Amongst [Ni(H_{2}O)_{6}]^{2+}, [Ni(PPh_{3})_{ | vedclass

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Question

(A) Species having unpaired electrons are paramagnetic.$Ni$ has atomic number $28$, electronic configuration $[Ar] 3d^{8} 4s^{2}$.$1$. $[Ni(H_{2}O)_{6

Vedclass · Accepted Answer

$[Ni(H_{2}O)_{6}]^{2+}$ and $[Ni(PPh_{3})_{2}Cl_{2}]$

Vedclass · Answer

(A) Species having unpaired electrons are paramagnetic.$Ni$ has atomic number $28$, electronic configuration $[Ar] 3d^{8} 4s^{2}$.$1$. $[Ni(H_{2}O)_{6}]^{2+}$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. $H_{2}O$ is a weak field ligand, so no pairing occurs. It has $2$ unpaired electrons (Paramagnetic).$2$. $[Ni(PPh_{3})_{2}Cl_{2}]$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. It is a tetrahedral complex. Due to the presence of $Cl^{-}$ ligands, it remains high spin with $2$ unpaired electrons (Paramagnetic).$3$. $[Ni(CO)_{4}]$: $Ni$ is in $0$ oxidation state $(3d^{8} 4s^{2})$. $CO$ is a strong field ligand, causing pairing of all electrons. It has $0$ unpaired electrons (Diamagnetic).$4$. $[Ni(CN)_{4}]^{2-}$: $Ni$ is in $+2$ oxidation state $(3d^{8})$. $CN^{-}$ is a strong field ligand, causing pairing of electrons. It has $0$ unpaired electrons (Diamagnetic).Thus, $[Ni(H_{2}O)_{6}]^{2+}$ and $[Ni(PPh_{3})_{2}Cl_{2}]$ are paramagnetic.

Amongst $[Ni(H_{2}O)_{6}]^{2+}$, $[Ni(PPh_{3})_{2}Cl_{2}]$, $[Ni(CO)_{4}]$ and $[Ni(CN)_{4}]^{2-}$, the paramagnetic species are

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