d
d
\(\text{CoC}{{\text{l}}_{3}}\cdot 6{{\text{H}}_{2}}\text{O}+\text{AgN}{{\text{O}}_{3}}\) \((\text{ excess })\to \text{AgCl}\)

\(100\, \mathrm{mL}, 0.1 \,\mathrm{M}\) \(1.2\, \times 10^{22} \text { ion }\)

\(\text {moles }\;\; 0.01 \,\mathrm{mol}\)   \(\frac{1.2 \times 10^{22}}{6 \times 10^{23}}=0.02\, \mathrm{mol}\)

since, \(0.01\, \mathrm{mol}\) of \(\mathrm{CoCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) produce \(0.02\,\mathrm{mol}\) of \(\mathrm{AgCl}\)

Therefore, \(1\,mol\) of \(\mathrm{CoCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) will produce \(2\, \mathrm{mol}\) of \(\mathrm{AgCl}\)

Thus, the complex is \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\)