Question

A coordination compound X gives pale yellow colour with $AgN{{O}_{3}}$ solution while its isomer Y gives white precipitate with $BaC{{l}_{2}}$. Two compounds are isomers of $CoBrS{{O}_{4}}.5N{{H}_{3}}$. What could be the possible formula of X and Y?
A) $X=[Co{{(N{{H}_{3}})}_{5}}S{{O}_{4}}]Br$ ,$Y=[Co{{(N{{H}_{3}})}_{5}}Br]S{{O}_{4}}$
B) $X=[Co{{(N{{H}_{3}})}_{5}}Br]S{{O}_{4}}$, $Y=[Co{{(N{{H}_{3}})}_{5}}S{{O}_{4}}]Br$
C) $X=[Co{{(N{{H}_{3}})}_{5}}BrS{{O}_{4}}]$, $Y=[CoBrS{{O}_{4}}{{(N{{H}_{3}})}_{5}}]$
D) $X=[Co{{(Br)}_{5}}N{{H}_{3}}S{{O}_{4}}]$, $Y=[CoBrS{{O}_{4}}(N{{H}_{3}})]$

Answer 1

Hint: The answer is dependent on the coordination chemistry part which involves the fact that the easily dissociable anion should be outside the brackets so that it can easily form isomers and that gives you the required answer.

Complete step by step answer:
In our lower classes of inorganic chemistry, we have studied about the coordination compounds and the various properties of the complexes.
Now let us see the definition and the terms which relates to it that will help us to lead to the required answer.
- Coordination compounds are those molecules that possess one or multiple metal centres that are bound to ligands. These coordination compounds or complexes can be neutral and also can be charged and when the complex is charged, it is stabilized by neighbouring counter ions.
- Ligands bound to the central atom are also called complexing agents.
- The isomers of a compound are those which have the same molecular formula and different arrangements of atoms. In coordination chemistry, easily dissociable ions are written outside the brackets which as per rule of writing the formula of coordinating molecules.
- Now, from the given options we know that the central metal ion is cobalt with the ligands ammonia, sulphate and bromine bound to it.
- Among these, bromine has greater tendency to be dissociated and hence the complex is $[Co{{(N{{H}_{3}})}_{5}}S{{O}_{4}}]Br$
- Therefore, the complex dissociation is given by,
$[Co{{(N{{H}_{3}})}_{5}}S{{O}_{4}}]Br\to {{[Co{{(N{{H}_{3}})}_{5}}S{{O}_{4}}]}^{+}}+B{{r}^{-}}$
- Now, this bromine ion dissociated reacts with the silver nitrate solution to form silver bromide which is pale yellow in colour.
- Now, barium chloride readily reacts with the sulphate ions and therefore, the sulphate ions dissociate in the complex easily which is having the sulphate ligand outside. The dissociation is as shown below,
$[Co{{(N{{H}_{3}})}_{5}}Br]S{{O}_{4}}\to {{[Co{{(N{{H}_{3}})}_{5}}Br]}^{2+}}+S{{O}_{4}}^{2-}$
Now, the released sulphate ion reacts with barium chloride to produce barium sulphate which is a white precipitate.

So, the correct answer is “Option A”.

Note: Before answering such types of questions, do not get confused about the same composition with different brackets. Make sure that you know the method to write the complexes as per the rules and this helps you to choose the correct answer.