In the substitution flow, Rh stays +3 and octahedral. Each arrow replaces two aqua ligands by one en. Count all geometrical and optical products, not just one selected pathway; ignore ring conformatio

Question

In the substitution flow, Rh stays +3 and octahedral. Each arrow replaces two aqua ligands by one en. Count all geometrical and optical products, not just one selected pathway; ignore ring conformations. Let $S_{B}$ , $S_{C}$ be total stereoisomer counts of B,C and $U_{A}$ the unpaired-electron count of low-spin A. Find $S_{B}$ + $S_{C}$ + $U_{A}$ .

TestHub · Accepted Answer

B is M(en)a₄, one achiral form. C is M(en)₂a₂, with achiral trans and two enantiomeric cis forms. A is low-spin d⁶, with no unpaired electrons. Hence 1+3+0 = 4. Counting only cis/trans misses an enantiomer.

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