Based VB theory explain why ` [Cr(NH_3)_6]^(3+)` is paramagnetic , while `[Ni(CN_4) ^(2-) ` is diamagnetic.

Explanation

(a) `[Cr(NH_(3))_(6)]^(3+)`
In this complex Cr is in the +3 oxidation state.
Electronic configuration of Cr atom
Electronic configuration of `Cr^(3+)` ion
Hybridisation and formation of `[Cr(NH_(3))_(6)]^(3+)` complex
* Due to the presence of three unpaired electrons in `[Cr(NH_(3))_(6)]^(3+)` it behaves as a paramagnetic substance
* The spin magnetic moment, `mu s= sqrt(3(3+2)) = sqrt15 = 3.87BM`
* `[Cr(NH_(3))_(6)]^(3+)` is an inner orbital octahedral complex.

(b) `[Ni(CN)_(4)]^(2-)`
In this complex Ni is the +2 oxidation state.
Electronic configuration of Ni atom
Electronic configuration of `Ni^(2+)` ion

Hybridisation and formation of `[Ni(CN)_(4)]^(2-)` Complex

Since `CN^(-)` is strong field ligand, hence the electrons in 3d orbitals for forced to pair up and there is no unpaired electron in `[Ni(CN)_(4)]^(2-)`, hence it should be diamagnetic substance.