Acidified KMnO4 reacts with just enough aqueous FeSO4: MnO4^- + 5Fe^2+ + 8H^+ -> Mn^2+ + 5Fe^3+ + 4H2O. Assuming high-spin aqua ions, the total number of unpaired electrons present (in the products) per mole of MnO4^- consumed is:

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Acidified $KMnO X_{4}$ reacts with just enough aqueous $FeSO X_{4}$ : $MnO X_{4} X^{-} + 5 Fe X^{2 +} + 8 H X^{+} Mn X^{2 +} + 5 Fe X^{3 +} + 4 H X_{2} O$ . Assuming high-spin aqua ions, the total number of unpaired electrons present (in the products) per mole of $MnO X_{4} X^{-}$ consumed is:

The equation balances (charge +17 on both sides). Products are 1

Mn X^{2 +}

(

d^{5}

, 5 unpaired electrons) and 5

Fe X^{3 +}

(each

d^{5}

, 5 unpaired electrons). Total

= 5 + 5 \times 5 = 30

unpaired electrons per mole of

MnO X_{4} X^{-}

, i.e.

30 N_{A}