[Enceph]

An element like Aluminum that has 3 valence electrons, can only get a maximum of 6 valence electrons through sharing with other atoms right?

Or since Aluminum needs 5 electrons to fill a full valence shield of 8, can it be 'given' electrons from other atoms, that are not being shared?

Because at the moment it kind of looks like this,

(in attachment), I know that some of the lines are crooked, but it was a quick paint job. 

There are still 2 electrons spots left in each Aluminum electron, does that mean that this bond does not work?  In my book it would appear that it does.  Or do these two Aluminum atoms stay at 6/8 in their valence shell?

[Alpha-Omega]

OH MY...that is very CREATIVE...let me think a little...you drew that youself...with paint?

[yosh]

i think he means with microsoft paint... on the computer.

[invisiblegs]

Aluminum oxide, aka alumina, is a difficult "molecule" to determine structurally.  If you go to Wiki, or the a chemical distributor's database, you will get the general formula Al₂O₃.  This general formula is used because it allows for charge balance, the Aluminum gives up 3 electrons (easier than acquiring 5), and the Oxygen receives 2 electrons (easier than giving up 6).  But as far as a discrete molecular structure goes, I don't think there is one in this case.  I seem to recall that since Aluminum is close to the semimetals in the periodic table its oxides may be similar.  Silicon dioxide (SiO₂) does not have a discrete molecular structure, the chemical formula is just used to represent a large matrix or network of bonded atoms.  In fact, I believe that aluminum is one of the metals used to dope or modify silicon semiconductors.  Your proposed molecule is very creative, but think of all those poor oxygen atoms with six bonds to them.  Oxygen atoms generally only like having 2 bonds (water), or three if they are positively charged.  I hope this helps you understand what is going on here and if you have further questions don't hesitate to ask.

[Alpha-Omega]

Yes good job with PAINT...use it all the time...OK:

Al2O3 (2 Al’s gives total of 6 e–‘s to 3 O’s to form 2Al3+ and 3O2–)

The geometry is octahedral.

The bonds are ionic (2 Al3+ cations to 3O2- anions).

Aluminum oxide exists in several different crystallographic forms, of which corundum is most common. Corundum is characterized by a high specific gravity (4.0), a high melting point (about 2,050° C, or 3,700° F), great insolubility, and hardness.

Element:  Al ;  % Composition:  52.93; Formal Oxidation State:  +3   Formal Electronic Structure:  [Ne]
Element: O; % Composition:  47.07; Formal Oxidation State:  -2 Formal Electronic Structure:  [He].2s2.2p6

A Group 3 metal + a Group 6 non-metal e.g. aluminium + oxygen ==> aluminium oxide Al2O3 or ionic formula (Al3+)2(O2-)3  In terms of electron arrangement, two aluminium atoms donate their three outer electrons to three oxygen atoms. This results in two triple positive aluminium ions to three double negative oxide ions. All the ions have the stable electronic structure of neon 2.8. Valencies, Al 3 and O 2.

2Al (2.8.3) + 3O (2.6) ==> 2Al3+(2. 3O2- (2.

can be summarized electronically as 2[2,8,3] + 3[2,6] ==> [2,8]3+2 [2,8]2-3

[Alpha-Omega]

Let me try sending that diagram again

[Alpha-Omega]

OK see attached document...should have dot structures...

[Enceph]

Wow the forum sure is fast today!

As always great, detailed help.   

Can someone though please answer this,

In this equation,

4Fe+3O2 -> 2Fe2O3

With Fe3+ and O2- as the molecules in question.  Why
does Oxygen (O) keep its subscript (2, before the reaction),
and Fe does not keep it's subscript (3, before the reaction)?
After the reaction it shows the subscripts that are produced, but
before the reaction it keeps Oxygen's prereaction subscript, but
does not show Fe's.  I understand that 4Fe3 wouldn't work, but you
aren't able to change the subscript in a formula.

[Alpha-Omega]

Really,  looks like O2  in reactants and O3 in products to me...TYPO perhaps?

[Enceph]

Really,  looks like O2  in reactants and O3 in products to me...

Yep that is correct.

Why is it like this though?  Where it shows Oxygen's reactant subscript, but doesn't show Iron's reactant subscript, but then shows their product subscripts.

[Alpha-Omega]

You have got to phrase those questions more clearly....you distincly ask  why the subscripts for O do not change...Here are your words:

4Fe+3O2 -> 2Fe2O3

With Fe3+ and O2- as the molecules in question. 

Why does Oxygen (O) keep its subscript (2, before the reaction),and Fe does not keep it's subscript (3, before the reaction)?

After the reaction it shows the subscripts that are produced, but before the reaction it keeps Oxygen's prereaction subscript, but does not show Fe's.  I understand that 4Fe3 wouldn't work, but you
aren't able to change the subscript in a formula.

I understand that 4Fe3 wouldn't work, but you aren't able to change the subscript in a formula.

Now by prereaction  am guessing you mean reactants or are you talking about the coefficient in front of the Fe???

Ok I am not seeing any of that in that equation...


4 Fe....  in reactants    and get 2Fe2 in products  ....charges balanced...right?? 4 = 2 x2 both sides equal right!!!!!!!!

Now O in reactant = 3 x 2 = 6 right....right!!!!!!!! and if I go to products...that 2 in front of the Fe is also associated with the O3....so 2 x 3 = 6  and so if 3 x 2 = 6 and 2 x 3 = 6....then we are all balanced right???

[LQ43]

Wow the forum sure is fast today!

As always great, detailed help.   

Can someone though please answer this,

In this equation,

4Fe+3O2 -> 2Fe2O3

With Fe3+ and O2- as the molecules in question.  Why
does Oxygen (O) keep its subscript (2, before the reaction),
and Fe does not keep it's subscript (3, before the reaction)?
After the reaction it shows the subscripts that are produced, but
before the reaction it keeps Oxygen's prereaction subscript, but
does not show Fe's.  I understand that 4Fe3 wouldn't work, but you
aren't able to change the subscript in a formula.

Fe³⁺ does not have a subscript, the 3+ is a superscript meaning charge 3+
same with the O^2- that is the superscript meaning charge 2-

however, that is all irrelevant.

Fe + O₂   before the reaction have NO charges, they are neutral and are elements

O₂ means diatomic oxygen with this 2 as a subscript

when they react in this redox reaction, electrons are transferred from Fe to O and to make the product Fe₂O₃

now , after the reaction in the product, Fe is Fe³⁺ and O is O^2-

with the proper subscripts is the balanced equation

4Fe  +  3O₂  -->   2Fe₂O₃

it takes a bit of time getting used to seeing all these without the sub and superscripts just because we all are too lazy to put them in.

[Alpha-Omega]

Is that what you were asking?

4 moles of elemental iron react with 3 moles of O2 gas (oxygen is diatomic in its natural state)  react to form 2 moles of Fe2O3...and it is all balanced....OK yes the charges are not usually written into the reaction equation...and ther has to be a charge and mass balance....the charges are usually assumed....well, bad choice of words...you can figure them out from state....

[Enceph]

Is that what you were asking?

Yeah pretty much.

So the diatomic oxygen, is a molecule with two oxygen atoms? (my question only refers to oxygen)

But how would I be able to tell that there are two oxygen atoms when putting together the equation, if all it asks me is
'Write the balanced chemical equation for the reaction between iron (its ion is Fe3+) and oxygen.

I understand how the product side works, the Fe2, O3.  But I still do not see how I can know that it would be Fe, and O2  on the reactant side.  (I was aware of the correct equation for the product side)

[LQ43]

'Write the balanced chemical equation for the reaction between iron (its ion is Fe3+) and oxygen.

Interpreting questions can be tricky (as we have seen). The two elements are given and the question assumes you know what their natural states are.

7 elements occur as diatomic molecules in their natural states

H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂

Its helpful to remember these

Fe³⁺ is given so that you will know what the charge on the iron will be in the product