$AgCl$ is dissolved in excess of each of ${NH}_{3}, KCN$ and ${Na}_{2} S_{2} O_{3}$ . The complex ions produced in each case are

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${[Ag {({NH}_{3})}_{2}]}^{+}, {[Ag (CN)_{2}]}^{+}$ and ${[Ag {(S_{2} O_{3})}_{2}]}^{3 -}$

${[Ag {({NH}_{3})}_{2}]}^{+}, {[Ag (CN)_{2}]}^{-}$ and ${[Ag {(S_{2} O_{3})}_{2}]}^{3 -}$

${[Ag {({NH}_{3})}_{2}]}^{2 +}, {[Ag (CN)_{2}]}^{3 -}$ and ${[Ag {(S_{2} O_{3})}_{2}]}^{2 -}$

${[Ag {({NH}_{3})}_{4}]}^{2 +}, {[Ag (CN)_{2}]}^{3 -}$ and ${[{Ag}_{2} {(S_{2} O_{3})}_{2}]}^{2 -}$

answer is D.

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Detailed Solution

$\begin{array}{l} AgCl + 2 {NH}_{4} OH ⟶ [Ag {({NH}_{3})}_{2}] Cl + 2 H_{2} O \\ AgCl + 2 KCN ⟶ K [Ag (CN)_{2}] + KCl \\ AgCl + {Na}_{2} S_{2} O_{3} ⟶ {Na}_{3} [Ag {(S_{2} O_{3})}_{2}] + NaCl \end{array}$

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