Animated Solution for Chemistry - Coordination Compounds: Among (A) - (D), the complexes that can display geometrical isomerism are (A) $[Pt(N{H}_3{)}_{3}Cl{]}^{+}$ (B) $[Pt(N{H}_3)C{l}_5{]}^{-}$ (C) $[Pt(N{H}_3{)}_{2}Cl(N{O}_2)]$ (D) $[Pt(N{H}_3{)}_{4}ClBr{]}^{2+}$

Geometrical isomerism is a fascinating phenomenon in coordination chemistry. It occurs when ligands can be arranged in different relative spatial positions around the central metal ion. This means that even with the exact same chemical formula, the 3D architecture of the molecule can vary, leading to distinct physical and chemical properties.

To determine if a complex exhibits geometrical isomerism, we must carefully analyze its coordination number, its geometry, and the types of ligands attached to it. Let's break down the four platinum complexes given in the problem.

Let's start by looking at complex (A), $[Pt(N{H}_3{)}_{3}Cl{]}^{+}$. This is a square planar complex with a coordination number of 4. We can classify it as an $M{a}_{3}b$ type complex, where 'M' is the metal, 'a' represents the three identical ammonia ligands, and 'b' is the single chloride ligand.

Because three out of the four ligands are identical, any attempt to swap their positions will simply result in the exact same spatial arrangement. You can rotate the molecule, but you cannot create a non-superimposable isomer. Therefore, complex (A) does not show geometrical isomerism.

The same logic applies to complex (B), $[Pt(N{H}_3)C{l}_5{]}^{-}$. This is an octahedral complex with a coordination number of 6, falling under the $Ma{b}_5$ category. With five identical chloride ligands, rearranging the single ammonia ligand relative to the chlorides will not yield a new geometrical isomer. Thus, complex (B) is also ruled out.

Complex C

Now, let's focus our attention on complex (C), $[Pt(N{H}_3{)}_{2}Cl(N{O}_2)]$. This is a square planar complex, but its ligand composition is different. It belongs to the $M{a}_{2}bc$ type.

Here, we have two identical ammonia ligands. This opens up a possibility! We can place these two ammonia ligands adjacent to each other, at a $90^{\circ }$ angle. This specific arrangement is known as the cis-isomer.

Alternatively, we can place the two ammonia ligands at opposite corners of the square plane, at a $180^{\circ }$ angle. This creates the trans-isomer. Because the cis and trans arrangements are spatially distinct, complex (C) definitely exhibits geometrical isomerism.

Complex D

Finally, let's examine complex (D), $[Pt(N{H}_3{)}_{4}ClBr{]}^{2+}$. With a coordination number of 6, this complex adopts an octahedral geometry. It is classified as an $M{a}_{4}bc$ type complex.

In this octahedral structure, we have four identical ammonia ligands and two unique ligands: Chlorine and Bromine. To find geometrical isomers, we look at the relative positions of these two unique ligands.

If we place the Chlorine and Bromine ligands adjacent to each other (at a $90^{\circ }$ angle), we form the cis-isomer. If we place them at opposite ends of the octahedron (at a $180^{\circ }$ angle), we form the trans-isomer. Therefore, complex (D) also successfully displays geometrical isomerism.

By systematically analyzing the geometry and ligand types of each complex, we have determined that complexes (A) and (B) cannot form geometrical isomers due to their high symmetry. However, complexes (C) and (D) can exist in distinct cis and trans forms.

Therefore, the complexes that can display geometrical isomerism are (C) and (D), making the correct option (b).