Question

$[NiC{l}_4{]}^{2-}$ is paramagnetic while $\left[Ni\right(CO{)}_4]$ is diamagnetic though both are tetrahedral. Why?

Answer 1

In $[NiC{l}_4{]}^{2-}$ oxidation state of $Ni$ is $+2$ and it have

$\upharpoonleft \downharpoonright \upharpoonleft \downharpoonright \upharpoonleft \downharpoonright \upharpoonleft \upharpoonleft$

$8$ electrons in d shell which is arranged as above, it has $2$ unpaired electrons, so it is paramagnetic.

In $\left[Ni\right(CO{)}_4]$ oxidation state of $Ni$ is $0$ and two electron from $4s$ is pushed into $3d$ oribtal.

$\upharpoonleft \downharpoonright \upharpoonleft \downharpoonright \upharpoonleft \downharpoonright \upharpoonleft \downharpoonright \upharpoonleft \downharpoonright$

It has fully filled d-orbital.

$\therefore \left[Ni\right(CO{)}_4]$ is diamagnetic