Answer
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Hint: The electronic configuration of the metal ions that are present in the complexes must first be determined. The pairing of electrons occurs depending on the strength of the ligands attached to the metal. If any unpaired electron remains in the complex, then it is paramagnetic in nature. The greater the number of unpaired electrons, the greater is the paramagnetism. However, if no unpaired electron remains in the complex, then it is diamagnetic in nature.
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) $[Ni{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}$
$N{{i}^{2+}}$ - $[Ar]3{{d}^{6}}$
$N{{H}_{3}}$ is a strong ligand. It induces a pairing of electrons as four $N{{H}_{3}}$ ligands approach the nickel metal. So, it is diamagnetic.
(C) $[Ni{{(N{{H}_{3}})}_{6}}]C{{l}_{2}}$
$N{{i}^{2+}}$ - $[Ar]3{{d}^{6}}$
$N{{H}_{3}}$ is a strong ligand. It induces a pairing of electrons as four $N{{H}_{3}}$ ligands approach the nickel metal. So, it is diamagnetic.
(D) $[Cu{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}$
$C{{u}^{2+}}$- $[Ar]3{{d}^{9}}$
$N{{H}_{3}}$ is a strong ligand. It induces a pairing of electrons as four $N{{H}_{3}}$ ligands approach the copper metal, but still, an electron remains unpaired. Hence, it is diamagnetic.
Thus, $[Cu{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}$has highest paramagnetism.
Correct Option: (D) $[Cu{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}$
Note: The coordination compound complexes have magnetic properties. A molecule's magnetic characteristics are determined by the number of unpaired electrons in it, and magnetism is produced by electronic spin. The complexes typically display the three types of magnetism: ferromagnetism, paramagnetism, and diamagnetism.
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) $[Ni{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}$
$N{{i}^{2+}}$ - $[Ar]3{{d}^{6}}$
$N{{H}_{3}}$ is a strong ligand. It induces a pairing of electrons as four $N{{H}_{3}}$ ligands approach the nickel metal. So, it is diamagnetic.
(C) $[Ni{{(N{{H}_{3}})}_{6}}]C{{l}_{2}}$
$N{{i}^{2+}}$ - $[Ar]3{{d}^{6}}$
$N{{H}_{3}}$ is a strong ligand. It induces a pairing of electrons as four $N{{H}_{3}}$ ligands approach the nickel metal. So, it is diamagnetic.
(D) $[Cu{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}$
$C{{u}^{2+}}$- $[Ar]3{{d}^{9}}$
$N{{H}_{3}}$ is a strong ligand. It induces a pairing of electrons as four $N{{H}_{3}}$ ligands approach the copper metal, but still, an electron remains unpaired. Hence, it is diamagnetic.
Thus, $[Cu{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}$has highest paramagnetism.
Correct Option: (D) $[Cu{{(N{{H}_{3}})}_{4}}]C{{l}_{2}}$
Note: The coordination compound complexes have magnetic properties. A molecule's magnetic characteristics are determined by the number of unpaired electrons in it, and magnetism is produced by electronic spin. The complexes typically display the three types of magnetism: ferromagnetism, paramagnetism, and diamagnetism.
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