A) \[[Cr\,{{(N{{H}_{3}})}_{4}}Br]\,C{{l}_{2}};[Cr\,(N{{H}_{3}})C{{l}_{2}}]\,Br\]
B) \[[Cr\,{{(N{{H}_{3}})}_{4}}BrCl]\,Cl;[Cr\,(N{{H}_{3}})C{{l}_{2}}Br]\,N{{H}_{3}}\]
C) \[[Cr\,{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]Br;[Cr\,(N{{H}_{3}})\,BrCl]\,Cl\]
D) \[[Cr\,{{(N{{H}_{3}})}_{4}}BrCl]\,Cl;[Cr\,{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]\,Br\]
Correct Answer: D
Solution :
[d]| Since, the compound A gives white precipitate with \[AgN{{O}_{3}},\] which is soluble in dilute ammonia solution, it contains at least one \[Cl\] atom outside the coordination sphere. Thus, its formula [a] is \[[Cr\,{{(N{{H}_{3}})}_{4}}BrCl]\,Cl.\] |
| \[[Cr\,{{(N{{H}_{3}})}_{4}}BrCl]\,Cl\xrightarrow{{}}{{[Cr\,{{(N{{H}_{3}})}_{4}}BrCl]}^{+}}C{{l}^{-}}\] |
| \[AgN{{O}_{3}}+C{{l}^{-}}\xrightarrow{{}}\underset{\text{White}\,\text{ppt}.}{\mathop{AgCl}}\,+NO_{3}^{-}\] |
| Further, the compound B on reaction with\[AgN{{O}_{3}}\]produces a pale yellow precipitate soluble in cone. \[N{{H}_{3}}\]solution. |
| Thus, \['B'\]is \[[Cr\,{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]Br\]\[[Cr\,{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]Br\xrightarrow{{}}{{[Cr{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]}^{+}}+B{{r}^{-}}\] |
| \[AgN{{O}_{3}}+B{{r}^{-}}\xrightarrow{{}}\underset{\text{Pale}\,\text{yellolw}\,\text{ppt}\text{.}}{\mathop{AgBr}}\,+NO_{3}^{-}\] |
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