# If 1 mole of MnO4 oxidizes 2.5 moles of M^x+ then the value of x is (a) 5 (b) 3 (c) 4 (d) 2

_{4}

^{-}instead of MnO

_{4}

$M{n}^{7+}+{M}^{x+}\to M{n}^{2+}+M\phantom{\rule{0ex}{0ex}}No.ofequivalentsofMnO{4}^{-}=No.ofequivalentsofM\phantom{\rule{0ex}{0ex}}moles\times valencyfactor=No.ofequivalents\phantom{\rule{0ex}{0ex}}so1\times 5=2.5\times valencyfactorofM\phantom{\rule{0ex}{0ex}}valencyfactorofM=2\phantom{\rule{0ex}{0ex}}ItmeanstheoxidationNo.changefor{M}^{x+}\to Mis2\phantom{\rule{0ex}{0ex}}SooptionDiscorrect.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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