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Question

How many unpaired electrons are present in the Brown Ring complex
[Fe(H2O)5(NO)]SO4?

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Solution

The spin of the compound is 3/2 as No+ is a paramagnetic substance.
So that unpaired electron it can give up to Fe to reduce it from Fe2+ to +1 and no becomes NO+
Hence, No. of unpaired electron present in [Fe(H2O)5(NO)]SO4 is 6.

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