Answer:
Option A,C,D
Explanation:
$K_{3}[Fe^{3+}(CN)_{6}]+KI(excess) $ $\rightarrow$
$K_{4}[Fe^{+2}(CN)_{6}]+KI_{3}$ (redxox)( Brownish yellow solution)
$K_{4}[Fe(CN)_{6}]+ZnSO_{4}$ $ \rightarrow$
$ K_{2}Zn_{3}[Fe(CN)_{6}]_{2}$ or $K_{2}Zn[Fe(CN)_{6}]$ (white ppt)
$I_{3}^{-}$(Brownish yellow filtrate)+$2Na_{2}S_{2}O_{3}$ $\rightarrow$ $Na_{2}S_{4}O_{6}$(clear solution)+2NaI+$I_{2}$ (Turns strach solution blue)
$K_{2}Zn[Fe(CN)_{6}]$ react with NaOH as
$K_{2}Zn[Fe(CN)_{6}]+NaOH$ $ \rightarrow$ $ [Zn(OH)_{4}]^{2-}+[Fe(CN)_{6}]^{4-}$
