Which of the following oxidation reactions are carried out by both K₂Cr₂O₇$\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ and KMnO₄$\mathrm{KMnO}_4$ in acidic medium? A. I^- arrow I₂$\mathrm{I}^- \rightarrow \mathrm{I}_2$ B. S²⁻ arrow S$\mathrm{S}^{2-} \rightarrow \mathrm{S}$ C. Fe²⁺ arrow Fe³⁺$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}$ D. I⁻ arrow IO₃⁻$\mathrm{I}^{-} \rightarrow \mathrm{IO}_{3}^{-}$ E. S₂O₃²⁻ arrow SO₄²⁻$\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow \mathrm{SO}_4^{2-}$ Choose the correct answer from the options given below:

Related Formula

Dissolution reaction pathway:

Core Logic

The reaction yields sodium tetrahydroxoplumbate(II), [Pb(OH)₄]²⁻$[\text{Pb(OH)}_4]^{2-}$. The charge of the complex species is -2$-2$ (dianionic), and it binds 4 hydroxo coordination ligands, matching a coordination number of four.

Chapter Mix

Class 12 Chemistry: d-and f-Block Elements Class 12 Chemistry: Coordination Compounds

Step 1: Production of Reddish Fumes

Reddish brown vapors (B) are chromyl chloride (CrO₂Cl₂$CrO_2Cl_2$).

Step 2: Conversion to Yellow Solution

Yellow solution (C) corresponds to sodium chromate (Na₂CrO₄$Na_2CrO_4$).

Pattern Recognition

Chloride detection signature: Cl^- arrow CrO₂Cl₂ (red-brown) arrow Na₂CrO₄ (yellow chromate).$\text{Cl}^- \rightarrow \text{CrO}_2\text{Cl}_2\text{ (red-brown)} \rightarrow \text{Na}_2\text{CrO}_4\text{ (yellow chromate)}.$

Chapter Mix

Class 12 Chemistry: d-and f-Block Elements

Core Logic

The noble gas configuration of Xenon (Z=54$Z=54$) provides the primary core layout. For Neodymium (Z=60$Z=60$), the 6 remaining valence electrons distribute into the inner 4f$4\text{f}$ orbital subshell rather than filling the 5d$5\text{d}$ subshell due to shielding effects. This results in an absolute atomic ground state electronic configuration of [Xe] 4f⁴ 6s²$\text{[Xe]} 4\text{f}^4 6\text{s}^2$.

Pattern Recognition

Lanthanide filling sequences generally bypass 5d$5d$ progression except for specific exceptions (La, Gd, Lu).

Chapter Mix

Class 12 Chemistry: d-and f-Block Elements

Core Logic

Potassium permanganate (KMnO₄$KMnO_4$) is a strong oxidizing agent. When heated to 513K$513\mathrm{K}$, it undergoes thermal decomposition to give potassium manganate (K₂MnO₄$K_2MnO_4$), manganese dioxide (MnO₂$MnO_2$), and oxygen gas (O₂$O_2$).

The balanced chemical equation is:

Step 1: Final Identification

The products formed along with O₂$O_2$ are K₂MnO₄$K_2MnO_4$ (green) and MnO₂$MnO_2$ (black).

Chapter Mix

Class 12 Chemistry: d and f Block Elements

Core Logic

The reaction sequence for the chromyl chloride test is:

The chromyl chloride gas is then passed through a basic medium (like NaOH$NaOH$) to form a yellow solution of chromate ions:

Acidification of the yellow CrO₄²⁻$CrO_4^{2-}$ solution followed by the addition of H₂O₂$H_2O_2$ and amyl alcohol yields a blue-colored organic layer due to the formation of chromium pentoxide (CrO₅$CrO_5$).

CrO₄²⁻ [yellow solution, 1. Acidification CrO₅ (blue compound)$$CrO_4^{2-} \xrightarrow[\text{yellow solution}, {1. \text{Acidification}} CrO_5 \text{ (blue compound)}$$

Step 1: Oxidation State Calculation

The structure of chromium pentoxide (CrO₅$CrO_5$) features a distinctive "butterfly" arrangement. It contains one double-bonded oxide oxygen (O²⁻$O^{2-}$) and four peroxide oxygens (O₂²⁻$O_2^{2-}$). Therefore, there are 2 peroxo linkages.

Let the oxidation state of Chromium be x$x$.

x - 2 - 4 = 0$x - 2 - 4 = 0$ x = +6$x = +6$

Thus, the oxidation state of Cr in CrO₅$CrO_5$ is +6$+6$.

Pattern Recognition

A classic oxidation state trap. Calculating simply via formula CrO₅$CrO_5$ yields x - 10 = 0 x = +10$x - 10 = 0 \implies x = +10$, which is impossible for Chromium (max +6). Whenever calculation exceeds the maximum group valency, peroxide bonds are present.

Chapter Mix

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