Explanation

(d)
(i) The sum of oxidation states of all atoms in a compound is zero.
(ii) Oxidation state of metal in carbonyl is zero.
(a) $K_{4}Fe(CN)_{6}$
Let, oxidation state of Fe in $K_{4}Fe(CN)_{6}$= x
$\therefore$ +4+ x +($-$1$\times$6)=0
$\therefore$ x =+2
(b) $K_{2}{FeO}_{4}$
Let, oxidation state of Fe in $K_{2}{FeO}_{4}$= x
$\therefore$ +1$\times$2+ x +($-$2$\times$4)=0
$\therefore$ x =+6
(c) ${Fe}_{2}2O_{3}$
Let, oxidation state of Fe in ${Fe}_{2}O_{3}$= x
$\therefore$ 2x +($-$2$\times$3)=0
or 2x =6
$\therefore$ x =$\frac{6}{2}$=+3
(d) $Fe(CO)_{5}$
Oxidation state of Fe in $Fe(CO)_{5}$=0
$\therefore$ Oxidation state of Fe is least in $Fe(CO)_{5}$_{.}