The reason behind relatively higher boiling point of water has to be explained. Concept Introduction: London forces: London(dispersion) forces happen because sometimes, when adjacent molecules line up so that they form temporary dipoles, there is a weak force of attraction between the end of one molecule where more of the electrons are accumulating at the moment and the end of a second molecule where the electrons are not staying at the moment. London(dispersion) forces are present between molecules of all molecular substances. Hydrogen bonding: Hydrogen bonding is a special form of dipole-dipole interaction. When a hydrogen atom is bonded to a small, electronegative atom (like N , O a n d F ), then the electronegative atom pulls electron density away from the hydrogen atom. This results a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom. The hydrogen atom on one molecule will be attracted to the electronegative atom on another molecule. Like this hydrogen bonding takes place. Dipole-dipole Interaction: This kind of attraction occurs between permanent dipoles in adjacent polar molecules or between two polar groups in the same large molecule. Molecules having permanent dipoles attract each other when the partial positive region of one is close to the partial negative region of another.

Question

Want to see the full answer?

Answer 1

Question

Chapter 9, Problem 55QRT

Interpretation Introduction

Interpretation:

The reason behind relatively higher boiling point of water has to be explained.

Concept Introduction:

London forces:

London(dispersion) forces happen because sometimes, when adjacent molecules line up so that they form temporary dipoles, there is a weak force of attraction between the end of one molecule where more of the electrons are accumulating at the moment and the end of a second molecule where the electrons are not staying at the moment. London(dispersion) forces are present between molecules of all molecular substances.

Hydrogen bonding:

Hydrogen bonding is a special form of dipole-dipole interaction. When a hydrogen atom is bonded to a small, electronegative atom (like N, O and F), then the electronegative atom pulls electron density away from the hydrogen atom. This results a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom. The hydrogen atom on one molecule will be attracted to the electronegative atom on another molecule. Like this hydrogen bonding takes place.

Dipole-dipole Interaction:

This kind of attraction occurs between permanent dipoles in adjacent polar molecules or between two polar groups in the same large molecule. Molecules having permanent dipoles attract each other when the partial positive region of one is close to the partial negative region of another.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

I

Draw the anti-Markovnikov product of the hydration of this alkene. this problem. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for esc esc ☐ Explanation Check F1 1 2 F2 # 3 F3 + $ 14 × 1. BH THE BH3 2. H O NaOH '2 2' Click and drag to start drawing a structure. F4 Q W E R A S D % 905 LL F5 F6 F7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility < & 6 7 27 8 T Y U G H I F8 F9 F10 F11 F12 9 0 J K L P + // command option Z X C V B N M H H rol option command

AG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, N