[getfirefox]

I'm working my way through a practise paper but have become stuck on a certain few questions. This is them in full. Can anyone help?

A sample of metallic americium can be made by reacting americium fluoride with barium metal at 1100 °C.

A) Suggest a formulae for the two different ions in americium fluoride.

B) Construct a balanced chemical equation for the reaction of barium with americium fluoride.

Could somebody please help me with those two questions? Many thanks to anybody that reads or posts a response.

[Astrokel]

Hi getfirefox, i wouldn't have known the most common oxidation state of Americium without googling. Were you taught on that?

[Scoobiedoobiedoo!]

I would say either Am4+ and F- making the compound AmF4 or
Am3+ and F- making the compound AmF3. AmF3 would be the most stable as it would complete the Am octete so I would go with that one. 

[Scoobiedoobiedoo!]

Oh and for the balanced equation; Ba + 2Am3F --> BaF2 + 6Am, but I am really not sure sorry I couldn't be of more help.

[sjb]

Oh and for the balanced equation; Ba + 2Am3F --> BaF2 + 6Am, but I am really not sure sorry I couldn't be of more help.

Although this is balanced in the sense of having the same elements on both sides of the equation, you seem to have contradicted yourself looking at the oxidation states of the metals and what you quoted earlier...

[Scoobiedoobiedoo!]

Yes sorry I should have removed my first choice of ions AmF3 is the one to go with. Sorry about confusion caused.