To solve this matching problem, we need to analyze each complex in Column I and determine its properties from Column II.

Analysis of Complex (A): $[Co(NH_3)_4(H_2O)_2]Cl_2$

1. Oxidation State of Metal Ion (S): Let the oxidation state of Cobalt (Co) be x. $NH_3$ and $H_2O$ are neutral ligands (charge 0). The chloride ions ($Cl^-$) outside the coordination sphere have a charge of -1 each. The overall charge of the compound is 0. The coordination sphere $[Co(NH_3)_4(H_2O)_2]$ has a charge of +2 to balance the two $Cl^-$ ions. $x + 4(0) + 2(0) = +2$ $x = +2$ So, the metal ion has a +2 oxidation state. This matches (S).

2. Magnetic Property (Q/R): The central metal ion is $Co^{2+}$. The atomic number of Co is 27, so its electronic configuration is $[Ar] 3d^7 4s^2$. The configuration of $Co^{2+}$ is $[Ar] 3d^7$. The complex is octahedral. In an octahedral field, the d orbitals split into $t_{2g}$ and $e_g$ sets. For a $d^7$ ion, there will always be unpaired electrons, regardless of whether it's a high-spin or low-spin complex.

   • High-spin (weak field): $t_{2g}^5 e_g^2$ (3 unpaired electrons)
   • Low-spin (strong field): $t_{2g}^6 e_g^1$ (1 unpaired electron) Since there are unpaired electrons, the complex is paramagnetic. This matches (Q).

3. Geometrical Isomers (P): The complex has the general formula $[MA_4B_2]$, where M = Co, A = $NH_3$, and B = $H_2O$. Octahedral complexes of this type exhibit cis-trans geometrical isomerism.

   • cis-isomer: The two $H_2O$ ligands are adjacent (90° apart).
   • trans-isomer: The two $H_2O$ ligands are opposite (180° apart). Therefore, it shows geometrical isomers. This matches (P).

Conclusion for (A): Matches (P), (Q), and (S).

Analysis of Complex (B): $[Pt(NH_3)_2Cl_2]$

1. Oxidation State of Metal Ion (S): Let the oxidation state of Platinum (Pt) be x. $NH_3$ is neutral (0) and Cl is -1. The overall complex is neutral. $x + 2(0) + 2(-1) = 0$ $$$$

ightarrow x = +2$$ So, the metal ion has a +2 oxidation state. This matches (S).

2. Magnetic Property (Q/R): The central metal ion is $Pt^{2+}$. Pt (Z=78) is a 5d transition metal. Its configuration is $[Xe] 4f^{14} 5d^9 6s^1$. The configuration of $Pt^{2+}$ is $[Xe] 4f^{14} 5d^8$. $d^8$ complexes of 4d and 5d series metals (like $Pt^{2+}$) are typically square planar and low-spin due to the large crystal field splitting energy. In a square planar field, the $d^8$ electrons fill the lower energy orbitals, resulting in a paired configuration: $(d_{xz})^2 (d_{yz})^2 (d_{z^2})^2 (d_{xy})^2$. There are no unpaired electrons. Therefore, the complex is diamagnetic. This matches (R).

3. Geometrical Isomers (P): The complex has the formula $[MA_2B_2]$ and a square planar geometry. This type of complex exhibits cis-trans isomerism (e.g., cisplatin and transplatin). Therefore, it shows geometrical isomers. This matches (P).

Conclusion for (B): Matches (P), (R), and (S).

Analysis of Complex (C): $[Co(H_2O)_5Cl]Cl$

1. Oxidation State of Metal Ion (S): Let the oxidation state of Co be x. $H_2O$ is neutral (0), Cl ligand is -1, and the counter-ion Cl is -1. The coordination sphere $[Co(H_2O)_5Cl]$ has a charge of +1. $x + 5(0) + 1(-1) = +1$ $$$$

ightarrow x = +2$$ So, the metal ion has a +2 oxidation state. This matches (S).

2. Magnetic Property (Q/R): The central ion is $Co^{2+}$ ($d^7$). The ligands $H_2O$ and $Cl^-$ are weak field ligands, so the complex is high-spin octahedral. The electronic configuration is $t_{2g}^5 e_g^2$, which has 3 unpaired electrons. Therefore, the complex is paramagnetic. This matches (Q).

3. Geometrical Isomers (P): The complex is of the type $[MA_5B]$. In an octahedral geometry, all positions are equivalent for the single 'B' ligand relative to the five 'A' ligands. Thus, it does not exhibit geometrical isomerism. Does not match (P).

Conclusion for (C): Matches (Q) and (S).

Analysis of Complex (D): $[Ni(H_2O)_6]Cl_2$

1. Oxidation State of Metal Ion (S): Let the oxidation state of Nickel (Ni) be x. $H_2O$ is neutral (0). The complex ion $[Ni(H_2O)_6]$ has a charge of +2. $x + 6(0) = +2$ $x = +2$ So, the metal ion has a +2 oxidation state. This matches (S).

2. Magnetic Property (Q/R): The central ion is $Ni^{2+}$. Ni (Z=28) has configuration $[Ar] 3d^8 4s^2$. $Ni^{2+}$ is $[Ar] 3d^8$. This is an octahedral complex. For a $d^8$ configuration in an octahedral field, the electrons are arranged as $t_{2g}^6 e_g^2$, irrespective of the ligand field strength. There are two unpaired electrons in the $e_g$ orbitals. Therefore, the complex is paramagnetic. This matches (Q).

3. Geometrical Isomers (P): The complex is of the type $[MA_6]$, where all six ligands are identical. Such complexes do not exhibit geometrical isomerism. Does not match (P).

Conclusion for (D): Matches (Q) and (S).

Summary of Matches

• (A) $[Co(NH_3)_4(H_2O)_2]Cl_2$ $\rightarrow$ (P), (Q), (S)
• (B) $[Pt(NH_3)_2Cl_2]$ $\rightarrow$ (P), (R), (S)
• (C) $[Co(H_2O)_5Cl]Cl$ $\rightarrow$ (Q), (S)
• (D) $[Ni(H_2O)_6]Cl_2$ $\rightarrow$ (Q), (S)

Comparing this with the given options:

• A: A - (p, q, s); B - (p, r, s); C - (q, s); D - (q, s) - This perfectly matches our findings.
• B: A is missing (p), C is missing (s).
• C: C is missing (s), D is missing (q).
• D: B is missing (p, r), D is missing (s).

The correct option is A.