[hpl912]

[FeBr3(H2O)]NO3

I need to find the geometry, metal oxidation state and the d configuration.

I don't get what the NO3 at the end means. Does it equal the overall charge (NO3= -1?) of the compound within the square brackets as [FeBr3(H2O)]^-1 ? Otherwise, how can we even get the oxidation state of the metal? ignore NO3?

where does the NO3 bind to? the whole [FeBr3(H2O)] which seems to be tetrahedral

[stewie griffin]

As you have written the chemical formula, the NO₃ group is a counterion. It is not bonded to the metal center. You can make an analogy to NaCl. The NO₃^- is just like the Cl^-. However you now have a "fancier sodium cation" here because the cation isn't simply an atom but a cationic complex of iron and some ligands.
You are incorrect when you ask if the iron compound in brackets has a charge of -1... in fact the complex in the brackets has a charge of +1.
Hopefully that helps get you started...

[hpl912]

thanks for the explanation!

but then how did you get +1? for me, my -1 charge i got from NO3 where one of the O has only one bond to N so it leaves with an extra electron, resulting the an overall charge of -1 for NO3.

if you saying the compound within the brackets is +1, then Fe oxidation state will be +4 since Br is -1 and H2O is neutral. The d configuration will be d4.

the geometry if as you said as a counterion, then the Fe will have 4 coordination, tetrahedral.

does my answers makes sense?

[stewie griffin]

Sorry if that wasn't clear. The +1 is for the compound in the brackets... I didn't mean to say that the charge or oxidation state of the iron was +1. This is simply because the nitrate anion is -1.
From that I would agree with your oxidation state of iron. Since each Br is going to be considered -1 and we have to end up with a +1 charge, then iron must be +4. Iron will also have a coordination number of 4. I'm pretty sure it will be tetrahedral. I know that Ni(CO)₄ is tetrahedral and I thought that for the "top row" of the d block that tetrahedral is preferred over square planar. Plus square planar usually only shows up for the d¹⁰ metals like Pd in Pd(PPh₃)₄.

[hpl912]

ah ok. got it now. thanks!