Identical Isomerism in Coordination Complex Pairs

To determine which pair(s) of coordination complexes or complex ions exhibit the identical type of isomerism, let us analyze each option individually:

1. Option (A):

   • $[\text{Cr}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2$: This is an octahedral complex of the type $\text{MA}_5\text{B}$. It cannot exhibit geometrical or optical isomerism. Furthermore, since the ligand inside the coordination sphere and the counter-ions are both $\text{Cl}^-$, it does not exhibit ionization isomerism either.
   • $[\text{Cr}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}$: This is an octahedral complex of the type $\text{MA}_4\text{B}_2$. It exhibits geometrical isomerism ($cis$- and $trans$-forms).
   • Since one complex does not show isomerism while the other does, they do not exhibit the same kind of isomerism.

2. Option (B):

   • $[\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+$: Cobalt is in the $+3$ oxidation state with coordination number $6$ (octahedral geometry, $\text{MA}_4\text{B}_2$ type). It exhibits geometrical isomerism ($cis$ and $trans$ isomers).
   • $[\text{Pt}(\text{NH}_3)_2(\text{H}_2\text{O})\text{Cl}]^+$: Platinum is in the $+2$ oxidation state ($5d^8$ configuration) with coordination number $4$, which always adopts a square planar geometry. Being of the type $\text{MA}_2\text{BC}$, it exhibits geometrical isomerism ($cis$ and $trans$ with respect to the two identical $\text{NH}_3$ ligands).
   • Therefore, both complexes exhibit geometrical isomerism.

3. Option (C):

   • $[\text{CoBr}_2\text{Cl}_2]^{2-}$: $\text{Co}^{2+}$ is a $3d^7$ ion with weak field halide ligands, forming a tetrahedral complex ($\text{MA}_2\text{B}_2$ type). Tetrahedral complexes of this type are spherically symmetric and do not show geometrical isomerism.
   • $[\text{PtBr}_2\text{Cl}_2]^{2-}$: $\text{Pt}^{2+}$ is a $5d^8$ ion that forms a square planar complex ($\text{MA}_2\text{B}_2$ type). It exhibits geometrical isomerism ($cis$ and $trans$ isomers).
   • Therefore, they do not exhibit the same kind of isomerism.

4. Option (D):

   • $[\text{Pt}(\text{NH}_3)_3(\text{NO}_3)]\text{Cl}$: This compound can exchange its coordinated nitrate ion ($\text{NO}_3^-$) with the chloride ion ($\text{Cl}^-$) present outside the coordination sphere to form the isomer $[\text{Pt}(\text{NH}_3)_3\text{Cl}]\text{NO}_3$. Hence, it exhibits ionization isomerism.
   • $[\text{Pt}(\text{NH}_3)_3\text{Cl}]\text{Br}$: Similarly, this compound can exchange its coordinated chloride ion ($\text{Cl}^-$) with the bromide ion ($\text{Br}^-$) present outside the coordination sphere to form $[\text{Pt}(\text{NH}_3)_3\text{Br}]\text{Cl}$. Hence, it also exhibits ionization isomerism.
   • Therefore, both complexes exhibit ionization isomerism.

Thus, the correct options are B and D.