MHTCET 16 May 2024 Shift 2 Paper

Concept:The oxidation number of an element in a polyatomic ion is calculated by equating the sum of all oxidation states to the overall ionic charge.Explanation:Let the oxidation number of Mn be $x$.In $\mathrm{MnO}_{4}^{-}$, oxygen has an oxidation number of $-2$.There are 4 oxygen atoms, so their total contribution is $4 \times (-2) = -8$.The sum of all oxidation numbers must equal the charge on the ion, which is $-1$.Thus, the equation becomes $x + (-8) = -1$.Simplifying, $x - 8 = -1$.Adding 8 to both sides gives $x = +7$.Hence, manganese in the permanganate ion has an oxidation number of $+7$.Answer:The oxidation number of Mn in $\mathrm{MnO}_{4}^{-}$ is $+7$.Correct option: D. 7