Nickel $(Z=28)$ combines with a uninegative monodentate ligand $\mathrm{X}^{-}$to form a paramagnetic…

Nickel $(Z=28)$ combines with a uninegative monodentate ligand $\mathrm{X}^{-}$to form a paramagnetic complex $\left[\mathrm{NiX}_4\right]^{2-}$. The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively
  1. one, tetrahedral
  2. two, tetrahedral
  3. one, square planar
  4. two, square planar

Solution

${ }_{28} \mathrm{Ni}: \ldots \ldots \ldots \mathrm{s}^2, 3 \mathrm{p}^6, 3 \mathrm{~d}^8, 4 \mathrm{~s}^2$ $\mathrm{Ni}^{2+}: 3 \mathrm{~s}^2, 3 \mathrm{p}^6, 3 \mathrm{~d}^8$

Asked in: JEE Main 2006

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