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	18	17	45

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

Q78 jee_main_2024_27_jan_morning Chromyl Chloride Test

NaCl

reacts with conc.

H₂SO₄

and

K₂Cr₂O₇

to give reddish fumes (B), which react with

NaOH

to give yellow solution (C). (B) and (C) respectively are;

A. $CrO₂Cl₂$ $CrO_2Cl_2$, $Na₂CrO₄$ $Na_2CrO_4$
B. $Na₂CrO₄$ $Na_2CrO_4$, $CrO₂Cl₂$ $CrO_2Cl_2$
C. $CrO₂Cl₂$ $CrO_2Cl_2$, $KHSO₄$ $KHSO_4$
D. $CrO₂Cl₂$ $CrO_2Cl_2$, $Na₂Cr₂O₇$ $Na_2Cr_2O_7$

Solution

Step 1: Production of Reddish Fumes

4NaCl + K₂Cr₂O₇ + 6H₂SO₄ arrow 2CrO₂Cl₂ + 2KHSO₄ + 4NaHSO₄ + 3H₂O

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

Step 2: Conversion to Yellow Solution

CrO₂Cl₂ + 4NaOH arrow Na₂CrO₄ + 2NaCl + 2H₂O

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

Pattern Recognition

Chloride detection signature: $Cl^- arrow CrO₂Cl₂ (red-brown) arrow Na₂CrO₄ (yellow chromate).$

Chapter Mix

Class 12 Chemistry: d-and f-Block Elements

Q80 jee_main_2024_27_jan_morning Lanthanide Configuration

The electronic configuration for Neodymium is: [Atomic Number for Neodymium 60]

A. $[Xe] 4f⁴ 6s²$ $\text{[Xe]} 4f^4 6s^2$
B. $[Xe] 5f⁴ 7s²$ $\text{[Xe]} 5f^4 7s^2$
C. $[Xe] 4f⁶ 6s²$ $\text{[Xe]} 4f^6 6s^2$
D. $[Xe] 4f¹ 5d¹ 6s²$ $\text{[Xe]} 4f^1 5d^1 6s^2$

Solution

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

Q jee_main_2024_29_jan_morning Potassium Permanganate

KMnO₄

decomposes on heating at

513K

to form

O₂

along with

A. $MnO₂ & K₂O₂$
B. $K₂MnO₄ & Mn$
C. $Mn & KO₂$ $\text{Mn} \text{ \& } \text{KO}_2$
D. $K₂MnO₄ & MnO₂$

Solution

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:

2KMnO₄ Δ K₂MnO₄ + MnO₂ + O₂

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

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 7

Solution

Related Formula

Core Logic

Step 1: Selection Verification

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Step 1: Production of Reddish Fumes

Step 2: Conversion to Yellow Solution

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Identification

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