Animated Solution for Chemistry - Coordination Compounds: The number of geometrical isomers possible in triamminetrinitrocobalt (III) is $X$ and in trioxalatochromate (III) is $Y$. Then, the value of $X+Y$ is ............... .

The $M{A}_3{B}_3$ System

Let's embark on this stereochemistry journey by analyzing our first coordination entity: triamminetrinitrocobalt (III). Its chemical formula is $[Co(N{O}_2{)}_3(N{H}_3{)}_3]$.

If we look closely at the ligands, we have three identical nitro groups ($N{O}_2$) and three identical ammine groups ($N{H}_3$). This perfectly matches the general formula of an $M{A}_3{B}_3$ type octahedral complex.

For an $M{A}_3{B}_3$ complex, the spatial arrangement of ligands is highly specific. We can arrange these ligands in exactly two distinct geometrical ways:

1. Facial (fac) Isomer: Imagine the eight triangular faces of an octahedron. If three identical ligands occupy the three corners of a single triangular face, they form a "face." This is the fac-isomer. 2. Meridional (mer) Isomer: If the three identical ligands are arranged around the perimeter of the octahedron—much like the meridian of a globe—they form the mer-isomer.

Because these are the only two unique spatial arrangements possible without simply rotating the molecule, the number of geometrical isomers for $[Co(N{O}_2{)}_3(N{H}_3{)}_3]$ is exactly $2$.

Therefore, we have our first value:

The $[M(AA{)}_3]$ System

Now, let's shift our focus to the second complex: trioxalatochromate (III), represented by the formula $[Cr(C_2{O}_4{)}_3{]}^{3-}$.

Here, the ligand is oxalate ($C_2{O}_4^{2-}$), which is a symmetrical bidentate ligand. It coordinates to the central chromium ion through two identical oxygen atoms. This makes our complex an $[M(AA{)}_3]$ type system.

Can we have geometrical isomers here? Think about the geometry. All three bidentate ligands are identical and symmetrical. No matter how you attach them to the six octahedral sites, the relative distances and angles between the donor oxygen atoms remain exactly the same. You cannot create a distinct cis or trans relationship between identical chelating rings.

Thus, it is impossible to create a distinct spatial arrangement that isn't just a 3D rotation of the original molecule. The number of geometrical isomers is zero.

Therefore, we have our second value:

We are asked to find the sum of $X$ and $Y$. Substituting the values we derived:

A Crucial Trap Warning: While the $[Cr(C_2{O}_4{)}_3{]}^{3-}$ complex has zero geometrical isomers, it is highly optically active! Because the arrangement of the three bidentate rings resembles a propeller, the molecule lacks any plane of symmetry. It exists as a pair of non-superimposable mirror images (enantiomers, often denoted as $\Delta$ and $\Lambda$ forms). Always read the question carefully to see if the examiner is asking for geometrical isomers, optical isomers, or total stereoisomers!