Question

Total number of geometrical isomers possible for the complexes [NiCl$_4$]$^{2-}$, [CoCl$_2$(NH$_3$)$_4$]$^+$, [Co(NH$_3$)$_3$(NO$_2$)$_3$] and [Co(NH$_3$)$_5$Cl]$^{2+}$ is

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

Memorize the common cases for geometrical isomerism in octahedral complexes: - [MA$_4$B$_2$]: 2 isomers (cis, trans) - [MA$_3$B$_3$]: 2 isomers (fac, mer) - [M(AA)$_2$B$_2$]: 2 isomers (cis, trans) where AA is a symmetric bidentate ligand. - [MA$_2$B$_2$C$_2$]: 5 isomers Tetrahedral [MA$_4$] and square planar [MA$_4$] do not show geometrical isomerism.

Answer 1

The Correct Option is C

We need to find the number of geometrical isomers for each complex and then sum them up.
1. [NiCl$_4$]$^{2-}$: This complex has a central Ni$^{2+}$ ion (d$^8$ configuration). With a weak field ligand like Cl$^-$, it has a tetrahedral geometry. Tetrahedral complexes of the type [MA$_4$] do not show geometrical isomerism because all positions are equivalent with respect to each other. Number of isomers = 0.
2. [CoCl$_2$(NH$_3$)$_4$]$^+$: This complex is of the type [MA$_4$B$_2$] and has an octahedral geometry. Complexes of this type can exist as two geometrical isomers: - cis-isomer: The two B ligands (Cl) are adjacent to each other (at a 90$^\circ$ angle). - trans-isomer: The two B ligands (Cl) are opposite to each other (at a 180$^\circ$ angle). Number of isomers = 2.
3. [Co(NH$_3$)$_3$(NO$_2$)$_3$]: This complex is of the type [MA$_3$B$_3$] and has an octahedral geometry. Complexes of this type can exist as two geometrical isomers: - facial (fac) isomer: The three identical ligands (e.g., NH$_3$) occupy the corners of one face of the octahedron. - meridional (mer) isomer: The three identical ligands occupy positions such that two are trans to each other, forming a 'meridian' around the central atom. Number of isomers = 2.
4. [Co(NH$_3$)$_5$Cl]$^{2+}$: This complex is of the type [MA$_5$B] and has an octahedral geometry. In this type, all positions of the five A ligands are equivalent relative to the single B ligand. No matter where the B ligand is placed, the resulting structure is the same. Number of isomers = 0.
Total number of geometrical isomers = (Isomers of complex 1) + (Isomers of complex 2) + (Isomers of complex 3) + (Isomers of complex 4).
Total number = $0 + 2 + 2 + 0 = 4$.