Question

4. Using the following data and crystal field theory, rationalise the observation that [Fe(bpy)31 and [Fe(bpy)3] have almost identical Fe-N bond lengths, but [Fe(H20)62 and [Fe(H2O)615 have significantly different Fe-O bond lengths. M-L bond length (A) [Fe(bpy)32 [Fe(bpy)]3 [Fe(OH2)6]2* [Fe(OH2)]* 1.96 1.98 2.10 1.99

Answer 1

Û [Fe (60), fe and [Fe tru.] in to I both Fe-n bonds lenght are same is due to both are strong field complexes and ti-acle aptar ligands, so fe-n bond lengtts are allonost same for Fet , Feln. CESE for See (Bry)] fet us 3d tigo ege a Ao = tagxo.4.2.0.68 eg = 6X0.4 = 2.440. CESE for (Fe (byty) e usado 40 = 5x0.4 = 2.0 40. * But in case of [e (H22] and he (4.2%) Fe-o'bandlengths are different in both complenus due to weak field complen and it is High spin complen, weak ligend. So here Fets have high lonic charge when compare to feel so, it too easy to attract the H₂O'ligand and so shows small feno' s and lengts. CASE: Fe ( 4 th us a dt tag egr 40 = 4x04 - 2x06 = 16-1.2-0.440 (Eelt2t, us sdf teglegt A = 3X0-4- 2x06 = 1.2-1270 EFLEnd zero for (Fe (Hr), t, so it has to len bond length between re-o.