• # question_answer AgCl is dissolved in excess of each of $\text{N}{{\text{H}}_{\text{3}}}\text{, KCN}$and$N{{a}_{2}}{{S}_{2}}{{O}_{3}}.$The complex ions produced in each case are A) ${{[Ag{{(N{{H}_{3}})}_{2}}]}^{2+}},{{[Ag{{(CN)}_{2}}]}^{3-}}$and${{[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]}^{2-}}$B) ${{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}},{{[Ag{{(CN)}_{2}}]}^{3-}}$and${{[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]}^{2-}}$C) ${{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}},{{[Ag{{(CN)}_{2}}]}^{-}}$and${{[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]}^{3-}}$D) ${{[Ag{{(N{{H}_{3}})}_{4}}]}^{2+}},{{[Ag{{(CN)}_{2}}]}^{3-}}$and${{[A{{g}_{2}}{{({{S}_{2}}{{O}_{3}})}_{2}}]}^{2-}}$

Correct Answer: C

Solution :

$AgCl+2N{{H}_{4}}OH\to [Ag{{(N{{H}_{3}})}_{2}}]Cl+2{{H}_{2}}O$ $AgCl+2KCN\to K[Ag{{(CN)}_{2}}]+KCl$ $AgCl+N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to N{{a}_{3}}[Ag{{({{S}_{2}}{{O}_{3}})}_{2}}]+NaCl$ Hence, the correct option is [c].

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