Question

Which kind of isomerism is exhibited by octahedral Co(NH${}_3$)${}_4$Br${}_2$Cl ?

• A. Geometrical and ionization
• B. Geometrical and optical
• C. Optical and ionization
• D. Geometrical only

Answer 1

Answer: (A)

Geometrical and ionization

$Co(N{H}_3{)}_{4}B{r}_{2}Cl$ is octahedral in shape.

It shows ionization and geometrical isomerism.

$\left[Co\right(N{H}_3{)}_{4}B{r}_2]Cl\rightleftharpoons [Co(N{H}_3{)}_{4}B{r}_2{]}^{+}+C{l}^{-}$

$\left[Co\right(N{H}_3{)}_{4}BrCl]Br\rightleftharpoons [Co(N{H}_3{)}_{4}BrCl{]}^{+}+B{r}^{-}$

The above two structures are geometrical isomers because they furnish different ions on dissociation.

$Co(N{H}_3{)}_{4}B{r}_{2}Cl$ is of $M{a}_4{b}_2$ type, and it shows geometrical isomerism as shown here.