What is the relationship between the melting point and the atomic radius?

Answer 1

The answer varies depending on whether or not you are dealing with an ionic compound or a pure substance. Other substances will not fit the following rules. As usual, transition metals often find exception.

As you go down a column in the Periodic Table, the atomic radius increases and melting point also increases. This is most obvious from the group 7A elements, the halogens. Caution, this is not only because of the atomic radius. It's also because we are comparing similar nonpolar molecules to one another. The melting and boiling points increase also because of random London dispersion forces. This is a type of force contributing most when you are comparing similar molecules. Iodine, I2, with a large radius, is solid at room temperature. It hasn't hit its melting point yet. Bromine, Br2, with a smaller radius, is a liquid at room temperature. Its melting point was lower than I2; it has already melted. Since chlorine, Cl2, with the smallest radius yet, is already gaseous at room temperature, it is safely assumed that its melting point is lower than either Br2 or I2. London dispersion forces also account for pure cesium's higher melting point vs. pure sodium, pure calcium's higher melting point vs. pure magnesium, etc.

When you consider compounds, however, the game changes. Radii and ionic charge must be considered. Magnesium oxide (MgO), for example has charges of 2+ and 2-, respectively. These ions are strongly attracted to each other and the radius between them is very small as a result. Strong ionic bonds and a small atomic radius mean a very high lattice energy, which is directly related to boiling and melting points.

Let's consider Na2O. Our anion is the same, oxygen, with charge 2-. Compared to the magnesium cation, sodium with charge 1+ does not attract as strongly to the oxygen and so the ionic bond is weaker. Also, the sodium ion radius is slightly larger than the magnesium ion radius because sodium has fewer protons to attract the same number of electrons. This makes the overall lattice energy much weaker compared to MgO! As we expect, the melting point of Na2O is much lower than that of MgO.

Answer 2

As you go down group 1 (the alkali metals) the melting point decreases as the atomic radius increases.[1] For example Lithium has an electronic configuration of 2,1 so it only has two shells so there are stronger forces of attraction between the shells which means more energy is required to break these forces. Whereas, Potassium has an electronic configuration of 2,8,8,1 thus it has 4 shells, therefore there is less force of attraction between the shells which means there is less energy required to break these forces of attraction between the shells.