The number of paramagnetic species from the f | vedclass

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Question

(C) To determine the number of paramagnetic species,we analyze the electronic configuration and magnetic behavior of each complex:$1. [Ni(CN)_4]^{2-}$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) To determine the number of paramagnetic species,we analyze the electronic configuration and magnetic behavior of each complex:$1. [Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons. It is diamagnetic.$2. [Ni(CO)_4]$: $Ni$ is $3d^8 4s^2$. $CO$ is a strong field ligand,forcing electrons to pair. It is diamagnetic.$3. [NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,no pairing occurs. It has $2$ unpaired electrons,so it is paramagnetic.$4. [Fe(CN)_6]^{4-}$: $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing. It is diamagnetic.$5. [Cu(NH_3)_4]^{2+}$: $Cu^{2+}$ is $3d^9$. It has $1$ unpaired electron,so it is paramagnetic.$6. [Fe(CN)_6]^{3-}$: $Fe^{3+}$ is $3d^5$. $CN^-$ is a strong field ligand,causing pairing. It has $1$ unpaired electron,so it is paramagnetic.$7. [Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ is $3d^6$. $H_2O$ is a weak field ligand,no pairing occurs. It has $4$ unpaired electrons,so it is paramagnetic.The paramagnetic species are $[NiCl_4]^{2-}, [Cu(NH_3)_4]^{2+}, [Fe(CN)_6]^{3-}$,and $[Fe(H_2O)_6]^{2+}$.Total number of paramagnetic species = $4$.

The number of paramagnetic species from the following is $............$.
$[Ni(CN)_4]^{2-}, [Ni(CO)_4], [NiCl_4]^{2-}$
$[Fe(CN)_6]^{4-}, [Cu(NH_3)_4]^{2+}$
$[Fe(CN)_6]^{3-}$ and $[Fe(H_2O)_6]^{2+}$

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The number of paramagnetic species from the following is $............$.$[Ni(CN)_4]^{2-}, [Ni(CO)_4], [NiCl_4]^{2-}$$[Fe(CN)_6]^{4-}, [Cu(NH_3)_4]^{2+}$$[Fe(CN)_6]^{3-}$ and $[Fe(H_2O)_6]^{2+}$