Question

Which of the following cannot show linkage isomerism?
A) NO$_2$ $^{-}$
B) SCN$^{-}$
C) CN$^{-}$
D) NH$_3$

Answer 1

Hint: Linkage isomerism is relatable with the ambidentate ligands. Ambidentate ligands can coordinate from more than one site. We are given with the four different ligands. So, identify the ambidentate ligands, and the ligand that doesn’t show linkage isomerism would be known.The correct option is often used as a rocket fuel, production of urea and an ingredient in fertilizers.

Complete answer:
> Now, we will define the linkage isomerism. As mentioned it is relatable with the ambidentate ligands. Thus, we can say that there will be more than one donor site which will bind with the central atom.
> Now, let us talk about the given options. The first we have nitro ion, it has more than one donor site. So, nitro ions can bind through the nitrogen atom as well as the oxygen atom. But the nitro ion cannot bind from both the atoms at once, so it is called a polydentate ligand too.
> Thus, the nitro ion represents the linkage isomerism, when binds through nitrogen it is nitro, whereas when it binds through oxygen it is known as nitrito.
> The second option we have thiocyanate, it also shows binding through the sulphur atom, and the nitrogen atom. It also represents the linkage isomerism.
> Now, if we see the third option is cyanide ion, it will also represent the linkage isomerism, as when it binds as NC it is known as isocyanide.
> The last option we have is ammine ligand; it represents the geometrical isomerism.
In the last we can conclude that NH$_3$ (ammine) ligand will not show linkage isomerism.

Hence, the correct option is (D).

Note: Don’t get confused between the different ligands given in the question. Just identify the ambidentate ligands. In the linkage isomerism, the representation like structure of the complex, properties and the naming of complexes differs; but the formula remains the same.