A) \[[Pt{{(N{{H}_{3}})}_{6}}]C{{l}_{4}}\]
B) \[[Pt{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]C{{l}_{2}}\]
C) \[[Pt{{(N{{H}_{3}})}_{5}}Cl]C{{l}_{3}}\]
D) \[[Pt{{(N{{H}_{3}})}_{3}}C{{l}_{3}}]\,Cl\]
Correct Answer: C
Solution :
The complex \[PtC{{l}_{4}}.5N{{H}_{3}}\]is designated as\[[Pt{{(N{{H}_{3}})}_{5}}Cl]\,C{{l}_{3}}\]which ionizes to\[{{[Pt{{(N{{H}_{3}})}_{5}}Cl]}^{3+}}\]\[+\,3C{{l}^{-}}\]ions. Thus total ions produced are four but three moles of\[AgCl\] are produced from\[3C{{l}^{-}}\] ions with\[AgN{{O}_{3}}\].
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