Type of isomerism which exists between $$[Pd(C_6H_5)_2(SCN)_2]$$ and $$[Pd(C_6H_5)_2(NCS)_2]$$ is :

The given compounds are $$[Pd(C_6H_5)_2(SCN)_2]$$ and $$[Pd(C_6H_5)_2(NCS)_2]$$. Both have palladium (Pd) as the central metal atom, two phenyl groups $$(C_6H_5)$$, and two thiocyanate groups. The molecular formula is identical: Pd, two C₆H₅, and two SCN groups.

The difference lies in the thiocyanate ligand $$(SCN^-)$$. Thiocyanate is an ambidentate ligand, meaning it can bind to the metal through either the sulfur atom (S) or the nitrogen atom (N). In $$[Pd(C_6H_5)_2(SCN)_2]$$, the notation SCN indicates bonding through sulfur (thiocyanato-S). In $$[Pd(C_6H_5)_2(NCS)_2]$$, the notation NCS indicates bonding through nitrogen (thiocyanato-N).

This variation in the bonding atom of the ambidentate ligand leads to isomerism. Specifically, this is linkage isomerism, where the same ligand coordinates to the metal through different donor atoms.

Now, evaluating the other options:

Coordination isomerism occurs in ionic complexes where both cation and anion are complex ions, and ligands are exchanged between them. Here, both compounds are neutral (Pd²⁺ charge balanced by two phenyl anions $$(C_6H_5^-)$$ and two thiocyanate anions $$(SCN^-)$$), so coordination isomerism does not apply.

Ionisation isomerism arises when isomers produce different ions in solution due to a ligand switching between being coordinated or being a counter ion. Both compounds have all ligands coordinated, with no counter ions, so ionisation isomerism is not present.

Solvate isomerism involves the exchange of solvent molecules (like water) between the coordination sphere and the solvent. Here, no solvent molecules are part of the ligands, so solvate isomerism is irrelevant.

Thus, the isomerism is linkage isomerism.

Hence, the correct answer is Option A.