General Chemistry

Question

The charges and sizes of the ions in an ionic compound affect the strength of the electrostatic attraction holding that compound together.

Based on ion charges and relative ion sizes, arrange these ionic compounds by their expected melting points.

NaF, MgO, MgF2, KBr

Answer 1

The stronger the attractions, the higher will be the melting point. So, the question is essentially asking us to arrange these compounds by their attractions.

The attractions will be based on two main variables: (1) charge and (2) size. The smaller the radii, the stronger the attraction. Looking first at charge, we have...

NaF: Na +1 and F -1

MgO: Mg +2 and O -2

MgF₂: Mg+2 and F -1

KBr: K +1 and Br-

Since both Mg compounds have a cation with a 2+ charge, the attractions will be stronger than for the NaF and KBr. So NaF and KBr will have lower melting points than MgO and MgF₂. Now, we turn to ionic size. Na is smaller than K and F is smaller than Br so NaF will have greater attractions than KBr and thus will have a higher melting point. So, to this point, we have melting points that rank NaF > KBr. Now we turn to MgO and MgF₂.

The cation is Mg in both compounds, so we look at the anion. F is smaller than O but recall that it has a charge of only 1-. Even though O is larger than F, it carries a charge of 2- and charge plays a greater role than size in determining the attraction. So, comparison of these two compounds predicts stronger attraction for MgO compared to MgF₂ and thus MgO would have a higher melting point.

In summary, we would have the following rank order for melting points:

MgO > MgF₂ > NaF > KBr