Past exam of the chemistry B course of the University of Cambridge / 2024 / Paper 2 / B7 / a / i

In the ionic assignment, both tetrahedral anions would require Mn(VII), $d^0$. Such a high oxidation state is strongly oxidizing and is best supported by strong bonding to oxygen. Short Mn–O bonds combine good orbital overlap with oxygen's resistance to further oxidation. This stabilizes permanganate relative to a hypothetical all-sulfide analogue.

sulfide is larger, more polarizable and much easier to oxidize than oxide. An attempted Mn(VII) sulfide would therefore combine an exceptionally oxidizing metal center with reducing ligands: internal redox can reduce manganese while oxidizing sulfur, for example toward S–S-bonded or elemental sulfur products. The larger sulfur donors also provide less favorable short multiple bonding to the small, highly charged manganese center. The oxygen compound supports Mn(VII); the sulfide combination is redox-incompatible. This predicts instability without requiring a specific unprovided decomposition product.