Calculate the number of metal cations which form octahedral complex with excess??????

Solution:
Cu+2 + 4CN-[Cu(CN)4]2-
Fe+2 + 6CN-[Fe(CN)6]4-
Fe+3 + 6CN-[Fe(CN)6]3-
Co+3 + 6CN-[Co(CN)6]3-
Ag+ + 2CN-[Ag(CN)2]-
Zn+2 + 4CN-[Zn(CN)4]2-
3 metal ion forms octahedral complexes with CN-

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