Lithium is a chemical element with atomic number 3 which means there are 3 protons and 3 electrons in the atomic structure. The chemical symbol for Lithium is Li. It is a soft, silvery-white alkali metal. Ni: Atomic Number: 28: Atomic Mass: 58.6934 atomic mass units: Number of Protons: 28: Number of Neutrons: 31: Number of Electrons: 28: Melting Point: 1453.0° C: Boiling Point: 2732.0° C: Density: 8.902 grams per cubic centimeter: Normal Phase: Solid: Family: Transition Metal.
[NiCl4]2− is paramagnetic, while [Ni(CO)4] is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28)
Why is [NiCl4]2– paramagnetic but [Ni(CO)4] is diamagnetic ? (At. nos. : Cr = 24, Co = 27, Ni = 28)
Solution 1
In [NiCl4]2−, Ni is in the +2 state. Cl− is a ligand which is a weak field ligand which does not cause pairing of unpaired 3d electrons. Hence, it is paramagnetic. In [Ni(CO)4], Ni has 0 oxidation state. CO is a strong field ligand, which causes pairing of unpaired 3d electrons. No unpaired electrons are present in this case. So, it is diamagnetic.
Solution 2
In [NiCl4]2−, the oxidation state of Ni is +2. Chloride is a weak field ligand and does not cause pairing up of electrons against the Hund's rule of maximum multiplicity. As a result, two unpaired electrons are present in the valence d-orbitals of Ni which impart paramagnetic character to the complex. On the other hand, carbonyl is a strong field ligand and causes pairing up of electrons against the Hund's rule of maximum multiplicity. As a result, no unpaired electrons are present and hence, the complex is diamagnetic.
Solution 3
[NiCl4]2− and [Ni(CO)4] both are tetrahedral. But their magnetic characters are different. This is due to difference in the nature of ligands.
Ni+2 = [Ar] 4s03d8
Ni+2 has 2 unpaired electrons hence, this complex is paramagnetic.
In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d84s2.
[Ni(CO)4]
Element With Atomic Number 86
Ni = [Ar] 4s2 3d8
But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization. Since, no unpaired electrons are present in this case, [Ni(CO)4] is diamagnetic.
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