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Sahay Sir > Question Answers > Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) > The magnetic moment of K3 [ Fe (CN ) 6 ] is found to be 1.7 B.M. How many unpaired electron(s) is /are present per molecule

The magnetic moment of K3 [ Fe (CN ) 6 ] is found to be 1.7 B.M. How many unpaired electron(s) is /are present per molecule

The magnetic moment of K3 [ Fe (CN ) 6 ] is found to be 1.7 B.M. How many unpaired electron(s) is /are present per molecule