Question

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

• A. $ [NiCl_4]^{2-} $ is diamagnetic while $ [Ni(CO)_4] $ is paramagnetic and both the complexes have square planar geometry
• B. $ [NiCl_4]^{2-} $ is paramagnetic while $ [Ni(CO)_4] $ is diamagnetic and both the complexes have tetrahedral geometry
• C. $ [NiCl_4]^{2-} $ is paramagnetic while $ [Ni(CO)_4] $ is diamagnetic and both the complexes have square planar geometry
• D. $ [NiCl_4]^{2-} $ is diamagnetic while $ [Ni(CO)_4] $ is paramagnetic and both the complexes have tetrahedral geometry

Answer 1

Answer: (B)

$[NiCl_4]^{2-}$
Configuration:

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

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