A) \[NiS{{O}_{4}}\]
B) \[Ba{{S}_{2}}{{O}_{3}}\]
C) \[Pb{{S}_{2}}{{O}_{3}}\]
D) \[CuS{{O}_{4}}\]
Correct Answer: B
Solution :
\[B{{a}^{+2}}+2C{{H}_{3}}CO{{O}^{-}}\to {{(C{{H}_{3}}COO)}_{2}}Ba\] \[Ba{{S}_{2}}{{O}_{3}}\]gives \[S{{O}_{2}}\]gas with dil.\[HCl\]and also yellow ppt. of Barium chromate. \[(C{{H}_{3}}COO)Ba+{{K}_{2}}Cr{{O}_{4}}\to 2C{{H}_{3}}COOK+\]\[\underset{\text{(yellow ppt}\text{.)}}{\mathop{BaCr{{O}_{4}}\downarrow }}\,\]
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