
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 $[PtB{r_2}{(N{H_3})_4}]C{l_2}$
Answer
577.2k+ views
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.
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.
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