In nitroprusside ion, the iron and `NO` exist as `Fe(II)` and `NO^(+)` rather than `Fe^(III)` and `NO`. These forms can be differentiated by

Options

A. estimating the concentration of iron
B. measuring the concentration of `CN^(-)`
C. measuring the solid state magnetic moment
D. thermally decomposing the compound

Answer

C

Explanation

In nitroprusside ion, Fe (II) has `d^(6)` configuration, i.e., 4 unpaired electrons which pair up for `d^(2)sp^(3)`-hybridisation and `NO^(+)` has also no unpaired electron. Hence, its magnetic moment is zero, i.e., it is diamagnetic. On the other hand, Fe (III) has `d^(5)` configuration and NO has unpaired electron. Even after pairing up, unpaired electrons are present and it would have net magnetic moment.