Among the following oxides of 3d elements, the number of mixed oxides are ____. Ti₂O₃, V₂O₄, Cr₂O₃, Mn₃O₄, Fe₃O₄, Fe₂O₃, Co₃O₄$$Ti_2O_3, V_2O_4, Cr_2O_3, Mn_3O_4, Fe_3O_4, Fe_2O_3, Co_3O_4$$

Related Formula

Core Logic

Acidified potassium dichromate (K₂Cr₂O₇$K_2Cr_2O_7$) acts as a strong oxidizing agent. In acidic medium, the dichromate ion (Cr₂O₇²⁻$Cr_2O_7^{2-}$, where Cr is in the +6$+6$ oxidation state) is reduced to the Chromium(III) ion (Cr³⁺$Cr^{3+}$).

Concurrently, iodide ions (I^-$I^-$ from KI) are oxidized to elemental iodine (I₂$I_2$). The full ionic equation is:

The final product containing chromium is the Cr³⁺$Cr^{3+}$ ion, meaning its oxidation state is +3$+3$.

Pattern Recognition

Always remember that in acidic media, Cr₂O₇²⁻$Cr_2O_7^{2-}$ (orange) universally reduces to Cr³⁺$Cr^{3+}$ (green). The oxidation state invariably goes from +6 arrow +3$+6 \rightarrow +3$.

Chapter Mix

Class 12 Chemistry: d and f Block Elements Class 11 Chemistry: Redox Reactions

Core Logic

The lightest element of Group 7 in the periodic table is Manganese (Mn$Mn$). The oxoanion of Mn$Mn$ where it resides in the +6 oxidation state is the manganate ion (MnO₄²⁻$MnO_4^{2-}$). The potassium salt of this oxoanion is potassium manganate (K₂MnO₄$K_2MnO_4$).

Step 1: Deduce Color

K₂MnO₄$K_2MnO_4$ (containing the MnO₄²⁻$MnO_4^{2-}$ ion) is known to have a green color. (Contrast with the +7 state in KMnO₄$KMnO_4$, which is intensely purple).

Pattern Recognition

Manganese oxoanions have signature colors: MnO₄⁻$MnO_4^{-}$ (Permanganate, +7) is Purple/Pink, while MnO₄²⁻$MnO_4^{2-}$ (Manganate, +6) is Green.

Chapter Mix

Class 12 Chemistry: The d- and f-Block Elements

Step 1: Evaluate Statement I

Color in transition metal ions arises from d-d transitions, which require unpaired d-electrons (d¹$d^1$ to d⁹$d^9$).\n$\bullet$ [Sc³⁺, Ti³⁺]$[Sc^{3+}, Ti^{3+}]$: Sc³⁺$Sc^{3+}$ is 3d⁰$3d^0$ (Colourless). Pair invalid.\n$\bullet$ [Mn²⁺, Cr²⁺]$[Mn^{2+}, Cr^{2+}]$: Mn²⁺$Mn^{2+}$ is 3d⁵$3d^5$ (Coloured), Cr²⁺$Cr^{2+}$ is 3d⁴$3d^4$ (Coloured). Pair valid.\n$\bullet$ [Cu²⁺, Zn²⁺]$[Cu^{2+}, Zn^{2+}]$: Zn²⁺$Zn^{2+}$ is 3d¹⁰$3d^{10}$ (Colourless). Pair invalid.\n$\bullet$ [Ni²⁺, Ti⁴⁺]$[Ni^{2+}, Ti^{4+}]$: Ti⁴⁺$Ti^{4+}$ is 3d⁰$3d^0$ (Colourless). Pair invalid.\nOnly ONE pair contains both colored ions. Statement I is false.

Step 2: Evaluate Statement II

Thorium (Th$Th$) exhibits a stable +4$+4$ oxidation state. Th⁴⁺$Th^{4+}$ has an empty shell (5f⁰ 6d⁰ 7s⁰$5f^0 6d^0 7s^0$) and cannot lose more electrons to act as a reducing agent (which requires getting oxidized further). Statement II is false.

Final Conclusion

Both Statement I and Statement II are false.

Pattern Recognition

d⁰$d^0$ and d¹⁰$d^{10}$ configurations never absorb visible light for d-d transitions, rendering them strictly colourless. Maximum group oxidation states cannot act as reducing agents.

Chapter Mix

Class 12 Chemistry: The d- and f-Block Elements

Core Logic

(A) Both MnO₄²⁻$MnO_{4}^{2-}$ (Manganate) and MnO₄⁻$MnO_{4}^{-}$ (Permanganate) have tetrahedral geometry utilizing d³s$d^3s$ hybridization. (Correct)

(B) The oxidation state of Mn in manganate (MnO₄²⁻$MnO_{4}^{2-}$) is +6 and in permanganate (MnO₄⁻$MnO_{4}^{-}$) is +7. The statement swaps these. (Incorrect)

(C) Mn²⁺ + S₂O₈²⁻ arrow MnO₄⁻$Mn^{2+} + S_{2}O_{8}^{2-} \rightarrow MnO_{4}^{-}$ (Permanganate ion), not manganate. (Incorrect)

(D) MnO₄⁻$MnO_{4}^{-}$ (Mn in +7, d⁰$d^0$) is diamagnetic. MnO₄²⁻$MnO_{4}^{2-}$ (Mn in +6, d¹$d^1$) is paramagnetic. (Correct)

(E) Acidified permanganate ion is an oxidizing agent, meaning it OXIDIZES oxalate, nitrite, and iodide ions; it does not reduce them. (Incorrect)

Step 1: Final Conclusion

Statements A and D are correct.

Pattern Recognition

Recall MnO₄^-$MnO_4^-$ is purple, diamagnetic, +7 state, powerful oxidizing agent. MnO₄²⁻$MnO_4^{2-}$ is green, paramagnetic, +6 state. Oxidizing agent means it reduces itself, thus oxidizes other substrates.

Chapter Mix

Class 12 Chemistry: The d- and f-Block Elements