Final answer:
CCl_4 would be more volatile than SiBr_4. This is because SiBr_4, with heavier bromine atoms, would likely exhibit stronger London dispersion forces than CCl_4, thus having a lower volatility.
Step-by-step explanation:
If carbon tetrachloride (CCl4) and silicon tetrabromide (SiBr4) were both liquids at the same temperature, the one that would be more volatile, meaning having a higher tendency to vaporize, depends on their respective boiling points and the strength of intermolecular forces. According to the information given, silicon tetrachloride (SiCl4) is a nonpolar, low-boiling liquid at 57 °C and is similar in nature to CCl4, which is also a nonpolar liquid. Although the question asks about SiBr4, we can assume it has properties similar to SiCl4 due to being in the same group. Both compounds would be subject to London dispersion forces, but because SiBr4 has heavier bromine atoms compared to the chlorine atoms in CCl4, SiBr4 would likely have larger London dispersion forces and hence a lower volatility. Following this logic, CCl4 would be more volatile of the two.
When comparing boiling points as an example of predicting volatility, ICI was noted to have a higher boiling point than Br2 due to ICI's polar nature resulting in dipole-dipole attractions in addition to London dispersion forces, whereas Br2 is nonpolar and has only London forces. Similarly, although both SiBr4 and CCl4 are nonpolar, the larger size and mass of SiBr4 would mean stronger London forces compared to CCl4, leading to a lower volatility for SiBr4.