Question

Which of the following does not show optical isomerism? (en = ethylenediamine)

• A. $[Co(en{)}_{2}C{l}_2{]}^{+}$
• B. $[Co(N{H}_3{)}_{3}C{l}_3{]}^0$
• C. $[Co(en)C{l}_2(N{H}_3{)}_2{]}^{+}$
• D. $[Co(en{)}_3{]}^{3+}$

Answer 1

Answer: (B)

Optical isomerism is exhibited by only those complexes in which elements of symmetry are absent. Octahedral complexes of the types $\left[M(aa{)}_3\right]$, $\left[M(aa)x_2,y_2\right]$ and $\left[M(aa{)}_2{x}_2\right]$ have absence of elements of symmetry, thus exhibit optical isomerism. Here, aa represents bidentate ligand, $x$ or $y$ represents monodentate ligand and $M$ represents central metal ion.
Hence, ${\left[Co{\left(N{H}_3\right)}_{3}C{l}_3\right]}^0$ due to presence of symmetry elements does not exhibit optical isomerism.
or
Octahedral complexes of $\left[M(AA{)}_2{B}_2\right]$ type, e.g.
${\left[Co(en{)}_{2}C{l}_2\right]}^{+},\left[M(AA)B_2{C}_2\right]$ type, e.g.
$\left[Co(en)C{l}_2{\left(N{H}_3\right)}_2\right]$ and $\left[M(AA{)}_3\right]$ type, e.g.
${[\text{ Co(en }{)}_3]}^{3+}$ show optical isomensim, whereas complexes of $\left[M{A}_3{B}_3\right]$ type, e.g. ${\left[Co\left(N{H}_3\right)C{l}_2\right]}^0$ do not show optical isomerism.