Correct Answer - B
`CuSO_(4) + K_(4) [Fe(CN)_(6)] rarr `no reaction
`4NH_(4)OH +CuSO_(4) rarr underset("Deep blue")([Cu(NH_(3))_(4)]SO_(4)) + 4H_(2)O`
`underset("Anhydrous")(CuSO_(4)) +5H_(2)O rarr underset(Blue)(CuSO_(4)).5H_(2)O`
`4FeCI_(3) +3Na[Fe(CN)_(6)] rarr underset(("Prussian blue"))underset("Ferric ferrocyanide")(Fe_(4)[Fe(CN)_(6)]+12 NaCI)`