Mastering Chemistry Chapter 13 Question (att. forces, energies, etc.)

[robert zellmer]

I got a question from someone about a homework problem.  I've changed it 
slightly.
"Why is the ion-solvent attractive force greater for a smaller ion, such 
as Mg^2+
compared to Ca^2+?"  Related to this would be "Which is more soluble in H2O,
NaCl or Fe2O3"?

This has to do with the form of the ion-ion interaction energy and the form
of the ion-dipole interaction energy.  The equations for the energy due 
to these
interactions are very similar.

LE (for ionic cmpds) = (Q+)(Q-)/d

This is lattice energy (energy req. to sep. ions).  This is the energy 
required to
separate the ions in order for them to dissolve in solution.

Ion-dipole energy (ions in polar solvent) = (Q+)(u)/d^2

This is the energy released when ions from an ionic solid are solvated 
(surrounded)
by the solvent.  This is due to the ion-solvent interaction.  This term 
(energy) is
negative since it's energy released when attractive forces are formed.

Q is charge on an ion
u is dipole moment of the polar molecule
d is the distance between the particles (measured center-to-center)

Look at both equations just in terms of these variables and what they
tell us about these energies.

If the value for Q (the charge on an ion) inc. then both types of 
energies inc.
The more polar the solvent (large dipole moment) the greater the ID energy.
The smaller the ions or molecules the shorter the distance between them. A
smaller distance means greater energy for both LE and ID energy.

Thus, as the charge on the ions inc. the greater the LE and ID energy.
The smaller the ions the greater the LE and ID energy.

In terms of solubility of an ionic cmpd dissolved in a polar solvent, both
terms increase as the charges on the ions inc. and the size of the ions
(and solvent molecules) dec.  This is the "charge density" in my notes and
I spoke about in class (charge per unit volume or charge/distance). So why
does the solubility of ionic cmpds decrease as charges inc. and the ions get
smaller if both terms dec.?  That's due to the form of the two eqns.  As
the charge density inc. (larger charges and smaller ions) the LE inc. more
than does the ID energy.  Thus, the delta(H)_soln becomes more positive
(more endothermic, i.e. the energy gap between the solute/solvent and
solution gets bigger).  As this occurs the delta(H)_soln can become so
positive that not enough disorder, entropy, can be created during the
solution formation to overcome this uphill energy difference.

Think about the energy diagrams I covered in lecture.  The first step
would be to separate the ions in the ionic solid from each other. The
energy required to do this is related to the lattice energy.  The third
step is due to the formation of attractive forces between the ions and
the solvent molecules (lets say H2O) and is the ion-dipole energy.
Both of these steps are relatively large.  The second step would be the
energy req. to separate the H2O molecules, del(H)2 (this is approx. zero,
very small compared to the energy of the first and third steps).

So lets say we're talking about comparing the solubility of two ionic cmpds
in which one of them has larger charges (to keep things simple we'll say
the sizes of the different ions remains about the same).  As the charges on
the ions inc. both the LE and ID energies increase for both ionic cmpds.
However, the inc. in del(H)1 is bigger than the inc. in del(H)3 (for example
del(H)1 inc. by a factor of 2 while del(H)3 inc. by a factor of 1.5.).
This would mean the del(H)soln for the cmpd with bigger charges is more
positive than del(H)soln for the cmpd with smaller charges.  This will
tend to make the one with larger charges less soluble.

In summary, for ionic cmpds the bigger the charges on the ions and the
smaller the ions (the greater the LE) the higher the melting pt. the less
soluble the ionic cmpd.  NaCl is very soluble in water while Fe2O3 is
insoluble.

In a similar vein, I stated discussed a diamond dissolving in a liquid.
Diamonds are a covalent-network solid (i.e. a crystal of carbon atoms
held together with covalent bonds). The first step would be separating
the C atoms from each other (breaking a huge number of covalent bonds).
A tremendous amount of energy is required to do this, even if you don't
break every last C-C bond (del(H)1 is huge).  Again, the energy req.
to separate the solvent molecules (such as water, C6H14, etc.) are very
small compared to the first step.  The third step, the mixing (solvation)
step would involve only London Forces between the C atoms/fragments
and the solvent molecules.  This del(H)3 would be very small compared
to del(H)1.  Thus, the del(H)soln would be very large (very positive).
Even though the del(S)soln would be positive (there's an inc. in disorder,
entropy, going from very ordered diamond to C atoms/fragments) not
enough entropy can be created to overcome this large positive del(H)soln
so that a solution would form spontaneously.  Diamonds are insoluble
even at relatively high temps (except in solvents in which they would
react with the solvent).

I hope this helps.

Dr. Zellmer