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The geometries of $\left[\mathrm{Ni}(\mathrm{CO})_4\right],\left[\mathrm{PtCl}_4\right]^{2-}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ respectively are
ChemistryCoordination CompoundsAP EAMCETAP EAMCET 2021 (24 Aug Shift 2)
Options:
  • A tetrahedral, tetrahedral and octahedral
  • B tetrahedral, square planar and square pyramidal
  • C square planar, square planar and octahedral
  • D tetrahedral, square planar and octahedral
Solution:
1102 Upvotes Verified Answer
The correct answer is: tetrahedral, square planar and octahedral
$\left[\mathrm{Ni}(\mathrm{CO})_4\right]$
Ground state electronic configuration $=3 d^8 4 s^2$
$\mathrm{CO}$ is strong field ligand, so it pushes $10 e^{-}$to $3 d$-orbital. The hybridisation of $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ is $s p^3$ and geometry is tetrahedral.


$\left[\mathrm{PtCl}_4\right]^{2-} \rightarrow$ Square planar. A strong repulsion with the electron and ligand take place in between platinum which leads to strong field splitting due to which square planar geometry is more stabilised than tetrahedral shape.

$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ is octahedral that involves $d^2 s p^3$ hybridisation.

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