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

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

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?
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:
$$\ce{H2SO4 + HBr -&gt; Br2 + SO2} \tag{unbalanced}$$
but
$$\ce{H2SO4 + HCl -&gt; X} \tag{does not occur}$$

$\ce{K2Cr2O7}$ 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 -&gt;&amp; Na2CrO4} \tag{yellow soln,unbalanced} \\
\ce{above + H+ -&gt;&amp; (Cr2O7)2-} \tag{orange solution} \\
\ce{Br2 + NaOH -&gt;&amp; NaBr + NaBrO3} \tag{light yellow soln}
\end{align*}