# $4FeCl_3+3K_4Fe(CN)_6 \longrightarrow (x)+12KCl$ the x formed :

$(a)\;Fe_4(Fe(CN)_36]_3 \\(b)\;Fe(Fe(CN)_6]_3\\(c)\; Fe(Fe(CN)_3]_6 \\(d)\;Fe(Fe(CN)_2]_6$

The x formed is $Fe_4(Fe(CN)_36]_3$
Hence a is the correct answer.