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Why is uranium converted from uranium(IV) to uranium(VI) during ion-exchange recovery?

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chemical structure of cation exchanger — ion-exchange reaction Figure 1: Chemical structure of cation exchanger. The exchangeable ions are marked ⁺. The whole structure is permeated by solvent molecules, usually water (not shown).

During the recovery of uranium, the conversion from uranium(IV) to uranium(VI) is necessary to facilitate the formation of soluble, negatively charged complex ions that can be captured by ion-exchange resins.12

In acidic leaching processes, uranium must be in the hexavalent state to react with sulfuric acid and form the uranyl sulfate complex anion, $[U O_{2} (S O_{4})_{3}]^{4 -}$ . Similarly, in alkaline leaching, the hexavalent uranium reacts to form the soluble uranyl carbonate complex ion, $[U O_{2} (C O_{3})_{3}]^{4 -}$ . These specific complex ions are required because the ion-exchange process relies on the selective absorption of these negatively charged species onto the resin.123

By ensuring the uranium is oxidized to the more stable hexavalent state, the process effectively separates the uranium from other metals, such as iron, which do not form these specific complexes under the same conditions.13