Explanation

(c)
$\left\lbrack Cu\left( {NH}_{3} \right)_{4} \right\rbrack{SO}_{4}$
Oxidation number of
Cu ⇒ x$\ + 4 \times 0 - 2 = 0$
x $-$2=0
x =+2
O.N of Cu=+2
O.N of pt in $\lbrack Pt\left( {NH}_{3} \right)_{2}{Cl}_{2}\rbrack$
x$\ + 2 \times 0 + 2 \times - 1 = 0$
x $-$2=0
x =+2
O.N of Ni in $\lbrack Ni(CO)_{4}\rbrack$
x +4⨯0=0
x =0
O.N of Fe in $K_{3}\lbrack Fe(CN)_{6}\rbrack$
3⨯(+1)+ x +6⨯$-$1=0
3+ x $-$6=0
x =+3
$\therefore\lbrack Ni(CO)_{4}\rbrack$ is zero valent compound.