can you please prove this theoritically
what is the oxidation state and no. of unpaired electrons of Fe in
[Fe(H2O)5NO]SO4
the answer for oxidation state is given +1 and no. of unpaired electrons is given 3
how can these both be simultaneously true
no of unpaired electrons in Fe([Ne]4s23d6) are 5
in +1 oxidation state ,after removing an electron from 4s2, no of unpaired electrons will be 6
plz somebody explain this
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4 Answers
The complex [Fe(H2O)5NO]2+ is formed in the brown ring test for
nitrates. The color is due to the charge transfer. This complex formally contains Fe(+1) and
NO+. Its magnetic moment is approximately 3.9 BM confirming the presence of three
unpaired electrons
hey pritish thats [Ar] 3d64s2 not [Ne]...also in the ground state it has 4 unpaired electrons
Fe2+ + NO → Fe+ + NO+
(3d7)
(there occurs charge transfer which increases one electron in Fe)
in 3d7 configuration it has sp3d2 hybridization and shows spin-only magnetic moment of √15 BM
[Fe(H2O)5NO]SO4
now for oxidation state
x+0+1-2=0
so x=1
this is oxidation state of iron
no of unpaired electrons in Fe([Ne]4s23d6) are 4 not 5....u hv written wrong
in Fe+1 there are 5 unpaired electrons.....4 in 3d subshell and one in 4s subshell
for sp3d2 hybridisation one unpaied electron will move to 3d subshell and pairing will occur.....so there will be 3 unpaied electrons..they won't pair coz water is a weak ligand and won't cause any more pairing of electrons......