Question

In nitroprusside ion, the iron and NO exist as Fe(II) and $N{O}^{+}$ rather than Fe(III) and NO. These forms can be distinguished by:

• (A) estimating the concentration of iron
• (B) measuring the concentration of $\mathrm{CN}$ -
• (C) measuring the solid-state magnetic moment
• (D) thermally decomposing the compound

Answer 1

The Correct Option is C

Explanation:
In nitroprusside ion, the iron and NO exist as Fe(II) and $N{O}^{+}$ rather than Fe(III) and NO. These forms can be distinguished by measuring the solid-state magnetic moment.Nitroprusside ion is ${[\mathrm{Fe}(\mathrm{CN}{)}_5\mathrm{NO}]}^{2-}$. If the central atom iron is present here in ${\mathrm{Fe}}^{2+}$ form, its effective atomic number will be $26-2+(6\times 2)=36$ and the distribution of electrons in valence orbitals (hybridised and unhybridized) of the ${\mathrm{Fe}}^{2+}$ will be

It has no unpaired electron. So this anionic complex is diamagnetic. If the nitroprusside ion has ${\mathrm{Fe}}^{3+}$ and $\mathrm{NO}$, the electronic distribution will be such that it will have one unpaired electron i.e. the complex will be paramagnetic.

Thus, magnetic moment measurement establishes that in nitroprusside ion, the Fe and NO exist as F(II) and $N{O}^{+}$ rather than Fe(III) and NO.Hence, the correct option is (C).