A mong the following options, select the option in which each complex in Set-I shows geometrical isomerism and the two complexes in Set-II are ionization isomers of each other.

[en = H₂NCH₂CH₂NH₂]

To determine the correct option, let us analyze the conditions for geometrical isomerism in Set-I and ionization isomerism in Set-II step-by-step.

1. Analysis of Set-I (Geometrical Isomerism):
Geometrical isomerism occurs when ligands have different spatial arrangements around the central metal atom/ion.
Let us analyze the two complexes in the correct option for Set-I:
- $\mathrm{[Co}\left({\mathrm{NH}}_3{)}_3\right({\mathrm{NO}}_2{)}_3]$:
This is an octahedral complex of the type $\left[{\mathrm{MA}}_3{B}_3\right]$. Such complexes exhibit geometrical isomerism, specifically existing as facial (fac) and meridional (mer) isomers. In the facial isomer, the three similar ligands occupy adjacent positions at the corners of an octahedral face. In the meridional isomer, the three similar ligands occupy positions around the meridian of the octahedron.
- $\mathrm{[Co}\left(\mathrm{en}{)}_2{\mathrm{Cl}}_2\right]$:
This is an octahedral complex of the type $\left[{\mathrm{M(AA)}}_2{B}_2\right]$, where $\mathrm{en}$ is a symmetrical bidentate ligand. This complex exists as cis and trans geometrical isomers. In the cis isomer, the two chloride ligands are adjacent to each other (at a 90-degree angle). In the trans isomer, the two chloride ligands are opposite to each other (at a 180-degree angle).
Thus, both complexes in Set-I show geometrical isomerism.

2. Analysis of Set-II (Ionization Isomerism):
Ionization isomers are complexes that have the same molecular formula but give different ions in solution. This happens because the counter-ion in the outer coordination sphere swaps positions with a ligand in the inner coordination sphere.
Let us analyze the two complexes in the correct option for Set-II:
- $\mathrm{[Co}\left({\mathrm{NH}}_3{)}_5\mathrm{Cl}\right]{\mathrm{SO}}_4$ and $\mathrm{[Co}\left({\mathrm{NH}}_3{)}_5\right({\mathrm{SO}}_4\left)\right]\mathrm{Cl}$:
These two compounds share the same chemical composition but differ in the placement of the chloride (${\mathrm{Cl}}^{-}$) and sulfate (${\mathrm{SO}}_4^{2-}$) ions.
- When $\mathrm{[Co}\left({\mathrm{NH}}_3{)}_5\mathrm{Cl}\right]{\mathrm{SO}}_4$ is dissolved in water, it dissociates to produce sulfate ions (${\mathrm{SO}}_4^{2-}$), which can be verified by the formation of a white precipitate with barium chloride (${\mathrm{BaCl}}_2$).
- When $\mathrm{[Co}\left({\mathrm{NH}}_3{)}_5\right({\mathrm{SO}}_4\left)\right]\mathrm{Cl}$ is dissolved in water, it dissociates to produce chloride ions (${\mathrm{Cl}}^{-}$), which can be verified by the formation of a white precipitate with silver nitrate (${\mathrm{AgNO}}_3$).
Therefore, these two complexes are ionization isomers of each other.

Hence, the correct option matching all these criteria is:
Set-I: $\mathrm{[Co}\left({\mathrm{NH}}_3{)}_3\right({\mathrm{NO}}_2{)}_3]$ and $\mathrm{[Co}\left(\mathrm{en}{)}_2{\mathrm{Cl}}_2\right]$
Set-II: $\mathrm{[Co}\left({\mathrm{NH}}_3{)}_5\mathrm{Cl}\right]{\mathrm{SO}}_4$ and $\mathrm{[Co}\left({\mathrm{NH}}_3{)}_5\right({\mathrm{SO}}_4\left)\right]\mathrm{Cl}$