Explanation

(c)
(a) $K_{3}\lbrack Fe(OH)_{6}\rbrack$
Let oxidation state of Fe in
$K_{3}\lbrack Fe(OH)_{6}\rbrack\$=x
(+1$\times$3)+ x +($-$1$\times$6)=0
$\therefore\$ x =+3
(b) $K_{2}\lbrack{FeO}_{4}\rbrack$
Let oxidation state of Fe in $K_{2}\lbrack{FeO}_{4}\rbrack$=x
(+1$\times$2)+x +(4$\times -$2)=0
x =+6
${FeSO}_{4}.\left( {NH}_{4} \right)_{2}{SO}_{4}\ .6H_{2}O$
Let oxidation state of Fe in
${FeSO}_{4}.\left( {NH}_{4} \right)_{2}{SO}_{4}\ .6H_{2}O$=x
$\therefore$ x +($-$2)+2+($-$2)=0
x =+2
(d) $\left\lbrack Fe(CN)_{6} \right\rbrack^{3 -}$
Let oxidation state of Fe in
$\left\lbrack Fe(CN)_{6} \right\rbrack^{3 -}$=x
$\therefore$ x +(6$\times -$1)=$\ -$3
$\therefore$ x =+3
${FeSO}_{4}.\left( {NH}_{4} \right)_{2}{SO}_{4}$ has Fe in lowest oxidation state.