Among [Ni(CO)_4],[NiCl_4]^{2-},[Co(NH_3)_4Cl_ | vedclass

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(C) Let us analyze the magnetic nature of each compound:$1$. $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand

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$4$

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(C) Let us analyze the magnetic nature of each compound:$1$. $[Ni(CO)_4]$: $Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons. All electrons are paired,so it is diamagnetic.$2$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is $3d^8$. $Cl^-$ is a weak field ligand,so it is paramagnetic (has $2$ unpaired electrons).$3$. $[Co(NH_3)_4Cl_2]Cl$: $Co^{3+}$ is $3d^6$. In an octahedral complex with $NH_3$ and $Cl^-$,$Co^{3+}$ is paramagnetic (has $4$ unpaired electrons).$4$. $Na_3[CoF_6]$: $Co^{3+}$ is $3d^6$. $F^-$ is a weak field ligand,so it is paramagnetic (has $4$ unpaired electrons).$5$. $Na_2O_2$: Contains peroxide ion $O_2^{2-}$,which has all electrons paired,so it is diamagnetic.$6$. $CsO_2$: Contains superoxide ion $O_2^-$,which has an odd number of electrons,so it is paramagnetic.The paramagnetic compounds are $[NiCl_4]^{2-}$,$[Co(NH_3)_4Cl_2]Cl$,$Na_3[CoF_6]$,and $CsO_2$. The total number is $4$.

Among $[Ni(CO)_4]$,$[NiCl_4]^{2-}$,$[Co(NH_3)_4Cl_2]Cl$,$Na_3[CoF_6]$,$Na_2O_2$ and $CsO_2$,the total number of paramagnetic compounds is

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