Asked by Melanie
How can i be accurate when assigning oxidation numbers?
By this i refer to elements that have more than 1 oxidation # .
For example:
I have been working with hydrides and i had this reaction in mind:
Nb + H
Shoud i use 3 or 5 for Nb, why?
By this i refer to elements that have more than 1 oxidation # .
For example:
I have been working with hydrides and i had this reaction in mind:
Nb + H
Shoud i use 3 or 5 for Nb, why?
Answers
Answered by
DrBob222
I looked up niobium hydride on Google and came up with NbH5; however, I think you have asked the wrong question. It appears to me that you aren't interested so much the oxidation state as in what product is formed.
The electronic structure is
[Kr]4d^4 5s^1. It would appear to me that 5 is the first choice.
And your H you know can't be just H. H2 perhaps? Or H^-; or MH?
The electronic structure is
[Kr]4d^4 5s^1. It would appear to me that 5 is the first choice.
And your H you know can't be just H. H2 perhaps? Or H^-; or MH?
Answered by
Melanie
Now I am a little bit puzzled.If the electronic structure of Fe is Ar 3d^6 4s^2 ,then what i can infer is that i would use 4,but the accurate number would be 3,plz make yourself clear.
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