intermolecular forces - Why does silicon tetrafluoride have a higher melting point than sulfur tetrafluoride?

So looking at the Wikipedia pages of sulfur tetrafluoride and silicon tetrafluoride, the melting points are −121 °C and −90 °C respectively, and so $\ce{SiF4}$ has the higher melting point. However, their boiling points are −38 °C and −86 °C, respectively, giving $\ce{SF4}$ the higher boiling point.

I can justify that $\ce{SF4}$ has the higher boiling point because it is more polar since its Lewis structure has a lone pair on the sulfur atom, and so it experiences greater dipole-dipole forces. However, why does $\ce{SF4}$ have a lower melting point?

Answer

High symmetry molecules fit into crystal lattices especially well (higher m.p.), but are volatile for having fewer van der Waals interactions (lower b.p.).

$$\begin{array}{lrr} \hline \text{Compound} & \text{m.p.}/\pu{°C} & \text{b.p.}/\pu{°C} \\ \hline \text{pentane} & −130 & 36.1 \\ \text{isopentane} & −160 & 27.2 \\ \text{neopentane} & −18 & 9.5 \\ \hline \text{octane} & −57 & 125 \\ \text{2-methylheptane} & −110 & 117 \\ \text{2,2,4-trimethylpentane} & −107 & 99 \\ \text{2,2,3,3-tetramethylbutane} & 101 & 107 \\ \text{cubane} & 131 & 133 \\ & & \text{(sub. at r.t.)*} \\ \hline \text{decane} & −30  & 174 \\ \text{isodecane} & −75 & 167 \\ \text{adamantane} & 270 & \text{sub.} \\ \hline \end{array}$$

* sub. — sublimes; r.t. — room temperature.