Animated Solution for Chemistry - Coordination Compounds: The total number of possible isomers for $[\text{Pt}({\text{NH}}_3{)}_4{\text{Cl}}_2]{\text{Br}}_2$ is ______ .

When we look at the coordination complex $[\text{Pt}({\text{NH}}_3{)}_4{\text{Cl}}_2]{\text{Br}}_2$, the first thing we need to do is decode its structure. We have a central Platinum (Pt) atom. Inside the square brackets, which denote the coordination sphere, we have four ammonia (${\text{NH}}_3$) ligands and two chloride (${\text{Cl}}^{-}$) ligands. This gives us a coordination number of 6, meaning the complex adopts an octahedral geometry.

Outside the brackets, in the ionization sphere, we have two bromide (${\text{Br}}^{-}$) counter ions. The presence of exchangeable anions both inside and outside the coordination sphere is a massive hint. It screams: Ionisation Isomerism!

Finding Ionisation Isomers

Ionisation isomerism occurs when ligands from the coordination sphere exchange places with ions from the ionization sphere. Let's systematically list all the possible combinations by swapping the halide ions.

1. Isomer I (The Original): $[\text{Pt}({\text{NH}}_3{)}_4{\text{Cl}}_2]{\text{Br}}_2$ 2. Isomer II (Full Swap): We swap both inner chlorides with both outer bromides to get $[\text{Pt}({\text{NH}}_3{)}_4{\text{Br}}_2]{\text{Cl}}_2$. 3. Isomer III (Partial Swap): We swap just one inner chloride with one outer bromide to get $[\text{Pt}({\text{NH}}_3{)}_4\text{BrCl}]\text{BrCl}$.

So, we have successfully identified three distinct ionisation isomers. But we are not done yet. We must dive deeper into the spatial arrangement of each of these isomers.

The Spatial Dance

For octahedral complexes, we must always check for geometrical isomerism, specifically cis-trans isomerism. Let's analyze our three ionisation isomers one by one.

For Isomer I: The coordination entity is $[\text{Pt}({\text{NH}}_3{)}_4{\text{Cl}}_2{]}^{2+}$. This matches the generic formula ${\text{MA}}_4{\text{B}}_2$. In an octahedral field, the two 'B' ligands (the chlorides) can be placed adjacent to each other at a $90^{\circ }$ angle, giving us the cis-isomer. Alternatively, they can be placed opposite each other at a $180^{\circ }$ angle, giving us the trans-isomer. Thus, Isomer I contributes 2 geometrical isomers.

For Isomer II: The coordination entity is $[\text{Pt}({\text{NH}}_3{)}_4{\text{Br}}_2{]}^{2+}$. This is also an ${\text{MA}}_4{\text{B}}_2$ type complex. Just like Isomer I, the two bromide ligands can be arranged in cis and trans positions. This gives us another 2 geometrical isomers.

For Isomer III: The coordination entity is $[\text{Pt}({\text{NH}}_3{)}_4\text{BrCl}{]}^{2+}$. This matches the generic formula ${\text{MA}}_4\text{BC}$. Even though the two halide ligands are different, they can still be placed adjacent ($90^{\circ }$) to form the cis-isomer, or opposite ($180^{\circ }$) to form the trans-isomer. This contributes the final 2 geometrical isomers.

To find the total number of possible isomers, we simply sum the geometrical isomers for each ionisation state:

A Note on Optical Isomerism: You might wonder why we didn't check for optical isomers. For a complex to be optically active, it must lack any plane of symmetry (it must be chiral). In all the cis and trans forms of ${\text{MA}}_4{\text{B}}_2$ and ${\text{MA}}_4\text{BC}$ complexes with monodentate ligands, there is always at least one plane of symmetry. Therefore, none of these 6 isomers are optically active.