Classify each of the following complexes as either high or low spin . Explain your answers.
(i) `[Co(H_(2)O)_(6)]^(2+) mu= 3.87 B.M. " " [Co(CN)_(6)]^(3-) mu=0.0 B.M.`
(ii) `[Fe(NO_(2))_(6)]^(4-) mu=0.0 B.M." "[Fe(H_(2)O)_(6)]^(3+) mu =5.94 B.M. `

Explanation

ntains `Co^(2+)` ion in an octahedral field. Obviously, the compound `[Co(H_(2)0)_(6)]^(2+)` is high spin having a magnetic moment value of 4.6 B.M. whereas `(Co(CN)_(6)]^(4-)` is low spin with a magnetic moment value of 1.9 B.M.
(ii) The complex `[Fe(NO_(2))_(6)]^(4-)` contains `Fe^(2+)` which belongs to a `d_(6)` system. As the magnetic moment of the complex is zero, it indicates that all the six d-electrons are paired up. Hence the compound is low spin. On the other hand, the magnetic moment of 5.94 B.M. for the complex `Fe(H_(2)O)_(6)]^(3+)` corresponds to five unpaired electrons in Fe. It indicates there is no pairing of d-electrons of `Fe^(3+)` ion, hence the complex is high spin.