1281371269
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Thermal Decomposition
My Chemistry textbook isn't explaining itself well:
It says that group 1 nitrates, with the exception of LiNO3, decompose to form nitrites whilst group 2 nitrates and LiNO3 decompose to form oxides.
It has stated that smaller ions make more stable lattices due to the smaller gap between them, increasing the strength of the attraction.
It then explains the difference in decomposition products by saying that for the relatively small group 1 cations, the relatively small change in size
from NO3- to NO2- is sufficient to achieve stability, whereas the larger group 2 cations need the change to the much larger O2- ion.
But Li is the smallest of the lot. So why does it form an oxide, not a nitrite?
But the explanation is even more flawed because group 2 metals actually form smaller ions. I'm completely confused. My only guess is that it means
that group 2 cations bond with two NO3 anions, thus have a larger overall size.
[Edited on 5-4-2010 by Mossydie]
[Edited on 5-4-2010 by Mossydie]
[Edited on 5-4-2010 by Mossydie]
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Arrhenius
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Hm... There's no way the nitrite anion is smaller than an oxide ion - remember that atomic radius decreases towards fluorine, and that anionic radii
are smaller than atomic radii. Firstly it's Li2O, so this may be causing some confusion. Likewise it's LiNO2.
I would also guess that the reaction of a group 2 metal losing NOx to form MO is entropically more favored than for group 1 - this is purely based on
reaction stoichiometry. As for lithium forming an oxide and not a nitrite, this is probably a result - like you said - of the cationic radius of
lithium being quite small. The nitrite decomposition product may form incipiently, but it may still be a rather large gegenion and thus undergo
further decomposition to a more stable lattice structure.
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DJF90
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It's been a while since I've done this, but I recall having to use Kapustinskii's equation and perhaps Gibb's free energy. If you work it out
correctly, you'll find that the decomposition of LiNO3 to LiNO2 is not half as favourable as that to Li2O (based on lattice enthalpies alone) - add to
that the entropic effect (using G=H-TS) and you're laughing. If you look at Kapustinskii's equation you can make some basic observations: Smaller ions
give a higher lattice enthalpy (i.e. more favourable), more highly charged ions also contribute to a larger lattice enthalpy, and more ion's in the
unit formula again contribute to a larger lattice enthalpy. You can also use this kind of rationale to explain why the alkali metals form different
oxides upon burning in oxygen (again, it is lattice enthalpy at work here).
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1281371269
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I tried to think it through logically and came up with this explanation, is it at all correct?
Group two form 2+ ions, thus are more stable with the oxide anion that has a 2- charge - with oxides the lattice can fit together better than with
nitrites, which would require two anions per group two metal cation. Group one is better off with the 1- NO2 ion for more or less the same reason,
except for Li who is so small that having to have two Li cations per oxide isn't a problem.
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DJF90
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You can't really say these ions fit in the lattice better than those ones as each compound may occupy a different structure type. For example, the
alkali metal oxides all exist in the antifluorite structure; CCP array of oxide anions with metal cations in all of the tetrahedral holes. Alkaline
metal oxides however (at least Mg and Ca) exist as the rocksalt structure; CCP array of oxide anions with metal cations in all of the octohedral
holes.
Like I said before, you can draw three pretty important conclusions from Kapustinskii's equation: Higher lattice enthalpy (more stable substance) is
favoured by - multiple ions, small ions, and multiply charged ions. Take Li2O stability over LiNO2. The ions are smaller in Li2O, there are more ions
in Li2O, and the oxygen anion is doubly charged. Why don't all the grp 1 metals form oxides on decomposition of nitrates I hear you ask? Try and work
through it in a logical way and I'll see if I can help you when you have an answer... I have little time atm to work it out myself I'm afraid.
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