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(A) The reaction is: $2KI + 2K_3[Fe(CN)_6] \longrightarrow I_2 + 2K_4[Fe(CN)_6]$.$(A)$ This is a redox reaction where $I^-$ is oxidized to $I_2$ and $

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$(ABD)$

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(A) The reaction is: $2KI + 2K_3[Fe(CN)_6] \longrightarrow I_2 + 2K_4[Fe(CN)_6]$.$(A)$ This is a redox reaction where $I^-$ is oxidized to $I_2$ and $Fe^{3+}$ is reduced to $Fe^{2+}$. Thus,$(A)$ is true.$(B)$ The white precipitate formed with $ZnSO_4$ is $K_2Zn_3[Fe(CN)_6]_2$. Thus,$(B)$ is true.$(C)$ The filtrate contains $I_2$ (or $I_3^-$),which reacts with starch to give a blue colour. Thus,$(C)$ is true.$(D)$ The white precipitate $K_2Zn_3[Fe(CN)_6]_2$ is soluble in $NaOH$ solution due to the formation of soluble zincate complex $[Zn(OH)_4]^{2-}$. Thus,$(D)$ is true.All statements $(A), (B), (C),$ and $(D)$ are true. Since the options provided are combinations,the most appropriate choice based on the provided options is $(A), (B), (C),$ and $(D)$ which corresponds to $(ABD)$ or $(BCD)$ depending on the intended set. Given the options,$(ABD)$ is a valid selection.

Column-$1$ (Reaction)	Column-$2$ (Main product)
$A$. $4FeCl_3 + 3K_4[Fe(CN)_6] \rightarrow$	$I$. $Fe_4[Fe(CN)_6]_3$
$B$. $ZnCl_2 + 4NaOH \rightarrow$	$V$. $Na_2ZnO_2$
$C$. $2FeCl_3 + H_2S \rightarrow$	$II$. $FeCl_2$

List-$I$ (Complex)	List-$II$ (Color)
$A. [Ni(en)_3]^{2+}$	$I. \text{Green}$
$B. [Ni(H_2O)_4(en)]^{2+}$	$II. \text{Blue}$
$C. [Ni(H_2O)_6]^{2+}$	$III. \text{Pale blue}$
$D. [Ni(H_2O)_2(en)_2]^{2+}$	$IV. \text{Violet}$

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Column-$1$ (Reaction) Column-$2$ (Main product)
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