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Chemistry Question on coordination compounds

For complexes [NiCl4]2−[NiCl_4]^{2-} and Ni(CO)4Ni(CO)_4 which one of the following statement is true

A

[NiCl4]2−[NiCl_4]^{2-} is diamagnetic while [Ni(CO)4][Ni(CO)_4] is paramagnetic and both the complexes have square planar geometry

B

[NiCl4]2−[NiCl_4]^{2-} is paramagnetic while [Ni(CO)4][Ni(CO)_4] is diamagnetic and both the complexes have tetrahedral geometry

C

[NiCl4]2−[NiCl_4]^{2-} is paramagnetic while [Ni(CO)4][Ni(CO)_4] is diamagnetic and both the complexes have square planar geometry

D

[NiCl4]2−[NiCl_4]^{2-} is diamagnetic while [Ni(CO)4][Ni(CO)_4] is paramagnetic and both the complexes have tetrahedral geometry

Answer

[NiCl4]2−[NiCl_4]^{2-} is paramagnetic while [Ni(CO)4][Ni(CO)_4] is diamagnetic and both the complexes have tetrahedral geometry

Explanation

Solution

[NiCl4]2−[NiCl_4]^{2-}
Configuration:

Oxidation state of metal =+2= + 2
Type of hybridisation =sp3= sp^3 (tetrahedral)
Number of unpaired electrons =2= 2
Magnetic nature == Paramagnetic
[Ni(CO)4][Ni(CO)_4]
Configuration:

Oxidation state of metal =O= O
Type of hybridisation =sp3= sp^3 (tetrahedral)
Number of unpaired electrons =O= O
Magnetic nature == Diamagnetic
Hence, [NiCl4]2−[NiCl_4]^{2-} is paramagnetic while [Ni(CO)4][Ni(CO)_4] is diamagnetic and both the complexes have tetrahedral geometry.