For each pair of compounds listed, check the box next to the one with the higher boiling point. compounds higher boiling point SiCl
SiI
SnH
GeH
CCl
CBr
X 5
The Correct Answer and Explanation is:
Here are the correct selections for the compounds with the higher boiling points:
- SiI₄ (higher boiling point than SiCl₄)
- SnH₄ (higher boiling point than GeH₄)
- CBr₄ (higher boiling point than CCl₄)
Explanation
Boiling points are influenced by the strength of intermolecular forces, molecular size, and polarizability. Let’s break down each pair:
1. SiCl₄ vs. SiI₄
- Both are tetrahalides of silicon with similar molecular geometry (tetrahedral, non-polar), and both experience London dispersion forces.
- Iodine atoms are larger and more polarizable than chlorine atoms, leading to stronger dispersion forces in SiI₄.
- Hence, SiI₄ has a higher boiling point due to stronger van der Waals interactions.
2. SnH₄ vs. GeH₄
- Both are group 14 hydrides, but SnH₄ (stannane) is heavier than GeH₄ (germane).
- As molecular weight increases down the group, London dispersion forces become stronger.
- Despite being gases at room temperature, SnH₄ has a higher boiling point than GeH₄ due to increased mass and polarizability.
3. CCl₄ vs. CBr₄
- Both are non-polar, symmetrical tetrahalides of carbon, so dispersion forces dominate.
- Bromine is larger and more polarizable than chlorine, making CBr₄ capable of stronger London dispersion forces.
- Thus, CBr₄ has a higher boiling point than CCl₄.
Summary Table:
| Compounds | Higher Boiling Point |
|---|---|
| SiCl₄ / SiI₄ | ✅ SiI₄ |
| SnH₄ / GeH₄ | ✅ SnH₄ |
| CCl₄ / CBr₄ | ✅ CBr₄ |
These trends are consistent with periodic trends in atomic size and intermolecular forces, particularly dispersion forces which are more pronounced in heavier, more polarizable molecules.
