Question

Amongst $Ni(CO{)}_4,[Ni(CN{)}_4{]}^{2-}$ and $NiC{l}_4^{2-}$

• A. $Ni(CO{)}_{4}andNiC{l}_4^{2-}$ are diamagnetic and $\left[Ni\right(CN)4{]}^{2-}$ is paramagnetic
• B. $NiC{l}_4^{2-}$ and $\left[Ni\right(CN{)}_4{]}^{2-}$ are diamagnetic and $Ni(CO{)}_4$ is paramagnetic
• C. $Ni(CO{)}_4$ and $\left[Ni\right(CN{)}_4{]}^{2-}$ are diamagnetic and
  $NiC{l}_4^{2-}$is paramagnetic
  
• D. $Ni(CO{)}_4$ is diamagnetic and $NiC{l}_4^{2-}$ and $\left[Ni\right(CN{)}_4{]}^{2-}$ is paramagnetic
  

Answer 1

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

$NiC{l}_4^{2-}$is paramagnetic

In $Ni(CO{)}_4$
Ni = $\left[Ar\right]3d^{8}4s^2$ hence it is diamagnetic


Hence it is diamagnetic
$Ni\left(CN{)}_4{]}^{2+}N{i}^{2+}{\to }_{[}Ar\right]3d^8$

It has 2 unpaired electron hence it is paramagnetic