Question

Which of the following isomers will give a white precipitate with $BaC{l}_2$ solution?

• 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. $\left[Co\right(N{H}_3{)}_4\left(S{O}_4{)}_2\right]Br$
• D. $\left[Co\right(N{H}_3{)}_{4}Br\left(S{O}_4\right)]$

Answer 1

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

$S{O}_4^{2-}$ will give a precipitate with $BaC{l}_2$ only if the sulphate ion is not present in the coordination sphere. Only then it will be able to dissociate as an anion in aqueous solution.

The complex $\left[Co\right(N{H}_3{)}_{5}Br]S{O}_4$ dissociates to give $\left[Co\right(N{H}_3{)}_{5}Br{]}^{2+}$ and $S{O}_4^{2-}$ ions. $S{O}_4^{2-}$ ion reacts with barium chloride to give white precipitate of $BaS{O}_4$.
$\left[Co\right(N{H}_3{)}_{5}Br]S{O}_4\rightleftharpoons [Co(N{H}_3{)}_{5}Br{]}^{2+}+S{O}_4^{2-}$
$S{O}_4^{2-}+BaC{l}_2\to BaS{O}_4\left(\text{white ppt.}\right)\downarrow +2C{l}^{-}$