Question

Among Ni(CO)${}_4$, [Ni(CN)${}_4$]${}^{2-}$, and NiCl${}_4^{2-}$:

• A. Ni(CO)${}_4$ and NiCl${}_4^{2-}$ are diamagnetic and [Ni(CN)${}_4$]${}^{2-}$ is paramagnetic
• B. NiCl${}_4^{2-}$ and [Ni(CN)${}_4$]${}^{2-}$ are diamagnetic and Ni(CO)${}_4$ is paramagnetic.
• C. Ni(CO)${}_4$ and [Ni(CN)${}_4$]${}^{2-}$ are diamagnetic and NiCl${}_4^{2-}$ is paramagnetic
• D. Ni(CO)${}_4$ is diamagnetic, and NiCl${}_4^{2-}$ and [Ni(CN)${}_4$]${}^{2-}$ are paramagnetic

Answer 1

The correct option is C Ni(CO)${}_4$ and [Ni(CN)${}_4$]${}^{2-}$ are diamagnetic and NiCl${}_4^{2-}$ is paramagnetic

$\left[Ni\right(CO{)}_4]$
The oxidation number of Ni is O.
Atomic number = 28
Ni = $\left[Ar\right]3d^{8}4s^2$
$s{p}^3$-Hybridization (tetrahedral)
There are no unpaired electrons, so the complex is diamagnetic.
Spin magnetic moment = zero
$\left[Ni\right(CN{)}_4{]}^{2-}$
The oxidation number of Ni is +2.
Atomic number = 28
Ni = $\left[Ar\right]3d^{8}4s^2$
$N{i}^{2+}$ = $\left[Ar\right]3d^8$
$ds{p}^2$-Hybridization (square planar)
There are no unpaired electrons so, the complex is diamagnetic.
Spin magnetic moment = zero
$[NiC{l}_4{]}^{2-}$
The oxidation number of Ni is +2 .
Atomic number = 28
Ni = $\left[Ar\right]3d^{8}4s^2$
$N{i}^{2+}$ = $\left[Ar\right]3d^8$
Chlorido is a weak field ligand, no pairing
$s{p}^3$-Hybridization (tetrahedral)
There are two unpaired electrons, so the complex is paramagnetic.
Spin magnetic moment
$\mu =\sqrt{n(n+2)}$ BM
$=\sqrt{2(2+2)}$ BM = $\sqrt{8}$ BM