# question on oxidation number

#### jamesq

10+ Year Member
Hello,

This is a question from AAMC 9 that im having trouble getting:

What is the oxidation state of aluminum in Na[Al(OH)4) that 4 should be a subscript. They say the answer is +3 but i cant wrap my head around it. Thanks!

#### rocuronium

10+ Year Member
Start with the (OH)4 part. Each oxygen supplies an oxidation number of -2. Therefore, when you multiply by the 4, the total for oxygen is -8. Each H supplies an oxidation number of +1, so the total for hydrogen is +4.

This means that the total oxidation number of the (OH)4 is -4.

If you look at the Na+, it has an oxidation number of 1+.

There is no charge on the molecule, so the oxidation numbers HAVE to add to zero. You know the oxidation numbers of everything except Al, so you can figure that out.

You need a +3 oxidation number on the Al to get a total of 0.

I hope that helps.

#### bluemonkey

10+ Year Member
7+ Year Member
Hello,

This is a question from AAMC 9 that im having trouble getting:

What is the oxidation state of aluminum in Na[Al(OH)4) that 4 should be a subscript. They say the answer is +3 but i cant wrap my head around it. Thanks!
[Al(OH)4] is a complex that Na is bound up to. Na forms a +1 cation when making ionic bonds. This means that the entire [Al(OH)4] has a -1 charge. Then, looking in this complex, you have Al and OH. OH has a -1 charge and there are four of them, so you have a next of -4 from hydroxide. Now the entire [Al(OH)4] complex needs to be net -1, so Al must be 3+ in order to balance out the -4 from hydroxides. Make sense?

#### bluemonkey

10+ Year Member
7+ Year Member
On an additional note, some elements only take on one oxidation state other than their ground (neutral) state. In example, Al & B only forms 3+ in compounds, Na only forms +1 in compounds, and so on...

#### rocuronium

10+ Year Member
Yeah it would probably be a good thing to remember certain common oxidation numbers. You can find charts of them in most chem texts.

OP
J

#### jamesq

10+ Year Member
thank you all for your help, it makes sense now. I knew the oxidation numbers, i just didnt know I had to include (OH)4 and just used OH instead and had the oxidation numbers for oxygen as -2, and for H as +1, but now I understand the concept.

#### freedom hater

10+ Year Member
5+ Year Member
I had problems grasping this same problem. I know this is an old post but this might help me with my lingering question.

The OP may be doing Formal Charge instead of Oxidation Number. Which are not the same thing.

The answers to the OP make sense when you consider what you 'know', that OH is a -1 ion. But if you do the formal charge formula, you get Al = +1 (Just like i did, maybe that's what you did), and the answer was Al = +3.

The question asks "what is the oxidation number", not "what is the formal charge".
If you did the formal charge on Al you would get: BaseValence - #covalent.bonds/2 - #free.electrons.pairs = +1, (Actually now that i type this, i cant remember if it is not supposed to be BaseValence - #covalentbonds - #free.electrons, which makes no sense as an answer, but is in notes online.)

How do you do formal charge? covalent.bonds/2 or not?

#### chemtopper

##### online organic/general chem/MCAT/DAT tutor
10+ Year Member
7+ Year Member
Formal charge is actually total valence electrons present with atom minus number of electrons used in lewis acid structure in bonding and as lone pair
so many ways of writing it
FC = VE- Bonds -2LP
FC = Group number - number of electrons in covalent bonding /2 - number of non bonding electrons.
I prefer to tell the first one because for every covalent bond that element is giving only one electron so you just count the number of bonds around that atom and for every lone pair atom has given two electrons in its lewis dot structure.