| \[Mn{{O}_{2}}+KOH+{{O}_{2}}\to \text{ }\,\,\,\underset{(green)}{\mathop{'B'}}\,\,\,+\,\,{{H}_{2}}O~~~~~~~~~~~\left( green \right)\] |
| \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\,\,\xrightarrow{\Delta }\,\,{{K}_{2}}Cr{{O}_{4}}+\,\,C{{r}_{2}}{{O}_{3}}\,+\,\,\,\,\,\underset{(Colourless\,\,gas)}{\mathop{'C'}}\,\]\[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\,\,+\,\,{{H}_{2}}S{{O}_{4}}+NaCl\text{ }\to \] |
| \[\begin{array}{*{35}{l}} {} \\ KHS{{O}_{4}}\,\,+\,\,NaHS{{O}_{4}}+{{H}_{2}}O \\ {} \\ \end{array}+\,\,\underset{\left( Reddish\text{ }brown\,\,vapour, \right)}{\mathop{'D'}}\,\] |
A) \[C{{r}^{3+}},\,\,{{K}_{2}}Mn{{O}_{4}},\,\,{{O}_{2}},\,\,C{{l}_{2}}\]
B) \[C{{r}_{2}}{{O}_{3}},\,\,{{K}_{2}}Mn{{O}_{4}},\,\,{{O}_{2}},\,\,Cr{{O}_{2}}C{{l}_{2}}\]
C) \[CrO{{\left( {{O}_{2}} \right)}_{2}},\text{ }KMn{{O}_{4}},\text{ }{{O}_{2}},\text{ }Cr{{O}_{2}}C{{l}_{2}}\]
D) \[Cr{{O}_{5}},\text{ }{{K}_{2}}Mn{{O}_{4}},{{O}_{2}},Cr{{O}_{2}}C{{l}_{2}}\]
Correct Answer: D
Solution :
\[C{{r}_{2}}{{O}_{7}}{{^{-}}^{2}}+{{H}^{+}}+{{H}_{2}}{{O}_{2}}\text{ }\to \text{ }Cr{{O}_{5}}+{{H}_{2}}O\] \[Mn{{O}_{2}}+KOH+{{O}_{2}}\to \,\,{{K}_{2}}Mn{{O}_{4}}+{{H}_{2}}O\] \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\,\,\xrightarrow{\Delta }{{K}_{2}}Cr{{O}_{4}}+C{{r}_{2}}{{O}_{3}}+{{O}_{2}}\] \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}+{{H}_{2}}S{{O}_{4}}+NaCl\text{ }\to \] \[KHS{{O}_{4}}+NaHS{{O}_{4}}+{{H}_{2}}O+Cr{{O}_{2}}C{{\text{l}}_{\text{2}}}\]
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