The boiling point of $$ \mathrm{SnH}_{4} $$ is higher than that of $$ \mathrm{CH}_{4} $$ because... ( a. The molecular weight of $$ \mathrm{SnH}_{4} $$ is higher than in $$ \mathrm{CH}_{4} $$ b. $$ \begin{array}{l} ext { The shape of } \mathrm{SnH}_{4} ext { is linear } \ ext { meanwhile } \mathrm{CH}_{4} ext { is not } \ ext { C. } \mathrm{SnH}_{4} ext { can form hydrogen bonds } \ ext { but } \mathrm{CH}_{4} ext { molecules cannot }\end{array} $$ d. $$ \begin{array}{l}\mathrm{SnH}_{4} ext { is a polar molecule but } \mathrm{CH}_{4} \ ext { is non nolar }\end{array} $$

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The boiling point of $$ \mathrm{SnH}_{4} $$ is higher than that of $$ \mathrm{CH}_{4} $$ because... ( a. The molecular weight of $$ \mathrm{SnH}_{4} $$ is higher than in $$ \mathrm{CH}_{4} $$ b. $$ \begin{array}{l}	ext { The shape of } \mathrm{SnH}_{4} 	ext { is linear } \ 	ext { meanwhile } \mathrm{CH}_{4} 	ext { is not } \ 	ext { C. } \mathrm{SnH}_{4} 	ext { can form hydrogen bonds } \ 	ext { but } \mathrm{CH}_{4} 	ext { molecules cannot }\end{array} $$ d. $$ \begin{array}{l}\mathrm{SnH}_{4} 	ext { is a polar molecule but } \mathrm{CH}_{4} \ 	ext { is non nolar }\end{array} $$

UpStudy AI · Accepted Answer

1. Both $$\mathrm{SnH}_4$$ and $$\mathrm{CH}_4$$ are tetrahedral molecules, so the difference in geometry (as mentioned in option b) is not a factor.

2. Neither molecule forms hydrogen bonds because hydrogen bonding requires highly electronegative atoms (such as $$\mathrm{O}$$, $$\mathrm{N}$$, or $$\mathrm{F}$$) bonded to hydrogen. Therefore, option c is incorrect.

3. Due to the tetrahedral symmetry, neither $$\mathrm{SnH}_4$$ nor $$\mathrm{CH}_4$$ has a net dipole moment. This makes option d incorrect.

4. The key difference is the molecular weight. The molar mass of $$\mathrm{SnH}_4$$ is significantly higher than that of $$\mathrm{CH}_4$$:

- For $$\mathrm{SnH}_4$$: 
     $$
     M(\mathrm{SnH}_4) \approx M(\mathrm{Sn}) + 4 \times M(\mathrm{H}) \approx 118.71 + 4 \approx 122.71\, \text{g/mol}
     $$
   - For $$\mathrm{CH}_4$$: 
     $$
     M(\mathrm{CH}_4) \approx M(\mathrm{C}) + 4 \times M(\mathrm{H}) \approx 12.01 + 4 \approx 16.01\, \text{g/mol}
     $$
   
   Higher molecular weight increases the strength of London dispersion forces (induced dipole-induced dipole interactions). Stronger intermolecular attractions require more energy (in the form of heat) to overcome, leading to a higher boiling point.

Thus, the correct explanation is given by option a.
The boiling point of $$\mathrm{SnH}_4$$ is higher than that of $$\mathrm{CH}_4$$ because $$\mathrm{SnH}_4$$ has a higher molecular weight, which increases the strength of intermolecular forces, requiring more energy to boil.

The boiling point of is higher than
that of because…
( a. The molecular weight of is
higher than in
b.
d.

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