Question

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

• A. $\left[NiCl_{4}\right]^{2-}, \left[Ni\left(H_{2}O\right)_{6}\right]^{2+},\left[Ni\left(PPh_{3}\right)_{2}Cl_{2}\right]$
• B. $\left[Ni\left(CO\right)_{4}\right],\left[Ni\left(PPh_{3}\right)_{2}Cl_{2}\right],\left[NiCl_{4}\right]^{2-}$
• C. $\left[Ni\left(CN\right)_{4}\right]^{2-}, \left[Ni\left(H_{2}O\right)_{6}\right]^{2+},\left[NiCl_{4}\right]^{2-}$
• D. $\left[Ni\left(PPh_{3}\right)_{2}Cl_{2}\right], \left[Ni\left(CO\right)_{4}\right],\left[Ni\left(CN\right)_{4}\right]^{2-}$

Answer 1

Answer: (A)

Species having unpaired electrons are paramagnetic
$Ni ^{2+}=[ Ar ] 3 d ^{8} $

(i) In $\left[ NiCl _{4}\right]^{2-}$ and $\left.N i \left( H _{2} O \right)_{6}\right]^{2+}$ ligands $Cl ^{-}$and
$H _{2} O$ are weak ligands, therefore no pairing will be possible. Thus, there are two unpained electrons.
(ii) In $\left[ Ni \left( Ph _{3}\right)_{2} Cl _{2}\right\}$ although $PPh _{3}$ has
d-acceptance nature but presence of $Cl$, makes electrons unpaired.