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Q.

Asked by Arushi Juyal 16th December 2017, 8:19 AM
Answered by Expert
Answer:
Highest space oxidation space state space of space Mn space is space present space in space KMnO subscript 4 space

For space KMnO subscript 4
Oxidation space number space of space space straight K space equals plus 1

Oxidation space number space of space space straight O equals negative 2

therefore left parenthesis plus 1 right parenthesis plus Mn plus open square brackets 4 open parentheses negative 2 close parentheses close square brackets space equals space 0

Mn equals left parenthesis negative 8 plus 1 right parenthesis
space space space space
space space space space space equals space left parenthesis negative 7 right parenthesis

For space straight K subscript 2 MnO subscript 4

left parenthesis 2 cross times 1 right parenthesis space plus Mn plus open square brackets 4 open parentheses negative 2 close parentheses close square brackets space equals 0

Mn equals space plus 6

For space Mn subscript 2 straight O subscript 3
2 Mn plus open parentheses 3 cross times open parentheses negative 2 close parentheses close parentheses equals 0

Mn equals space plus 3

For space MnO subscript 2

Mn plus open square brackets 2 open parentheses negative 2 close parentheses close square brackets equals space 0
Mn equals space plus space 4
Answered by Expert 16th December 2017, 11:36 PM
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