The diamagnetic species is

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(C) To determine the magnetic property,we calculate the number of unpaired electrons in each complex:$1$. In $[Co(H_2O)_6]^{2+}$,$Co$ is in $+2$ oxida

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$[Co(NH_3)_6]^{3+}$

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(C) To determine the magnetic property,we calculate the number of unpaired electrons in each complex:$1$. In $[Co(H_2O)_6]^{2+}$,$Co$ is in $+2$ oxidation state $(d^7)$. $H_2O$ is a weak field ligand,so it has $3$ unpaired electrons.$2$. In $[Ni(H_2O)_6]^{2+}$,$Ni$ is in $+2$ oxidation state $(d^8)$. $H_2O$ is a weak field ligand,so it has $2$ unpaired electrons.$3$. In $[Co(NH_3)_6]^{3+}$,$Co$ is in $+3$ oxidation state $(d^6)$. $NH_3$ is a strong field ligand,causing pairing of electrons. The configuration becomes $t_{2g}^6 e_g^0$,resulting in $0$ unpaired electrons. Thus,it is diamagnetic.$4$. In $[Ni(NH_3)_6]^{2+}$,$Ni$ is in $+2$ oxidation state $(d^8)$. $NH_3$ is a strong field ligand,but for $d^8$ configuration,it still has $2$ unpaired electrons.

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