Ni(CO)_4 and [Ni(NH_3)_4]^{2+} do not differ | vedclass

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(A) For $Ni(CO)_4$: The oxidation state of $Ni$ is $0$. The configuration is $3d^{10} 4s^0$. The $EAN = 28 + 2(4) = 36$. It undergoes $sp^3$ hybridiza

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(A) For $Ni(CO)_4$: The oxidation state of $Ni$ is $0$. The configuration is $3d^{10} 4s^0$. The $EAN = 28 + 2(4) = 36$. It undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry. It has $0$ unpaired electrons,so the magnetic moment is $0 \ BM$.For $[Ni(NH_3)_4]^{2+}$: The oxidation state of $Ni$ is $+2$. The configuration is $3d^8$. The $EAN = 28 - 2 + 2(4) = 34$. It undergoes $dsp^2$ hybridization,resulting in a square planar geometry. It has $0$ unpaired electrons,so the magnetic moment is $0 \ BM$.Since both complexes have $0$ unpaired electrons,they do not differ in their magnetic moment.

$Ni(CO)_4$ and $[Ni(NH_3)_4]^{2+}$ do not differ in

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