Answer to Question #84456 in Inorganic Chemistry for Aishwarya

Question

guve the no of unpaired electrons in [Cu(Nh3)4]So4

Answer 1

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 SO₄^2- 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 : t_2g⁶e_g³

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