Questions: For each pair of compounds listed, check the box next to the one with the higher boiling point. compounds higher boiling point CI4 0 CBr4 (Circle) GeCl4 GeF4 (Circle) SiH4 (Circle) SiI4 (Circle)

For each pair of compounds listed, check the box next to the one with the higher boiling point.

compounds higher boiling point

CI4 0

CBr4 (Circle)

GeCl4

GeF4 (Circle)

SiH4 (Circle)

SiI4 (Circle)

Transcript text: For each pair of compounds listed, check the box next to the one with the higher boiling point. \begin{tabular}{|c|c|} \hline compounds & higher boiling point \\ \hline $\mathrm{CI}_{4}$ & 0 \\ $\mathrm{CBr}_{4}$ & $\bigcirc$ \\ \hline $\mathrm{GeCl}_{4}$ \\ $\mathrm{GeF}_{4}$ & $\bigcirc$ \\ \hline $\mathrm{SiH}_{4}$ & $\bigcirc$ \\ $\mathrm{SiI}_{4}$ & $\bigcirc$ \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Understanding Boiling Points

Boiling points are influenced by molecular weight, intermolecular forces, and molecular structure. Generally, larger molecules with stronger intermolecular forces have higher boiling points.

Step 2: Analyzing the First Pair: $\mathrm{CI}_{4}$ vs. $\mathrm{CBr}_{4}$

$\mathrm{CI}_{4}$ (Carbon tetraiodide) and $\mathrm{CBr}_{4}$ (Carbon tetrabromide) are both tetrahalides of carbon.
$\mathrm{CBr}_{4}$ has a higher boiling point than $\mathrm{CI}_{4}$ because bromine is less massive than iodine, but $\mathrm{CBr}_{4}$ has stronger van der Waals forces due to better packing efficiency.

Step 3: Analyzing the Second Pair: $\mathrm{GeCl}_{4}$ vs. $\mathrm{GeF}_{4}$

$\mathrm{GeCl}_{4}$ (Germanium tetrachloride) and $\mathrm{GeF}_{4}$ (Germanium tetrafluoride) are both tetrahalides of germanium.
$\mathrm{GeCl}_{4}$ has a higher boiling point than $\mathrm{GeF}_{4}$ because chlorine is larger than fluorine, leading to stronger van der Waals forces.

Step 4: Analyzing the Third Pair: $\mathrm{SiH}_{4}$ vs. $\mathrm{SiI}_{4}$

$\mathrm{SiH}_{4}$ (Silane) and $\mathrm{SiI}_{4}$ (Silicon tetraiodide) are both silicon compounds.
$\mathrm{SiI}_{4}$ has a higher boiling point than $\mathrm{SiH}_{4}$ because iodine is much larger and heavier than hydrogen, resulting in stronger van der Waals forces.

Final Answer

$\mathrm{CBr}_{4}$ has a higher boiling point than $\mathrm{CI}_{4}$: $\boxed{\mathrm{CBr}_{4}}$
$\mathrm{GeCl}_{4}$ has a higher boiling point than $\mathrm{GeF}_{4}$: $\boxed{\mathrm{GeCl}_{4}}$
$\mathrm{SiI}_{4}$ has a higher boiling point than $\mathrm{SiH}_{4}$: $\boxed{\mathrm{SiI}_{4}}$

Was this solution helpful?