Question

One cationic complex has two isomers A and B. Each has one $C{o}^{3+}$, five $N{H}_3$, one $B{r}^{-}$, and one $S{O}_4^{2-}$ stoi-chiometrically. A gives white ppt with $BaC{l}_2$ whle B give yellow ppt with $AgN{O}_3$.
B can be:

• A. $\left[Co\right(N{H}_3{)}_{5}S{O}_4]Br$
• B. $\left[Co\right(N{H}_3{)}_{5}Br]S{O}_4$
• C. $\left[Co\right(N{H}_3{)}_{3}Br\left(S{O}_4\right)].2N{H}_3$
• D. None of these

Answer 1

The correct option is A

$\left[Co\right(N{H}_3{)}_{5}S{O}_4]Br$

This question combines the knowledge of both coordination compounds as well as colours of salts (qualitative inorganic salt(s) analysis).

To start with, the isomers should definitely be octahedral and the complex is said to be cationic! This means that outside the coordination sphere - given the colour of the precipitates - either $S{O}_4^{2-}$ or $B{r}^{-}$ are present. Now we want to figure out what the compound 'B' is:

The yellow precipitate has to be $AgBr$ while the white precipitate should be $BaS{O}_4$.

Hence the answer is $\left[Co\right(N{H}_3{)}_{5}S{O}_4]Br$!