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For example, how do the bonds in NaCl form?

I understand that the elements need to obtain an octet structure to become more stable, by either accepting or losing electrons.

In NaCl, from my understanding, the sodium atom loses an electron and gives it to the chlorine atom, in turn gaining a +1 charge and then causing chlorine to have a -1 charge.

The new compound formed is NaCl. How does it form ? Do i need to place the respective atoms close to each other? What is needed to them to bond? Also, what is needed for bonding for other types such as covalent and metalic? Does it happen naturally ?

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closed as too broad by Klaus-Dieter Warzecha, Todd Minehardt, ron, Jan, jerepierre Jan 19 '17 at 20:58

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Bonding surely does happen naturally, in that it did exist before we humans came along and started placing atoms this way and that. $\endgroup$ – Ivan Neretin Jan 19 '17 at 12:53
  • $\begingroup$ Your second and third sentences go a long way toward answering your question. "Nature" wants to be in it's most stable (low energy) state, and swapping/sharing electrons so that everyone gets a complete octet is one way in which the most stable chemical state is achieved. $\endgroup$ – airhuff Jan 19 '17 at 17:18
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The collision theory aims to explain how reactions happen.

In the theory, when the particles of the suitable reactant collide with each other, the particles with the right conditions (orientation, velocity, etc.) will react and form new chemical elements.

It is quite difficult to form $\ce{NaCl}$ from sodium metal and chlorine gas alone. If you have to produce $\ce{NaCl}$ from sodium metal and chlorine gas (which you would not), you would do it in an aqeuous environment:

$$\ce{Na(s) + H2O(l) -> NaOH(aq) + H2(g) ^}$$ $$\ce{Cl2(g) + H2O(l) -> HCl(aq) + HOCl(aq)}$$ $$\ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O}$$

But in this method $\ce{NaOCl}$ is produced as a side product, so I do not recommend using this method.

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    $\begingroup$ It is quite easy to make $\ce{NaCl}$ from sodium and chlorine. It is neither safe nor cost-effective. $\endgroup$ – Ben Norris Jan 19 '17 at 13:48
  • $\begingroup$ I mean directly in an anhydrous environment. $\endgroup$ – DHMO Jan 19 '17 at 13:49
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    $\begingroup$ Drop sodium metal into chlorine gas. After the dust settles, the shrapnel will be coated with salt. The reaction is spontaneous and violent. The process is not cost-effective because both elements are manufactured by electrolysis of sodium chloride. $\endgroup$ – Ben Norris Jan 19 '17 at 13:53
  • $\begingroup$ @BenNorris Are you sure that it is not because of the oxide layer? $\endgroup$ – DHMO Jan 19 '17 at 14:44
  • $\begingroup$ Certain. The oxide layer would slow this reaction down. $\endgroup$ – Ben Norris Jan 19 '17 at 19:13

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