Chapter 14: Problem 121
Which of the following order is incorrect? (1) \(\mathrm{PH}_{3}<\mathrm{AsH}_{3}<\mathrm{SbH}_{3}<\mathrm{NH}_{3}\) (Melting point) (2) \(\mathrm{PH}_{3}<\mathrm{AsH}_{3}<\mathrm{NH}_{3}<\mathrm{SbH}_{3}\) (Boiling point) (3) \(\mathrm{NH}_{3}>\mathrm{PH}_{3}>\mathrm{AsH}_{3}>\mathrm{SbH}_{3}>\mathrm{BiH}_{3}\) (Dipole moment) (a) only 1 (b) only 2 (c) both (1) and (2) (d) none is incorrect here
Short Answer
Step by step solution
Understanding the Physical Properties
Analyzing Melting Points
Evaluating Statement 1
Analyzing Boiling Points
Evaluating Statement 3
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Melting Point
For compounds like ammonia (\( ext{NH}_3\)) and the hydrogen chalcogenides (\( ext{PH}_3, ext{AsH}_3, ext{SbH}_3\)), this means that although ammonia might have a lower molecular mass, its strong hydrogen bonds significantly raise its melting point. Compare that to larger but less interactive molecules like \( ext{SbH}_3\), where mass contributes more dominantly than hydrogen bonding.
- This explains why ammonia’s melting point is actually higher than that of \( ext{SbH}_3\), despite its smaller size.
Boiling Point
In the case of \( ext{NH}_3\), its ability to form hydrogen bonds plays a substantial role. Hydrogen bonds make it more energy-intensive to break molecular forces and transition to the gaseous state. Therefore, ammonia, despite having a lower molecular mass than compounds like \( ext{SbH}_3\), exhibits a high boiling point due to this bonding.
- It's why the trend in statement 2, where \( ext{NH}_3\) comes before \( ext{SbH}_3\), holds true and showcases hydrogen bonding's influence.
Dipole Moment
Take \( ext{NH}_3\) (ammonia), for example, it has a pronounced dipole moment due to its trigonal pyramidal shape and significant electronegativity of nitrogen. The shape causes the electrons to focus more on one side, creating a distinct positive and negative end. Other molecules like \( ext{PH}_3, ext{AsH}_3\), and \( ext{SbH}_3\) lack this combination of shape and electronegativity interplay.
- The order from the problem reflects these concepts well, acknowledging ammonia's exceptional dipole moment compared to the heavier and less electronegative counterparts.