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In KMnO₄ (Potassium Permanganate), K = +1 O = -2 (each oxygen) Let x be the oxidation number of Mn: (+1)+(x)+4(−2)=0(+1) + (x) + 4(-2) = 0 (+1)+(x)+4(−2)=01+x−8=01 + x - 8 = 0 1+x−8=0x=+7x = +7 x = +7 Hence, the oxidation state of Mn in KMnO₄ is +7.
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