Answer to Question #119462 in Inorganic Chemistry for Bernard Asant

Question

Predict whether these complexes would be labile or inert and explain your choices. The magnetic moment is given in Bohr magnetons (µB) after each complex. Ammonium oxopentachlorochromate(V) 1.82 Potassium hexaiodomanganate(IV) 3.82 Potassium hexacyanoferrate(III) 2.40 Hexammineiron(II) chloride

Answer 1

Ammonium oxopentachlorochromate(V) 1.82 is labile, Cr(V) has one d electron and configuration in octahedral split shell is t_2g¹e_g⁰, it is prone to Jahn-Teller distortion and hence complex is labile.

Potassium hexaiodomanganate(IV) 3.82 is inert. Mn(IV) has 3 electrons and configuration t_2g³e_g⁰, it is not prone to Janh-Teller effect and has no antibonding e_g electrons and hence is inert.

Potassium hexacyanoferrate(III) 2.40 is labile, since Fe(III) in a strong field of cyanoligands has one unpaired electrons and two electron pairs, configuration is t_2g⁵e_g⁰, which makes complex prone to a weak Janh-Teller distortion, and hence is labile.

Hexammineiron(II) chloride is labile, since Fe(II) in ammineligand field has one electron pair and four unpaired electrons, configuration is t_2g⁴e_g², complex is labile because this configuration has electrons on antibonding e_g orbitals and undergoes a weak Jahn-Teller distortion because of asymmetric filling of t_2g shell