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