For YBa2Cu3O7-d where d=0.1, it says that each cation is in its highest ordinary valence state. Assuming oxygen takes -2 so, -2x6.1= -12.2, How do I
balance that 0.2 charge. I am a rookie in this. I have to write a defect reaction for removing an oxygen vacancy. So first step I need to find out the
oxidation states each element takes.
Please help. woelen - 22-2-2013 at 11:10
The only element in this compound which easily takes multiple oxidation states is copper. Y has oxidation state +3, Ba has oxidation state +2, O has
oxidation state -2.
By controlling the amount of oxygen (varying the parameter d), apparently it is possible to change the average oxidation state of all copper ions.
The equation for total oxidation states is 1*3 + 2*2 + 3*x - (7-d)*2 = 0, where x is the average oxidation state of copper.
So, the oxidation state (on average) of the copper is x = (7-2d)/3. Copper ions have oxidation state +1, +2, and rarely +3. With d equal to 0.1, the
average oxidation state of the copper is 6.8/3, which is between 2 and 3. So, most of the copper will be in oxidation state +2 and some will be in
oxidation state +3.
I do not know the precise structure of this compound, it may well be that indeed the copper has fractional oxidation state and that electronic charge
is distributed over multiple copper atoms. The number 6.8/3 is only an average number, knowledge about the precise oxidation state of the copper atoms
requires knowledge about the specific structure and chemistry of this type of compounds (which I do not have).
[Edited on 22-2-13 by woelen]emily123 - 22-2-2013 at 11:18
Even Y can take oxidation states of +1/+2/+3 right? Then why do we fixate on integral oxidation state for Y and not Cu? I don't know much about non
stoichiometric equations.Mixell - 22-2-2013 at 11:43
Yes, Yttrium indeed exhibits those oxidation states, but you mentioned "highest ordinary valence state", so I'd go with yttrium being in the +3
oxidation state.
That leaves us only copper to mess around with.
Even Y can take oxidation states of +1/+2/+3 right? Then why do we fixate on integral oxidation state for Y and not Cu? I don't know much about non
stoichiometric equations.
Yttrium is not likely to have a +1 oxidation state in a simple compound. The only transition metals that commonly form simple +1 ions are the coinage
metals (which have an electron configuration of ns1 (n-1)d10). It's also a very early transition metal (left-hand side of the periodic table), so
it's expected to have a low ionization energy, and being a second-row transition metal, will favour its higher oxidation states.
Now, it's not impossible that yttrium will form some weird compound with a lower oxidation states (Greenwood's Chemistry fo the Elements mentions some
reduced compounds in which Sc is +2 or in mixed oxidation states), but if you're going to be reducing things in this compound, copper(II) is by far
the stronger oxidizing agent. emily123 - 22-2-2013 at 12:49
