The user wants to identify the option that satisfies two conditions:

1. Set-I: Each complex in the set must exhibit geometrical isomerism.
2. Set-II: The pair of complexes must be ionization isomers of each other.

Let's analyze each option step-by-step.

Condition 1: Geometrical Isomerism

Geometrical isomerism (cis-trans, fac-mer) arises from different spatial arrangements of ligands. Key points to remember:

• Tetrahedral complexes (Coordination Number = 4) do not show geometrical isomerism as all positions are equivalent.
• Square planar complexes (CN = 4) of type $Ma_2b_2$ and Mabcd show geometrical isomerism.
• Octahedral complexes (CN = 6) of type $Ma_6$ and $Ma_5b$ do not show geometrical isomerism.
• Octahedral complexes of type $Ma_4b_2$, $Ma_3b_3$, and $M(AA)_2b_2$ (where AA is a bidentate ligand) show geometrical isomerism.

Condition 2: Ionization Isomerism

Ionization isomers are compounds with the same molecular formula but yield different ions in solution. This occurs when a ligand within the coordination sphere and a counter-ion outside the sphere are exchanged.

Evaluation of Options

Option A:

• Set-I: $[Ni(CO)_4]$ and $[PdCl_2(PPh_3)_2]$.
  • $[Ni(CO)_4]$ has a tetrahedral geometry. Tetrahedral complexes do not exhibit geometrical isomerism.
  • Thus, Set-I does not satisfy the condition.
• Conclusion: Option A is incorrect.

Option B:

• Set-I: $[Co(en)(NH_3)_2Cl_2]$ and $[PdCl_2(PPh_3)_2]$.
  • $[Co(en)(NH_3)_2Cl_2]$ is an octahedral complex of the type $M(AA)b_2c_2$. The relative positions of the $NH_3$ and Cl ligands can be different, leading to geometrical isomers.
  • $[PdCl_2(PPh_3)_2]$ is a square planar complex of the type $Ma_2b_2$. It shows cis-trans isomerism.
  • Set-I satisfies the condition.
• Set-II: $[Co(NH_3)_6][Cr(CN)_6]$ and $[Cr(NH_3)_6][Co(CN)_6]$.
  • These isomers differ by the exchange of ligands between the cationic and anionic coordination entities. This is known as coordination isomerism, not ionization isomerism.
  • Thus, Set-II does not satisfy the condition.
• Conclusion: Option B is incorrect.

Option C:

• Set-I: $[Co(NH_3)_3(NO_2)_3]$ and $[Co(en)_2Cl_2]$.
  • $[Co(NH_3)_3(NO_2)_3]$ is an octahedral complex of the type $Ma_3b_3$. It exhibits facial (fac) and meridional (mer) geometrical isomerism.
  • $[Co(en)_2Cl_2]$ is an octahedral complex of the type $M(AA)_2b_2$. It exhibits cis-trans geometrical isomerism.
  • Set-I satisfies the condition.
• Set-II: $[Co(NH_3)_5Cl]SO_4$ and $[Co(NH_3)_5(SO_4)]Cl$.
  • The first complex ionizes to give $[Co(NH_3)_5Cl]^{2+}$ and $SO_4^{2-}$ ions.
  • The second complex ionizes to give $[Co(NH_3)_5(SO_4)]^+$ and $Cl^-$ ions.
  • These have the same overall formula but produce different ions in solution due to the exchange of the ligand $Cl^-$ and the counter-ion $SO_4^{2-}$. They are a classic example of ionization isomers.
  • Set-II satisfies the condition.
• Conclusion: Option C satisfies both conditions and is the correct answer.

Option D:

• Set-I: $[Cr(NH_3)_5Cl]Cl_2$ and $[Co(en)(NH_3)_2Cl_2]$.
  • $[Cr(NH_3)_5Cl]Cl_2$ is an octahedral complex of the type $Ma_5b$. It does not show geometrical isomerism as all positions for the five $NH_3$ ligands are equivalent relative to the single Cl ligand.
  • Thus, Set-I does not satisfy the condition.
• Conclusion: Option D is incorrect.