When we look at ionization energy for this group, everything is straightfoward. IE decreases as you move down.
But other trends don't seem to make sense to me.
<b>Hypothesis:</b> There should be a general constant trend where enthalpy of formation of these salts should decrease from Li to Cs
because of the ionization energies.
<b>Data (from General Chemistry by Linus Pauling)</b>
Enthalpy of formations:
--------------------------------
For M+(aq)
Li < K < Cs < Rb < Na
For M2O(s)
Li < Na < K < Rb < Cs
For MH(s)
Li < Cs < Rb < Na = K
MF
Li < Na < K < Rb < Cs
MCl
K < Cs < Rb < Na < Li
MBr
Cs < K < Rb < Na < Li
MI
Cs < Rb < K < Na < Li
--------------
<b>Conclusion:</b>
1) In aqueous solution, it is more thermodynamically favorable to form Li+ than all the others, probably due to the fact that is is more easily
hydrated due to its small size (but what happened to Na+).
2) Almost the same trend is observed when forming oxides, the order is the exact opposite of what was expected by looking at ionization energies.
Could be explained by the fact that there is more electrostatic attraction between smaller ions and oxygen.
3)No conclusion (Need help)
4) Follows same trend as oxides.
5,6,7) Start to reverse the trend of oxides and fluorides. (WHY? help needed)
-----------
So can anyone help me try and figure oout what is going on here? Thanks.
[Edited on 7/13/2006 by guy]12AX7 - 12-7-2006 at 19:16
Are these molecules or solids? Solids have crystalline binding energy involved, too (which I believe is mostly what keeps SiC, WC and such together,
since they have such massive melting points (fractional eV average energy!), but little liquid range before vaporizing; also compare formation of
C(4+) carbide ion!).
Timguy - 12-7-2006 at 19:33
Well I can see a correlation between electronegativity of the anion with the order of the enthalpy values. For F and O, the enthalpy values depend
only on the size of the cation, because a smaller cation will lead to stronger electrostatic forces in the crystal. The less electronegative the
anions, the more dependent they are on the ionization energies of the metal because they depend more one gaining the electron rather than taking it,
and their ionic character depends on it. So, in conclusion the enthaply of formations for salts with highly electronegative anions depends on the
electrostatic interaction in the crystal, and the salts with less electronegative anions depend more on the ionic character of the salts.
For aqueous solutions, the smaller the cation, the lower the enthalpy of hydration leading to a more stable ion. Still, why is Na+ the lowest one?
The trend in the hydrides is still kind of weird.
[Edited on 7/13/2006 by guy]JohnWW - 13-7-2006 at 09:30
Those enthalpies of formation are not always consistent with electronegativities of the alkali mmetals. Also, what about francium, or is it too
unstable (half-lives of longest-lived isotopes 19 and 21 minutes) for its compounds to be examined in detail?franklyn - 13-7-2006 at 10:57
Quote:
Originally posted by guy
other trends don't seem to make sense to me.
What more proof do you need that chemistry is not a science
.guy - 13-7-2006 at 17:37
Attached is a chart comparing Lattice energy and Electronegativity difference(ionic characer) with Enthalpy of formation.
From the chart it is apparent that the enthalpy of formations for fluorides and oxides(not in chart) are more dependent on lattice energy. All the
other halides are more dependent on ionic character.
In the boxed area, it shows an inverse relationship between lattice energy and enthalpy of formation, but it is not caused by it.
The second graph shows increase in the slopes, showing that as electronegativity of the anion decrease, the enthalpy of formation is dependent on the
cation. The fluorides seem to be less stable by increase in electronegativity difference, but this is not the cause, so it is not being determined by
ionic character.
