Questions
Hybridization of Fe in [Fe(H2O)5NO]SO4 (brown ring complex) is
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a
dsp2
b
sp3d
c
sp3d2
d
d2sp3
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Correct option is C
- The oxidation state of Fe in the brown ring complex is +1 and the electronic configuration thus becomes 3d64s1. Then the weak field ligand water and strong field ligand attack the metal ion and change the configuration to 3d74s0 and therefore the complex is left with 3 unpaired electrons. Also, since the ligands donate electrons in the vacant 4s,4p,4d orbitals, therefore, the hybridization is sp3d2.
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