The diamagnetic species is

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(A) To determine the magnetic property,we look at the electronic configuration of the central metal ion and the nature of the ligand:$1$. In $[Ni(CN)_

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$[Ni(CN)_4]^{2-}$

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(A) To determine the magnetic property,we look at the electronic configuration of the central metal ion and the nature of the ligand:$1$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in the $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of electrons in the $3d$ orbitals,resulting in no unpaired electrons. Thus,it is diamagnetic.$2$. In $[NiCl_4]^{2-}$,$Ni^{2+}$ $(3d^8)$ is bonded to $Cl^-$ (a weak field ligand),which does not cause pairing,resulting in two unpaired electrons. Thus,it is paramagnetic.$3$. In $[CoCl_4]^{2-}$,$Co^{2+}$ $(3d^7)$ is bonded to $Cl^-$ (a weak field ligand),resulting in three unpaired electrons. Thus,it is paramagnetic.$4$. In $[CoF_6]^{2-}$,$Co^{4+}$ $(3d^5)$ is bonded to $F^-$ (a weak field ligand),resulting in five unpaired electrons. Thus,it is paramagnetic.Therefore,the correct option is $A$.

Similar Questions

Arrange the following in the increasing order of the number of unpaired electrons present in the central metal ion:
$I.$ $[MnCl_6]^{3-}$
$II.$ $[FeF_6]^{3-}$
$III.$ $[Mn(CN)_6]^{3-}$
$IV.$ $[Fe(CN)_6]^{3-}$

Among the following complexes,the one that can exist as facial $(fac)$ and meridional $(mer)$ isomers is

Determine the magnetic property of $[CoF_6]^{3-}$.

Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers: $Mn = 25$; $Fe = 26$)
$(A)$ $[FeF_{6}]^{3-}$
$(B)$ $[Fe(CN)_{6}]^{3-}$
$(C)$ $[MnCl_{6}]^{3-}$ (high spin)
$(D)$ $[Mn(CN)_{6}]^{3-}$

The tetrahedral $[CoI_4]^{2-}$ and square planar $[PdBr_4]^{2-}$ complex ions are respectively:

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