The correct arrangement of the compounds provided in the question in decreasing order of their boiling points is:
\( \text{H}_2\text{O} > \text{HF} > \text{NH}_3 > \text{PH}_3 \)
Let's understand why this order is correct, considering the intermolecular forces and molecular structures involved:
- Understanding Intermolecular Forces:
- Hydrogen Bonding: This is a strong type of dipole-dipole interaction that significantly elevates the boiling points of substances where it occurs. It is found in compounds like water (\(\text{H}_2\text{O}\)), hydrogen fluoride (\(\text{HF}\)), and ammonia (\(\text{NH}_3\)).
- Van der Waals Forces: Present in all molecules, but particularly significant in nonpolar ones like phosphine (\(\text{PH}_3\)). Van der Waals forces are generally weaker than hydrogen bonds.
- Boiling Points of the Given Compounds:
- Water (\(\text{H}_2\text{O}\)): Exhibits extensive hydrogen bonding because each molecule can form multiple hydrogen bonds. This results in a very high boiling point relative to its molecular weight.
- Hydrogen Fluoride (\(\text{HF}\)): Also exhibits hydrogen bonding, but each molecule is only able to form hydrogen bonds at one site (as compared to water's multiple bonding sites), giving it a slightly lower boiling point than water.
- Ammonia (\(\text{NH}_3\)): Forms hydrogen bonds as well, but not as extensively as water or HF. This gives it a moderate boiling point in the group.
- Phosphine (\(\text{PH}_3\)): Does not partake in hydrogen bonding and has weaker van der Waals forces, resulting in the lowest boiling point among the compounds mentioned.
- Conclusion: Based on the above analysis, the boiling points decrease in the sequence: \(\text{H}_2\text{O}\) (highest due to multiple hydrogen bonds) > \(\text{HF}\) (high due to hydrogen bonding) > \(\text{NH}_3\) (moderate due to limited hydrogen bonding) > \(\text{PH}_3\) (lowest, due to lack of hydrogen bonding).
Therefore, the correct answer is Option 2: \(\text{H}_2\text{O} > \text{HF} > \text{NH}_3 > \text{PH}_3\).