Question

Give reason for the statement. $‘[Ni(CN)_{4}]^{2-}$ is diamagnetic while $[NiCl_{4}]^{2-}$ is paramagnetic in nature.’

• A. In $[NiCl_{4}]^{2-}$, no unpaired electrons are present while in $[Ni(CN)_{4}]^{2-}$ two unpaired electrons are present
• B. In $[Ni(CN)_{4}]^{2-}$, no unpaired electrons are present while in $[NiCl_{4}]^{2-}$ two unpaired electrons are present
• C. $[NiCl_{4}]^{2-}$ shows $dsp^{2}$ hybridisation hence it is paramagnetic
• D. $[Ni(CN)_{4}]^{2-}$ shows $sp^{3}$ hybridisation hence it is diamagnetic

Answer 1

Answer: (B)

In $[Ni(CN)_{4}]^{2-}$ there is no unpaired electrons because $CN^{-}$ is a strong field ligand. Therefore it is diamagnetic in nature.

In $[NiCl_{4}]^{2-}$, there are two unpaired electrons because $Cl^{-}$ is a weak field ligand. Therefore, it is paramagnetic in nature.