Question
$\left[ Cr \left( NH _3\right)_6\right]^{3+}$ is paramagnetic while [Ni(CN)4]2- is diamagnetic. Explain why?
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Answer
In [Cr(NH3)6]3+, Cr is in +3 oxidation state whose electronic configuration is 3d3, which has three unpaired electrons.
These unpaired electrons remain present even after the hybridization (d2sp3) because only vacant d-orbitals are used in hybridization, hence [Cr(NH3)6]3+ is paramagnetic but in $\left[ Ni ( CN )_4\right]^{2-}, Ni$ is in $Ni ^{2+}$ state whose configuration is 3d8. Due to strong ligand CN-, its unpaired electrons becomes paired at the time of hybridization, hence it is diamagnetic.
These unpaired electrons remain present even after the hybridization (d2sp3) because only vacant d-orbitals are used in hybridization, hence [Cr(NH3)6]3+ is paramagnetic but in $\left[ Ni ( CN )_4\right]^{2-}, Ni$ is in $Ni ^{2+}$ state whose configuration is 3d8. Due to strong ligand CN-, its unpaired electrons becomes paired at the time of hybridization, hence it is diamagnetic.
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