Answer to Question #119462 in Inorganic Chemistry for Bernard Asant

Question #119462
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
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Expert's answer
2020-06-04T10:31:02-0400

Ammonium oxopentachlorochromate(V) 1.82 is labile, Cr(V) has one d electron and configuration in octahedral split shell is t2g1eg0, 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 t2g3eg0, it is not prone to Janh-Teller effect and has no antibonding eg 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 t2g5eg0, 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 t2g4eg2, complex is labile because this configuration has electrons on antibonding eg orbitals and undergoes a weak Jahn-Teller distortion because of asymmetric filling of t2g shell


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