A) \[CoC{{l}_{3}}.3N{{H}_{3}}\]
B) \[CoC{{l}_{3}}.\,4N{{H}_{3}}\]
C) \[CoC{{l}_{3}}.5N{{H}_{3}}\]
D) \[CoC{{l}_{3}}.6N{{H}_{3}}\]
Correct Answer: A
Solution :
\[[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}\to {{[Co{{(N{{H}_{3}})}_{6}}]}^{3+}}+3C{{l}^{-}}\] \[[Co{{(N{{H}_{3}})}_{3}}\text{C}{{\text{l}}_{\text{3}}}]\to [Co{{(N{{H}_{3}})}_{2}}C{{l}_{3}}]\] \[[Co{{(N{{H}_{3}})}_{4}}\text{C}{{\text{l}}_{2}}]Cl\to {{[Co{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]}^{+}}\text{C}{{\text{l}}^{\text{-}}}\] \[[Co{{(N{{H}_{3}})}_{5}}Cl]C{{l}_{2}}\to {{[Co{{(N{{H}_{3}})}_{5}}Cl]}^{2+}}+2C{{l}^{-}}\]So, \[[Co{{(N{{H}_{3}})}_{3}}C{{l}_{3}}]\] does not ionize so does not give test for chloride ions.
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