The correct combination is :

The correct combination is :
  1. $\left[\mathrm{NiCl}_4\right]^{2-}$ - square-planar; $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ - paramagnetic
  2. $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$-tetrahedral; $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ - paramagnetic
  3. $\left[\mathrm{NiCl}_4\right]^{2-}$- paramagnetic; $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ - tetrahedral
  4. $\left[\mathrm{NiCl}_4\right]^{2-}$- dimagnetic; $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ -square-planar

Solution

$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is square planar, diamagnetic $(0$ unpaired electrons) with $d s p^2$ hybridisation. $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ - is tetrahedral,diamagnetic ( $0$ unpaired electrons) with $s p^3$ hybridisation. $\left[\mathrm{NiCl}_4\right]^{2-}$ is tetrahedral, paramagnetic ($2$ unpaired electrons) with $s p^3$ hybridisation. Hence, the option (c) is the correct answer.

Asked in: JEE Main 2018 (15 Apr Shift 1 Online)

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