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at depression temperatures, loftier pressures, or both (Tabular array \(\PageIndex{2}\)).

What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German language physicist who afterward worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could issue in the formation of short-lived instantaneous dipole moments, which produce attractive forces called London dispersion forces between otherwise nonpolar substances.

Table \(\PageIndex{2}\): Normal Melting and Boiling Points of Some Elements and Nonpolar Compounds
Substance	Molar Mass (g/mol)	Melting Indicate (°C)	Boiling Indicate (°C)
Ar	40	−189.four	−185.9
Xe	131	−111.8	−108.1
N₂	28	−210	−195.8
O₂	32	−218.8	−183.0
F₂	38	−219.7	−188.1
I₂	254	113.7	184.4
CH₄	16	−182.5	−161.5

Consider a pair of next He atoms, for example. On boilerplate, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in abiding motion, however, their distribution in 1 atom is likely to be asymmetrical at whatever given instant, resulting in an instantaneous dipole moment. As shown in function (a) in Figure \(\PageIndex{3}\), the instantaneous dipole moment on 1 atom can collaborate with the electrons in an adjacent cantlet, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The internet issue is that the get-go atom causes the temporary formation of a dipole, called an induced dipole, in the 2nd. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing altitude. London was able to testify with quantum mechanics that the attractive free energy betwixt molecules due to temporary dipole–induced dipole interactions falls off as one/r ⁶. Doubling the distance therefore decreases the attractive energy by ii⁶, or 64-fold.

alt — Figure \(\PageIndex{3}\): Instantaneous Dipole Moments. The formation of an instantaneous dipole moment on one He cantlet (a) or an H₂ molecule (b) results in the germination of an induced dipole on an adjacent atom or molecule.

Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances similar Xe. This consequence, illustrated for 2 H_ii molecules in part (b) in Effigy \(\PageIndex{3}\), tends to go more pronounced as diminutive and molecular masses increase (Table \(\PageIndex{two}\)). For example, Xe boils at −108.one°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given cantlet can be perturbed. In small atoms such equally He, the two 1s electrons are held close to the nucleus in a very modest book, and electron–electron repulsions are strong enough to foreclose pregnant asymmetry in their distribution. In larger atoms such every bit Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an cantlet or molecule is called its polarizability. Because the electron distribution is more than easily perturbed in large, heavy species than in small-scale, light species, we say that heavier substances tend to be much more polarizable than lighter ones.

For similar substances, London dispersion forces get stronger with increasing molecular size.

The polarizability of a substance too determines how it interacts with ions and species that possess permanent dipoles. Thus London dispersion forces are responsible for the general trend toward higher humid points with increased molecular mass and greater surface expanse in a homologous series of compounds, such as the alkanes (part (a) in Effigy \(\PageIndex{4}\)). The strengths of London dispersion forces also depend significantly on molecular shape considering shape determines how much of i molecule tin interact with its neighboring molecules at any given time. For case, part (b) in Figure \(\PageIndex{4}\) shows 2,ii-dimethylpropane (neopentane) and due north-pentane, both of which take the empirical formula C_vH₁₂. Neopentane is near spherical, with a small expanse for intermolecular interactions, whereas n-pentane has an extended conformation that enables it to come up into close contact with other north-pentane molecules. As a result, the boiling indicate of neopentane (nine.5°C) is more 25°C lower than the boiling betoken of n-pentane (36.1°C).

alt — Figure \(\PageIndex{4}\): Mass and Surface Area Touch the Strength of London Dispersion Forces. (a) In this serial of four simple alkanes, larger molecules have stronger London forces between them than smaller molecules and consequently higher boiling points. (b) Linear *northward*-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.

All molecules, whether polar or nonpolar, are attracted to one some other past London dispersion forces in improver to any other bonny forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate.

Case \(\PageIndex{2}\)

Arrange northward-butane, propane, 2-methylpropane [isobutene, (CH₃)₂CHCH_three], and n-pentane in order of increasing boiling points.

Given: compounds

Asked for: order of increasing boiling points

Strategy:

Determine the intermolecular forces in the compounds and then arrange the compounds co-ordinate to the forcefulness of those forces. The substance with the weakest forces will have the lowest boiling betoken.

Solution:

The iv compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, then propane should have the lowest boiling indicate and n-pentane should have the highest, with the 2 butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n-butane has the more extended shape. Consequently, we expect intermolecular interactions for n-butane to be stronger due to its larger surface area, resulting in a higher humid point. The overall order is thus equally follows, with actual boiling points in parentheses: propane (−42.i°C) < ii-methylpropane (−11.vii°C) < n-butane (−0.5°C) < n-pentane (36.one°C).

Exercise \(\PageIndex{2}\)

Conform GeH_iv, SiCl₄, SiH₄, CH₄, and GeCl₄ in order of decreasing boiling points.

Reply: GeCl₄ (87°C) > SiCl_four (57.6°C) > GeH_iv (−88.v°C) > SiH₄ (−111.8°C) > CH_four (−161°C)

Dipole–Dipole Interactions

Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.east., the two bonded atoms generate a dipole). If the structure of a molecule is such that the individual bond dipoles exercise not abolish one another, and so the molecule has a internet dipole moment. Molecules with net dipole moments tend to align themselves and then that the positive end of one dipole is near the negative stop of some other and vice versa, equally shown in Effigy \(\PageIndex{1a}\).

alt — Figure \(\PageIndex{1}\): Attractive and Repulsive Dipole–Dipole Interactions. (a and b) Molecular orientations in which the positive end of one dipole (δ⁺) is near the negative cease of another (δ⁻) (and vice versa) produce attractive interactions. (c and d) Molecular orientations that juxtapose the positive or negative ends of the dipoles on next molecules produce repulsive interactions.

These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (Figure \(\PageIndex{1c}\)). Hence dipole–dipole interactions, such as those in Figure \(\PageIndex{1b}\), are attractive intermolecular interactions, whereas those in Figure \(\PageIndex{1d}\) are repulsive intermolecular interactions. Because molecules in a liquid motility freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, every bit shown in Figure \(\PageIndex{ii}\). On average, yet, the bonny interactions boss.

alt — Effigy \(\PageIndex{2}\): Both Attractive and Repulsive Dipole–Dipole Interactions Occur in a Liquid Sample with Many Molecules

Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between ii ions, each of which has a accuse of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In add-on, the attractive interaction between dipoles falls off much more apace with increasing distance than do the ion–ion interactions. Recall that the bonny energy between two ions is proportional to 1/r, where r is the distance between the ions. Doubling the distance (r → iir) decreases the attractive energy past one-half. In dissimilarity, the energy of the interaction of two dipoles is proportional to ane/r ⁱⁱⁱ, so doubling the distance between the dipoles decreases the strength of the interaction by two³, or eight-fold. Thus a substance such equally \(\ce{HCl}\), which is partially held together by dipole–dipole interactions, is a gas at room temperature and one atm pressure, whereas \(\ce{NaCl}\), which is held together by interionic interactions, is a loftier-melting-point solid. Inside a series of compounds of similar tooth mass, the strength of the intermolecular interactions increases equally the dipole moment of the molecules increases, as shown in Table \(\PageIndex{one}\).

Table \(\PageIndex{i}\): Relationships between the Dipole Moment and the Humid Point for Organic Compounds of Similar Molar Mass
Chemical compound	Molar Mass (grand/mol)	Dipole Moment (D)	Boiling Point (K)
C_iiiH_half-dozen (cyclopropane)	42	0	240
CH₃OCH_iii (dimethyl ether)	46	1.30	248
CH₃CN (acetonitrile)	41	3.9	355

The attractive free energy between two ions is proportional to 1/r, whereas the attractive free energy between two dipoles is proportional to one/r6.

Example \(\PageIndex{1}\)

Adapt ethyl methyl ether (CH_iiiOCH₂CH₃), 2-methylpropane [isobutane, (CH_iii)_twoCHCH₃], and acetone (CH₃COCH₃) in order of increasing boiling points. Their structures are equally follows:

alt

Given: compounds

Asked for: order of increasing boiling points

Strategy:

Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points.

Solution:

The three compounds have substantially the same molar mass (58–60 thousand/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds.

The beginning compound, ii-methylpropane, contains but C–H bonds, which are not very polar because C and H take similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point.

Ethyl methyl ether has a structure similar to H₂O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bail dipoles partially reinforce i another and generate a meaning dipole moment that should give a moderately high boiling point.

Acetone contains a polar C=O double bail oriented at well-nigh 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling signal.

Thus nosotros predict the following lodge of humid points:

2-methylpropane < ethyl methyl ether < acetone.

This result is in good understanding with the actual data: 2-methylpropane, boiling point = −xi.seven°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = seven.iv°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D.

Practise \(\PageIndex{1}\)

Arrange carbon tetrafluoride (CF₄), ethyl methyl sulfide (CH₃SC₂H₅), dimethyl sulfoxide [(CH₃)₂S=O], and 2-methylbutane [isopentane, (CH_iii)_iiCHCH₂CH_iii] in guild of decreasing boiling points.

Answer: dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (humid indicate = 67°C) > 2-methylbutane (boiling point = 27.viii°C) > carbon tetrafluoride (boiling betoken = −128°C)

Hydrogen Bonds

When hydrogen is covalently bonded to a small-scale, highly electronegative chemical element (N, O and F) in a molecule, the molecules are capable of interacting through a particularly strong dipole-dipole interaction. The large deviation in electronegativity between the hydrogen atom and North, O or F results in a highly polar bond with the hydrogen end of the bond partially positive and the N, O or F stop partially negative. Additionally, since the atoms involved in this bond are so small, they tin can approach the comparable atoms in another molecule closely. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds, as shown for ice in Effigy \(\PageIndex{six}\). A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the hydrogen bail donor) and the atom that has the lone pair of electrons (the hydrogen bond acceptor). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can exist formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of side by side water molecules. The bridging hydrogen atoms are not equidistant from the two oxygen atoms they connect, withal. Instead, each hydrogen cantlet is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to ii H atoms at the shorter altitude and ii at the longer distance, corresponding to 2 O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from side by side water molecules, respectively. The resulting open, cagelike construction of ice ways that the solid is actually slightly less dumbo than the liquid, which explains why ice floats on water rather than sinks.

alt — Effigy \(\PageIndex{6}\): The Hydrogen-Bonded Structure of Ice.

Each water molecule accepts two hydrogen bonds from 2 other water molecules and donates 2 hydrogen atoms to form hydrogen bonds with two more than water molecules, producing an open, cagelike structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion.

Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor.

Because ice is less dense than liquid h2o, rivers, lakes, and oceans freeze from the meridian downward. In fact, the ice forms a protective surface layer that insulates the rest of the h2o, assuasive fish and other organisms to survive in the lower levels of a frozen lake or sea. If water ice were denser than the liquid, the ice formed at the surface in cold conditions would sink equally fast as information technology formed. Bodies of water would freeze from the bottom upward, which would exist lethal for most aquatic creatures. The expansion of water when freezing likewise explains why automobile or boat engines must be protected past "antifreeze" and why unprotected pipes in houses break if they are allowed to freeze.

Example \(\PageIndex{3}\)

Considering CH_threeOH, C₂H₆, Xe, and (CH₃)_threeN, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.

Given: compounds

Asked for: formation of hydrogen bonds and structure

Strategy:

Identify the compounds with a hydrogen atom attached to O, N, or F. These are probable to be able to deed as hydrogen bond donors.
Of the compounds that tin act as hydrogen bail donors, identify those that likewise contain lone pairs of electrons, which allow them to exist hydrogen bail acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, describe a structure showing the hydrogen bonding.

Solution:

A Of the species listed, xenon (Xe), ethane (C₂H_{half dozen}), and trimethylamine [(CH₃)_threeNorthward] practice not contain a hydrogen atom attached to O, North, or F; hence they cannot act every bit hydrogen bond donors.

B The i compound that can act as a hydrogen bond donor, methanol (CH_iiiOH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two alone pairs of electrons on O (making it a hydrogen bond acceptor); methanol tin can thus course hydrogen bonds past acting every bit either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is equally follows:

alt

Practise \(\PageIndex{3}\)

Because CH_iiiCO_twoH, (CH_iii)_iiiN, NH₃, and CH_threeF, which tin can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.

Answer

CH₃CO₂H and NH₃;

alt

Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of but 15–25 kJ/mol, they have a meaning influence on the concrete properties of a compound. Compounds such every bit HF can course but 2 hydrogen bonds at a time as can, on average, pure liquid NH_iii. Consequently, fifty-fifty though their molecular masses are similar to that of water, their humid points are significantly lower than the boiling indicate of water, which forms four hydrogen bonds at a fourth dimension.

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

Arrange C₆₀ (buckminsterfullerene, which has a cage construction), 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 chemical compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the everyman boiling point.

Solution:

Electrostatic interactions are strongest for an ionic chemical compound, and then nosotros expect NaCl to have the highest humid point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to grade hydrogen bonds, and their tooth mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so information technology should have the everyman boiling bespeak. Argon and N_twoO have very similar tooth masses (forty and 44 g/mol, respectively), merely N_iiO is polar while Ar is non. Consequently, Due north₂O should take a college humid point. A C₆₀ molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or Due north₂O. Because the boiling points of nonpolar substances increase speedily with molecular mass, C_sixty should eddy at a higher temperature than the other nonionic substances. The predicted guild is thus as follows, with bodily boiling points in parentheses:

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

Do \(\PageIndex{four}\)

Arrange two,4-dimethylheptane, Ne, CS₂, Cl_ii, and KBr in order of decreasing humid points.

Answer: KBr (1435°C) > ii,four-dimethylheptane (132.9°C) > CS₂ (46.6°C) > Cl₂ (−34.6°C) > Ne (−246°C)

Example \(\PageIndex{5}\):

Identify the well-nigh significant intermolecular strength in each substance.

C₃H₈
CH₃OH
H₂Southward

Solution

a. Although C–H bonds are polar, they are only minimally polar. The most significant intermolecular forcefulness for this substance would exist dispersion forces.

b. This molecule has an H atom bonded to an O atom, and then it will experience hydrogen bonding.

c. Although this molecule does not experience hydrogen bonding, the Lewis electron dot diagram and VSEPR indicate that it is aptitude, so information technology has a permanent dipole. The almost significant force in this substance is dipole-dipole interaction.

Exercise \(\PageIndex{6}\)

Identify the most significant intermolecular force in each substance.

Answer a: hydrogen bonding
Answer b: dipole-dipole interactions

Summary

Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds. Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that agree the atoms together inside molecules and polyatomic ions. Transitions between the solid and liquid or the liquid and gas phases are due to changes in intermolecular interactions but do non touch on intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively every bit van der Waals forces), and hydrogen bonds. Dipole–dipole interactions ascend from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to one/r³ , where r is the distance between dipoles. London dispersion forces are due to the formation of instantaneous dipole moments in polar or nonpolar molecules as a result of brusque-lived fluctuations of electron charge distribution, which in plow crusade the temporary formation of an induced dipole in adjacent molecules. their energy falls off as ane/r ^six. Larger atoms tend to be more polarizable than smaller ones because their outer electrons are less tightly jump and are therefore more than easily perturbed. Hydrogen bonds are especially strong dipole–dipole interactions between molecules that accept hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H cantlet on one molecule (the hydrogen bond donor) can interact strongly with a lonely pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the hydrogen bond acceptor). Because of strong O⋅⋅⋅H hydrogen bonding between water molecules, water has an unusually high boiling signal, and water ice has an open, cagelike construction that is less dense than liquid h2o.