Animated Solution for Chemistry - Coordination Compounds: The number of stereoisomers possible for $[Co(ox{)}_2(Br)(N{H}_3){]}^{2-}$ is ......... $[ox=\text{oxalate}]$.

Coordination chemistry is a fascinating world where molecules build intricate 3D architectures. Today, we are going to dissect the complex $[Co(ox{)}_2(Br)(N{H}_3){]}^{2-}$ and uncover all its hidden stereoisomers.

First, let's break down the components of our complex. We have a central cobalt ion ($C{o}^{3+}$) surrounded by a specific set of ligands. There are two oxalate ions ($o{x}^{2-}$), which are symmetrical bidentate ligands. This means each oxalate bites onto the cobalt atom at two different spots. We also have two monodentate ligands: one bromide ion ($B{r}^{-}$) and one ammonia molecule ($N{H}_3$).

Because the total number of coordinate bonds formed with the central metal is six ($2\times 2+1+1=6$), the complex adopts an octahedral geometry. In general terms, this complex fits the formula $[M(A-A{)}_{2}BC]$, where $M$ is the metal, $A-A$ is the symmetrical bidentate ligand, and $B$ and $C$ are the monodentate ligands.

Cis and Trans

Stereoisomers are molecules that have the same bonds but differ in the spatial arrangement of their atoms. The first type of stereoisomerism we look for is geometrical isomerism.

We can arrange the two monodentate ligands ($B{r}^{-}$ and $N{H}_3$) in two distinct ways: 1. Trans-isomer: We place the $B{r}^{-}$ and $N{H}_3$ ligands exactly opposite to each other (at a $180^{\circ }$ angle). The two oxalate ligands will then occupy the equatorial plane. 2. Cis-isomer: We place the $B{r}^{-}$ and $N{H}_3$ ligands adjacent to each other (at a $90^{\circ }$ angle). The oxalate ligands will wrap around the remaining positions.

The Mirror Test

Now comes the crucial part: checking for optical activity. A molecule is optically active (chiral) if it cannot be superimposed on its mirror image. The easiest way to check this is to look for a plane of symmetry.

Let's examine the trans-isomer. Imagine a plane slicing vertically through the $B{r}^{-}$, the $C{o}^{3+}$, and the $N{H}_3$, cutting right between the two oxalate ligands. The left side of the molecule perfectly reflects the right side. Because it possesses this plane of symmetry, the trans-isomer is achiral and optically inactive. It only counts as one stereoisomer.

Now, let's look at the cis-isomer. Try as you might, you cannot find a plane that slices this molecule into two identical, reflecting halves. Because it lacks any plane of symmetry, the cis-isomer is chiral. This means its mirror image is a completely different, non-superimposable molecule. Therefore, the cis-isomer exists as an enantiomeric pair (often referred to as the $d$ and $l$ forms). This gives us two optically active stereoisomers.

To find the total number of stereoisomers, we simply add them up:

So, the complex $[Co(ox{)}_2(Br)(N{H}_3){]}^{2-}$ can exist in exactly 3 different stereoisomeric forms.