Animated Solution for Chemistry - Coordination Compounds: The species that can have a trans-isomer is (en = ethane -1, 2-diamine, ox = oxalate)

Welcome to a fascinating exploration of geometrical isomerism in coordination chemistry! Our mission is to identify which of the given complexes can exist as a trans-isomer.

To solve this, we must visualize the 3D geometry of each complex. This depends entirely on the coordination number of the central metal ion and the denticity of the attached ligands.

Let's begin by examining the complexes with a coordination number of four.

Option (a), $[Pt(en)C{l}_2]$, is a platinum(II) complex that forms a square planar geometry. Option (d), $[Zn(en)C{l}_2]$, is a zinc(II) complex, which typically adopts a tetrahedral geometry.

Here is the critical constraint: ethylenediamine ('en') is a bidentate ligand. In a square planar complex, the carbon chain of 'en' is simply too short to reach across the metal atom and occupy opposite ($180^{\circ }$) positions. It is physically restricted to binding at adjacent ($90^{\circ }$) positions.

Therefore, square planar complexes of the type $M(a-a)b_2$ are exclusively cis and can never form trans-isomers. Furthermore, tetrahedral complexes do not exhibit geometrical isomerism at all because all positions are adjacent to one another!

Next, let's evaluate Option (b), the chromium complex $[Cr(en{)}_2(ox){]}^{+}$.

This is an octahedral complex featuring two 'en' ligands and one oxalate ('ox') ligand. Notice a pattern? All three of these are bidentate ligands!

For a trans-isomer to exist in this setup, one of these bidentate ligands would have to stretch across the central metal to connect two opposite axial positions. As we established earlier, their carbon chains are far too short for such a stretch. Consequently, this complex is exclusively cis and cannot show trans-isomerism.

This brings us to Option (c): $[Pt(en{)}_{2}C{l}_2{]}^{2+}$.

This is an octahedral complex with two bidentate 'en' ligands and two monodentate chloride ligands. It fits the general formula $M(a-a{)}_2{b}_2$. Because the chloride ligands are monodentate, they are not tethered to each other and can independently occupy any available position.

If we place the two chloride ligands adjacent to each other at a $90^{\circ }$ angle, the two 'en' ligands will occupy the remaining positions, giving us the cis-isomer.

However, if we place the two chloride ligands exactly opposite to each other at a $180^{\circ }$ angle, the two 'en' ligands can comfortably occupy the four equatorial positions. This perfectly forms the trans-isomer!

Therefore, the only species among the choices that can exhibit a trans-isomer is $[Pt(en{)}_{2}C{l}_2{]}^{2+}$.

A Peek into Optical Activity

As a bonus thought, consider the optical activity of these isomers.

The trans-isomer possesses a beautiful plane of symmetry, making it optically inactive. In contrast, the cis-isomer lacks this symmetry and exists as a pair of non-superimposable mirror images, making it optically active.

Always keep an eye out for these hidden layers of symmetry in coordination chemistry!