Indicate the complex/complex ion which did not show any geometrical isomerism.
(a) [CoCl2(en)2]
(b)

Q: Why doesn't $MA_5B$ show geometrical isomerism?

Because all 5 'A' ligands are identical. No matter where you place the single 'B' ligand, you can always rotate the molecule to make it look exactly the same. There is no distinct "cis" or "trans" relative to another 'B' ligand.

Q: What about linkage isomerism in option (b)?

The question specifically asks for *geometrical* isomerism. While $[Co(CN)_5(NC)]^{3-}$ and $[Co(CN)_6]^{3-}$ are linkage isomers, a single molecule of $[Co(CN)_5(NC)]^{3-}$ does not have geometrical isomers.

Q: Does $M(AA)_2B_2$ always show optical isomerism?

The *cis* form of $M(AA)_2B_2$ is optically active because it lacks a plane of symmetry, but the *trans* form is optically inactive. However, both show geometrical isomerism.

Q: What are facial and meridional isomers?

They occur in $MA_3B_3$ complexes. If the three identical ligands occupy one face of the octahedron, it's 'facial' (fac). If they form an arc around the meridian, it's 'meridional' (mer).

Q: Can a complex with a coordination number of 4 show geometrical isomerism?

Yes, square planar complexes of type $MA_2B_2$ show cis-trans isomerism. However, tetrahedral complexes do not, because all positions are adjacent to each other at $109.5^\circ$.

Question

Indicate the complex/complex ion which did not show any geometrical isomerism.

Options:

(A) $[CoCl_2(en)_2]$

(B) $[Co(CN)_5(NC)]^{3-}$

(C) $[Co(NH_3)_3(NO_2)_3]$

(D) $[Co(NH_3)_4Cl_2]^+$

Sigma Percentile · Accepted Answer

### The Essence of Geometrical Isomerism

Geometrical isomerism in coordination compounds is all about spatial real estate. It occurs when we can arrange the same set of ligands around a central metal atom in different relative positions, creating distinct 3D structures.

Imagine you are building a model. If swapping the positions of two different ligands creates a molecule that cannot be superimposed on the original, you have found a geometrical isomer!

However, there is a strict mathematical constraint for octahedral complexes. To have geometrical isomers, you must have at least two different types of ligands, and at least one of them must be present in a quantity of two or more.

This means that complexes of the type $MA_6$ or $MA_5B$ are completely symmetrical in terms of relative positioning. They **cannot** show geometrical isomerism.

### Analyzing the Suspects

Let's put our options to the test.

First, we have option (a), $[CoCl_2(en)_2]$. This is a classic $M(AA)_2B_2$ type complex. Because it has two distinct chlorine ligands, we can place them adjacent to each other at a $90^\circ$ angle to form the **cis** isomer, or opposite to each other at a $180^\circ$ angle to form the **trans** isomer.

Next is option (c), $[Co(NH_3)_3(NO_2)_3]$. This is an $MA_3B_3$ type complex. This is a favorite concept for JEE! It exhibits **facial (fac)** and **meridional (mer)** isomerism depending on whether the three identical ligands occupy the corners of one octahedral face or form an arc around the meridian.

Then we have option (d), $[Co(NH_3)_4Cl_2]^+$. This is an $MA_4B_2$ type complex. Just like option (a), the presence of two chlorine ligands allows it to easily form **cis** and **trans** isomers.

### The $MA_5B$ Anomaly

Finally, let's look at option (b), $[Co(CN)_5(NC)]^{3-}$. There is a catch here!

At first glance, you see two different ligands: cyanide ($CN^-$) and isocyanide ($NC^-$). However, this is an $MA_5B$ type complex. It contains five identical cyanide ligands and only one isocyanide ligand.

Think about the geometry. No matter which of the six octahedral vertices you choose to place the single isocyanide ligand, the remaining five positions will always be filled by cyanide.

If you rotate the molecule, every single arrangement is identical. No new spatial arrangement is formed. Therefore, it **will not show geometrical isomerism**, making it our correct answer!

Animated Solution for Chemistry - Coordination Compounds: Indicate the complex/complex ion which did not show any geometrical isomerism.

Select Answer:

Visualized Solution

Geometrical Isomerism

Condition for Octahedral Complexes

Option (a): $[C o C l_{2} (e n)_{2}]$

Option (c): $[C o (N H_{3})_{3} (N O_{2})_{3}]$

Option (d): $[C o (N H_{3})_{4} C l_{2}]^{+}$

Option (b): $[C o (C N)_{5} (N C)]^{3 -}$

Conclusion

The Sigma Insight: Nomenclature, Isomerism, Importance and Werner's Theory

The Essence of Geometrical Isomerism

Analyzing the Suspects

The $M A_{5} B$ Anomaly

Similar Questions

Which one of the following complex ions has geometrical isomers?

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

Which one of the following complexes shows optical isomerism?

Among (A) - (D), the complexes that can display geometrical isomerism are (A) $[P t (N H_{3})_{3} C l]^{+}$ (B) $[P t (N H_{3}) C l_{5}]^{-}$ (C) $[P t (N H_{3})_{2} C l (N O_{2})]$ (D) $[P t (N H_{3})_{4} C l B r]^{2 +}$

Which of the following compounds shows optical isomerism ?

Which of the following has an optical isomer?

The number of geometrical isomers found in the metal complexes $[P tC l_{2} (N H_{3})_{2}]$ , $[N i (C O)_{4}]$ , $[R u (H_{2} O)_{3} C l_{3}]$ and $[C o C l_{2} (N H_{3})_{4}]^{+}$ respectively, are

The one that is not expected to show isomerism is

The number of geometric isomers possible for the complex $[CoL_{2} Cl_{2}]^{-}$ ( $L = H_{2} NCH_{2} CH_{2} O^{-}$ ) is