MC test: Intermolecular forces

Multiple choice test on 4.4 Intermolecular forces

Use the following 'quiz' to test your knowledge and understanding of this sub-topic. You will need access to a periodic table (Section 6 of the IB data booklet).

If you get an answer wrong, read through the explanation carefully to learn from your mistakes.

The following substances all possess intermolecular hydrogen bonding. Which one has the highest boiling point?

methanol, CH₃OH

water, H₂O

ammonia, NH₃

hydrogen fluoride, HF

The boiling points of NH₃, HF and CH₃OH are −33.3, +19.5 and +64.7 ^oC respectively. Water has the highest boiling point as it has two non bonding pairs of electrons, the polarity of the O atom is 2δ⁻ and each molecule of water can form hydrogen bonds to two other water molecules.

Which molecule will be polar?

tetrachloromethane (carbon tetrachloride), CCl₄

carbon dioxide, CO₂

dibromomethane, CH₂Br₂

boron trichloride, BCl₃

Although they all contain polar bonds the resultant dipole in CO₂, CCl₄ and BCl₃ is zero.

Which are classified as van der Waals' forces of attraction?

I. London (dispersion) forces

II. Dipole-dipole forces

III. Dipole-induced dipole forces

I and II only

I, II and III

II and III only

I and III only

All of these are relatively weak attractive forces and all come under the heading of van der Waals' forces. It is debatable whether hydrogen bonding should also come under this heading. In past IB multiple choice questions the IB has signified that hydrogen bonding is not included but many text books and web sites etc. would disagree with this.

Which is the correct order of increasing forces of attraction between atoms or molecules (weakest first)?

London forces < hydrogen bonding < dipole-dipole < covalent bonding

covalent bonding < London forces < dipole < dipole < hydrogen bonding

covalent bonding < hydrogen bonding < London forces < dipole < dipole

London forces < dipole-dipole < hydrogen bonding < covalent bonding

Covalent bonding is much stronger than the other three and London forces are the weakest. Hydrogen bonding is stronger than dipole-dipole.

Butane H₃CCH₂CH₂CH₃, propanone H₃CCOCH₃ and propan-1-ol, H₃CCH₂CH₂OH all have a similar molar mass. Which is the correct order of their increasing boiling points (lowest first)?

propan-1-ol < propanone < butane

butane < propan-1-ol < propanone

butane < propanone < propan-1-ol

propanone < butane < propan-1-ol

Butane molecules have London forces between them, propanone molecules have dipole-dipole as the strongest intermolecular force between them and propan-1-ol molecules can form hydrogen bonds with other propan-1-ol molecules.

Which explains why the boiling points of the halogens increase down the group from fluorine to iodine?

The ionization energy of the halogens decreases with increasing atomic number.

The dipole-dipole forces between the molecules increases with increasing atomic number.

The London (dispersion) forces between the molecules increases with increasing atomic number.

The covalent bonds between the molecules increases with increasing atomic number.

Because all the halogens are simple diatomic molecules they are all non-polar. The only attractive forces between them are weak London (dispersion) forces.

Butan-1-ol, H₃CCH₂CH₂CH₂OH and butan-2-ol, H₃CCH₂CH(OH)CH₃ both have the same mass. The boiling point of butan-1-ol is 117.7 ^oC and the boiling point of butan-2-ol is 99.0 ^oC. Which best explains why butan-1-ol has a higher boiling point than butan-2-ol.

The London forces are greater in butan-1-ol due to the longer hydrocarbon chain.

The δ− charge on the oxygen atom is less in butan-2-ol than in butan-1-ol because the oxygen atom is joined to the second carbon atom.

Only butan-1-ol can form hydrogen bonds

Both molecules can form hydrogen bonds but the intermolecular bonding is weaker in butan-2-ol as it is more spherical in shape so there is less surface area contact between the molecules.

The polarity of the O−H bond is the same (or at least very similar) in both molecules but the intermolecular forces of attraction also depend upon the contact surface area between the molecules.

Which best describes all the forces of attraction present between butanal (H₃CCH₂CH₂CHO) molecules?

Dipole-dipole forces of attraction

London dispersion forces

London (dispersion) forces and dipole-dipole attractions

London (dispersion) forces and dipole-dipole attractions and hydrogen bonding

There are always London (dispersion) forces between any two molecules. Butanal is polar so there will also be dipole-dipole forces of attraction but as the hydrogen atom is bonded to carbon not oxygen there will be no hydrogen bonding.

Methoxymethane, H₃COCH₃ and ethanol, C₂H₅OH are isomers. Which best explains why ethanol has a considerably higher boiling point than methoxymethane?

Because of its linear shape methoxymethane is not polar so only weak London (dispersion) forces of attraction exist between the molecules.

Methoxymethane is more spherical than ethanol so there is less contact surface area between the molecules.

There is hydrogen bonding between ethanol molecules but only dipole-dipole interactions between methoxymethane molecules.

The C−O bond in methoxymethane is much weaker than the O−H bond in ethanol.

Because H is bonded directly to O in ethanol there will be hydrogen bonding between ethanol molecules but in methoxymethane the O is bonded to C, not H, so no hydrogen bonding is possible.

The solubility of alcohols in water decreases in the order methanol > ethanol> propan-1-ol > butan-1-ol with methanol being the most soluble. Why does the solubility of alcohols in water follow this order?

As the length of the non-polar carbon chain increases there is increased London forces of attraction between the alcohol molecules and they tend to attract each other more than water molecules.

The polarity of the O−H bond decreases as the length of the carbon chain increases.

There is more contact between the alcohol molecules and water as the length of the carbon chain increases .

The polarity of the O−H bond is spread over the whole of the molecule and as the length of the carbon chain increases the charge density of the polarity decreases.

The hydrocarbon chain is non-polar so tends to attract the non-polar hydrocarbon chain of other alcohol molecules through London forces of attraction. This attraction increases as the chain length increases.

Total Score:

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MC test: Intermolecular forces

Multiple choice test on 4.4 Intermolecular forces

MC test: Intermolecular forces