Question

Based on VB theory explain why \[\ce{[Cr(NH3)6]^3+}\] is paramagnetic, while \[\ce{[Ni(CN)4]^2-}\] is diamagnetic.

Answer 1

1. \[\ce{[Cr(NH3)6]^3+}\]

In this complex, Cr is in the +3 oxidation state. Electronic configuration of Cr atom. Electronic configuration of Cr³⁺ ion 

Hybridisation and formation of [Cr(NH₃)₆]³⁺ Complex

Due to the presence of three unpaired electrons in [Cr(NH₃)₆]³⁺ it behaves as a paramagnetic substance.

The spin magnetic moment,

µ_s = `"g" sqrt(3(3 + 2)) = "g" sqrt(15)` = 3.87 BM

[Cr(NH₃)₆]³⁺ is an inner orbital octahedral complex.

2. \[\ce{[Ni(CN)4]^2-}\]

In this complex, Ni is in the +2 oxidation state. Electronic configuration of Ni atom. Electronic configuration of Ni²⁺ ion. Hybridisation and formation of \[\ce{[Ni(CN)4]^2-}\] Complex

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