Matching Coordination Complexes to Magnetic and Isomeric Properties

To determine the correct matching between List-I and List-II, let us analyze the magnetic nature and isomerism exhibited by each complex:

1. Complex P: $[\text{Cr}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}$

   • Oxidation State & Configuration: $\text{Cr}^{3+} \implies [\text{Ar}]\,3d^3$ It has 3 unpaired electrons in the $t_{2g}$ orbitals ($t_{2g}^3 e_g^0$), meaning it is paramagnetic.
   • Isomerism: The complex is of the octahedral type $[\text{M}a_4b_2]$. Complexes of this type exhibit $cis\text{-}trans$ (geometrical) isomerism, with the two chlorine ligands placed either adjacent ($cis$) or opposite ($trans$) to each other. It cannot exhibit ionisation isomerism because the counter-ion is identical to the ligand ($\text{Cl}^-$).
   • Hence, P corresponds to 3 (Paramagnetic and exhibits $cis\text{-}trans$ isomerism).

2. Complex Q: $[\text{Ti}(\text{H}_2\text{O})_5\text{Cl}](\text{NO}_3)_2$

   • Oxidation State & Configuration: $\text{Ti}^{3+} \implies [\text{Ar}]\,3d^1$ It possesses 1 unpaired electron, making it paramagnetic.
   • Isomerism: It is of the type $[\text{M}a_5b]$, which does not show geometrical ($cis\text{-}trans$) isomerism. However, exchange of the coordinated $\text{Cl}^-$ ligand with the ionisable $\text{NO}_3^-$ counter-ion gives the isomer $[\text{Ti}(\text{H}_2\text{O})_5(\text{NO}_3)]\text{Cl}(\text{NO}_3)$. Thus, it exhibits ionisation isomerism.
   • Hence, Q corresponds to 1 (Paramagnetic and exhibits ionisation isomerism).

3. Complex R: $[\text{Pt}(\text{en})(\text{NH}_3)\text{Cl}]\text{NO}_3$

   • Oxidation State & Configuration: $\text{Pt}^{2+} \implies 5d^8$ For $5d$ transition metals, the crystal field splitting ($\Delta$) is very large, leading to low-spin square planar geometry with all electrons paired. Thus, it is diamagnetic.
   • Isomerism: The bidentate ethylenediamine ($\text{en}$) ligand can only coordinate at adjacent ($cis$) positions. As a result, the other two coordination sites occupied by $\text{NH}_3$ and $\text{Cl}^-$ are fixed relative to each other, preventing $cis\text{-}trans$ isomerism. However, exchange between the coordinating $\text{Cl}^-$ and the counter-ion $\text{NO}_3^-$ produces $[\text{Pt}(\text{en})(\text{NH}_3)(\text{NO}_3)]\text{Cl}$, demonstrating ionisation isomerism.
   • Hence, R corresponds to 4 (Diamagnetic and exhibits ionisation isomerism).

4. Complex S: $[\text{Co}(\text{NH}_3)_4(\text{NO}_3)_2]\text{NO}_3$

   • Oxidation State & Configuration: $\text{Co}^{3+} \implies 3d^6$ Cobalt(III) in an octahedral field with strong-field ligands like $\text{NH}_3$ forms a low-spin complex ($t_{2g}^6 e_g^0$) with zero unpaired electrons, making it diamagnetic.
   • Isomerism: Being an octahedral complex of the type $[\text{M}a_4b_2]$, it exhibits $cis\text{-}trans$ isomerism. It cannot exhibit ionisation isomerism as the counter-ion ($\text{NO}_3^-$) is identical to the inner anionic ligands.
   • Hence, S corresponds to 2 (Diamagnetic and exhibits $cis\text{-}trans$ isomerism).

Matching summary: $\text{P-3, Q-1, R-4, S-2}$

Therefore, the correct option is B.