
Which of the following complexes is diamagnetic? [RPMT 1997]
(A) $[Ni{{(CO)}_{4}}]$
(B) ${{[NiC{{l}_{4}}]}^{2-}}$
(C) ${{[NiB{{r}_{4}}]}^{2-}}$
(D) $[NiC{{l}_{2}}].4{{H}_{2}}O$
Answer
164.1k+ views
Hint: Firstly, we have to determine the electronic configuration of metal ions present in the complexes. Then, according to the strength of the ligands attached to the metal, the pairing of the electrons occurs. If in a complex, all the electrons are paired up, then the complex is diamagnetic.
Complete answer:The electronic configurations of the metal ions present in the solution are as shown:
(A) $[Ni{{(CO)}_{4}}]$
$N{{i}^{{}}}$- $[Ar]3{{d}^{8}}$
$CO$ is a strong ligand. It induces a pairing of electrons as four $CO$ ligands approach the nickel metal. So, it is diamagnetic.
(B) ${{[NiC{{l}_{4}}]}^{2-}}$
$N{{i}^{2+}}$ - $[Ar]3{{d}^{6}}$
$C{{l}^{-}}$ is a weak ligand. It does not induce the pairing of electrons as ligands approach the nickel metal. So, it is paramagnetic.
(C) ${{[NiB{{r}_{4}}]}^{2-}}$
$N{{i}^{2+}}$ - $[Ar]3{{d}^{6}}$
$B{{r}^{-}}$ is a weak ligand. It does not induce the pairing of electrons as ligands approach the nickel metal. So, it is paramagnetic.
(D) $[NiC{{l}_{2}}].4{{H}_{2}}O$
$N{{i}^{2+}}$ - $[Ar]3{{d}^{6}}$
$C{{l}^{-}}$ is a weak ligand. It does not induce the pairing of electrons as ligands approach the nickel metal. So, it is paramagnetic.
Hence, among the following complexes, $[Ni{{(CO)}_{4}}]$ is diamagnetic.
Correct Option: (A) $[Ni{{(CO)}_{4}}]$
Note: Complexes of the coordination compounds exhibit magnetic characteristics. The quantity of unpaired electrons in a molecule determines its magnetic properties, and the electronic spin produces magnetism. There are primarily three forms of magnetism exhibited by the complexes: ferromagnetism, paramagnetism, and diamagnetism. If any unpaired electrons are present in the complex, it is said to be paramagnetic.
Complete answer:The electronic configurations of the metal ions present in the solution are as shown:
(A) $[Ni{{(CO)}_{4}}]$
$N{{i}^{{}}}$- $[Ar]3{{d}^{8}}$
$CO$ is a strong ligand. It induces a pairing of electrons as four $CO$ ligands approach the nickel metal. So, it is diamagnetic.
(B) ${{[NiC{{l}_{4}}]}^{2-}}$
$N{{i}^{2+}}$ - $[Ar]3{{d}^{6}}$
$C{{l}^{-}}$ is a weak ligand. It does not induce the pairing of electrons as ligands approach the nickel metal. So, it is paramagnetic.
(C) ${{[NiB{{r}_{4}}]}^{2-}}$
$N{{i}^{2+}}$ - $[Ar]3{{d}^{6}}$
$B{{r}^{-}}$ is a weak ligand. It does not induce the pairing of electrons as ligands approach the nickel metal. So, it is paramagnetic.
(D) $[NiC{{l}_{2}}].4{{H}_{2}}O$
$N{{i}^{2+}}$ - $[Ar]3{{d}^{6}}$
$C{{l}^{-}}$ is a weak ligand. It does not induce the pairing of electrons as ligands approach the nickel metal. So, it is paramagnetic.
Hence, among the following complexes, $[Ni{{(CO)}_{4}}]$ is diamagnetic.
Correct Option: (A) $[Ni{{(CO)}_{4}}]$
Note: Complexes of the coordination compounds exhibit magnetic characteristics. The quantity of unpaired electrons in a molecule determines its magnetic properties, and the electronic spin produces magnetism. There are primarily three forms of magnetism exhibited by the complexes: ferromagnetism, paramagnetism, and diamagnetism. If any unpaired electrons are present in the complex, it is said to be paramagnetic.
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