What Holds A Molecule Together

iv.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 betwixt those of gases and solids, but are more like to solids. In contrast to intramolecular forces, such equally 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 by and large much weaker than covalent bonds. For instance, information technology requires 927 kJ to overcome the intramolecular forces and interruption both O–H bonds in 1 mol of water, only it takes but most 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly depression value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds.

The properties of liquids are intermediate betwixt those of gases and solids but are more like to solids.

Intermolecular forces decide bulk properties such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal 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 bonny and repulsive components. Because electrostatic interactions fall off quickly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are shut together.

In this section, we explicitly consider 3 intermolecular forces of attraction between neutral molecules: London dispersion, dipole-dipole and hydrogen bonding.

London Dispersion Forces

Thus far we take considered only interactions betwixt polar molecules, but other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature, and others, such as iodine and naphthalene, are solids. Even the noble gases can exist liquefied or solidified at low temperatures, high pressures, or both (Table \(\PageIndex{2}\)).

What kind of attractive forces tin can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German language physicist who later worked in the U.s.. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could consequence 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 Humid Points of Some Elements and Nonpolar Compounds

Substance	Molar Mass (g/mol)	Melting Point (°C)	Humid Bespeak (°C)
Ar	40	−189.4	−185.nine
Xe	131	−111.8	−108.1
N2	28	−210	−195.eight
O2	32	−218.8	−183.0
Ftwo	38	−219.7	−188.i
I2	254	113.vii	184.iv
CH4	16	−182.5	−161.5

Consider a pair of adjacent He atoms, for example. On average, the 2 electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in i atom is likely to be asymmetrical at whatsoever given instant, resulting in an instantaneous dipole moment. Every bit shown in part (a) in Effigy \(\PageIndex{3}\), the instantaneous dipole moment on one atom can interact with the electrons in an side by side atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net upshot is that the outset atom causes the temporary formation of a dipole, called an induced dipole, in the 2d. Interactions betwixt these temporary dipoles crusade atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to evidence with quantum mechanics that the attractive free energy between molecules due to temporary dipole–induced dipole interactions falls off equally one/r ^{half dozen}. Doubling the altitude therefore decreases the attractive energy by ii^six, or 64-fold.

Effigy \(\PageIndex{three}\): Instantaneous Dipole Moments. The formation of an instantaneous dipole moment on 1 He atom (a) or an H₂ molecule (b) results in the formation of an induced dipole on an adjacent atom or molecule.

Instantaneous dipole–induced dipole interactions betwixt nonpolar molecules tin produce intermolecular attractions just every bit they produce interatomic attractions in monatomic substances like Xe. This issue, illustrated for ii H₂ molecules in part (b) in Figure \(\PageIndex{3}\), tends to become more pronounced as atomic and molecular masses increase (Table \(\PageIndex{ii}\)). For example, Xe boils at −108.i°C, whereas He boils at −269°C. The reason for this trend is that the force of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In pocket-sized atoms such as He, the two 1s electrons are held close to the nucleus in a very pocket-size volume, and electron–electron repulsions are strong enough to foreclose significant asymmetry in their distribution. In larger atoms such every bit Xe, nevertheless, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a upshot, it is relatively like shooting fish in a barrel to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability. Considering the electron distribution is more hands perturbed in large, heavy species than in small-scale, low-cal species, nosotros say that heavier substances tend to exist much more than polarizable than lighter ones.

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

The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus London dispersion forces are responsible for the full general trend toward college boiling points with increased molecular mass and greater surface area in a homologous serial of compounds, such as the alkanes (part (a) in Effigy \(\PageIndex{4}\)). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of 1 molecule tin collaborate with its neighboring molecules at any given time. For example, office (b) in Figure \(\PageIndex{4}\) shows 2,2-dimethylpropane (neopentane) and n-pentane, both of which have the empirical formula C₅H₁₂. Neopentane is almost spherical, with a pocket-size surface area for intermolecular interactions, whereas n-pentane has an extended conformation that enables it to come into close contact with other due north-pentane molecules. Every bit a result, the boiling point of neopentane (nine.5°C) is more than than 25°C lower than the boiling point of n-pentane (36.1°C).

Effigy \(\PageIndex{4}\): Mass and Surface Area Bear on the Force of London Dispersion Forces. (a) In this serial of iv simple alkanes, larger molecules have stronger London forces between them than smaller molecules and consequently college boiling points. (b) Linear n-pentane molecules take a larger expanse 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 another past London dispersion forces in improver to any other attractive forces that may be nowadays. In general, even so, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the sometime predominate.

Case \(\PageIndex{2}\)

Arrange due north-butane, propane, 2-methylpropane [isobutene, (CH₃)₂CHCH₃], and due north-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 and then arrange the compounds co-ordinate to the strength of those forces. The substance with the weakest forces volition have the lowest boiling point.

Solution:

The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n-pentane should have the highest, with the two butane isomers falling in betwixt. Of the two butane isomers, 2-methylpropane is more meaty, and n-butane has the more extended shape. Consequently, we expect intermolecular interactions for n-butane to be stronger due to its larger area, resulting in a college boiling betoken. The overall social club is thus as follows, with bodily humid points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.vii°C) < n-butane (−0.5°C) < due north-pentane (36.ane°C).

Exercise \(\PageIndex{2}\)

Arrange GeH₄, SiCl₄, SiH₄, CH_four, and GeCl₄ in order of decreasing humid points.

Reply

GeCl₄ (87°C) > SiCl_four (57.6°C) > GeH_four (−88.5°C) > SiH₄ (−111.8°C) > CH₄ (−161°C)

Dipole–Dipole Interactions

Polar covalent bonds behave as if the bonded atoms take localized fractional charges that are equal simply opposite (i.e., the two bonded atoms generate a dipole). If the construction of a molecule is such that the individual bond dipoles do not cancel ane another, then the molecule has a net dipole moment. Molecules with internet dipole moments tend to align themselves so that the positive terminate of one dipole is near the negative end of another and vice versa, as shown in Figure \(\PageIndex{1a}\).

Figure \(\PageIndex{one}\): Attractive and Repulsive Dipole–Dipole Interactions. (a and b) Molecular orientations in which the positive end of ane dipole (δ⁺) is near the negative end of another (δ⁻) (and vice versa) produce bonny 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 2 positive or two negative ends are adjacent (Effigy \(\PageIndex{1c}\)). Hence dipole–dipole interactions, such as those in Figure \(\PageIndex{1b}\), are attractive intermolecular interactions, whereas those in Effigy \(\PageIndex{1d}\) are repulsive intermolecular interactions. Because molecules in a liquid motion freely and continuously, molecules ever experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure \(\PageIndex{2}\). On average, however, the bonny interactions dominate.

Figure \(\PageIndex{2}\): 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 charge of at to the lowest degree ±ane, or between a dipole and an ion, in which one of the species has at least a full positive or negative accuse. In addition, the attractive interaction between dipoles falls off much more than chop-chop with increasing distance than do the ion–ion interactions. Recall that the bonny energy betwixt two ions is proportional to one/r, where r is the distance between the ions. Doubling the distance (r → 2r) decreases the attractive energy by 1-one-half. In contrast, the energy of the interaction of two dipoles is proportional to ane/r ³, then doubling the distance between the dipoles decreases the force of the interaction by 2³, or 8-fold. Thus a substance such as \(\ce{HCl}\), which is partially held together by dipole–dipole interactions, is a gas at room temperature and i atm pressure, whereas \(\ce{NaCl}\), which is held together past interionic interactions, is a loftier-melting-point solid. Within a series of compounds of similar molar mass, the force of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(\PageIndex{1}\).

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

Chemical compound	Molar Mass (1000/mol)	Dipole Moment (D)	Boiling Point (M)
CthreeH6 (cyclopropane)	42	0	240
CH3OCH3 (dimethyl ether)	46	1.30	248
CH3CN (acetonitrile)	41	3.9	355

The attractive energy betwixt ii ions is proportional to ane/r, whereas the bonny energy betwixt two dipoles is proportional to ane/r6.

Example \(\PageIndex{1}\)

Accommodate ethyl methyl ether (CH₃OCH₂CH₃), 2-methylpropane [isobutane, (CH₃)₂CHCH₃], and acetone (CH₃COCH_iii) in guild of increasing boiling points. Their structures are as follows:

Given: compounds

Asked for: order of increasing boiling points

Strategy:

Compare the molar masses and the polarities of the compounds. Compounds with higher tooth masses and that are polar volition take the highest humid points.

Solution:

The three compounds have substantially the same molar mass (58–60 g/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 offset compound, 2-methylpropane, contains just C–H bonds, which are non very polar because C and H take like electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point.

Ethyl methyl ether has a construction similar to H₂O; it contains two polar C–O unmarried bonds oriented at nigh a 109° angle to each other, in add-on to relatively nonpolar C–H bonds. As a result, the C–O bail dipoles partially reinforce 1 some other and generate a significant dipole moment that should give a moderately high boiling bespeak.

Acetone contains a polar C=O double bail oriented at nearly 120° to 2 methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should issue in both a rather large dipole moment and a high boiling point.

Thus we predict the following lodge of boiling points:

2-methylpropane < ethyl methyl ether < acetone.

This result is in good understanding with the actual information: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, humid point = vii.four°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D.

Exercise \(\PageIndex{1}\)

Arrange carbon tetrafluoride (CF₄), ethyl methyl sulfide (CH₃SC₂H₅), dimethyl sulfoxide [(CH₃)_iiS=O], and 2-methylbutane [isopentane, (CH₃)₂CHCH₂CH₃] in order of decreasing boiling points.

Reply

dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (humid betoken = 67°C) > ii-methylbutane (humid bespeak = 27.8°C) > carbon tetrafluoride (boiling point = −128°C)

Hydrogen Bonds

When hydrogen is covalently bonded to a small, highly electronegative element (Due north, O and F) in a molecule, the molecules are capable of interacting through a particularly strong dipole-dipole interaction. The large deviation in electronegativity betwixt the hydrogen cantlet and N, O or F results in a highly polar bond with the hydrogen finish of the bond partially positive and the N, O or F cease partially negative. Additionally, since the atoms involved in this bail are then pocket-sized, they can approach the comparable atoms in another molecule closely.  The combination of big bail dipoles and short dipole–dipole distances results in very potent dipole–dipole interactions chosen hydrogen bonds, as shown for ice in Figure \(\PageIndex{6}\). A hydrogen bail is unremarkably indicated past a dotted line betwixt the hydrogen cantlet 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 ii hydrogen atoms and 2 lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that class bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are non equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom 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 distance and 2 at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dumbo than the liquid, which explains why ice floats on water rather than sinks.

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

Each water molecule accepts 2 hydrogen bonds from two other h2o molecules and donates two hydrogen atoms to class hydrogen bonds with ii more water molecules, producing an open up, cagelike construction. The structure of liquid water is very similar, simply in the liquid, the hydrogen bonds are continually broken and formed considering of rapid molecular motion.

Hydrogen bail germination requires both a hydrogen bond donor and a hydrogen bond acceptor.

Considering ice is less dumbo than liquid h2o, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the residual of the water, assuasive fish and other organisms to survive in the lower levels of a frozen lake or bounding main. If ice were denser than the liquid, the water ice formed at the surface in cold weather would sink as fast equally information technology formed. Bodies of water would freeze from the bottom upwardly, which would exist lethal for most aquatic creatures. The expansion of h2o when freezing also explains why car or boat engines must be protected by "antifreeze" and why unprotected pipes in houses break if they are immune to freeze.

Example \(\PageIndex{3}\)

Considering CH₃OH, C_iiH₆, Xe, and (CH₃)_threeN, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.

Given: compounds

Asked for: germination of hydrogen bonds and structure

Strategy:

1. Identify the compounds with a hydrogen cantlet attached to O, Due north, or F. These are likely to exist able to act every bit hydrogen bond donors.
2. Of the compounds that can act as hydrogen bail donors, place those that likewise incorporate solitary pairs of electrons, which allow them to be hydrogen bail acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding.

Solution:

A Of the species listed, xenon (Xe), ethane (C₂H₆), and trimethylamine [(CH_three)₃Northward] exercise non contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors.

B The one compound that tin human action as a hydrogen bond donor, methanol (CH_threeOH), contains both a hydrogen cantlet attached to O (making information technology a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded construction of methanol is as follows:

Do \(\PageIndex{3}\)

Considering CH₃CO_iiH, (CH_three)₃N, NH₃, and CH_threeF, which tin form hydrogen bonds with themselves? Depict the hydrogen-bonded structures.

Answer

CH₃CO₂H and NH₃;

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 concrete properties of a compound. Compounds such every bit HF can grade only two hydrogen bonds at a time every bit can, on average, pure liquid NH_iii. Consequently, fifty-fifty 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.

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

Arrange C₆₀ (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N_twoO in order of increasing humid points.

Given: compounds

Asked for: order of increasing boiling points

Strategy:

Place the intermolecular forces in each chemical compound and then arrange the compounds co-ordinate to the force of those forces. The substance with the weakest forces will have the everyman boiling signal.

Solution:

Electrostatic interactions are strongest for an ionic compound, so we look NaCl to have the highest boiling point. To predict the relative humid points of the other compounds, nosotros must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their tooth mass (for London dispersion forces). Helium is nonpolar and past far the lightest, so it should have the lowest boiling betoken. Argon and Northward₂O have very similar tooth masses (forty and 44 g/mol, respectively), but N₂O is polar while Ar is not. Consequently, N₂O should have a higher humid betoken. A C_threescore molecule is nonpolar, but its tooth mass is 720 grand/mol, much greater than that of Ar or Due north₂O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C₆₀ should eddy 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.vii°C) < N₂O (−88.5°C) < C₆₀ (>280°C) < NaCl (1465°C).

Practise \(\PageIndex{four}\)

Arrange 2,four-dimethylheptane, Ne, CS₂, Cl_ii, and KBr in social club of decreasing boiling points.

Respond

KBr (1435°C) > ii,4-dimethylheptane (132.9°C) > CS₂ (46.six°C) > Cl_two (−34.vi°C) > Ne (−246°C)

Example \(\PageIndex{five}\):

Place the most significant intermolecular force in each substance.

1. C ₃ H ₈
2. CH ₃ OH
3. H ₂ S

Solution

a. Although C–H bonds are polar, they are only minimally polar. The virtually pregnant intermolecular strength for this substance would exist dispersion forces.

b. This molecule has an H cantlet bonded to an O atom, so it volition experience hydrogen bonding.

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

Practice \(\PageIndex{half dozen}\)

Identify the near significant intermolecular forcefulness in each substance.

1. HF
2. HCl

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 hold 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 just do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these ii are oftentimes referred to collectively as 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 force is proportional to the magnitude of the dipole moment and to 1/r^three , where r is the distance betwixt dipoles. London dispersion forces are due to the formation of instantaneous dipole moments in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an induced dipole in adjacent molecules. their energy falls off as one/r ^half-dozen. Larger atoms tend to be more polarizable than smaller ones because their outer electrons are less tightly bound and are therefore more hands perturbed. Hydrogen bonds are particularly strong dipole–dipole interactions betwixt molecules that have 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 lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the hydrogen bond acceptor). Considering of strong O⋅⋅⋅H hydrogen bonding between water molecules, water has an unusually high boiling bespeak, and ice has an open up, cagelike structure that is less dumbo than liquid water.

What Holds A Molecule Together,

Source: https://chem.libretexts.org/Courses/Harper_College/CHM_110:_Fundamentals_of_Chemistry/04:_Water/4.04:_What_makes_molecules_stick_together_--_Intermolecular_Forces

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