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Q.

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

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a

d2sp3

b

sp3d2

c

dsp2

d

sp3d

answer is C.

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Detailed Solution

  • 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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