Animated Solution for Chemistry - Coordination Compounds: The complex(es), which can exhibit the type of isomerism shown by $[Pt(N{H}_3{)}_{2}B{r}_2]$, is(are) $[en=H_{2}NC{H}_{2}C{H}_{2}N{H}_2]$

Square Planar Geometry

Imagine you are an architect, but instead of buildings, you are designing molecules. The question hands us a reference blueprint: the diamine dibromo platinum(II) complex, $[Pt(N{H}_3{)}_{2}B{r}_2]$. Because $P{t}^{2+}$ has a $d^8$ electron configuration, it strongly prefers a square planar geometry.

In this flat, square arrangement, the two identical ammonia ligands and the two identical bromo ligands can be placed in two distinct ways. If the identical ligands are placed right next to each other ($90^{\circ }$ apart), we get the cis isomer. If they are placed diagonally opposite to each other ($180^{\circ }$ apart), we get the trans isomer. This spatial dance is known as geometrical isomerism. Our mission is to find which of the given options can also perform this dance.

The Bidentate Constraint

Let's look at option A: $[Pt(en)(SCN{)}_2]$. Here, we have a square planar platinum center, but one of the ligands is ethylenediamine ('en'). This is a bidentate ligand, meaning it has two 'teeth' (nitrogen atoms) that bite onto the metal simultaneously.

However, there is a physical constraint. The carbon chain connecting the two nitrogen atoms in 'en' is relatively short. It forms a tight 5-membered chelate ring with the platinum atom. Because of this ring strain, the two nitrogen atoms can only reach adjacent (cis) positions. They simply cannot stretch across the metal to occupy opposite (trans) positions. Therefore, this complex is locked into a single geometry and cannot exhibit cis-trans isomerism.

The Tetrahedral Trap

Moving to option B, we encounter $[Zn(N{H}_3{)}_{2}C{l}_2]$. Zinc(II) is a $d^{10}$ metal ion. With a completely full d-subshell, it doesn't benefit from the crystal field stabilization energy that drives square planar geometries. Instead, it adopts a tetrahedral geometry to minimize ligand-ligand repulsion.

In a perfect tetrahedron, every position is exactly $109.5^{\circ }$ away from every other position. There is no such thing as 'opposite' in a tetrahedron; all positions are adjacent. Because you cannot place two ligands opposite to each other, tetrahedral complexes never exhibit geometrical isomerism.

The Octahedral Playground

Option C presents $[Pt(N{H}_3{)}_{2}C{l}_4]$. Platinum(IV) has a coordination number of 6, which means it forms an octahedral complex. This is an $M{A}_4{B}_2$ type complex.

An octahedron is like a square bipyramid. It has plenty of room for spatial variations. We can place the two ammonia ligands adjacent to each other (at a $90^{\circ }$ angle) to form the cis isomer. Alternatively, we can place them at the top and bottom poles (at a $180^{\circ }$ angle) to form the trans isomer. Thus, this complex definitely exhibits geometrical isomerism.

Asymmetry in Octahedral Complexes

Finally, we examine option D: $[Cr(en{)}_2(H_{2}O)(S{O}_4){]}^{+}$. This is a chromium(III) complex, which also forms an octahedral geometry. It is an $M(AA{)}_{2}BC$ type complex, featuring two bidentate 'en' ligands and two different monodentate ligands ($H_{2}O$ and $S{O}_4^{2-}$).

Even though the monodentate ligands are different, the logic remains the same. We can place the water and sulfate ligands adjacent to each other to create the cis form. Or, we can place them opposite to each other to create the trans form. The two bidentate 'en' ligands will happily fold into the remaining adjacent positions in either case. Therefore, this complex also exhibits geometrical isomerism.

By carefully analyzing the geometry and the physical constraints of the ligands, we can confidently conclude that both the octahedral platinum(IV) complex in option C and the octahedral chromium(III) complex in option D can exhibit cis-trans isomerism, just like our reference molecule. The correct choices are (C) and (D).