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Q. Statement 1: $[Fe(H_{2}O)_{5}NO]SO_{4}$ is paramagnetic.
Statement 2:The $Fe$ in $[Fe(H_{2}O)_{5}NO]SO_{4}$ has three impaired electrons.

Coordination Compounds

Solution:

Iron in $[Fe(H_2O )5NO]SO_4$ exists in +2 oxidation state having electronic configuration $(3d)^6$. The unpaired electron of $NO$ is shifted to $Fe^{2+}$ changing it to $Fe^+$ and also changes $NO$ to $NO^+$. The electronic configuration of iron in the complex thus becomes $(3d)^7$ with three unpaired electrons as in the complex, $sp^3d^2$ hybridization exists.