Hydrogen Bonding (H-FON Bonding) A very special kind of dipole-dipole force occu
ID: 977589 • Letter: H
Question
Hydrogen Bonding (H-FON Bonding) A very special kind of dipole-dipole force occurs in a bulk sample of hydrogen fluoride. Consider the following schematic. This type of intermolecular force exists in samples of compounds that have O H and N H bonds as well. Since all of them involve an atom that is bound to a hydrogen atom, these dipole-dipole forces are (unfortunately) called "hydrogen bonds". Hydrogen bonds, typically on the order of 30 kJ/mol, are many times stronger than the ordinary dipole-dipole force (and, thus, the reason they are classified as a distinct type of intermolccular force). What are two special characteristics of the atoms of H, N, O, and F that sets them apart from other elements? Use Coulomb's Law (F Q_+ Q/r^2) to explain why hydrogen bonding only occurs in compounds with N H, O H, and F H covalent bonds. Consider the boiling points of the hydrides of the Group VIA elements: Describe the intermolecular forces that exist in a bulk sample of each compound. Which intermolecular forces predominate (i.e., are the strongest)? Explain why H_2O has the highest boiling point. Explain the trend in boiling point in going from H_2S to H_2Se to H_2Te.Explanation / Answer
7 a) only these three element N, O , F can form hydrogen bonding with H atom, all are having similar size and electronegativity.
b) force of attraction is directly propotional product of charges and inversely propotional to distance between them.
So first row element N, O, F will have smaller size , so smaller distance , hence effective attraction, in their heviar analogue charge will be similar but distance between them will be higher so there will be no H-bonding with P, S, Cl.
c) In H2O , there will be two H-bonding per molecule so it exists as a liquid and has higher boiling point , whereas there will be no H-bonding in remaining heviar analogue in their family, so they all exists as a vapour and hence low boiling point.
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