Answer:
Option A
Explanation:
$S^{2-}+CuCl_{2}\rightarrow CuS\downarrow$ (black ppt.)
$SO_{4}^{2-}+CuCl_{2}\rightarrow Soluble, $ thus
(a) $CuCl_{2}$ selectively precipitates $S^{2-}$
(b) $S^{2-} (soluble)+BaCl_{2}\rightarrow BaS\downarrow$
$SO_{4}^{2-}(white ppt.)+BaCl_{2}\rightarrow BaSO_{4}\downarrow$
(b) precipitates $SO_{4}^{2-}$ and not $S^{2-}$
(c) $S^{2-} (black .ppt)+Pb^{2+}\rightarrow PbS\downarrow$
$SO_{4}^{2-} (white.ppt)+Pb^{2+} \rightarrow PbSO_{4}\downarrow$
$S^{2-} $ and $SO_{4}^{2-}$, both are precipitated.
(d) $ S^{2-}+Na_{2}[Fe(CN)]_{5}NO\rightarrow Na _{4}[Fe(CN)]_{5}NOS]$
Sodium nitroprusside (purple colour)
But no colour with $SO_{4}^{2-}$