A) \[Mn{{O}_{2}}\]
B) \[HMn{{O}_{4}}\]
C) \[M{{n}_{3}}{{O}_{4}}\]
D) \[PbMn{{O}_{4}}\]
Correct Answer: B
Solution :
On boiling \[MnS{{O}_{4}}\] with \[Pb{{O}_{2}}\] and cone. \[HN{{O}_{3}}\], permanganic acid is obtained, due to which the colour of solution turns deep blue. \[Pb{{O}_{2}}+2HN{{O}_{3}}(conc)\xrightarrow{{}}\] \[Pb{{(N{{O}_{3}})}_{2}}+{{H}_{2}}O+[O]\] \[2MnS{{O}_{4}}+3{{H}_{2}}O+5[O]\xrightarrow{{}}\underset{\begin{smallmatrix} permanganic \\ acid \end{smallmatrix}}{\mathop{2HMn{{O}_{4}}}}\,\] \[+2{{H}_{2}}S{{O}_{4}}\]
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