Question

[NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is  diamagnetic though both are tetrahedral. Why? (Atomic no. Ni = 28) 

Answer 1

[NiCl₄]^2- 
Ni is in + 2 oxidation state.
3d⁸ configuration 

Cl^- is weak field ligand. So, pairing doesnot occur. 
sp3 hybridised orbitals of 2 Ni²⁺

[ ] NiCl  is paramagnetic as n = 2
4 [ ( ) ] Ni CO Ni is in ‘0’ oxidation state

In presence of ‘CO’ pairing of e– takes place ‘CO’ is strong field ligand. So, with ‘CO
[Ni(CO)4]  →



^{[Ni(CO)4] →}

So, [Ni(CO)₄] is diamagnetic as n = 0.