Question

The CORRECT statement(s) about \([Ni(CN)_4]^{2-}\), \([Ni(CO)_4]\) and \([NiCl_4]^{2-}\) is(are):
(Given: Atomic number of Ni = 28)

• A. Both \([Ni(CN)_4]^{2-}\) and \([Ni(CO)_4]\) are square planar complexes.
• B. \([Ni(CN)_4]^{2-}\) is diamagnetic and \([NiCl_4]^{2-}\) is paramagnetic.
• C. Both \([Ni(CO)_4]\) and \([NiCl_4]^{2-}\) are paramagnetic.
• D. \([Ni(CN)_4]^{2-}\) is square planar and \([NiCl_4]^{2-}\) is tetrahedral in shape.

Hint:

The geometry and magnetic property of a complex depend on the oxidation state of the metal and whether the ligand is a strong or weak field ligand. Use Crystal Field Theory and hybridization to justify your answer.

Answer 1

The Correct Option is B, D

Nickel has atomic number 28, and its ground-state electron configuration is \([Ar]\,3d^8\,4s^2\). In its complexes, the oxidation state and ligand field strength determine geometry and magnetic properties.
1. For \([Ni(CN)_4]^{2-}\): The oxidation state of Ni is +2. CN\(^{-}\) is a strong field ligand and causes pairing of electrons. The complex adopts a square planar geometry due to dsp\(^2\) hybridization. Since all electrons pair up, the complex is diamagnetic.
2. For \([NiCl_4]^{2-}\): Ni is in the +2 oxidation state again, but Cl\(^{-}\) is a weak field ligand. There is no pairing of electrons, leading to tetrahedral geometry with sp\(^3\) hybridization. The presence of unpaired electrons makes it paramagnetic.
3. For \([Ni(CO)_4]\): The Ni is in the 0 oxidation state. CO is a strong field ligand and causes pairing of electrons. The complex forms a tetrahedral geometry with sp\(^3\) hybridization. It is diamagnetic because there are no unpaired electrons.
Hence, \([Ni(CN)_4]^{2-}\) is diamagnetic and square planar, while \([NiCl_4]^{2-}\) is paramagnetic and tetrahedral.