The reagent(s) that can selectively precipitate S ^ 2 - from a...

(A)

$\begin{cases}CuC{l}_{2} & + & S^{2 -} \\ \left(sol^{n}\right) & & \left(sol^{n}\right)\end{cases}$\begin{cases}\to & CuS ↓ & + \\ & \left(Blackppt .\right) & \end{cases}$$\begin{array}{ll}2Cl^{-} \\ \left(Sol^{n}\right)\end{array}$$
$\begin{cases}CuC{l}_{2} & + & S{O}_{4}^{2 -} \\ \left(sol^{n}\right) & & \left(sol^{n}\right)\end{cases}$\begin{cases}\to & Noppt . \\ & \end{cases}<br />(B)</p><p>$\begin{cases}BaC{l}_{2} & + & S^{2 -} \\ \left(sol^{n}\right) & & \left(sol^{n}\right)\end{cases}$\begin{cases}\to & BaS & + \\ & \left(Noppt .\right) & \end{cases}\begin{array}{ll}2Cl^{-} \\ \left(Sol^{n}\right)\end{array}<br />$\begin{cases}BaC{l}_{2} & + & {SO}_{4}^{2 -} \\ \left(sol^{n}\right) & & \left(sol^{n}\right)\end{cases}$\begin{cases}\to & BaS{O}_{4} ↓ & + \\ & \left(Whiteppt .\right) & \end{cases}\begin{array}{ll}2Cl^{-} \\ \left(Sol^{n}\right)\end{array}<br />(C)</p><p>$\begin{cases}Pb{\left(OOCC{H}_{3}\right)}_{2} & + & S^{2 -} \\ \left(sol^{n}\right) & & \left(sol^{n}\right)\end{cases}$\begin{cases}\to & PbS ↓ & + \\ & \left(Blackppt .\right) & \end{cases}\begin{array}{ll}2C{H}_{3}COO^{-} \\ \left(Sol^{n}\right)\end{array}<br />$\begin{cases}Pb{\left(OOCC{H}_{3}\right)}_{2} & + & {SO}_{4}^{2 -} \\ \left(sol^{n}\right) & & \left(sol^{n}\right)\end{cases}$\begin{cases}\to & PbS{O}_{4} ↓ & + \\ & \left(Whiteppt .\right) & \end{cases}\begin{array}{ll}2C{H}_{3}COO^{-} \\ \left(Sol^{n}\right)\end{array}<br />(D)</p><p>$\begin{cases}N{a}_{2}\left[Fe{\left(CN\right)}_{5}NO\right] & + & S^{2 -} \\ \left(sol^{n}\right) & & \left(sol^{n}\right)\end{cases}$\begin{cases}\to & N{a}_{4}\left[Fe{\left(CN\right)}_{5}NOS\right] \\ & ( Purplecoloursolution )\end{cases}
$\begin{cases}N{a}_{2}\left[Fe{\left(CN\right)}_{5}NO\right] & + & {SO}_{4}^{2 -} \\ \left(sol^{n}\right) & & \left(sol^{n}\right)\end{cases}$\begin{cases}\to & Noppt . \\ & \end{cases}<br />Note: \left\begin{array}{ll}PbS{O}_{4}Ksp = 2.5 \times {10}^{- 8} \\ PbSKsp = 3 \times {10}^{- 28}\end{array}$\right\}Whicharenotgiveninquestion
As in question selective precipitation is asked $\text{P}\text{b}\text{S}$ will be precipitate much easier than $PbS{O}_{4}$ though both are insoluble. Hence, answer should be $\text{P}\text{b}{( \text{O}\text{O}\text{C} - \text{C}{H}_{3} )}_{2}$ also along with \text{C}\text{u}\text{C}{l}_{2}.