
Which of the following exhibits linkage isomerism?
(A)- \[[Co{{(N{{H}_{3}})}_{5}}S{{O}_{4}}]Br\]
(B)- \[[Co{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}\]
(C)- \[[Co{{(N{{H}_{3}})}_{5}}(N{{O}_{2}})]C{{l}_{2}}\]
(D)- \[[Co{{(N{{H}_{3}})}_{6}}][Cr{{(CN)}_{6}}]\]
Answer
123.9k+ views
Hint: The coordination complexes that show linkage isomerism must have one special type of ligand which has two donor atoms, but only one of the donor atoms can bind to the central metal atom at a time.
Complete step by step solution:
- The coordination complexes that show linkage isomerism have ambidentate ligand.
- Ambidentate ligands are those which have two donor atoms, but only one of the donor atoms can bind to the central metal atom at a time.
- ‘Ambi’ means both and dentate means the denticity of the ligand.
- Denticity is the number of the donor atoms of the ligand that binds to the central metal atom in the coordination complex.
- Here, \[[Co{{(N{{H}_{3}})}_{5}}(N{{O}_{2}})]C{{l}_{2}}\]exhibits linkage isomerism.
\[N{{O}_{2}}\] is the ambidentate ligand in this coordination complex. This ligand is the reason the coordination complex shows the linkage isomerism.
When the central metal atom is bonded with nitrogen, the ligand is\[N{{O}_{2}}\]. When the metal atom is bonded with oxygen, the ligand is\[ONO\].
- Other examples of the ambidentate ligand are \[C{{N}^{-}}\]and\[N{{C}^{-}}\], \[SC{{N}^{-}}\]and\[NC{{S}^{-}}\].
- \[[Co{{(N{{H}_{3}})}_{5}}(N{{O}_{2}})]C{{l}_{2}}\]and \[[Co{{(N{{H}_{3}})}_{5}}(ONO)]C{{l}_{2}}\]are the linkage isomers.
So, option C is the correct option.
Additional Information:
Ligands can be monodentate, bidentate, and multidentate as well depending upon the number of the donor atoms of the ligand binding to the central metal atom.
Note \[N{{O}_{2}}\]is called a nitro ligand. When the donation is from the nitrogen to the metal center, it is nitro complex. The linkage of the ligand is as follow

\[ONO\]is called nitrito ligand. When the donation is from the nitrogen to the metal centre, it is nitrito complex. The linkage is as follows:

The complexes are octahedral in shape.
Complete step by step solution:
- The coordination complexes that show linkage isomerism have ambidentate ligand.
- Ambidentate ligands are those which have two donor atoms, but only one of the donor atoms can bind to the central metal atom at a time.
- ‘Ambi’ means both and dentate means the denticity of the ligand.
- Denticity is the number of the donor atoms of the ligand that binds to the central metal atom in the coordination complex.
- Here, \[[Co{{(N{{H}_{3}})}_{5}}(N{{O}_{2}})]C{{l}_{2}}\]exhibits linkage isomerism.
\[N{{O}_{2}}\] is the ambidentate ligand in this coordination complex. This ligand is the reason the coordination complex shows the linkage isomerism.
When the central metal atom is bonded with nitrogen, the ligand is\[N{{O}_{2}}\]. When the metal atom is bonded with oxygen, the ligand is\[ONO\].
- Other examples of the ambidentate ligand are \[C{{N}^{-}}\]and\[N{{C}^{-}}\], \[SC{{N}^{-}}\]and\[NC{{S}^{-}}\].
- \[[Co{{(N{{H}_{3}})}_{5}}(N{{O}_{2}})]C{{l}_{2}}\]and \[[Co{{(N{{H}_{3}})}_{5}}(ONO)]C{{l}_{2}}\]are the linkage isomers.
So, option C is the correct option.
Additional Information:
Ligands can be monodentate, bidentate, and multidentate as well depending upon the number of the donor atoms of the ligand binding to the central metal atom.
Note \[N{{O}_{2}}\]is called a nitro ligand. When the donation is from the nitrogen to the metal center, it is nitro complex. The linkage of the ligand is as follow

\[ONO\]is called nitrito ligand. When the donation is from the nitrogen to the metal centre, it is nitrito complex. The linkage is as follows:

The complexes are octahedral in shape.
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