Question

How many unpaired electrons are present in brown ring complex?

Answer 1

Answer:

The spin of the compound is $$3/2$$ as $$N{o^ + }$$ is a paramagnetic substance$$.$$

So that unpaired electron it can give up to $$Fe$$ to reduce it from $$F{e^{2 + }}$$ to $$+1$$ and no becomes $$N{O^ + }$$

Hence$$,$$ No$$.$$ of unpaired electron present in $$\left[ {Fe{{\left( {{H_2}O} \right)}_5}\left( {NO} \right)} \right]S{O_4}$$ is $$6.$$