A) formation of red vapours
B) formation of lead chromate
C) formation of chromyl chloride
D) liberation of chlorine
Correct Answer: D
Solution :
Chromyl chloride test is used for\[C{{l}^{-}}\]ions. Chlorine is not liberated in this test. \[4NaCl+{{K}_{2}}C{{r}_{2}}{{O}_{7}}+3{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}{{K}_{2}}S{{O}_{4}}\] \[+2N{{a}_{2}}S{{O}_{4}}+\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,\,chromyl\,\,chloride \\ (reddish\,\,brown\,\,vapour) \end{smallmatrix}}{\mathop{2Cr{{O}_{2}}C{{l}_{2}}}}\,\uparrow +3{{H}_{2}}O\] \[Cr{{O}_{2}}C{{l}_{2}}+4NaOH\xrightarrow{{}}2NaCl+N{{a}_{2}}Cr{{O}_{4}}\] \[+2{{H}_{2}}O\] \[N{{a}_{2}}Cr{{O}_{4}}+{{(C{{H}_{3}}COO)}_{2}}Pb\xrightarrow{{}}2C{{H}_{3}}COONa\] \[+\underset{lead\,\,chromate}{\mathop{PbCr{{O}_{4}}}}\,\]
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