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Depends on the oxidation state Vanadium is in. If it is the Vanadium (III) state you will have a [V(CN)7]4- salt such as K4[V(CN)7] which can be prepared from VCl3 and KCN in aqueous solution to form a pentagonal bi-pyramidal complex. You can then reduce this complex in liquid NH3 to form the Vanadium(II) complex K4[V(CN)6]

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