Compound $ Cr{O_2}C{l_2} $ is formed while testing:
(A) $ NO_3^ - $
(B) $ C{l^ - } $
(C) $ C{r^{3 + }} $
(D) $ F{e^{3 + }} $
Answer
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Hint: Qualitative analysis is a branch of chemistry that deals with the identification of elements and ions present in a particular sample. The tests and verification process varies for each element and ion. The given compound $ Cr{O_2}C{l_2} $ is named by IUPAC norms as Chromyl chloride and it is a dark red blood colour liquid.
Complete Step by Step answer
We know that qualitative analysis is a branch of chemistry that deals with the verification tests to identify the presence of ions and elements in a compound. The compound or a sample of the compound is put through a series of tests to identify the presence of an ion or element in it based on the reaction the compound shows to a particular substrate.
Similarly chromyl chloride is a compound that is formed as a by-product that indicates the presence of $ C{l^ - } $ ions in the compound tested. This test is called chromyl chloride and it is carried out in the presence of sulphuric acid $ \left( {{H_2}S{O_4}} \right) $ and potassium dichromate $ \left( {{K_2}C{r_2}{O_7}} \right) $ . Take a look at an example where a chlorine containing salt is subjected to chromyl chloride test.
$ {K_2}C{r_2}{O_7} + 4NaCl + 6{H_2}S{O_4}\xrightarrow{{}}2Cr{O_2}C{l_2} + 2KHS{O_4} + 4NaHS{O_4} + 3{H_2}O $
You can see that the product contains the compound $ Cr{O_2}C{l_2} $ which is indicated by the release of red fumes from the reaction environment, once the reaction is completed. Although the reaction mechanism seems spontaneous on heating, it is carried out in a number of steps.
Thus we know that the compound $ Cr{O_2}C{l_2} $ (Chromyl chloride) is a by-product of the chromyl chloride qualitative analysis that indicates the presence of $ C{l^ - } $ ions in a compound.
Hence option B is correct.
Note
Although not all compounds having the presence of chlorine shows the chromyl chloride test. Only the ones that form a $ C{l^ - } $ ionic bond in the compound show the test, the rest show no changes when the chromyl chloride test is carried out on them.
Complete Step by Step answer
We know that qualitative analysis is a branch of chemistry that deals with the verification tests to identify the presence of ions and elements in a compound. The compound or a sample of the compound is put through a series of tests to identify the presence of an ion or element in it based on the reaction the compound shows to a particular substrate.
Similarly chromyl chloride is a compound that is formed as a by-product that indicates the presence of $ C{l^ - } $ ions in the compound tested. This test is called chromyl chloride and it is carried out in the presence of sulphuric acid $ \left( {{H_2}S{O_4}} \right) $ and potassium dichromate $ \left( {{K_2}C{r_2}{O_7}} \right) $ . Take a look at an example where a chlorine containing salt is subjected to chromyl chloride test.
$ {K_2}C{r_2}{O_7} + 4NaCl + 6{H_2}S{O_4}\xrightarrow{{}}2Cr{O_2}C{l_2} + 2KHS{O_4} + 4NaHS{O_4} + 3{H_2}O $
You can see that the product contains the compound $ Cr{O_2}C{l_2} $ which is indicated by the release of red fumes from the reaction environment, once the reaction is completed. Although the reaction mechanism seems spontaneous on heating, it is carried out in a number of steps.
Thus we know that the compound $ Cr{O_2}C{l_2} $ (Chromyl chloride) is a by-product of the chromyl chloride qualitative analysis that indicates the presence of $ C{l^ - } $ ions in a compound.
Hence option B is correct.
Note
Although not all compounds having the presence of chlorine shows the chromyl chloride test. Only the ones that form a $ C{l^ - } $ ionic bond in the compound show the test, the rest show no changes when the chromyl chloride test is carried out on them.
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