Question

When ${\text{C}}{{\text{O}}_{\text{2}}}$ is passed into aqueous:
From the following which options are correct?
(This question has multiple correct options).
A.${\text{N}}{{\text{a}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}$ solution, its yellow colour changes to orange
B.${{\text{K}}_{\text{2}}}{\text{Mn}}{{\text{O}}_{\text{4}}}$ solution, it disproportionate into ${\text{KMn}}{{\text{O}}_{\text{4}}}$and ${\text{Mn}}{{\text{O}}_2}$
C.${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}$ solution, its orange colour changes to green
D.${\text{KMn}}{{\text{O}}_{\text{4}}}$ solution, its pink colour changes to green

Answer 1

When carbon dioxide gas is dissolved in water, it exists in equilibrium with carbonic acid.
${\text{C}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$
Thus, carbon dioxide in aqueous medium becomes acidic in nature.

Complete step by step answer:
In acidic medium, the chromate ions get converted into dichromate ions.
${\text{2Cr}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}\left( {{\text{aq}}} \right){\text{ + 2}}{{\text{H}}^{\text{ + }}}\left( {{\text{aq}}} \right) \to {\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{{\text{2 - }}}\left( {{\text{aq}}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right)$
Now, the chromate ions impart yellow color to the solution and the dichromate ions impart orange – red color to the solution. So, when carbon dioxide is passed into aqueous sodium chromate solution, its yellow colour changes to orange. So, option A is correct.
In aqueous salts, manganate ion undergoes disproportionation very easily into permanganate ion and manganese dioxide. The balanced equation for the disproportionation reaction is shown below.
$3{{\text{K}}_{\text{2}}}{\text{Mn}}{{\text{O}}_{\text{4}}}{\text{ + 4C}}{{\text{O}}_2}{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}} \to 2{\text{KMn}}{{\text{O}}_{\text{4}}}{\text{ + Mn}}{{\text{O}}_2} + 4{\text{KHC}}{{\text{O}}_3}$
So, option B is also correct.
Passing carbon dioxide into aqueous dichromate solution, the reaction doesn’t have any effect. So, when carbon dioxide is passed into aqueous sodium dichromate solution, its orange colour does not change to green. So, option C is not correct.
When carbon dioxide is passed into aqueous potassium permanganate solution, it doesn’t have any effect. So, the pink color of potassium permanganate solution will not change into green. So, option D is also not correct.

Hence option A and B are correct.

Note:
Unlike carbon dioxide gas, sulphur dioxide gas can turn an acidified orange coloured potassium dichromate solution green due to the formation of chromium ions.
${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{{\text{2 - }}}{\text{ + 2}}{{\text{H}}^{\text{ + }}}{\text{ + 3S}}{{\text{O}}_{\text{2}}} \to {\text{C}}{{\text{r}}^{{\text{3 + }}}}{\text{ + 3S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
Sulphur dioxide gas can also turn lime water milky just like carbon dioxide gas due to the formation of calcium sulphate. The reaction is as follows.
${\text{Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}}\left( {{\text{aq}}} \right) + {\text{S}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{CaS}}{{\text{O}}_3}\left( {\text{s}} \right) + {{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right)$
The corresponding reaction with carbon dioxide gas is shown below.
${\text{Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}}\left( {{\text{aq}}} \right) + {\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{CaC}}{{\text{O}}_3}\left( {\text{s}} \right) + {{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right)$