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

We have to count only geometrical (cis-trans / fac-mer) arrangements, ignoring optical and linkage effects.

First consider triamminetrinitrocobalt (III): $$[\,\text{Co}(NH_3)_3(NO_2)_3\,]$$. This fits the general octahedral type $$[MA_3B_3]$$ where $$A=NH_3$$ and $$B=NO_2^-$$.

For an octahedral complex of the form $$[MA_3B_3]$$ the well-known result is that exactly two geometrical isomers exist:

$$\text{facial (fac)}:\;$$ the three identical ligands occupy one face of the octahedron, all mutually cis.
$$\text{meridional (mer)}:\;$$ two identical ligands are trans while the third is cis to both of them, lying on the same meridian.

Thus $$X=2$$ for $$[\,\text{Co}(NH_3)_3(NO_2)_3\,]$$.

Now take trioxalatochromate (III): $$[\,\text{Cr}(\text{C}_2\text{O}_4)_3\,]^{3-}$$. Each oxalate $$(\text{C}_2\text{O}_4)^{2-}$$ is a symmetrical bidentate ligand. In an octahedral field three identical bidentate chelates must occupy all six coordination sites pairwise; there is only one possible spatial arrangement—every position is equivalent, so no cis-trans or fac-mer choice arises.

Therefore no separate geometrical isomer exists, giving $$Y=0$$.

Adding the two counts we obtain $$X+Y = 2 + 0 = 2$$.

Hence, the correct answer is Option 2.