Abstract [Au9L8]3+ reacts with L to give [Au8L8]2+ via the intermediate [Au8L7]2+. The reaction of [Au9L8]3+ with SCN− and Cl− results in the formation of [Au11L8(SCN2]+ and [Au11L8Cl2]+, respectively. [Au8L8]2+ acts as an intermediate in this conversion of Au9 into Au11. The 2+ charge on the [Au8L8]2+ cluster and its diamagnetic nature can be rationalized by a simple Hückel MO calculation.
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Vollenbroek et al. (1980) studied this question.
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