Melting and Boiling Points

In the last three sections, we have mostly focused our attention solely on molecular substances, learning how to determine their shape, molecular polarity, and type of IMFs they experience. We will now zoom back out to interpret what we have learned in the context of network substances again.

When substances melt or boil, the attractive forces holding particles together must be broken. Sometimes, those forces are actual chemical bonds: ionic, covalent, or metallic, as shown in the first three rows of the table below. Because these bonds are relatively strong, it takes lots of heat to melt these kinds of substances.

*Note: there are some exceptions. For example, mercury experiences a particularly weak form of metallic bonding, causing its melting point to be exceptionally low, -38 °C. However, most metals have melting points of at least several hundred degrees Celsius.

When molecular substances melt, it is not bonds that must be broken but IMFs. Since IMFs are much weaker than bonds, the temperatures required to melt these are much lower, as you can see. In addition, you can observe that substances with the weakest IMFs have the lowest melting points among the molecular substances.

The takeaway here, and what you will be expected to do with this knowledge on homework and tests, is fairly simple. Melting and boiling points are higher in substances with stronger forces holding them together. So you should be able to identify the IMFs present in a molecular substances, and know which are stronger and weaker. You should also be able to identify network substances, and know that they are held together by chemical bonds. Finally, you should be able to apply this knowledge to predict, for a given set of substances, which will have the lowest melting points and which will have the highest.

Practice

You can practice on the following sets of substances. For each one, order them from lowest melting point to highest melting point and identify the forces holding each one together.

1. NCl₃, Cl₂, MgCl₂

2. C, CH₄, CH₂O

3. HF, NF₃, F₂

Answers

1. Cl₂ would be lowest, since it is non-polar (dispersion forces). Next would be NCl₃, a polar substance (dipole-dipole forces). MgCl₂ would be much higher than both these (ionic bonding)

2. CH₄ would be lowest, since it is non-polar (dispersion forces). Next would be CH₂O, a polar substance (dipole-dipole forces). C would be highest, since elemental carbon (diamond) is a covalent network (covalent bonding)

3. F₂ would be lowest, since it is non-polar (dispersion forces). Next would be NF₃ , a polar substance (dipole-dipole forces). HF would be highest (hydrogen bonding).