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Intermolecular Forces

Dr. Melinda Oliver

IntermolecularForcesDr. Melinda Oliver

Why do substances exist as gases or liquids or solids at room temperature?

The answer is the forces of attraction between particles determines whether a substance will be a solid, liquid or gas AT room temperature

The attractions between molecules are not nearly as strong as the intramolecular "force" such as the covalent bond in the example below.

These intermolecular attractions are, however, strong enough to control physical properties, such as boiling and melting points, vapor pressures, surface tension, and viscosities.

An important exception to intra-vs-inter is in ionic compounds. The interparticle force is the same as the intermolecular force: the ionic bond and it is the strongest of the interparticle forces.

With the exception of ionic compounds, the words polar and nonpolar only pertain to MOLECULAR substances and the difference between them dictates the strength and type of IMF.

Polar & Non-Polar Molecules: Crash Course Chemistry #23

3 Main IMF's (for molecular substances) explained below

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Dipole-dipole

Think polar bonds and molecules. The polar bond is a dipole and if there is a polar bond that ends up making a molecule overall polar, then dipole-dipole is the IMF and this one can be pretty strong.
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Hydrogen bonding

A special case of dipole-dipole is H-bonding and that just means a polar bond with H at one end and either N, O, or F at the other end (only these 3 highly electronegative atoms bonded to H will give H-bonding). Of the 3 discussed here, this is the strongest IMF.
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London Dispersion (van der waals, induced dipole)

This type of IMF is present in almost every substance. The key to understanding London dispersion forces is to think of nonpolar molecules in particular as like water balloons that can be "squishy". The squishiness increases with size and the same is true for nonpolar substances. The larger the nonpolar molecule, the "squishier" the electron density (science word = polarizability) and the stronger the attraction between molecules. Another way to think about it is this IMF is like velcro. Small molecule = small piece of velcro = weaker London dispersion force. Large molecule = larger piece of velcro = larger London dispersion force.
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Factors affecting London Dispersion forces

  • The shape of the molecule affects the strength of dispersion forces: long, skinny molecules (like n-pentane) tend to have stronger dispersion forces than short, fat ones (like neopentane).This is due to the increased surface area in n-pentane.

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Find the molecules with the higher dipole moments in the figure above.

Obviously, molecules that are more polar, have stronger attractions and, therefore, higher boiling points. Why use boiling vs melting??

How do we explain this diagram?

Account for the differences between the groups (think size, polarity, and geometry). For example, why is group 4 below all of the other groups? Another question is to account for the big deviation for NH3, HF and H2O. The dotted line drawn for the red curve is an interpolation of the red curve-where water would be expected to be.

The extent of Hydrogen bonding depends on the atom H is attached to

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Each water molecule can form 4 hydrogen bonds

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Each ammonia molecule can form 2 hydrogen bonds

Two hydrogens are "wasted" due to a lack of lone pairs.
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Each HF molecule can form 2 hydrogen bonds

Two lone pairs are "wasted" due to lack of hydrogens.

More detailed explanation:

If temperature changes, the kinetic energy of the particles changes and so can the phase of matter. How much the kinetic energy changes is directly related to the strength of the IMF between the molecules.

In the heating curve below, why is the amount of heat different for boiling vs melting?

Practice

1. True or False:

(a) For molecules with similar molecular weights, the dispersion forces become stronger as the molecules become more polarizable.

(b) For the noble gases the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table.

(c) In terms of the total attractive forces for a given substance dipole-dipole interactions, when present, are always larger than dispersion forces.

(d) All other factors being the same, dispersion forces between linear molecules are greater than dispersion forces between molecules whose shapes are nearly spherical.


2. Rationalize the difference in boiling points in each pair: (a) HF (20°C) and HCl (-85°C); (b) CHCl3 (61°C) and CHBr3 (150°C); (c) Br2 (59°C) and ICl (97°C).


3. Identify the type of types of intermolecular forces present in each substance and then select the substance in each pair that has the higher boiling point: (a) propane C3H8 or n-butane C4H10; (b) diethyl ether CH3CH2OCH2CH3 or 1-butanol CH3CH2CH2CH2OH; (c) sulfure dioxide (SO2) or sulfur trioxide (SO3); (d) phosgene Cl2CO or formaldehyde H2CO.


4. Arrange substances Ga, Ne, and Br2 in order of increasing boiling point.


5. Which type of intermolecular force accounts for each of these differences: (a) CH3OH boils at 65 C; CH3SH boils at 6 C (b) Xe is a liquid at atmospheric pressure and 120 K, whereas Ar is a gas under the same conditions. (c) Kr, atomic weight 84, boils at 120.9 K, whereas Cl2, molecular weight about 71, boils at 238 K. (d) acetone boils at 56 C, whereas 2-methylpropane boils at -12 C.

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Acetone

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2-methylpropane

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Isopropanol

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Practice

6. Hydrazine (H2NNH2). hydrogen peroxide (HOOH), and water (H2O) all have exceptionally high surface tensions compared with other substances of comparable molecular weights. (a) Draw the Lewis structures for these three compounds (formulas are given as structural formulas so you know how they are connected). (b) What structural property do these substances have in common, and how might that account for the high surface tensions?


7. Would you expect the viscosity of n-pentane (CH3CH2CH2CH2CH3) to be larger or smaller than the viscosity of n-hexane (6 C's instead of 5)? (b) If you compared their viscosities at 270 K, would you expect the viscosity of neopentane (CH3)4C, to be smaller or larger than n-pentane.


8. True or False:

(a) CBr4 is more volatile than CCl4 (b) CBr4 has a higher boiling point than CCl4

(c) CBr4 has a weaker intermolecular forces than CCl4

(d) CBr4 has a higher vapor pressure at the same temperature than CCl4

What can you do to help your understanding?

Chemistry is a subject that must be practiced everyday if possible. Take advantage of the practice in Mastering Chemistry to give you the practice you need to be successful. DO NOT PROCRASTINATE! Check announcements on Blackboard everyday. Keep a printed copy of the most recent course calendar (included in syllabus) next to your work area. Email me with questions!!
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Dr. Melinda Oliver