Ferrous ion changes to the ion X on reacting with acidified hydrogen peroxide. The number of d-electrons in X and its spin-only magnetic moment (in BM) are, respectively:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 5 and 5.92
Acidified hydrogen peroxide is an oxidising agent. It oxidises the ferrous ion, \( \ce{Fe^2+} \), to the ferric ion, so X is \( \ce{Fe^3+} \). In the table of standard electrode potentials, \( E^{\circ}_{\ce{H2O2/H2O}}=+1.78\,\mathrm{V} \) (in acid) is much higher than \( E^{\circ}_{\ce{Fe^3+/Fe^2+}}=+0.77\,\mathrm{V} \), so \( \ce{H2O2} \) takes electrons from \( \ce{Fe^2+} \). The half-reactions and the balanced ionic equation:
Oxidation: \( \displaystyle \ce{Fe^2+ -> Fe^3+ + e^-} \) (× 2)
Reduction: \( \displaystyle \ce{H2O2 + 2H^+ + 2e^- -> 2H2O} \)
Ionic equation: \(\displaystyle \ce{2Fe^2+ + H2O2 + 2H^+ -> 2Fe^3+ + 2H2O} \)
Configuration: Iron (Z = 26) is [Ar]3d64s2.
Electrons removed: removing the two 4s electrons and one 3d electron gives \( \ce{Fe^3+} \): [Ar]3d5, so X has 5 d-electrons.
Unpaired electrons: by Hund’s rule the five electrons occupy the five 3d orbitals singly, so n = 5 unpaired electrons.
The spin-only magnetic moment:
\( \displaystyle \mu=\sqrt{n(n+2)}\,\mathrm{BM} \)
\( \displaystyle \mu=\sqrt{5(5+2)}=\sqrt{35} \)
\( \displaystyle \mu=5.92\,\mathrm{BM} \)
So X has 5 d-electrons and a magnetic moment of 5.92 BM.