Which of the following complex species is not expected to exhibit optical isomerism?

First, recall the basic condition for a coordination compound to be optically active. A molecule will show optical isomerism only when it is chiral, that is, when it cannot be super-posed on its mirror image. In symmetry language, a chiral molecule must lack (i) a plane of symmetry, (ii) a centre of symmetry and (iii) an improper rotation axis $$S_n\;(n\ge 2)$$. Whenever any one of these symmetry elements is present, the complex becomes achiral and therefore optically inactive.

Bidentate ligands such as ethane-1,2-diamine (abbreviated “en”) usually induce chirality in an octahedral complex because each ligand itself ties two adjacent positions together, generating an asymmetric three-dimensional arrangement. On the other hand, if only monodentate ligands are present, it is much easier for the molecule to possess a symmetry plane or a centre, so such complexes are very often optically inactive.

Now we examine the four given species one by one.

Option A has the formula $$[Co(NH_3)_3Cl_3]$$. The ligands are all monodentate, three are $$NH_3$$ and three are $$Cl^-$$. For an octahedral complex of the general type $$MA_3B_3$$ two geometrical possibilities exist:

1. Facial (fac): three identical ligands occupy one triangular face of the octahedron.
2. Meridional (mer): three identical ligands lie in the same plane passing through the metal centre.

We test each form for symmetry elements.

• The mer isomer possesses a plane that bisects the octahedron and exchanges two $$NH_3$$ ligands with two $$Cl^-$$ ligands; it also has a centre of symmetry. Either element is enough to make the molecule achiral, so mer-$$[Co(NH_3)_3Cl_3]$$ is optically inactive.

• The fac isomer belongs to the point group $$C_{3v}$$ because it has one $$C_3$$ axis and three vertical mirror planes. The presence of mirror planes again destroys chirality, so the fac form is also optically inactive.

Since both possible geometrical isomers are achiral, the complex $$[Co(NH_3)_3Cl_3]$$ cannot exhibit optical isomerism.

Option B is $$[Co(en)(NH_3)_2Cl_2]^+$$. Here the bidentate ligand $$en$$ creates an asymmetric five-membered chelate ring. If the two $$Cl^-$$ ligands are cis to each other, the molecule has no plane or centre of symmetry and becomes chiral. Thus at least one geometrical form of this complex is optically active.

Option C is $$[Co(en)_3]^{3+}$$. The three bidentate $$en$$ ligands wrap completely around the metal ion. This situation is the classic “tris-chelate” case, analogous to $$[Cr(ox)_3]^{3-}$$, and inevitably gives two non-superposable mirror images designated $$\Delta$$ and $$\Lambda$$. Therefore the complex is optically active.

Option D is $$[Co(en)_2Cl_2]^+$$. When the two $$en$$ ligands are arranged so that the two $$Cl^-$$ ligands are cis, the molecule lacks any symmetry plane or inversion centre, producing a pair of optical enantiomers. Consequently this species can also exhibit optical isomerism.

Summarising, the only species among the four that is incapable of becoming chiral under any geometrical arrangement is $$[Co(NH_3)_3Cl_3]$$.

Hence, the correct answer is Option A.