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Author: Subject: Scandium: why only 3+?
Polverone
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[*] posted on 12-5-2003 at 11:13
Scandium: why only 3+?


The setup:

"The 2+ oxidation state occurs commonly for nearly all metals in the first transition series. A significant exception is scandium. Explain why scandium is an exception."

My response:

"Scandium forms the extremely stable Ar electron configuration when it loses 3 electrons, so the 3+ state is strongly favored."

My score: 7/10, with the query "but why not Sc 2+?"

The really pedantic answer, after consulting some inorganic references, is that scandium does form a handful of 2+ oxidation state compounds. However, I'm positive that's not the answer she was looking for, and I am still unsure why scandium cannot (as a rule) retain its 3d1 electron and lose the 4s2 electrons. Even the advanced inorganic textbook I consulted mentioned that scandium appeared almost exclusively in the 3+ state, without really explaining why.

Am I missing something so obvious that no text feels a need to explain it?

Or is there really no handy explanation, and the professor just wanted me to say something different?
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rikkitikkitavi
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[*] posted on 12-5-2003 at 11:36


solve the schroedinger and it shall be revealed :)

no, but seriously, in which compounds , more specific, with what elements does scandium form 2+ ions?

could have to do with electronegativity, but it is late and I am tired...

/rickard
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[*] posted on 13-5-2003 at 15:24
Mystery solved


I finally went and asked the professor about it. Why didn't I get full credit? Because I did not answer the question as asked.

I explained that scandium is extremely prone to enter the 3+ state. For full credit I should have mentioned that being in the 3+ state (as it is wont to do) precludes it from being in the 2+ state. So, -3 for failing to mention the blindingly obvious.

Sometimes I really do feel like this professor is from a different planet. Since I was in there to ask about the scandium question anyway, I made the mistake of asking why we didn't earlier learn/apply the band model of metals, using the simpler "electron sea" model. Is the band model only for transition metals, or could it equally well explain (for example) the conductivity of sodium? She said that the band model can be used for any substance, but you don't usually think of sodium as a conductor because you can only use it in an inert atmosphere. :o :o

I usually don't think of bathrobes as clothing because you can't wear them to the opera. :P

Also, I asked why scandium couldn't retain a lone D electron when titanium and vanadium can. She said the energies involved were different, and you couldn't predict them just by looking at the electron structure. So I needed "GAUSSIAN: TI-89 edition" to answer the question if I hadn't previously memorized that "scandium is 3+"?!

In other words, even the answer she wanted was just a handy justification or mnemonic rather than a derivable consequence of known facts available to Us, the Test-Takers. The explanation she gave prior to the exam is no explanation at all.

I don't usually care to whine about things; thank you for indulging this personal venting. We now return you to your regularly scheduled chemistry fun.
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[*] posted on 14-5-2003 at 00:30


Quote:
Originally posted by Polverone
Sometimes I really do feel like this professor is from a different planet. Since I was in there to ask about the scandium question anyway, I made the mistake of asking why we didn't earlier learn/apply the band model of metals, using the simpler "electron sea" model. Is the band model only for transition metals, or could it equally well explain (for example) the conductivity of sodium? She said that the band model can be used for any substance, but you don't usually think of sodium as a conductor because you can only use it in an inert atmosphere. :o :o


That is bad! The electron sea model is a special case of the band model in which the spatial variation of the lattice potential is zero. This causes the bands to merge into the single parabolic band of the electron sea model. The band model works for group 1 metals, but there is little point in using it, as the spatial variation of the lattice potential is small and so the electron sea model works well.




1f `/0u (4|\\| |234d 7|-|15, `/0u |234||`/ |\\|33d 70 937 0u7 /\\/\\0|23.
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Polverone
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[*] posted on 14-5-2003 at 11:50
thank you!


Neither my professor nor the book we're using made that connection between the models. In fact, the book we're using presents the band model in a way so vague as to be almost worthless. I get the feeling that its authors are trying to keep it Dumb 'N Simple for first year students.
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Organikum
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[*] posted on 15-5-2003 at 04:51
It´s for your best only !


They want to lead you smoothly to the hard and evil truth behind chemistry:
(I shouldn´t tell but I can´t behave)
Chemistry has no valid theory. It´s an empirical science (if such a thing exists at all) and is from the theoretical background far behind statistics (what lies on the same level as astrology - no typo). This may explain the asked high suicide rate by chemists too. Now you know it. Be strong! It could have come harder, also I can´t imagine how..... ;)

Did you never got a odd feeling when thinking on Lewis and Broenstedt? Is this normal in a natural science to have two excluding theories which both don´t cover the topic at all? Tz, tz... Organic Chemistry is the crown, a world where a Parr-Shaker is regarded as a sophiticated device must have an surreal touch. :)






;)
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Polverone
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[*] posted on 15-5-2003 at 09:09
I can handle the truth!


Actually, I've had this conversation with PrimoPyro too, about how chemistry really has no overarching "theory of everything." It is especially telling, I think, that ab initio quantum mechanical calculations generally fare worse at predicting observed properties than semi-empirical calculations. I've often wondered if this is because of limitations in the numerical accuracy/extent of simulations, or if the theory simply isn't adequate.

On the downside, this means you can't take a fistful of fundamental equations and derive most of the important other equations for your subject from them (as in physics). On the upside, when everybody stumbles around in the dark even non-specialists may discover interesting things. And I am more of a chef than a mathematician anyway. ;)

But if chemistry is a pseudo-science, what does that make biology, economics, sociology?! :o
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[*] posted on 15-5-2003 at 10:13


The fundamental principles behind chemistry are very well understood. However we lack the computing power to usefully apply them to complex systems.



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Organikum
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[*] posted on 15-5-2003 at 11:16
Oh! Fine! I obviously missed something.


And now please be so kind to give me, or point me at least, to a consistent definition of acid - base as this must exist for a computabel algorithm.

Will ya? ;)


Polverone: PrimoPyro with the Klingon knifes? Who is at the AirForce now? Who can make acetaldehyde from ethyleneglycol so easily? The "we at the HIVE" PP? Yes?
Never heard of such a person, sorry :cool:

Qantum mechanics is widely statistics of course. And I hope the ironic undertone in my previous post was recognized for I don´t want to start a piefight on the honor of chemical science. As there exists a part I like quite well (and where most innovations come from) - chemical engineering. But thats my personal point of view which I ask for being accepted as I accept the views of others.

The next joke is on my costs! :D
ok?
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[*] posted on 15-5-2003 at 12:50


Quote:
Originally posted by Organikum
And now please be so kind to give me, or point me at least, to a consistent definition of acid - base as this must exist for a computabel algorithm.

Will ya? ;)


The contradictory definitions of 'acid' are merely a result of the word being ambiguously defined. The flaw isn't in the chemistry, but in the language used to describe it.




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[*] posted on 15-5-2003 at 14:37


Lewis acid base theory is pretty consistant, and very widely aplicable, but it doesnt need to be computable. 'Acids' are mearly a human interpretation of what is going on.

About scandium.
There is an additional energy gain associated with having an empty, half full, or full d orbital.

A similar reason is behind silver, Ag, being [Kr] 4d10 5s1, and not [Kr] 4d9 5s2.
And why it forms Ag+ easily, and Ag(2+) only on rare occations under force.

How much energy is required for each state, determines how strongly oxidising or reducing a species is, and from that what will be most common/stable in normal use.

I suspect what was being looked for was 'increased stability of empty d orbital vursus extra ionisation energy'.

[Edited on 15-5-2003 by Marvin]
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