There are four complexes of Ni. Select the co | vedclass

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(D) Paramagnetic complexes are attracted by a magnetic field.$1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pair

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$II$ and $IV$

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(D) Paramagnetic complexes are attracted by a magnetic field.$1$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. It is diamagnetic.$2$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,no pairing occurs. It has $2$ unpaired electrons,so it is paramagnetic.$3$. $[Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing. It is diamagnetic.$4$. $[Ni(H_2O)_6]^{2+}$: $Ni^{2+}$ is $3d^8$. $H_2O$ is a weak field ligand,no pairing occurs. It has $2$ unpaired electrons,so it is paramagnetic.Thus,complexes $(II)$ and $(IV)$ are paramagnetic and will be attracted by a magnetic field.

There are four complexes of $Ni$. Select the complex(es) which will be attracted by a magnetic field:
$(I) \ [Ni(CN)_4]^{2-}, \ (II) \ [NiCl_4]^{2-}, \ (III) \ [Ni(CO)_4]$
$(IV) \ [Ni(H_2O)_6]^{2+}$

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There are four complexes of $Ni$. Select the complex(es) which will be attracted by a magnetic field:$(I) \ [Ni(CN)_4]^{2-}, \ (II) \ [NiCl_4]^{2-}, \ (III) \ [Ni(CO)_4]$$(IV) \ [Ni(H_2O)_6]^{2+}$