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Which statement is incorrect : -
A
$Ni(CO)_4$ – Tetrahedral, paramagnetic
B
$[Ni(CN)_4]^{2-}$ – Square planar, diamagnetic
C
$Ni(CO)_4$ – Tetrahedral, diamagnetic
D
$[NiCl_4]^{2-}$ – Tetrahedral, paramagnetic
Detailed Solution
$Ni(CO)_4$: Ni(0) is $3d^8\,4s^2$. CO is a strong field ligand, so the 4s electrons pair into 3d ($3d^{10}$). $sp^3$ hybridisation gives a tetrahedral, diamagnetic complex.
So '$Ni(CO)_4$ – tetrahedral, paramagnetic' is incorrect, and '$Ni(CO)_4$ – tetrahedral, diamagnetic' is correct.
$[Ni(CN)_4]^{2-}$: $Ni^{2+}$, $3d^8$ with strong field $CN^-$; all electrons pair, $dsp^2$, square planar and diamagnetic; correct.
$[NiCl_4]^{2-}$: $Ni^{2+}$, $3d^8$ with weak field $Cl^-$; two unpaired electrons, $sp^3$, tetrahedral and paramagnetic; correct.
So the incorrect statement is '$Ni(CO)_4$ – tetrahedral, paramagnetic'.
So '$Ni(CO)_4$ – tetrahedral, paramagnetic' is incorrect, and '$Ni(CO)_4$ – tetrahedral, diamagnetic' is correct.
$[Ni(CN)_4]^{2-}$: $Ni^{2+}$, $3d^8$ with strong field $CN^-$; all electrons pair, $dsp^2$, square planar and diamagnetic; correct.
$[NiCl_4]^{2-}$: $Ni^{2+}$, $3d^8$ with weak field $Cl^-$; two unpaired electrons, $sp^3$, tetrahedral and paramagnetic; correct.
So the incorrect statement is '$Ni(CO)_4$ – tetrahedral, paramagnetic'.