The number of geometrical isomers possible in triamminetrinitrocobalt(III) is X and in trioxalatochromate(III) is Y. Then the value of X+Y is _________.
Show Hint
For octahedral complexes:
$[MA_3B_3]$ $\rightarrow$ 2 geometrical isomers (fac, mer)
$[M(AA)_3]$ $\rightarrow$ 0 geometrical but 2 optical isomers
Always check whether the question asks for geometrical or optical isomerism.
We are asked to find the number of geometrical isomers in two coordination compounds. (I) Triamminetrinitrocobalt(III)
Chemical formula:
\[
[Co(NH_3)_3(NO_2)_3]
\]
This is an octahedral complex of the type:
\[
[MA_3B_3]
\]
Octahedral complexes of the type $[MA_3B_3]$ show facial (fac) and meridional (mer) geometrical isomerism.
- fac-isomer: All three identical ligands occupy adjacent positions on one face of the octahedron.
- mer-isomer: The three identical ligands lie in a plane containing the metal ion.
Hence, the number of geometrical isomers is:
\[
X = 2
\]
(Neither fac nor mer is optically active.)
(II) Trioxalatochromate(III)
Chemical formula:
\[
[Cr(ox)_3]^{3-}
\]
where oxalate $(ox^{2-})$ is a bidentate ligand.
This is an octahedral complex of the type:
\[
[M(AA)_3]
\]
In $[M(AA)_3]$ complexes:
- All ligands are identical
- All coordination positions are equivalent
Therefore, no geometrical isomerism is possible.
(The complex does show optical isomerism, but this is not asked.)
Hence:
\[
Y = 0
\]
Final Calculation
\[
X + Y = 2 + 0 = \boxed{2}
\]
