Preparation of potassium permanganate from MnO2 involves two step process in which the 1st step is a reaction with KOH…

Q: Preparation of potassium permanganate from MnO2 involves two step process in which the 1st step is a reaction with KOH and KNO3 to produce

The standard preparation of potassium permanganate begins with the oxidative fusion of pyrolusite ore (MnO2). Fusing the solid reactant directly along an alkaline base payload (KOH) combined explicitly with an oxidizing carrier (KNO3) yields the intermediate green product, **potassium manganate** (K2MnO4). Step 1 yields the +6 green compound (K2MnO4); the subsequent Step 2 steps oxidize this intermediate to synthesize the target deep purple +7 agent (KMnO4). Class 12 Chemistry: The d-and f-Block

Year	2026	2025	2024	Total
Questions	10	16	17	43

Q jee_main_2025_29_jan_morning Preparation and Properties of Potassium Dichromate

The molar mass of the water insoluble product formed from the fusion of chromite ore

(FeCr₂O₄)

with

Na₂CO₃

in presence of

O₂

is ________

g mol⁻¹

.

Numerical Answer. Answer: 160 to 160

Solution

Related Formula

Balanced fusion reaction process description

Core Logic

Write the balanced chemical equation for the industrial preparation stage of chromate salts:

4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ arrow 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

Evaluating the solubilities of the products:

$Na₂CrO₄$ $\mathrm{Na_2CrO_4}$ is highly soluble in water.
$Fe₂O₃$ $\mathrm{Fe_2O_3}$ (Iron(III) oxide) is water-insoluble.

Molar Mass of $Fe₂O₃$ $\mathrm{Fe_2O_3}$:

M = (2 \cdot 55.85) + (3 \cdot 16.0) (2 \cdot 56) + (3 \cdot 16) = 112 + 48 = 160 ~g/mol

Pattern Recognition

Transition metal oxides in high oxidation states with minimal ionic breakdown parameters reliably act as insoluble precipitates in water.

Chapter Mix

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

Q jee_main_2024_01_february_morning Oxidising Properties

In acidic medium,

K₂Cr₂O₇

shows oxidising action as represented in the half reaction

Cr₂O₇²⁻ + XH^+ + Ye^- arrow 2A + ZH₂O

X, Y, Z and A are respectively are:

A. $8, 6, 4 and Cr₂O₃$ $8, 6, 4 \text{ and } Cr_2O_3$
B. $14, 7, 6 and Cr³⁺$ $14, 7, 6 \text{ and } Cr^{3+}$
C. $8, 4, 6 and Cr₂O₃$ $8, 4, 6 \text{ and } Cr_2O_3$
D. $14, 6, 7 and Cr³⁺$ $14, 6, 7 \text{ and } Cr^{3+}$

Solution

Core Logic

The balanced half-reaction for the dichromate ion acting as an oxidising agent in an acidic medium is:

Cr₂O₇²⁻ + 14H^+ + 6e^- arrow 2Cr³⁺ + 7H₂O

Step 1: Compare with Given Equation

Comparing this with the given equation $Cr₂O₇²⁻ + XH^+ + Ye^- arrow 2A + ZH₂O$ :

$X = 14$ $X = 14$ $Y = 6$ $Y = 6$ $Z = 7$ $Z = 7$ $A = Cr³⁺$ $A = Cr^{3+}$

Pattern Recognition

In acidic medium, dichromate ( $Cr₂O₇²⁻$ $Cr_2O_7^{2-}$) always requires $14H^+$ $14H^+$ to balance $7O$ $7O$ atoms, forming $7H₂O$ $7H_2O$. Chromium reduces from +6 to +3 state, taking $6e^-$ $6e^-$ overall.

Chapter Mix

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

Q73 jee_main_2024_29_january_evening Lanthanoid Oxidation States

Which of the following acts as a strong reducing agent? (Atomic number : Ce = 58, Eu = 63, Gd = 64, Lu = 71)

A. $Lu³⁺$ $\mathrm{Lu}^{3+}$
B. $Gd³⁺$ $\mathrm{Gd}^{3+}$
C. $Eu²⁺$ $\mathrm{Eu}^{2+}$
D. $Ce⁴⁺$ $\mathrm{Ce}^{4+}$

Solution

Related Formula

Electronic configuration of Eu = [Xe] 4f⁷ 6s²

Core Logic

The most common and stable oxidation state for lanthanoids is $+3$ $+3$. In the case of Europium:

Eu²⁺ = [Xe] 4f⁷

This configuration possesses a highly stable half-filled $f$ $f$-subshell. However, because the $+3$ $+3$ state is universally favored by thermodynamics in solution, $Eu²⁺$ $\text{Eu}^{2+}$ readily undergoes oxidation to lose one more electron:

Eu²⁺ arrow Eu³⁺ + 1e^-

By releasing an electron to stabilize into the $+3$ $+3$ state, it behaves as a potent reducing agent.

Step 1: Evaluation

Conversely, $Ce⁴⁺$ $\text{Ce}^{4+}$ acts as a powerful oxidizing agent to return to $+3$ $+3$, while $Lu³⁺$ $\text{Lu}^{3+}$ and $Gd³⁺$ $\text{Gd}^{3+}$ are already perfectly configured at their native stable limits.

Pattern Recognition

Europium( $II$ $II$) has a stable half-filled $f⁷$ $f^7$ configuration, yet easily loses an electron to attain the highly stable $+3$ $+3$ state typical of lanthanoids, making it a strong reducing agent.

Chapter Mix

Class 12 Chemistry: d and f Block Elements

Q75 jee_main_2024_29_january_evening Properties of Zinc, Cadmium and Mercury

Which of the following statements are correct about Zn, Cd and

Hg

? A. They exhibit high enthalpy of atomization as the d-subshell is full. B. Zn and Cd do not show variable oxidation state while Hg shows

+I

and

+II

. C. Compounds of Zn, Cd and Hg are paramagnetic in nature. D. Zn, Cd and Hg are called soft metals. Choose the most appropriate from the options given below:

A. B, D only
B. B, C only
C. A, D only
D. C, D only

Solution

Related Formula

General configuration of Group 12: (n-1)d¹⁰ ns²

Core Logic

Analyzing each statement based on inorganic chemistry principles:

Statement A is false: Because their $d$ $d$-subshell is completely full ( $d¹⁰$ $d^{10}$), these elements do not form strong metallic bonds. As a result, they exhibit the lowest enthalpy of atomization in their respective periods.
Statement B is true: $Zn$ $\text{Zn}$ and $Cd$ $\text{Cd}$ show only a stable $+2$ $+2$ oxidation state, whereas $Hg$ $\text{Hg}$ exhibits variable states forming both $+1$ $+1$ (as $Hg₂²⁺$ $\text{Hg}_2^{2+}$) and $+2$ $+2$.
Statement C is false: With a fully paired $d¹⁰$ $d^{10}$ subshell, their compounds lack unpaired electrons and are explicitly diamagnetic.
Statement D is true: Due to weak metallic bonds, these elements have low melting points and are classified as soft metals.

Step 1: Selection Verification

Statements B and D are true, matching choice (1).

Pattern Recognition

Group 12 metals have a full $d¹⁰$ $d^{10}$ subshell, leading to exceptionally weak metallic bonding, low enthalpies of atomization, and diamagnetic characteristics.

Chapter Mix

Class 12 Chemistry: d and f Block Elements

Q75 jee_main_2024_27_jan_morning Qualitative Analysis of Lead

Yellow compound of lead chromate gets dissolved on treatment with hot

NaOH

solution. The product of lead formed is a :

A. Tetraanionic complex with coordination number six
B. Neutral complex with coordination number four
C. Dianionic complex with coordination number six
D. Dianionic complex with coordination number four

Solution

Related Formula

Dissolution reaction pathway:

PbCrO₄ + 4NaOH (hot excess) arrow Na₂[Pb(OH)₄] + Na₂CrO₄

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

Preparation of potassium permanganate from MnO₂$\mathrm{MnO}_2$ involves two step process in which the 1st$1^{\text{st}}$ step is a reaction with KOH and KNO₃$\mathrm{KNO}_3$ to produce

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More The d-and f-Block Elements Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Compare with Given Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Selection Verification

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

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Preparation of potassium permanganate from $MnO₂$ $\mathrm{MnO}_2$ involves two step process in which the $1st$ $1^{\text{st}}$ step is a reaction with KOH and $KNO₃$ $\mathrm{KNO}_3$ to produce