Hybridization of Fe in [$$Fe(H2O)5NO$$]$$SO4$$ (brown$$ ring $$complex) is

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 $$sp3d2$$.

Questions

Hybridization of Fe in [Fe(H₂O)₅NO]SO₄ (brown ring complex) is

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dsp²

sp³d

sp³d²

d²sp³

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detailed solution

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.