d and f block elements

(A) $\mathrm{CuCl}_2+\mathrm{S}^{2-}\rightarrow \mathrm{CuS}\downarrow +2\mathrm{Cl}^-$       $(\mathrm{Sol}^n)\;(\mathrm{Sol}^n)\;(\mathrm{Black\;ppt})\;(\mathrm{sol}^n)$      $\mathrm{CuCl}_2+\mathrm{SO}_4^{2-}\rightarrow$No ppt.      $(\mathrm{Sol}^n)\;(\mathrm{Sol}^n)$(B) $\mathrm{BaCl}_2+\mathrm{S}^{2-}\rightarrow \mathrm{BaS}+2\mathrm{Cl}^-$       $(\mathrm{Sol}^n)\;(\mathrm{Sol}^n)\;(\mathrm{No\;ppt.})\;(\mathrm{Sol}^n)$      $\mathrm{Bacl}_2+\mathrm{SO}_4^{2-}\rightarrow \mathrm{BaSO}_4\downarrow +2\mathrm{Cl}^-$      $(\mathrm{Sol}^n)\;(\mathrm{Sol}^n)\;(\mathrm{White\;ppt.})\;(\mathrm{Sol}^n)$(C)        $(\mathrm{Sol}^n)\;(\mathrm{Sol}^n)\;(\mathrm{Black\;ppt.})\;(\mathrm{Sol}^n)$           $(\mathrm{Sol}^n)\;(\mathrm{Sol}^n)\;(\mathrm{White\;ppt.})\;(\mathrm{Sol}^n)$(D)         $(\mathrm{Sol}^n)\;(\mathrm{Sol}^n)\;$(Purple colour solution ppt.)        $\mathrm{Na}_2[\mathrm{Fe(C)}_5\mathrm{NO}]+\mathrm{SO}_4^{2-}$→No ppt.        $(\mathrm{Sol}^n)\;(\mathrm{Sol}^n)\;$Note: $\mathrm{PbSO}_4\ \mathrm{Ksp}=2.5\times 1.^{-8}$$\mathrm{PbS}\ \mathrm{Ksp}=3\times 10^{-28}$ which are not given in questionAs in question selective precipitation is asked PbS will be precipitate much easier than $\mathrm{PbSO}_4$ though both are insoluble. Hence answer should be (C) also along with (A)