MnO42-, in acidic medium, disproportionates to : — jee_main_2026_21_jan

Q: MnO42-, in acidic medium, disproportionates to :

Manganate ion (MnO42-), where Mn is in +6 oxidation state, is unstable in acidic medium and undergoes disproportionation. It oxidizes to Permanganate (MnO4^-, +7 state) and reduces to Manganese dioxide (MnO2, +4 state). Manganate (green, +6) disproportionates in acid to Permanganate (purple, +7) and MnO2 (brown/black precipitate, +4). Class 12 Chemistry: d and f Block Elements

Year	2026	2025	2024	Total
Questions	10	18	17	45

More d and f Block Elements Previous-Year Questions

Q75 jee_main_2026_21_jan_morning Chromyl Chloride Test and Chromate Chemistry

Consider the following reactions:

NaCl + K₂Cr₂O₇ + H₂SO₄ arrow A + KHSO₄ + NaHSO₄ + H₂O

A + NaOH arrow B + NaCl + H₂O

B + H₂SO₄ + H₂O₂ arrow C + Na₂SO₄ + H₂O

In the product 'C', 'X' is the number of

O₂²⁻

units, 'Y' is the total number oxygen atoms present and 'Z' is the oxidation state of Cr. The value of

X + Y + Z

is ____.

Numerical Answer. Answer: 13 to 13

Solution

Core Logic

The first reaction is the classical Chromyl Chloride Test: $4NaCl + K₂Cr₂O₇ + 6H₂SO₄ arrow 2CrO₂Cl₂ (A) + 2KHSO₄ + 4NaHSO₄ + 3H₂O$ Product A is Chromyl chloride ( $CrO₂Cl₂$ $\mathrm{CrO_2Cl_2}$), a red-orange gas.

When Chromyl chloride gas is passed into NaOH solution, it forms a yellow solution of sodium chromate (B): $CrO₂Cl₂ (A) + 4NaOH arrow Na₂CrO₄ (B) + 2NaCl + 2H₂O$

Acidifying the sodium chromate solution with $H₂SO₄$ $H_2SO_4$ and adding $H₂O₂$ $H_2O_2$ yields a deep blue solution of Chromium(VI) peroxide, $CrO₅$ $CrO_5$ (C): $Na₂CrO₄ (B) + H₂SO₄ + 2H₂O₂ arrow CrO₅ (C) + Na₂SO₄ + 3H₂O$

Structure of $CrO₅$ $CrO_5$:

Chromyl Chloride Test diagram for Q75 - JEE Main 2026 Morning

It has a butterfly structure.
Number of peroxy units ( $O₂²⁻$ $O_2^{2-}$), $X = 2$ $X = 2$.
Total number of oxygen atoms, $Y = 5$ $Y = 5$.
Oxidation state of Cr, $Z = +6$ $Z = +6$.

Sum: $X + Y + Z = 2 + 5 + 6 = 13$ $X + Y + Z = 2 + 5 + 6 = 13$.

Step 1: Final Calculation

$X + Y + Z = 13$ $X + Y + Z = 13$

Pattern Recognition

Chromyl chloride test $arrow$ $\rightarrow$ $CrO₂Cl₂$ $CrO_2Cl_2$ (red gas). Absorbed in NaOH $arrow$ $\rightarrow$ $Na₂CrO₄$ $Na_2CrO_4$ (yellow). Tested with $H₂O₂/H^+$ $H_2O_2/H^+$ $arrow$ $\rightarrow$ $CrO₅$ $CrO_5$ (butterfly structure, blue, two peroxy links, Cr in +6).

Chapter Mix

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

Q66 jee_main_2026_21_jan_evening Oxides of Manganese and Properties

Given below are some of the statements about

Mn

and

Mn₂O₇

. Identify the correct statements: A. Mn forms the oxide

Mn₂O₇

in which Mn is in its highest oxidation state. B. Oxygen stabilizes the Mn in higher oxidation states by forming multiple bonds with Mn. C.

Mn₂O₇

is an ionic oxide. D. The structure of

Mn₂O₇

consists of one bridged oxygen. Choose the correct answer from the options given below:

A. $(1) A, B, C and D$ $(1) \text{ A, B, C and D}$
B. $(2) A, B and D Only$ $(2) \text{ A, B and D Only}$
C. $(3) A, C and D Only$ $(3) \text{ A, C and D Only}$
D. $(4) A, B and C Only$ $(4) \text{ A, B and C Only}$

Solution

Core Logic

A is correct: $Mn₂O₇$ $\text{Mn}_2\text{O}_7$ features Mn in +7 state (its highest oxidation state).
B is correct: Oxygen stabilizes high oxidation states via multiple bonding.
C is incorrect: $Mn₂O₇$ $\text{Mn}_2\text{O}_7$ is a covalent green oil/oxide, not ionic.
D is correct: Structure consists of two $MnO₄$ $\text{MnO}_4$ tetrahedra sharing one bridging oxygen atom ( $O₃Mn-O-MnO₃$ ).

Step 1: Final Conclusion

Statements A, B and D are correct, matching option (2).

Pattern Recognition

Sees: Properties and bonding of transition metal oxides like $Mn₂O₇$ $\text{Mn}_2\text{O}_7$. Trap: Assuming high oxidation state oxides of transition metals are ionic.

Chapter Mix

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

Q58 jee_main_2026_22_january_morning Reactions of Transition Metals

A first row transition metal (M) does not liberate

H₂

gas from dilute HCl. 1 mol of aqueous solution of

MSO₄

is treated with excess of aqueous KCN and then

H₂S(g)

is passed through the solution. The amount of MS (metal sulphide) formed from the above reaction is ____ mol.

A. $2$ $\text{2}$
B. $1$ $\text{1}$
C. $3$ $\text{3}$
D. $0$ $\text{0}$

Solution

Related Formula

Cu²⁺ + 4CN⁻ arrow [Cu(CN)₄]³⁻ (after redox with CN^-)

Core Logic

The first-row transition metal that does not liberate $H₂$ $H_{2}$ gas from dilute HCl is Copper (Cu), because its standard reduction potential is positive ( $E°Cu²⁺/Cu = +0.34 V$ ).

When $CuSO₄$ $CuSO_{4}$ is treated with excess KCN, it forms a very stable soluble cyano complex:

CuSO₄ + 2KCN arrow Cu(CN)₂ + K₂SO₄

$2Cu(CN)₂ arrow 2CuCN + (CN)₂$ $CuCN + 3KCN arrow K₃[Cu(CN)₄]$

The complex ion $[Cu(CN)₄]³⁻$ $[Cu(CN)_{4}]^{3-}$ is highly stable (a perfect complex). When $H₂S$ $H_{2}S$ is passed through this solution, it does not yield sufficient $Cu⁺$ $Cu^{+}$ ions to exceed the solubility product ( $Kₛₚ$ $K_{sp}$) of $Cu₂S$ $Cu_{2}S$.

Step 1: Final Conclusion

Since no copper sulphide precipitates, the amount of MS formed is 0 moles.

Pattern Recognition

Cu and Cd separation: $Cu²⁺$ $Cu^{2+}$ forms a very stable cyanide complex that does not precipitate with $H₂S$ $H_2S$, whereas $Cd²⁺$ $Cd^{2+}$ forms a less stable complex that does precipitate as $CdS$ $CdS$.

Chapter Mix

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

Q54 jee_main_2026_22_january_evening Ionization Enthalpy Trends in Transition Metals

Given below are two statements: Statement-I: The first ionization enthalpy of Cr is lower than that of Mn. Statement-II: The second and third ionization enthalpies of Cr are higher than those of Mn. In the light of the above statements, choose the correct answer from the options given below:

A. Both Statement-I and Statement-II are false.
B. Statement-I is true but Statement-II is false.
C. Both Statement-I and Statement-II are true.
D. Statement-I is false but Statement-II is true.

Solution

Related Formula

Electronic Configurations: Cr = [Ar]3d⁵ 4s¹, Mn = [Ar]3d⁵ 4s²

Core Logic

Step 1: Compare $IE₁$ $IE_1$:

Cr: 4s¹ Removal of single 4s electron requires less energy than removing 4s² in Mn.

Hence, $IE₁(Cr) < IE₁(Mn)$ $IE_1(\text{Cr}) < IE_1(\text{Mn})$ (Statement-I is TRUE).

Step 2: Compare $IE₂$ $IE_2$ and $IE₃$ $IE_3$:

Cr^+ = 3d⁵ stable half-filled configuration d⁵, so IE₂(Cr) > IE₂(Mn)

For IE₃, Mn²⁺ = 3d⁵ removing electron from stable 3d⁵ in Mn²⁺ requires more energy than Cr²⁺ (3d⁴).

Hence, $IE₃(Cr) < IE₃(Mn)$ $IE_3(\text{Cr}) < IE_3(\text{Mn})$. Thus Statement-II is FALSE.

Pattern Recognition

Sees: Ionization enthalpy comparison of Cr and Mn. Shortcut: Stable $3d⁵$ $3d^5$ configuration in $Cr^+$ $\text{Cr}^+$ makes $IE₂$ $IE_2$ very high, whereas $3d⁵$ $3d^5$ in $Mn²⁺$ $\text{Mn}^{2+}$ makes $IE₃$ $IE_3$ of Mn higher than Cr.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements Class 11 Chemistry: Classification of Elements and Periodicity in Properties

MnO₄²⁻$MnO_{4}^{2-}$, in acidic medium, disproportionates to :

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More d and f Block Elements Previous-Year Questions

Solution

Core Logic

Step 1: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More d and f Block Elements Questions — jee_main_2026_21_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

$MnO₄²⁻$ $MnO_{4}^{2-}$, in acidic medium, disproportionates to :