Before a sodium atom and a chlorine atom combine together to form the ionic compound $NaCl$, they are just neutral atoms. However, when an electron is transferred from the outer energy level of the sodium atom to the chlorine atom, both of these gain a charge on them; the sodium atom or ion has a charge of $1+$ and the chlorine ion has a charge of $1-$. Both of these ions attract each other and an ionic bond is formed between them.

But what happens when an atom of an element say $Ca$ combines with a polyatomic ion say $OH^-$. Is this calcium atom an ion, is it already a charged atom when it interacts with the hydroxide ion or does it gain the charge by any transfer of electrons to the $OH^-$ ion and becomes $Ca^{2+}$ to bond with the $OH^-$ ion?

I think the polyatomic ions say $OH^-$ is already an ion and not an uncharged compound of elements that gains charge as in the case of $NaCl$, and the bond between an ion and a polyatomic ion is due to the attractive forces between them if already they have opposite charges.

  • $\begingroup$ Does $2Na+Cl_2$ give really $2NaCl$? For the second question I'm quite sceptic if you can us $Ca$ without a reaction with the water of the solution in which $OH^-$ is present... However theoretically is a good question! $\endgroup$ – G M Nov 8 '13 at 23:00
  • $\begingroup$ @GM I guess for the second you could imagine a reaction with the molten metal hydroxide. $\endgroup$ – Nicolau Saker Neto Nov 8 '13 at 23:14
  • $\begingroup$ What about the answer to my question? $\endgroup$ – Samama Fahim Nov 8 '13 at 23:16
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    $\begingroup$ Your question comes from an odd angle so it's a bit hard to answer. You can get $NaCl$ by reacting $Na$ and $Cl_2$, so neither "start out" with charge. Also, you can get polyatomic ions such as $OH^-$ from a neutral polyatomic compound, like in the reaction $\ce{Ca + 2H_2 O → Ca(OH)_2 + H_2}$. There's no relation between "starting out with charge" or not and a species being polyatomic or not. $\endgroup$ – Nicolau Saker Neto Nov 8 '13 at 23:46
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    $\begingroup$ As I commented above, you could consider $Na$ vapour reacting with molten $CsOH$, where the sodium is directly exposed to hydroxide ions. However, the reaction actually happens between $Cs^+$ and $Na^0$, yielding the exchange reaction $\ce{Na^0 + CsOH → Cs^0 + NaOH}$. The hydroxide is merely a spectator ion, and you could well write the equation simply as $\ce{Na^0 + Cs^+ → Cs^0 + Na^+}$. You could also attempt to react $Na$ with $Cl^-$ directly in molten $CsCl$, but again all that happens is the exchange $\ce{Na^0 + CsCl → Cs^0 + NaCl}$ with no direct participation of $Cl^-$ $\endgroup$ – Nicolau Saker Neto Nov 9 '13 at 0:02

I risk not directly answering the question as I have not perfectly understood it.

In your first example, you consider the reaction between two neutral atoms, $\ce{Na}$ and $\ce{Cl}$. An electron is transferred generating $\ce{NaCl}$ where the species are now charged ($\ce{Na^+ \ Cl^-}$). Then you follow up with a hypothetical reaction between a neutral atom ($\ce{Ca}$) and an already charged species ($\ce{OH^-}$). The two reactions are fundamentally different, but not because only the latter has polyatomic ions; the difference is in the very nature of the reactions you constructed (two neutral species reacting in the former, compared to one neutral species and an ion reacting in the latter).

The most direct way to obtain $\ce{NaCl}$ is the reaction between sodium metal (composed of neutral atoms, $\ce{Na^0}$) and chlorine gas (composed of neutral diatomic molecules, $\ce{Cl2}$). One then has:

$$\ce{2Na^0_{(s)} + Cl2_{(g)} → 2NaCl_{(s)}}$$

To obtain $\ce{Ca(OH)2}$, a similarly simple reaction would be that between calcium metal (composed of neutral atoms, $\ce{Ca^0}$) and water (composed of neutral molecules, $\ce{H_2O}$), where the water serves as a source of hydroxyl ions according to the equation:

$$\ce{Ca^0_{(s)} + 2H2O_{(l)} → Ca(OH)2_{(s)} + H2_{(g)}}$$

Now imagine you want to force a reaction between a neutral atom and an ion. There are many cases of this, but we can stick to something similar to what you wanted. In order to expose a substance directly to ions, a conceptually easy way is to add it to a molten salt. For now, let us consider pumping sodium vapour into molten caesium hydroxide ($\ce{CsOH}$). You seem to expect some interaction between the neutral $\ce{Na^0}$ and charged $\ce{OH^-}$, but in fact it is $\ce{Cs^+}$ that reacts with the sodium, according to the following equation:

$$\ce{Na^0_{(g)} + CsOH_{(l)} → NaOH_{(l)} + Cs^0_{(g)}}$$

Here we have simple exchange reaction. The hydroxide ion is actually only a spectator, and can be omitted without loss of understanding in an ionic equation:

$$\ce{Na^0_{(g)} + Cs^+_{(l)} → Na^+_{(l)} + Cs^0_{(g)}}$$

If we consider the reaction between sodium and molten caesium chloride instead, where chloride ions would be present, we have again a simple exchange reaction:

$$\ce{Na^0_{(g)} + CsCl_{(l)} → NaCl_{(l)} + Cs^0_{(g)}}$$

Indeed, it can be described by the same ionic equation above.

(Disclaimer: I actually am not sure the reactions between sodium and caesium compounds works as above. I don't have any understanding of the relative electrochemical potentials in molten salts, though I believe I've read that the reaction with molten chloride is a legitimate way to obtain metallic caesium. But that is mostly beside the point, either way. I avoided exemplifying with calcium and calcium hydroxide because I think the reaction wouldn't work, and the hydroxide is thermally unstable)

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