Looking for classes? Ksquare Career Institute, Bengaluru →
Of the following complex ions, which is diamagnetic in nature?
A
$[CoF_6]^{3-}$
B
$[NiCl_4]^{2-}$
C
$[Ni(CN)_4]^{2-}$
D
$[CuCl_4]^{2-}$
Detailed Solution
$[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand and pairs up the 3d electrons, leaving one 3d orbital empty; the hybridisation is $dsp^2$ (square planar) and there are no unpaired electrons, so it is diamagnetic.
$[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand, no pairing occurs; hybridisation is $sp^3$ (tetrahedral) with 2 unpaired electrons, so it is paramagnetic.
$[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand; hybridisation is $sp^3d^2$ with 4 unpaired electrons, so it is paramagnetic.
$[CuCl_4]^{2-}$: $Cu^{2+}$ is $3d^9$ with 1 unpaired electron, so it is paramagnetic.
Hence the diamagnetic ion is $[Ni(CN)_4]^{2-}$.
$[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand, no pairing occurs; hybridisation is $sp^3$ (tetrahedral) with 2 unpaired electrons, so it is paramagnetic.
$[CoF_6]^{3-}$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand; hybridisation is $sp^3d^2$ with 4 unpaired electrons, so it is paramagnetic.
$[CuCl_4]^{2-}$: $Cu^{2+}$ is $3d^9$ with 1 unpaired electron, so it is paramagnetic.
Hence the diamagnetic ion is $[Ni(CN)_4]^{2-}$.