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.
-----------------------
"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.