Isomerism of Coordination Compounds isomerism;: NEET MCQs with Answers

Explanation: Platinum(II) with coordination number four gives a square-planar complex of the type [Ma₂b₂], which has exactly two geometrical isomers. In the cis isomer the two chlorido ligands occupy adjacent corners (90° apart); this isomer is cisplatin, used as an anticancer drug. In the trans isomer the two chlorido ligands occupy opposite corners (180° apart); this isomer is biologically inactive. If the complex were tetrahedral instead, all four positions would be adjacent and no geometrical isomerism would be possible, which is how this pair of isomers helped establish the square-planar geometry of Pt(II).

Explanation: This is an octahedral complex of the type [MA₃B₃], and such complexes have exactly two geometrical isomers. In the facial (fac) isomer the three identical ligands occupy the three corners of one triangular face of the octahedron, so all three are mutually cis (90° apart). In the meridional (mer) isomer the three identical ligands lie along a meridian, occupying a plane that contains the metal, so two of them are trans to each other. No further arrangements are possible, so the answer is two.

Explanation: Platinum(II) with a coordination number of four is square planar, and here all four ligands are different, giving a complex of the type [Mabcd]. For a square-planar [Mabcd] complex the number of geometrical isomers is three. Fixing one ligand, each of the remaining three can in turn be placed trans to it, and each choice gives a distinct arrangement: Cl trans to Py, NH₃ trans to Py, and Br trans to Py. Square-planar complexes are planar and so always possess a plane of symmetry; none of the three is optically active.

Explanation: With two bidentate en ligands and two chlorides in an octahedron, the chlorides may be adjacent or opposite, giving cis and trans geometrical isomers. There is no ambidentate ligand, no complex counter ion and no ionisable group exchange.

Explanation: The ion exists as trans and cis geometrical isomers, and the cis form is chiral and therefore splits into a pair of enantiomers. Counting the achiral trans form plus the two optical forms of cis gives three isomers.

Explanation: Both ions are complex, so coordination isomerism is possible: the ammine and chlorido ligands can be shared out between copper and platinum in every allowed way. The four resulting compounds are: [Cu(NH₃)₄][PtCl₄] [Cu(NH₃)₃Cl][Pt(NH₃)Cl₃] [Pt(NH₃)₃Cl][Cu(NH₃)Cl₃] [Pt(NH₃)₄][CuCl₄] Each keeps four ligands on each metal and preserves the overall composition, so there are four isomers in all.

Explanation: [Cr(SCN)₂(NH₃)₄]⁺ is an octahedral [MA₄B₂] complex with cis and trans forms. In addition, the ambidentate thiocyanate can bind through S or N, so linkage isomerism multiplies the possible forms; none of the other complexes contains an ambidentate ligand.

Explanation: In [MA₅B] there is only one B ligand. All six corners of a regular octahedron are equivalent, so wherever that single B is placed the resulting structure is identical — it can always be rotated onto any other placement. With only one possible arrangement there can be no geometrical isomerism. The others do show it: [MA₄B₂] and the equivalent [MA₂B₄] give cis and trans forms according to whether the two B (or two A) ligands are 90° or 180° apart. [MA₃B₃] gives facial (fac) and meridional (mer) forms.

Explanation: Tris(ethylenediamine)cobalt(III) bromide is [Co(en)₃]Br₃. The complex cation carries three bidentate chelate rings arranged in a propeller fashion about the octahedral cobalt. It has no plane and no centre of symmetry, so it exists as a resolvable pair of enantiomers — this is the classic textbook example of a chiral coordination compound. The others are all achiral: [Co(NH₃)₅(NO₂)]I₂ is of the type [MA₅B] and highly symmetric; [Pt(NH₃)₂Cl₂] is square planar and therefore planar; and the trans isomer of [Cr(en)₂(CN)₂]⁺ has a plane of symmetry (it is only the cis form of such complexes that is chiral).

Explanation: Only cis-[Co(en)₂Cl₂]⁺ is chiral. In this arrangement the two en rings are disposed about the metal in a way that leaves no plane of symmetry and no centre of symmetry, so the ion and its mirror image cannot be superimposed and it resolves into d and l forms. The trans isomer of the same complex has the two chlorides on an axis and a plane of symmetry containing both en rings, so it is optically inactive. Both platinum complexes are square planar. A square-planar complex is by definition planar, and the molecular plane is itself a plane of symmetry, so square-planar complexes are essentially never optically active.

Explanation: Take the three possibilities in turn. Linkage isomerism: NO₂⁻ is ambidentate, binding through N (nitro) or through O (nitrito), giving [Co(NH₃)₄(NO₂)₂]Cl and [Co(NH₃)₄(ONO)₂]Cl. ✔ Ionisation isomerism: the chloride outside can exchange with a nitrite inside, giving [Co(NH₃)₄(NO₂)Cl]NO₂. ✔ Geometrical isomerism: the ion is octahedral [MA₄B₂], so the two nitro groups may be cis or trans. ✔ Optical isomerism: both the cis and the trans forms of an [MA₄B₂] complex possess a plane of symmetry, so neither is chiral. ✘ The correct combination is therefore linkage, ionisation and geometrical isomerism.

Explanation: Optical isomerism requires the complex to be chiral, that is to have no plane of symmetry and no centre of symmetry, so that it is not superimposable on its mirror image. [Co(NH₃)₃Cl₃] is of the type [MA₃B₃] and exists as fac and mer forms, but both of these possess a plane of symmetry. The molecule is therefore achiral and shows no optical isomerism. The other three all contain at least one chelating en ring, which twists the octahedron and destroys the symmetry planes: cis-[Co(en)₂Cl₂]⁺, cis-[Co(en)Cl₂(NH₃)₂]⁺ and [Co(en)₃]³⁺ are all resolvable into enantiomers.

Explanation: [Co(NH₃)₄Cl₂]⁺ is octahedral of the type [MA₄B₂]. The two chlorido ligands may sit adjacent to each other (cis, 90°) or opposite each other (trans, 180°), giving two distinct geometrical isomers. Because the two arrangements have different ligand fields around the cobalt, they absorb at different wavelengths and so differ in colour: the cis form is violet and the trans form is green. Linkage isomerism needs an ambidentate ligand, coordination isomerism needs two complex ions, and ionisation isomerism needs an exchangeable counter ion — none of which is available here.

Explanation: Nickel(II) with four monodentate ligands and weak-field donors forms a tetrahedral complex, so [Ni(NH₃)₂Cl₂] is tetrahedral of the type [Ma₂b₂]. In a tetrahedron every pair of positions is adjacent — there is no cis or trans distinction — so no geometrical isomerism is possible. And an [Ma₂b₂] tetrahedral complex still has a plane of symmetry, so it is not chiral either. Hence no isomerism at all. The others do show isomerism: [Ni(en)₃]²⁺ is octahedral with three chelate rings and is optically active; [Ni(NH₃)₄(H₂O)₂]²⁺ is octahedral [MA₄B₂] and shows cis–trans isomerism; [Pt(NH₃)₂Cl₂] is square planar and shows the familiar cis–trans pair.

Explanation: In both compounds the cation and the anion are each complex ions. Going from the first compound to the second, the ammine ligands and the cyanido ligands have been interchanged between the two metal centres — cobalt and chromium have swapped ligand sets. This redistribution of ligands between a complex cation and a complex anion is precisely the definition of coordination isomerism. The essential requirement is that both ions must be complex, which is satisfied here. Ionisation isomerism would require exchange between the coordination sphere and a simple counter ion, and linkage isomerism would require an ambidentate ligand — neither applies.