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

[Zellmer, Robert]

I got a question from someone dealing with sections 13.1 and 13.3.  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.



The lattice energy (LE) is the energy required to separate the ions in order

for them to dissolve in solution.



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



Q is charge on an ion

d is the distance between the ions (measured center-to-center)



The ion-dipole (ID) energy is the energy released when ions from an ionic

solid dissolve and 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.



ID energy (ions in polar solvent) = (Q+)(u)/d^2



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

inc.?  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, LE.  The third step is due to the formation

of attractive forces between the ions and the solvent molecules (let's say H2O)

and is the ion-dipole, ID, energy and energy is released in this step.  Both 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

energies of the first and third steps).



So, let's say we're talking about comparing the solubility of two ionic cmpds in

which one of them has larger charges.  For simplicity I'll use an ionic cmpd with

charges of +1 and -1 and an ionic cmpd with charges of +2 and -2 and sizes

of the different ions about the same.  I'll also assume the dipole moment of the

solvent is 0.5 ("charge" due to the dipole moment).



Remember, del(H_1) is the LE, del(H_2) is approximately zero (very small

compared to the first and third steps) and del(H_3) is the solvation energy

(also called the del(H_mix).  The delta H of solution is given by,

del(H_soln) = del(H_1) + del(H_2) + del(H_3) = del(H_1) + del(H_3) since

del(H_2) ~ 0.



As the charges on the ions inc. from +1/-1 to +2/-2 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).  The del(H_1) inc. by a factor of 4 while

del(H_3) inc. by a factor of 2.  This would mean the del(H_soln) for the

cmpd with bigger charges is more positive (greater) than the del(H_soln)

for the cmpd with smaller charges.  This will tend to make the one with

larger charges less soluble.



In general, for ionic cmpds, the bigger the charges on the ions and the smaller

the ions (the greater the LE) the higher the melting pt. and the less soluble the

ionic cmpd.  NaCl is very soluble in water while Fe2O3 is insoluble.  Also, in

general, the process of an ionic solid dissolving in H2O is endothermic,

del(H_soln) > 0.  That means an increase in entropy is necessary for the

cmpd. to dissolve.  This isn't always the case.  There are cases of ionic cmpds

dissolving in water which have del(H_soln) < 0, exothermic soln formation

(which would mean an inc in entropy is not necessary for the soln to form).



In a similar vein, I 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.) is 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 in just about any solvent even at relatively high temps,

except in solvents in which they would react with the solvent to form a soluble

product.  In class I stated it could be oxidized by concentrated HNO3. That actually

wouldn't even happen except maybe at elevated temps.  There are a few very

strong oxidizing acids or combinations of oxidizing acids with other powerful

oxidizing agents, which might dissolve a diamond due to a reaction.  Diamonds

are almost indestructible unless they're under extreme conditions because of the

very strong covalent bonds between the C atoms.



I hope this helps.



Dr. Zellmer
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