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Question: Answered & Verified by Expert
Amongst $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}, \quad\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2} ]\right.$, $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right.]$ and $\left[\mathrm{Ni}\left(\mathrm{CN}_{4}\right)\right]^{2-}$ the paramagnetic species are
ChemistryCoordination CompoundsJEE Main
Options:
  • A $\left[\mathrm{NiCl}_{4}\right]^{2-},\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \cdot\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$
  • B $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right],\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right],\left[\mathrm{NiCl}_{4}\right]^{2-}$
  • C $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-} \cdot\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+},\left[\mathrm{NiCl}_{4}\right]^{2-}$
  • D $\left[\mathrm{Nl}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right\}\left[\mathrm{Ni}(\mathrm{CO})_{4}\right\}\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$
Solution:
2021 Upvotes Verified Answer
The correct answer is: $\left[\mathrm{NiCl}_{4}\right]^{2-},\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \cdot\left[\mathrm{Ni}\left(\mathrm{PPh}_{3}\right)_{2} \mathrm{Cl}_{2}\right]$
Species having unpaired electrons are paramagnetic
$$
\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{d}^{8}
$$

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

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