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Chapter 7: Problem 4

Based on the strength of their molecular dipole moments, which compound should have the higher boiling point, \(\mathrm{H}_{2} \mathrm{~S}\) or \(\mathrm{H}_{2} \mathrm{O}\) ? Explain.

Short Answer

Expert verified
Based on the electronegativity differences, H₂O has a stronger molecular dipole moment (1.24) than H₂S (0.38). Therefore, H₂O should have a higher boiling point than H₂S due to its stronger molecular dipole moment.

Step by step solution

01

Determine the electronegativity of each involved atom

To evaluate the molecular dipole moments of the compounds, first, we need to understand the electronegativity of the involved atoms. Electronegativity values can be found in various references, such as an electronegativity table or the periodic table. In general, electronegativity increases as you move across the periodic table from left to right and decreases as you move down a group. Electronegativity values for atoms involved: - Hydrogen (H): 2.20 - Oxygen (O): 3.44 - Sulfur (S): 2.58
02

Calculate the electronegativity differences in H₂S and H₂O

Now that we have the electronegativity values for the involved atoms, we can calculate the electronegativity differences in the two compounds: For H₂S: Electronegativity difference: |2.58 (S) - 2.20 (H)| = 0.38 For H₂O: Electronegativity difference: |3.44 (O) - 2.20 (H)| = 1.24
03

Analyze the molecular dipole moments based on electronegativity differences

As we can see, the electronegativity difference in H₂O (1.24) is greater than that in H₂S (0.38). This suggests that the molecular dipole moment in H₂O is stronger than in H₂S. Recall that a greater molecular dipole moment results in a higher boiling point.
04

Conclusion

Based on the strength of their molecular dipole moments determined by the electronegativity differences, H₂O should have a higher boiling point than H₂S due to its stronger molecular dipole moment (1.24 > 0.38).

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Most popular questions from this chapter

Which of the molecules below would not form hydrogen bonds? (a) \(\mathrm{CH}_{3} \mathrm{OH}\) (b) \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{COOH}\) (d) \(\mathrm{NH}_{3}\) (e) None of these would form hydrogen bonds.

Which of the following would you expect to have the highest boiling point? Justify your choice. (a) Propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\) (b) Carbon dioxide, \(\mathrm{CO}_{2}\) (c) Ethyl alcohol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (d) Methyl fluoride, \(\mathrm{CH}_{3} \mathrm{~F}\)

How is it possible for a nonpolar molecule to have a higher boiling point than a polar one?

Would you expect \(\mathrm{CCl}_{4}\) or \(\mathrm{CBr}_{4}\) to have the higher boiling point? Explain your answer.

Draw a picture that shows how three polar \(\mathrm{HBr}\) molecules in the gas phase would attract one another. What kind of intermolecular force is involved?
See all solutions

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