- A$[Cu{(N{H_3})_4}]S{O_4}$
- B$[Pt{(N{H_3})_2}C{l_2}]$
- ✓$[Ni{(CO)_4}]$
- D${K_3}[Fe{(CN)_6}]$
Oxidation numbers of $C u:$
$x+4 \times 0-2=0$
$x-2=0$
$x=+2$
(b) Oxidation number of $P t$ in $\left[P t\left(N H_{3}\right)_{2} C l_{2}\right]$ $x+2 \times 0+2 \times-1=0$
$x-2=0$
$x=+2$
(c) Oxidation number of $N i$ in $\left[N i(C O)_{4}\right]$
$x+4 \times 0=0$
$x=0$
(d) Oxidation number of $F e \operatorname{in} K_{3}\left[F e(C N)_{6}\right]$
$3 \times(+1)+x+6 \times-1=0$
$3+x-6=0$
$x=+3$
$\therefore\left[N i(C O)_{4}\right]$ is a zero-valent compound.
Hence, option $C$ is correct.
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$\mathrm{NaOH}+\mathrm{Cl}_{2} \rightarrow(\mathrm{A})+$ side products
(hot and conc.)
$\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{Cl}_{2} \rightarrow(\mathrm{B})+$ side products
$(dry)$