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