`[Ni(CN)_(4) ]^(2-)` diamagnetic, whereas `[NiCI_(4)]^(2-)` is paramagnetic. Explain.

[Cu(NH_3)_4]^(2+) is square planar, where as [NiCl_(4)]^(2-) is tetrahedral. Explain.

[Fe F_6]^(4-) is paramagnetic because

Amongst Ni(CO)_(4), [Ni(CN)_(4)]^(2-) and [NiCl_(4)]^(2-) :

The geometry of [Ni(CN)_4]^(2-) is

Give reason: Cu^(2+) is paramagnetic and Cu^(+) is diamagnetic.

The hybridization states of [Ni(CO)_(4)], [Ni(CN)_(4)]^(2-) and [NiCl_(4)]^(2-) species are respectively

Why is [NiCl_4]^(2-) paramagnetic but [Ni(CO)_4] is diamagnetic ? (At. Nos. : Cr= 24 , Co= 27 , Ni= 28)

A solution of [Ni(H_2 O)_6]^(2+) is green but a solution of [Ni(CN)_4]^(2-) is colourless . Explain.

Both [Ni(CO)_(4)] and [Ni(CN)_(4)]^(2-) are diamagnetic. The hybridisations of nickel in these complexes respectively are :

The complex [Ni(CN)_(4)]^(2-) is diamagnetic and it involves the following hybridisation of nickel :