Among Ni(CO)_4, [Ni(CN)_4]^ 2- and [Ni(Cl)_4]^ 2- - Vidyadip

MCQ

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

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
✓
$Ni(CO)_4$ 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

Correct option: C.

$Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic and $[NiCl_4]^{2-}$ is paramagnetic

c
$\left[ Ni ( CO )_4\right]$

The oxidation number of $Ni$ is $O$.

Atomic number $=28$

$Ni =[ Ar ] 3 d ^8 4 s ^2$

$sp ^3$-Hybridization (tetrahedral)

There are no unpaired electrons, so the complex is diamagnetic.

Spin magnetic moment = zero

$\left[ Ni ( CN )_4\right]^{2-}$

The oxidation number of $Ni$ is $+2$.

Atomic number $=28$

$Ni =[ Ar ] 3 d ^8 4 s ^2$

$Ni ^{2+}=[ Ar ] 3 d ^8$

$dsp ^2$-Hybridization (square planar)

There are no unpaired electrons so, the complex is diamagnetic.

Spin magnetic moment $=$ zero

$\left[ NiCl _4\right]^{2-}$

The oxidation number of $Ni$ is $+2$.

Atomic number $=28$

$Ni =[ Ar ] 3 d ^8 4 s ^2$

$Ni ^{2+}=[ Ar ] 3 d ^8$

Chlorido is a weak field ligand, no pairing

$sp ^3$-Hybridization (tetrahedral)

There are two unpaired electrons, so the complex is paramagnetic.

Spin magnetic moment

$\mu=\sqrt{ n ( n +2)}\,BM$

$=\sqrt{2(2+2)} BM =\sqrt{8}\,BM$

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