Question

To determine the percentage of iron in magnetite, a sample of the ore was dissolved in acid, thus reducing the iron to $Fe^{2+}$ completely. The solution was then titrated with aqueous acidic $KMnO_4$. The correct statement(s) is (are)

• A. The number of electrons involved in the redox reaction during titration is 5.
• B. The paramagnetism of the solution increases during the reaction of $Fe^{2+}$ with $KMnO_4$.
• C. Dissolution of magnetite in sulfuric acid reduces the iron from +3 to +2.
• D. During the titration, the manganese containing species changes its geometry from square planar to tetrahedral.

Answer 1

Answer: (A), (B)

(A), (B)

Answer 2

The question asks for the correct statement(s) regarding the process of determining the percentage of iron in magnetite ($Fe_3O_4$) by titration with acidic $KMnO_4$. The process involves dissolving the ore in acid and ensuring all iron is in the $Fe^{2+}$ state, followed by titration with $KMnO_4$.

Let's analyze each statement:

(A) The number of electrons involved in the redox reaction during titration is 5.

The titration reaction involves the oxidation of $Fe^{2+}$ to $Fe^{3+}$ and the reduction of $MnO_4^-$ to $Mn^{2+}$ in acidic solution.

The oxidation half-reaction is: $Fe^{2+} \rightarrow Fe^{3+} + e^-$

The reduction half-reaction is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

To balance the electrons, we multiply the oxidation half-reaction by 5:

$5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$

Now, add the two half-reactions:

$MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$

In this balanced reaction, 5 electrons are transferred from $5Fe^{2+}$ ions to one $MnO_4^-$ ion. The number of electrons involved in the redox reaction (per mole of $MnO_4^-$ or per 5 moles of $Fe^{2+}$) is 5.

Thus, statement (A) is correct.

(B) The paramagnetism of the solution increases during the reaction of $Fe^{2+}$ with $KMnO_4$.

Paramagnetism is due to the presence of unpaired electrons. We need to consider the number of unpaired electrons in the relevant species: $Fe^{2+}$, $Fe^{3+}$, $MnO_4^-$, and $Mn^{2+}$.

Electron configuration of Fe: $[Ar] 3d^6 4s^2$.

$Fe^{2+}$: $[Ar] 3d^6$. In aqueous solution (high spin), it has 4 unpaired electrons ($t_{2g}^4 e_g^2$).

$Fe^{3+}$: $[Ar] 3d^5$. In aqueous solution (high spin), it has 5 unpaired electrons ($t_{2g}^3 e_g^2$).

Electron configuration of Mn: $[Ar] 3d^5 4s^2$.

$MnO_4^-$: Mn is in +7 oxidation state, $[Ar] 3d^0$. It has 0 unpaired electrons and is diamagnetic.

$Mn^{2+}$: $[Ar] 3d^5$. In aqueous solution (high spin), it has 5 unpaired electrons ($t_{2g}^3 e_g^2$).

The reaction consumes $Fe^{2+}$ (4 unpaired electrons) and $MnO_4^-$ (0 unpaired electrons) and produces $Fe^{3+}$ (5 unpaired electrons) and $Mn^{2+}$ (5 unpaired electrons).

Consider the change in the number of unpaired electrons for the stoichiometric reaction unit:

$5Fe^{2+} + MnO_4^- \rightarrow 5Fe^{3+} + Mn^{2+}$

Initial state (reactants): $5 \times (\text{unpaired e}^- \text{ in } Fe^{2+}) + 1 \times (\text{unpaired e}^- \text{ in } MnO_4^-) = 5 \times 4 + 1 \times 0 = 20$ unpaired electrons.

Final state (products): $5 \times (\text{unpaired e}^- \text{ in } Fe^{3+}) + 1 \times (\text{unpaired e}^- \text{ in } Mn^{2+}) = 5 \times 5 + 1 \times 5 = 25 + 5 = 30$ unpaired electrons.

As the reaction proceeds, species with fewer unpaired electrons ($Fe^{2+}$ and $MnO_4^-$) are replaced by species with more unpaired electrons ($Fe^{3+}$ and $Mn^{2+}$). The total number of unpaired electrons in the solution increases. Since paramagnetism is proportional to the number of unpaired electrons, the paramagnetism of the solution increases during the reaction.

Thus, statement (B) is correct.

(C) Dissolution of magnetite in sulfuric acid reduces the iron from +3 to +2.

Magnetite ($Fe_3O_4$) is a mixed oxide containing both $Fe^{2+}$ and $Fe^{3+}$ ions in a 1:2 ratio ($FeO \cdot Fe_2O_3$). When magnetite dissolves in a non-reducing acid like sulfuric acid, the ions are simply released into solution:

$Fe_3O_4(s) + 4H_2SO_4(aq) \rightarrow FeSO_4(aq) + Fe_2(SO_4)_3(aq) + 4H_2O(l)$

or in ionic form:

$Fe_3O_4(s) + 8H^+(aq) \rightarrow Fe^{2+}(aq) + 2Fe^{3+}(aq) + 4H_2O(l)$

Dissolution in sulfuric acid itself does not reduce the $Fe^{3+}$ ions present in magnetite to $Fe^{2+}$. The problem states that the ore was dissolved in acid, "thus reducing the iron to $Fe^{2+}$ completely". This implies that a reduction step was performed, but dissolution in sulfuric acid alone does not achieve this reduction. A separate reducing agent would be required to convert all $Fe^{3+}$ (both original and from dissolution) to $Fe^{2+}$.

Thus, statement (C) is incorrect.

(D) During the titration, the manganese containing species changes its geometry from square planar to tetrahedral.

The manganese containing species involved are $MnO_4^-$ and $Mn^{2+}$.

$MnO_4^-$ (permanganate ion): Mn is in +7 oxidation state. It is bonded to four oxygen atoms. The structure is tetrahedral (Tétrahedral geometry, like sulfate or phosphate ions).

$Mn^{2+}$ (manganese(II) ion): In aqueous solution, $Mn^{2+}$ ions are hydrated, forming complex ions like $[Mn(H_2O)_6]^{2+}$. These hydrated metal ions typically have octahedral geometry.

The statement claims the geometry changes from square planar to tetrahedral. This is incorrect. $MnO_4^-$ is tetrahedral, not square planar. $Mn^{2+}$ is octahedral (when hydrated), not tetrahedral.

Thus, statement (D) is incorrect.

Based on the analysis, statements (A) and (B) are correct.