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This is the very famous chromyl chloride test, used for detection of chloride ion in solution -

$$\ce{K2Cr2O7 + 4NaCl + 6H2SO4 -> 2CrO2Cl2 + 2KHSO4 + 4NaHSO4 +3H2O}$$

When a mixture containing chloride ion is heated with potassium dichromate and concentrated sulphuric acid, deep orange-red fumes of chromyl chloride are formed. My question is, how?

I want to know the proper mechanism for this reaction, particularly for chromyl chloride formation. I couldn't find it in several texts (and online), such as JD Lee and Atkins.

Also, if instead of chloride ion, bromide ion (or iodide) is used then 'chromyl bromide' is not formed. Instead, bromine gas is liberated, and chromium (III) sulphate is formed along with potassium (I) sulphate. Why? What is so special about the chloride ion?

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    $\begingroup$ Under those conditions you will have some amount of H2CrO4 in the mixture so I would start from there. Oxygen protonation and double OH2+ displacement with chloride should provide the product. $\endgroup$
    – EJC
    Commented Apr 15, 2018 at 14:48
  • $\begingroup$ Are you sure about the mechanism? In fact, please post in the form of an answer with proper 'arrow pushing' steps. It'd be great. Thank you. $\endgroup$
    – stoic-santiago
    Commented Apr 15, 2018 at 14:49
  • $\begingroup$ I didn't want to post an answer as I am not sure, it's my guess. Couldn't find any proof. $\endgroup$
    – EJC
    Commented Apr 15, 2018 at 15:17
  • $\begingroup$ >proper mechanism || Unlikely to happen, as the mechanisms are usually very hard to prove. But as a general idea, Cr(VI)O2X fragments are fairly stable, so as a first guess you can assume that the reaction proceeds through a series of such intermediates. $\endgroup$
    – permeakra
    Commented Apr 15, 2018 at 17:49
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    $\begingroup$ >What is so special about the chloride ion || Two reasons actually. First, chloride is a lot smaller and thus better binds with ultra-small Cr(VI). Second, bromide is a lot easier to oxidize, see there electrode potentials. CrO2F2 forms just as easy (or even easier) than CrO2Cl2 $\endgroup$
    – permeakra
    Commented Apr 15, 2018 at 17:52

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The answer may (and does) lie in the oxidation potential of the halogens. As we move down the group 17 the tendency for oxidation increases.

Refer: $$\ce{H2SO4 + HBr -> Br2 + SO2} \tag{unbalanced}$$ but $$\ce{H2SO4 + HCl -> X} \tag{does not occur}$$

$\ce{K2Cr2O7}$ is not able to oxidise $\ce{Cl-}$ to $\ce{Cl2}$ but it does so with $\ce{Br2}$, thus we observe reddish brown fumes of $\ce{Br2}$ as if $\ce{Br-}$ had been present, which are then passed through $\ce{NaOH}$ to confirm the presence of chloride or bromide.

Refer:

\begin{align*} \ce{CrO2Cl2 + NaOH ->& Na2CrO4} \tag{yellow soln,unbalanced} \\ \ce{above + H+ ->& (Cr2O7)2-} \tag{orange solution} \\ \ce{Br2 + NaOH ->& NaBr + NaBrO3} \tag{light yellow soln} \end{align*}

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  • $\begingroup$ hey guys if this is not the answer then why doesnt any body who downvotes suggest the correct one ?.Are u guys too busy to answer thaan down vote? $\endgroup$
    – Aashish Kohli
    Commented Apr 16, 2018 at 13:56
  • $\begingroup$ perhaps you wud like to refer (E=RC-RA) E is electrode potential rc reduction potential at cathode and RA reduction potential at anode $\endgroup$
    – Aashish Kohli
    Commented Apr 16, 2018 at 14:08
  • $\begingroup$ It does not provide an answer to the main mechanism question. The selective oxidation question is quite easy. $\endgroup$
    – Poutnik
    Commented Aug 11, 2019 at 11:38

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