The geometries of Ni(CO) $_{4}$ and Ni(PPh $_{3}$ ) $_{2}$ Cl $_{2}$ are:

Both square planar

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Tetrahedral and square planar, respectively

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Both tetrahedral

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Square planar and tetrahedral, respectively

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Solution

The correct option is D Both tetrahedral
$N i (C O)_{4}$
Oxidation number = 0
Ni = $1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{8}, 4 s^{2}$
$N i^{2 +} = 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{1} 0$
$s p^{3}$ -Hybridization
Geometry of $N i (C O)_{4}$ is tetrahedral
$N i (P P h_{3})_{2} C l_{2}$
Oxidation number = +2
$N i^{2 +} = 1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{1} 0$
$s p^{3}$ -Hybridization
Geometry of $N i (P P h_{3})_{2} C l_{2}$ , is tetrahedral.

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