d and f Block Elements Part 3

Concept:The manganate ion is $\mathrm{MnO}_4^{2-}$ and the permanganate ion is $\mathrm{MnO}_4^{-}$.Both have a tetrahedral geometry due to $\mathrm{sp}^3$ hybridization of Mn.Explanation:Statement A is correct. In both ions, Mn is bonded to four oxygen atoms, giving a tetrahedral shape.Statement B is false. For permanganate ($\mathrm{MnO}_4^{-}$): $x + 4(-2) = -1 \Rightarrow x = +7$.For manganate ($\mathrm{MnO}_4^{2-}$): $x + 4(-2) = -2 \Rightarrow x = +6$.The statement reverses the oxidation states (says +7 for manganate and +6 for permanganate).Statement C is false. Oxidation of $\mathrm{Mn}^{2+}$ by peroxodisulphate ($\mathrm{S}_2\mathrm{O}_8^{2-}$) gives permanganate ($\mathrm{MnO}_4^{-}$), not manganate.Statement D is correct. In $\mathrm{MnO}_4^{-}$, Mn is $+7$ ($3d^0$), no unpaired electrons – diamagnetic.In $\mathrm{MnO}_4^{2-}$, Mn is $+6$ ($3d^1$), one unpaired electron – paramagnetic.Statement E is false. Acidified permanganate is a strong oxidizing agent. It oxidizes oxalate, nitrite, and iodide ions, not reduces them.Answer:Correct statements are A and D only. Option A.