Question

The complex(es), which can exhibit the type of isomerism shown by [Co(en)$_3$]Cl$_3$, is(are) [en = H$_2$NCH$_2$CH$_2$NH$_2$]

• A. [Pt(en)(SCN)$_2$]
• B. [Zn(gly)$_2$]
• C. [Co(NH$_3$)$_4$Cl$_2$]$^+$
• D. [Cr(en)$_2$(H$_2$O)(SO$_4$)]$^+$

Answer 1

Answer: (B), (D)

B, D

Answer 2

The complex [Co(en)$_3$]Cl$_3$ is an octahedral complex with three bidentate ligands (en). Complexes of the type M(AA)$_3$, where AA is a symmetric bidentate ligand, are chiral and exhibit optical isomerism. They exist as a pair of enantiomers (delta and lambda forms). The question asks which of the given complexes can exhibit the same type of isomerism, i.e., optical isomerism.

Let's examine each option:

(A) [Pt(en)(SCN)$_2$]: Platinum(II) complexes are typically square planar. In a square planar complex with one bidentate ligand (en) and two monodentate ligands (SCN), the formula is M(AA)B$_2$. The en ligand occupies two adjacent coordination sites. The two SCN ligands occupy the remaining two adjacent sites, forming a cis arrangement relative to each other and the en ligand. A square planar M(AA)B$_2$ complex (cis isomer) has a plane of symmetry passing through the metal center, bisecting the angle between the two SCN ligands and the en ligand. Thus, it is achiral and does not exhibit optical isomerism.

(B) [Zn(gly)$_2$]: Zinc(II) complexes with coordination number 4 are typically tetrahedral. Glycinate (gly = H$_2$NCH$_2$COO$^-$) is an unsymmetric bidentate ligand coordinating through N and O atoms. The complex is of the type M(AB)$_2$, where AB represents the unsymmetric bidentate ligand. A tetrahedral complex of the type M(AB)$_2$ is chiral because it lacks a plane of symmetry or a center of inversion. The arrangement of the two unsymmetric ligands creates a non-superimposable mirror image. Therefore, [Zn(gly)$_2$] exhibits optical isomerism.

(C) [Co(NH$_3$)$_4$Cl$_2$]$^+$: This is an octahedral complex of the type MA$_4$B$_2$. Such complexes exhibit geometrical isomerism (cis and trans). The trans isomer (Cl ligands opposite) has a plane of symmetry and a center of inversion. The cis isomer (Cl ligands adjacent) also has a plane of symmetry. Neither isomer is chiral. Thus, [Co(NH$_3$)$_4$Cl$_2$]$^+$ does not exhibit optical isomerism.

(D) [Cr(en)$_2$(H$_2$O)(SO$_4$)]$^+$: This is an octahedral complex of the type M(AA)$_2$BC, where AA is en, B is H$_2$O, and C is SO$_4$. Such complexes can exhibit both geometrical and optical isomerism. The two en ligands can be arranged in a cis or trans manner relative to each other. If the two en ligands are in a trans position, the complex has a plane of symmetry and is achiral. If the two en ligands are in a cis position, the complex has a structure similar to cis-M(AA)$_2$B$_2$ or cis-M(AA)$_2$BC. The cis arrangement of the two bidentate ligands creates a chiral environment, similar to the M(AA)$_3$ complex. The cis-M(AA)$_2$BC isomer is generally chiral and exists as a pair of enantiomers. Therefore, [Cr(en)$_2$(H$_2$O)(SO$_4$)]$^+$ can exhibit optical isomerism (specifically, the cis-en isomer is chiral).

The complexes that exhibit optical isomerism, similar to [Co(en)$_3$]Cl$_3$, are [Zn(gly)$_2$] and [Cr(en)$_2$(H$_2$O)(SO$_4$)]$^+$.