1.

Prussian blue is

Answer»

`K_(3)[Fe(CN)_(6)]`
`K_(4) [Fe(CN)_(6)]`
`KFE[Fe(CN)_(6)]`
`Fe_(4)[Fe(CN)_(6)]`

Solution :Prussian BLUE in `K overset(III)(Fe)[overset(II) (Fe) (CN)_(6)]`


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