S^{2-} and SO_3^{2-} can be distinguished by | vedclass

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(C) $S^{2-}$ reacts with lead acetate $(CH_3COO)_2Pb$ to form a black precipitate of $PbS$,whereas $SO_3^{2-}$ does not form a black precipitate.$S^{2

Vedclass · Accepted Answer

both $(A)$ and $(B)$

Vedclass · Answer

(C) $S^{2-}$ reacts with lead acetate $(CH_3COO)_2Pb$ to form a black precipitate of $PbS$,whereas $SO_3^{2-}$ does not form a black precipitate.$S^{2-} + (CH_3COO)_2Pb 	o PbS (black \downarrow) + 2CH_3COO^{-}$$S^{2-}$ reacts with sodium nitroprusside $Na_2[Fe(CN)_5NO]$ to form a violet-colored complex,whereas $SO_3^{2-}$ does not give this test.$S^{2-} + [Fe(CN)_5NO]^{2-} 	o [Fe(CN)_5NOS]^{4-} (violet)$Therefore,both reagents can be used to distinguish $S^{2-}$ from $SO_3^{2-}$.

$S^{2-}$ and $SO_3^{2-}$ can be distinguished by using

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