I’m trying to learn about ions. There is a slide with examples about combining cations with anions. It goes like this:

$$\text{For} \ \ \ce{Al^3+} \ \ \text{ and } \ \ \ce{O^2-} \ \ \text{you get} \ \ \ce{Al2O3}$$

Wait what? So you have an atom of aluminum (in this case a cation because it loses three electrons) and an atom of oxygen (an anion since it receives two electrons). You combine them. Why do you now have TWO atoms of aluminum and THREE atoms of oxygen? Why does combining a cation and an anion suddenly spawn new atoms? And what happened to their charges?

The next example is this:

$$\text{For} \ \ \ce{Ca^2+} \ \ \text{ and } \ \ \ce{Br-} \ \ \text{you get} \ \ \ce{CaBr2}$$

Alright. I see a pattern. For some reason, if the charge of an atom is positive or negative, such number will define the number of atoms for the other element. So since calcium has $2{+}$, there would be two bromine. And since bromine has $-1$ there will be one calcium. I don’t get why, though.

And a third example,

$$\text{For} \ \ \ce{Na+} \ \ \text{ and } \ \ \ce{CO3^2-} \ \ \text{you get} \ \ \ce{Na2CO3}$$

Ok so this is a bit weird since there are now three elements involved.

But according to the pattern, since sodium has charge $+1$, there should be just one $\ce{CO3}$ molecule. And since $\ce{CO3^2-}$ has a charge of $-2$, there should be two of sodium.

Great, I suppose it’s an useful pattern, but my questions are:

  • Why? Why does losing/gaining an electron cause the other element to gain/lose atoms?
  • Does this pattern apply only when you combine a cation with an anion? What if it is the other way around? (An anion with a cation, or does it make no sense at all?)
  • After combining a cation with an anion, do both atoms lose their charge? Observe how the results no longer have $+$ or $-$ superscripts.
  • 1
    $\begingroup$ Ah, I see what you did there. Don't get mixed with arbitrary ways of finding the empirical formula for the ionic compounds. Try to understand them better instead. For example, note that ionic compounds are billions of cations and anions together and we only tend to demonstrate the 'smallest' relation between them. You could say $\ce{Na50(CO3)25}$. It would be correct, but non-standard. $\endgroup$ – M.A.R. ಠ_ಠ Jun 16 '15 at 19:22
  • $\begingroup$ @M.A.Ramezani hm... ok I got an idea which might be completely wrong: let's see... when people say $Al_2O_3$ they don't mean that there are two Aluminum atoms and three Oxygen atoms, but they actually mean that there is a 2:3 relation of Aluminum:Oxygen if you were to combine $Al^{3+}$ with $O^{2-}$ because one $Al^{3+}$ will attract $3$ Oxygens whereas one $O^{2-}$ will attract 2 Aluminums therefore the relation is $2:3$ and hence the end result is written as $Al_2O_3$? $\endgroup$ – Voldemort Jun 16 '15 at 20:06
  • $\begingroup$ Yes, kinda like that. For each three $\ce{O^2-}$ ions you get two $\ce{Al^3+}$. Don't get the impression that this is literally happening in the unit cells though. Ionic compounds tend to form special crystal structures. (e.g. fcc) $\endgroup$ – M.A.R. ಠ_ಠ Jun 16 '15 at 20:09
  • $\begingroup$ @M.A.Ramezani But in that case, if $Al^{3+}$ "attracts" $3$ Oxygens and $O^{2-}$ "attracts" $2$ Aluminums, shouldn't the result be like $Al_2O_3Al^{3+}O^{2-}$? I mean, why do $Al^{3+}O^{2-}$ vanish? Why don't they just stay? $\endgroup$ – Voldemort Jun 16 '15 at 20:20
  • $\begingroup$ Why do compounds form bonds? I've got three words for you: Less potential energy. I don't remember a case of positive lattice enthalpies$\ldots$ $\endgroup$ – M.A.R. ಠ_ಠ Jun 16 '15 at 20:21

I really like your process of trying to work it out! Trying to find patterns is a great way to look at this (and all science really). I think all you're really missing is some nomenclature, as this appears to be primarily a question of understanding of what's actually going on.

The 3+ alumnium ion ($\ce{Al^{3+}}$) and 2- oxygen ion ($\ce{O^{2-}}$) form a crystal lattice in solid form.

(A crystal lattice is a $3D$ arrangement of ions in a repeating pattern, in which each ion is interacting with more than one other ion)

The total charges need to balance out per "unit cell" (an intermediate-level chemistry concept that means the "smallest repeating subunit of a crystal lattice.") So this means that you need to balance the total charges by changing the number of ($\ce{Al^{3+}}$) and ($\ce{O^{2-}}$) ions to get the correct chemical formula.

In this case: $(3+) *(2) = (6+)$ balances $(2-) *(3) = (6-)$, giving $\ce{2 Al^{3+}}$ and $\ce{3 O^{2-}}$.

Ionic compounds that have an overall neutral charge are written without the charge superscripts, thus, $\ce{Al^{3+}_2O^{2-}_3}$ is written as $\ce{Al_2O_3}$

If you're actually looking at precipitation reactions (where the solid salt forms from a solution), you have to balance the equations.

Thus to form $\ce{Al_2O_3}$ the correct ratio would be: $\ce{2 Al^{3+} + 3 O^{2-}} -> Al_2O_3$

Similarly for $\ce{CaBr_2}$, the correct ratio would be: $\ce{Ca^{2+} + 2 Br^-} -> CaBr_2$.

For $\ce{Na_2CO_3}$ you're applying the exact same concept, except you're dealing with an anionic (negative charged) molecule instead of an atom.

Thus, to form $\ce{Na_2CO_3}$ the correct ratio would be: $\ce{2 Na^+ + CO^{2-}_3} -> Na_2CO_3$.


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