A) \[KlC{{O}_{4}}\]
B) \[KlC{{O}_{3}}\]
C) \[KlC{{O}_{2}}\]
D) \[KlCO\]
Correct Answer: B
Solution :
When chlorine is passed through hot concentrated solution of KOH, potassium chlorate is formed. \[3\,C{{l}_{2}}+\underset{\begin{smallmatrix} hot, \\ cone. \end{smallmatrix}}{\mathop{6KOH}}\,\xrightarrow{{}}5KCl+\,\,\,\underset{\begin{smallmatrix} potassium \\ chlorate \end{smallmatrix}}{\mathop{KCl{{O}_{3}}}}\,\,\,+3{{H}_{2}}O\] Note - When chlorine ne is passed into cold dilute solution of KOH, potassium hypochlorite is formed. \[C{{l}_{2}}+2KOH\xrightarrow{{}}KCl+\,\,\,\underset{\begin{smallmatrix} potassium \\ hypochlorite \end{smallmatrix}}{\mathop{KClO}}\,\,\,\,+{{H}_{2}}O\]
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