Question

Oxidation no of (fe(h2o)5(no)+)so4

Answer 1

number of SO4= -2

number of H2O= 0

number of NO= -1

Therefore ON of Fe= -(-1+0–2)=+3

The correct structure is [FeIII(H2O)5(NO−)]2+

For many years it was thought that iron was reduced to FeI and NO oxidized to NO+ ,based upon an observed magnetic moment suggestive of three unpaired electrons, however, the current thinking is that high spin FeIII (S=5/2) antiferromagnetically couples with NO-(S=1) for an observed spin of S=3/2.