Question

Which of the following statements is not true for the complex $Co(N{H}_3{)}_{4}B{r}_{2}N{O}_2$?

• A. It shows ionization, linkage and geometrical isomerism
• B. It does not show optical isomerism
• C. Its ionization isomers cannot be differentiated by silver nitrate solution
• D. It forms inner orbital complex with a low spin and has the hybridization $d^{2}s{p}^3$

Answer 1

The correct option is C Its ionization isomers cannot be differentiated by silver nitrate solution

(a): $Co(N{H}_3{)}_{4}B{r}_{2}N{O}_2$ can show ionisation and geometrical isomerism. Ionisation isomers are $\left[Co\right(N{H}_3{)}_{4}B{r}_2]N{O}_2$ and $\left[Co\right(N{H}_3{)}_{4}BrN{O}_2]Br$
This can also show linkage isomerism as $N{O}_2$ can either bind with $N$ and $O$.

(b): Due to the presence of mirror symmetry, it does not show optical isomerism.

(c): $\left[Co\right(N{H}_3{)}_{4}B{r}_2]N{O}_2$ and $\left[Co\right(N{H}_3{)}_{4}BrN{O}_2]Br$ can be differentiated by $AgN{O}_3$ solution as the latter will give a yellow ppt of $AgBr$.

(d): $Co$ is in the +3 oxidation state, here $N{H}_3$ acts as a strong field ligand and facilitates pairing. Hence, a low spin inner orbital complex will be formed with $d^{2}s{p}^3$ hybridisation.