does hydration energy become less exothermic down group 2? is it because of the weaker ion-dipole attractions due to lower charge densities?
will compounds with hydrogen bonding have pd pd also?
Originally posted by ohnoez!:does hydration energy become less exothermic down group 2? is it because of the weaker ion-dipole attractions due to lower charge densities?
will compounds with hydrogen bonding have pd pd also?
Both the endothermic lattice dissociation enthalpies and the exothermic hydration enthalpies are decreased in magnitude as you go down Group II, for the same reason : decrease in charge density results in weaker ionic bonds and weaker ion-dipole interactions.
If the anion is small (eg. OH-), the rate of decrease in endothermic lattice dissociation enthalpy outweighs the rate of decrease of hydration enthalpy, as you go down Group II; consequently, solution enthalpy becomes more exothermic (or less endothermic) as you go down Group II. Therefore solubility of Group II hydroxides increase down the group.
If the anion is large (eg. CO3 2- or SO4 2-), the rate of decrease of exothermic hydration enthalpy outweighs the rate of decrease of endothermic lattice dissociation enthalpy, as you go down Group II; consequently, solution enthalpy becomes more endothermic (or less exothermic) as you go down Group II. Therefore solubility of Group II carbonates & sulfates decrease down the group.
However, this is only half of the full picture. The other half is entropy. Gibbs free energy completes the picture. Near the top of the group (ie. high charge density), the decrease in entropy during the hydration process outweighs the increase in entropy during the lattice dissociation process. Near the bottom of the group the opposite occurs.
The result, is a complex balance between enthalpy versus entropy, lattice dissociation versus hydration. Group II carbonates decrease in solubility from Be to Sr (because thus far, enthalpy effect outweighs entropy effect), thereafter solubility increases from Sr to Ba (because from here on, entropy effect outweighs enthalpy effect).
For 'A' level H2 Chemistry, usually discussing enthalpy effect will suffice. Only in cases where enthalpy effect does not adequately explain the solubility trend, then bring in entropy, effect. As a further example, the solution or dissolving process of ammonium salts are endothermic (ie. endothermic lattice dissociation enthalpy outweighs exothermic hydration enthalpy), yet all ammonium salts are soluble. This is due to the favourable (ie. positive) entropy effect outweighing the unfavourable (ie. endothermic) enthalpy effect.
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"Permanent Dipole - Permanent Dipole" Van der Waals interactions (or more specifically, forces of attraction) are between polar molecules, which are not capable of hydrogen bonding.
"Permanent Dipole - Induced Dipole" Van der Waals interactions (or more specifically, forces of attraction) are between polar molecules and non-polar molecules.
"Instantaneous Dipole - Induced Dipole" Van der Waals interactions (or more specifically, forces of attraction) are between non-polar molecules. (Equivalently, terms such as "London forces" or "Dispersion forces" or "temporary dipoles" or "induced dipoles" or "induced dipole-dipoles" or "induced dipole - induced dipole forces", or simply "Van der Waals forces" may be used to refer to such interactions between non-polar molecules).
Hydrogen bonding, refers to the particularly strong permanent dipole - permanent dipole interactions that exist between molecules capable of hydrogen bonding. These are electrostatic attractions (with some degree of covalent nature) between a partially positively charged H atom (for the partial positive charge to be of sufficient magnitude for H-bonding, the H atom must be covalently bonded to an electronegative N, O or F atom) and a partially or formally negatively charged N, O or F atom (if the N, O or F atom has a partial or formal positive charge, no hydrogen bonding is possible with a partially positively charged H atom, because positive-positive like-charges repel) with at least one lone pair available.
Ion-dipole interactions are similar in strength to hydrogen bonding; usually slightly stronger, though this varies depending on individual charge densities of the ions involved. There is some overlap between some ion-dipole interactions and hydrogen bonding.
For instance, when aqueous sodium hydroxide is added to phenol, the phenol ppt dissolves. This is because when phenol is deprotonated, the resulting solute-solvent interactions between the phenoxide ion and water solvent, are stronger than the limited hydrogen bonding between phenol and water. These stronger interactions may be thought of as ion-dipole interactions, or as particularly strong hydrogen bondings (there is an upgrade in strength of hydrogen bonding because the partially negatively charged O atom in phenol has been upgraded to a formally negatively charged O atom in phenoxide ion. The O atom in phenoxide ion has one bond pair and 3 lone pairs, conferring on the O atom a negative formal charge. Yes, there is delocalization by resonance of the negative charge into the benzene ring, specifically the ortho and para carbon atoms; but because carbon is not as electronegative as oxygen, most of the negative charge in the resonance hybrid is still borne by the O atom).
Similarly do hydroxide ions have ion-dipole or hydrogen bonding with water? The interaction between the partially positively charged H atom of OH-, and water, is obviously hydrogen bonding. But the interaction between the negatively formally charged O atom in OH-, and water, may be thought of, or argued to be, either ion-dipole or hydrogen bonding. Both ways of looking at this, are acceptable by Cambridge at 'A' levels.
The interaction between NH4+ and H2O though, should be more accurately regarded as hydrogen bonding rather than ion-dipole. This is because the positively formal charged N atom does not have the opportunity for ion-dipole interaction with water. Why? Because of the 4 partially positively charged H atoms arranged tetrahedrally about the N atom. Before any water molecule can get close enough to the central positively charged N atom, these tetrahedrally arranged H atoms will seize the opportunity to hydrogen bond with the water molecule, denying the N atom the chance for any such interaction.
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