Which of the following pairs represents linkage isomers?
(A) $[Cu{(N{H_3})_4}][PtC{l_4}]$ and $[Pt(N{H_3})][CuC{l_4}]$
(B) $[Pd{(PP{h_3})_2}{(NCS)_2}]$ and $[Pd{(PP{h_3})_2}{(SCN)_2}]$
(C) $[Co{(N{H_3})_5}N{O_3}]S{O_4}$ and $[Co{(N{H_3})_5}S{O_4}]N{O_3}$
(D) $[PtC{l_2}{(N{H_3})_4}]B{r_2}$ and $[PtB{r_2}{(N{H_3})_4}]C{l_2}$

Hint: Any metal complex that has ambidentate ligands may show linkage isomerism. When a ligand is bound to metal with two different atoms in different complexes, then both the complexes are called linkage isomers.

Complete step by step solution:
We are given four pairs of metal complexes and we need to find which of the pair shows linkage isomerism. Let’s know what linkage isomerism is.
- Isomers are the compounds that have the same chemical formula but have different properties. We know that isomerism is broadly divided into two categories; structural isomerism and stereoisomerism. Linkage isomerism is a type of a structural isomerism.
- Linkage isomerism is seen in the complexes which involves ambidentate ligand. Ambidentate ligand is a ligand which can donate the electrons to metal from more than one atom.
- The ligands $C{l^ - },B{r^ - },PP{h_3},N{H_3},N{O_3}^ - {\text{ and S}}{{\text{O}}_4}^{2 - }$ can only donate electrons from one site. So, in option (A), (C) and (D) the ligands present are not ambidentate. So, they cannot show linkage isomerism.
- In option (B), one of the ligands are $NC{S^ - }{\text{ and SC}}{{\text{N}}^ - }$. These ligands are ambidentate because they can donate electrons from S-atoms as well as N-atoms.
- So, we can see that in one complex, the bond is like M-SCN and in another complex the bond is like M-NCS. So, it is clear that the same ligand is donating electrons from different atoms. Therefore linkage isomerism is present in this pair.

So, the correct answer to this question is (B).

Note: Do not get confused with coordination isomerism. It actually involves interchange of ligands between anionic and cationic species of different metal ions present in the metal complex.