11
$\begingroup$

If oxygen is more electronegative why doesn't it replace chlorine in compounds?

Example:

$\ce{2NaBr + Cl2 -> 2NaCl + Br2}$

works, while

$\ce{4NaCl + O2 -> 2Na2O + 2Cl2}$

doesn't.

$\endgroup$

2 Answers 2

2
$\begingroup$

You can displace chlorine from hydrogen chloride.

With sodium chloride, if you did form sodium oxide it would be a strong base. As such it extracts a $\ce{Cl^+}$ moiety from any potential chlorine molecules. Sodium chloride is oxidized by oxygen, but the oxidized chlorine ends up as chlorine oxyanions such as hypochlorite (i.e., oxide coordinated to chlorine).

$\endgroup$
5
  • $\begingroup$ It's redox not acid base. $\endgroup$
    – Mithoron
    Feb 10, 2018 at 17:37
  • $\begingroup$ Why would it not be both? $\endgroup$ Feb 10, 2018 at 17:38
  • $\begingroup$ I'm still not sure what you even mean in this. Proper answer is that O2 has to low redox potential. $\endgroup$
    – Mithoron
    Feb 10, 2018 at 17:43
  • $\begingroup$ Then say so in your own answer. $\endgroup$ Feb 10, 2018 at 18:45
  • $\begingroup$ Re: "chlorine gets oxidized to oxyanions such hypochlorite" So, the products formed from $\ce{Na2O\+Cl2}$ are $\ce{NaCl\+NaOCl}$? Thanks! $\endgroup$ Feb 13, 2018 at 4:43
1
$\begingroup$

In order to know if a reaction is spontaneous or not (if it will happen towards the direction you have drawn it) you must look at its Gibbs energy, ΔG. Electronegativity of one element in a reaction is one of the many factors that contribute to it. In this case you are dealing with a redox reaction (because the oxidation states of the elements involved change) and therefore you need to use the relationship: ΔG=-nFE$^0$cell with E$^0$cell calculated from the relevant half reactions. For more details read an introduction to electrochemistry from some inorganic chemistry book.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.