Answer to Question #84456 in Inorganic Chemistry for Aishwarya

Question #84456
guve the no of unpaired electrons in [Cu(Nh3)4]So4
1
Expert's answer
2019-01-24T05:25:13-0500

Complexes with coordination number 4 can have tetrahedral or square planar structure. The choice depends on the strength of ligands. Strong field ligands lead to formation of square planar complexes whereas weak field ligands lead to formation of tetraherdal complexes.

Spectrochemical series:

(weak) I < Br < S2− < SCN < Cl < NO3 < N3− < F < OH< C2O42− ≈ H2O <

NCS < CH3CN < py < NH3 < en < bipy < phen < NO2− < PPh3 < CN ≈ CO (strong)

Strong field ligands repel the electrons of the d orbitals. As a result unpaired electrons of d orbitals get paired. Pairing of electrons creates empty d orbiatls.

The electron configuration of Copper is : [Ar] 3d104s1

Oxidation state of Copper in [Cu(NH3)4]SO4 is Cu2+, as ligands NH3 have no charge and SO42- has charge -2

As atom of Cu loses two electrons to become an ion Cu2+ , it's electronic configuration is d9

The complex [Cu(NH3)4]2+ has a square planar shape, as NH3 is a strong field ligand, but the appropriate crystal field diagram shows that only one configuration is possible irrespective of the strength of the ligand field : t2g6eg3

So, there is only one unpaired electron in this complex what makes this complex paramagnetic.

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