Question

for each pair of compounds listed, check the box next to the one with the higher boiling point. compounds higher boiling point rn ne ch₃ch₃ c₂h₄ sncl₄ gecl₄

Explanation:

Step1: Analyze Rn vs Ne

Rn (Radon) and Ne (Neon) are noble gases. For noble gases (non - polar molecules with only London dispersion forces), the boiling point is related to the molar mass. The molar mass of Rn ($M_{Rn}$) is greater than the molar mass of Ne ($M_{Ne}$). Since London dispersion forces are stronger for larger (more massive) molecules, Rn has a higher boiling point.

Step2: Analyze $CH_3CH_3$ vs $C_2H_4$

$CH_3CH_3$ (ethane) and $C_2H_4$ (ethene) are both non - polar hydrocarbons with London dispersion forces. The molar mass of $CH_3CH_3$ ($M_{CH_3CH_3}=30\ g/mol$) and $C_2H_4$ ($M_{C_2H_4} = 28\ g/mol$). Also, ethane has a more spherical (less unsaturated) structure compared to ethene, and the molar mass of ethane is slightly higher. So, $CH_3CH_3$ has a higher boiling point.

Step3: Analyze $SnCl_4$ vs $GeCl_4$

$SnCl_4$ (tin(IV) chloride) and $GeCl_4$ (germanium(IV) chloride) are both non - polar molecules (tetrahedral geometry, symmetric) with London dispersion forces. The molar mass of $SnCl_4$ ($M_{SnCl_4}$) is greater than the molar mass of $GeCl_4$ ($M_{GeCl_4}$) because Sn is below Ge in the periodic table (higher atomic mass). Stronger London dispersion forces in $SnCl_4$ lead to a higher boiling point.

Answer:

• For the pair Rn and Ne: Check the box next to Rn.
• For the pair $CH_3CH_3$ and $C_2H_4$: Check the box next to $CH_3CH_3$.
• For the pair $SnCl_4$ and $GeCl_4$: Check the box next to $SnCl_4$.