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Topic: calculating transition metals valences  (Read 1819 times)

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Offline summoner

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calculating transition metals valences
« on: September 22, 2014, 04:09:26 PM »
I've been beating my head against this for a while, so hopefully someone can help me out here.  I really don't understand what determines the common oxidation numbers for some transition metals.

Looking at Cobalt, the 3d subshell is partially filled ( :spinpaired: :spinpaired: :spinpaired: :spinup: :spinup: :spinup:), and the 4s subshell is completely filled ( :spinpaired:).  So I guess I can understand the oxidation number of +2; the 4th shell is incomplete, so you lose the two electrons from the highest (4s) energy subshell.  But why +3?  You lose the two 4s electrons, but then you lose 1 electron from 3d?  Why do you stop at 1 electron from 3d?  Why is it more likely to lose 1 than to lose 0, or 4 (to leave 3d half-filled which I thought was more stable)?

Am I missing something obvious here?  Is it just one of those weird rules of transition metals that just some combinations of incomplete subshells are more stable than others?

Offline Corribus

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Re: calculating transition metals valences
« Reply #1 on: September 22, 2014, 05:13:19 PM »
Transition metals are strange.

Not helpful?
It's hard to generalize properties of transition metals (and don't forget, the spin state and ligand sphere also can affect transition metal electron configurations), but maybe these links will help.

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/trans.php
http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/d-Block_Elements/Electron_Configuation_of_Transition_Metals/Oxidation_States_of_Transition_Metals
http://www.chemguide.co.uk/inorganic/transition/features.html
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline kriggy

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Re: calculating transition metals valences
« Reply #2 on: September 24, 2014, 06:15:53 AM »
Well, the D shell orbitals are not same in energy, it depends on structure of the compound but they lose the degeneracy and some of them are lower in energy, some of them are higher in energy as you can see here
http://en.wikipedia.org/wiki/Crystal_field_theory#Crystal_field_splitting_diagrams

If you look at octahedral diagram you can see that in Co3+ the bottom 3 orbitals are fully filled while the upper two are each partialy filled. Given the fact that the upper two are antibonding in this case it is more stable to have only 2 electrons there instead of 3, thats why the Co3+ is also stable

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