Match the complex species given in List-I with the possible isomerism given in List-II
Choose the correct answer from the options given below: |
(A)-(I), (B)-(II), (C)-(IV), (D)-(III) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (A)-(IV), (B)-(I), (C)-(III), (D)-(II) (A)-(IV), (B)-(I), (C)-(II), (D)-(III) |
(A)-(IV), (B)-(I), (C)-(II), (D)-(III) |
The correct answer is Option (4) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(A) $[Co(NH_3)_4Cl_2]^+$ $\rightarrow$ (IV) Geometrical This complex follows the $MA_4B_2$ formula for octahedral geometry. It exhibits geometrical isomerism because the two chloride ($Cl^-$) ligands can be arranged in two different spatial positions:
(B) cis-$[Co(en)_2Cl_2]^+$ $\rightarrow$ (I) Optical While the trans-isomer of this complex is symmetrical and optically inactive, the cis-isomer lacks a plane of symmetry. This makes it chiral, allowing it to exist as two non-superimposable mirror images known as enantiomers ($d$ and $l$ forms). Therefore, it exhibits optical isomerism. (C) $[Co(NH_3)_5(NO_2)]Cl_2$ $\rightarrow$ (II) Ionization Although this complex contains an ambidentate ligand ($NO_2^-$) and is a classic example of linkage isomerism, in the context of the provided options, it is linked to ionization isomerism. This occurs when the ligand inside the coordination sphere ($NO_2^-$) can be interchanged with the counter-ion outside ($Cl^-$), resulting in isomers like $[Co(NH_3)_5Cl](NO_2)Cl$ that produce different ions in solution. Note: Many textbooks use this specific complex to illustrate linkage isomerism ($M-NO_2$ vs $M-ONO$), but for this matching set, it serves the role of the species capable of ion exchange. (D) $[Co(NH_3)_6][Cr(CN)_6)]$ $\rightarrow$ (III) Coordination This isomerism occurs in compounds where both the cation and the anion are complex ions. Coordination isomerism arises from the interchange of ligands between the two metal centers. For example, its isomer would be $[Cr(NH_3)_6][Co(CN)_6]$. |