Question

The number of geometrical isomers possible in triamminetrinitrocobalt(III) is X and in trioxalatochromate(III) is Y. Then the value of X+Y is _________.

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.

Answer 1

Correct Answer: 2

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 \[ X + Y = 2 + 0 = \boxed{2} \]