In chromyl chloride test for confirmation of Cl^- ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and 10% H2O2 turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is

The reaction sequence for the chromyl chloride test is: The chromyl chloride gas is then passed through a basic medium (like NaOH) to form a yellow solution of chromate ions: Acidification of the yellow CrO42- solution followed by the addition of H2O2 and amyl alcohol yields a blue-colored organic layer due to the formation of chromium pentoxide (CrO5). The structure of chromium pentoxide (CrO5) features a distinctive "butterfly" arrangement. It contains one double-bonded oxide oxygen (O2-) and four peroxide oxygens (O22-). Therefore, there are 2 peroxo linkages. Let the oxidation state of Chromium be x. x - 2 - 4 = 0 x = +6 Thus, the oxidation state of Cr in CrO5 is +6. A classic oxidation state trap. Calculating simply via formula CrO5 yields x - 10 = 0 x = +10, which is impossible for Chromium (max +6). Whenever calculation exceeds the maximum group valency, peroxide bonds are present. Class 12 Chemistry: d and f Block Elements Class 11 Chemistry: Redox Reactions

In chromyl chloride test for confirmation of Cl^- ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and 10% H2O2 turns organic layer blue indicating format

The reaction sequence for the chromyl chloride test is: The chromyl chloride gas is then passed through a basic medium (like NaOH) to form a yellow solution of chromate ions: Acidification of the yellow CrO42- solution followed by the addition of H2O2 and amyl alcohol yields a blue-colored organic layer due to the formation of chromium pentoxide (CrO5). The structure of chromium pentoxide (CrO5) features a distinctive "butterfly" arrangement. It contains one double-bonded oxide oxygen (O2-) and

In chromyl chloride test for confirmation of Cl^- ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and 10% H2O2 turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is

The reaction sequence for the chromyl chloride test is: The chromyl chloride gas is then passed through a basic medium (like NaOH) to form a yellow solution of chromate ions: Acidification of the yellow CrO42- solution followed by the addition of H2O2 and amyl alcohol yields a blue-colored organic layer due to the formation of chromium pentoxide (CrO5). The structure of chromium pentoxide (CrO5) features a distinctive "butterfly" arrangement. It contains one double-bonded oxide oxygen (O2-) and four peroxide oxygens (O22-). Therefore, there are 2 peroxo linkages. Let the oxidation state of Chromium be x. x - 2 - 4 = 0 x = +6 Thus, the oxidation state of Cr in CrO5 is +6. A classic oxidation state trap. Calculating simply via formula CrO5 yields x - 10 = 0 x = +10, which is impossible for Chromium (max +6). Whenever calculation exceeds the maximum group valency, peroxide bonds are present. Class 12 Chemistry: d and f Block Elements Class 11 Chemistry: Redox Reactions

In chromyl chloride test for confirmation of Cl^- ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and 10% H2O2 turns organic layer blue indicating format

The reaction sequence for the chromyl chloride test is: The chromyl chloride gas is then passed through a basic medium (like NaOH) to form a yellow solution of chromate ions: Acidification of the yellow CrO42- solution followed by the addition of H2O2 and amyl alcohol yields a blue-colored organic layer due to the formation of chromium pentoxide (CrO5). The structure of chromium pentoxide (CrO5) features a distinctive "butterfly" arrangement. It contains one double-bonded oxide oxygen (O2-) and

In chromyl chloride test for confirmation of Cl^- ion, a yellow solution is obtained. Acidification of the solution and…

Year	2026	2025	2024	Total
Questions	10	18	17	45

More d and f Block Elements Previous-Year Questions — Page 9

Q66 jee_main_2024_30_jan_morning Lanthanoids

Diamagnetic Lanthanoid ions are:

A. $Nd³⁺ and Eu³⁺$ $Nd^{3+}\text{ and }Eu^{3+}$
B. $La³⁺ and Ce⁴⁺$ $La^{3+}\text{ and }Ce^{4+}$
C. $Nd³⁺ and Ce⁴⁺$ $Nd^{3+}\text{ and }Ce^{4+}$
D. $Lu³⁺ and Eu³⁺$ $Lu^{3+}\text{ and }Eu^{3+}$

Solution

Core Logic

An ion is diamagnetic if all its electrons are paired (i.e., zero unpaired electrons). Let's write the electronic configuration for the elements in question.

Step 1: Checking configurations

Cerium ( $Ce$ $Ce$, Z=58): $[Xe] 4f¹ 5d¹ 6s²$ $[Xe] 4f^1 5d^1 6s^2$ $arrow Ce⁴⁺: [Xe] 4f⁰$ $\rightarrow Ce^{4+}: [Xe] 4f^0$ (0 unpaired electrons $arrow$ $\rightarrow$ Diamagnetic)

Lanthanum ( $La$ $La$, Z=57): $[Xe] 4f⁰ 5d¹ 6s²$ $[Xe] 4f^0 5d^1 6s^2$ $arrow La³⁺: [Xe] 4f⁰$ $\rightarrow La^{3+}: [Xe] 4f^0$ (0 unpaired electrons $arrow$ $\rightarrow$ Diamagnetic)

Pattern Recognition

Ions with an empty f-subshell ( $f⁰$ $f^0$, e.g., $La³⁺, Ce⁴⁺$ $La^{3+}, Ce^{4+}$) or a completely filled f-subshell ( $f¹⁴$ $f^{14}$, e.g., $Lu³⁺, Yb²⁺$ $Lu^{3+}, Yb^{2+}$) are invariably diamagnetic.

Chapter Mix

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

Q76 jee_main_2024_30_jan_morning Transition Elements

Match List-I with List-II.

List-I (Species)	List-II (Electronic distribution)
(A) $Cr⁺²$ $Cr^{+2}$	(I) $3d⁸$ $3d^8$
(B) $Mn^+$ $Mn^+$	(II) $3d⁵4s¹$ $3d^54s^1$
(C) $Ni⁺²$ $Ni^{+2}$	(III) $3d⁴$ $3d^4$
(D) $V^+$ $V^+$	(IV) $3d³4s¹$ $3d^34s^1$

Choose the correct answer from the options given below:

A. $(A)-(I), (B)-(II), (C)-(III), (D)-(IV)$
B. $(A)-(III), (B)-(II), (C)-(I), (D)-(IV)$
C. $(A)-(IV), (B)-(III), (C)-(I), (D)-(II)$
D. $(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$

Solution

Core Logic

Let's determine the electronic configuration for each species by first writing the neutral atom's configuration, and then removing electrons starting from the outermost $4s$ $4s$ orbital.

(A) $Cr$ $Cr$ (Z=24): $[Ar] 3d⁵ 4s¹$ $[Ar] 3d^5 4s^1$ $arrow$ $\rightarrow$ $Cr²⁺$ $Cr^{2+}$: $[Ar] 3d⁴$ $[Ar] 3d^4$ (B) $Mn$ $Mn$ (Z=25): $[Ar] 3d⁵ 4s²$ $[Ar] 3d^5 4s^2$ $arrow$ $\rightarrow$ $Mn^+$ $Mn^+$: $[Ar] 3d⁵ 4s¹$ $[Ar] 3d^5 4s^1$ (C) $Ni$ $Ni$ (Z=28): $[Ar] 3d⁸ 4s²$ $[Ar] 3d^8 4s^2$ $arrow$ $\rightarrow$ $Ni²⁺$ $Ni^{2+}$: $[Ar] 3d⁸$ $[Ar] 3d^8$ (D) $V$ $V$ (Z=23): $[Ar] 3d³ 4s²$ $[Ar] 3d^3 4s^2$ $arrow$ $\rightarrow$ $V^+$ $V^+$: $[Ar] 3d³ 4s¹$ $[Ar] 3d^3 4s^1$

Step 1: Match execution

A $arrow$ $\rightarrow$ III B $arrow$ $\rightarrow$ II C $arrow$ $\rightarrow$ I D $arrow$ $\rightarrow$ IV

Chapter Mix

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

Q jee_main_2024_31_jan_evening Properties of Transition Metal Oxides

Choose the correct statements from the following A.

Mn₂O₇

is an oil at room temperature B.

V₂O₄

reacts with acid to give

VO₂²⁺

CrO

is a basic oxide D.

V₂O₅

does not react with acid Choose the correct answer from the options given below:

A. $A, B and D only$ $\text{A, B and D only}$
B. $A and C only$ $\text{A and C only}$
C. $A, B and C only$ $\text{A, B and C only}$
D. $B and C only$ $\text{B and C only}$

Solution

Core Logic

(A) $Mn₂O₇$ $Mn_2O_7$ is a covalent oxide and exists as a green oil at room temperature. (Correct) (B) $V₂O₄$ $V_2O_4$ dissolves in acids to give $VO²⁺$ $VO^{2+}$ (vanadyl) salts, not $VO₂²⁺$ $VO_2^{2+}$. (Incorrect) (C) $CrO$ $CrO$ has chromium in the $+2$ $+2$ oxidation state. Lower oxidation state metal oxides are typically basic in nature. (Correct) (D) $V₂O₅$ $V_2O_5$ is an amphoteric oxide; it reacts with both acids as well as bases. (Incorrect)

Step 1: Final Selection

Only statements A and C are correct, which corresponds to option (2).

Chapter Mix

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

Q86 jee_main_2024_31_jan_evening Chromyl Chloride Test

In the reaction of potassium dichromate, potassium chloride and sulfuric acid (conc.), the oxidation state of the chromium in the product is (+) ________

Numerical Answer. Answer: 6 to 6

Solution

Related Formula

K₂Cr₂O₇(s) + 4KCl(s) + 6H₂SO₄(conc.) arrow 2CrO₂Cl₂(g) + 6KHSO₄ + 3H₂O

Core Logic

This reaction represents the Chromyl Chloride test used to detect the presence of chloride ions. When potassium dichromate is heated with a metal chloride in concentrated sulfuric acid, red vapors of chromyl chloride ( $CrO₂Cl₂$ $CrO_2Cl_2$) are evolved.

Step 1: Oxidation State Calculation

In chromyl chloride ( $CrO₂Cl₂$ $CrO_2Cl_2$): Let the oxidation state of Chromium be $x$ $x$. Oxygen is typically $-2$ $-2$ and Chlorine is $-1$ $-1$.

x + 2(-2) + 2(-1) = 0

$x - 4 - 2 = 0$ $x - 4 - 2 = 0$ $x = +6$ $x = +6$ Thus, the oxidation state of Chromium in the product is $6$ $6$.

Chapter Mix

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

Q71 jee_main_2024_31_jan_morning Potassium Dichromate and Permanganate

Identify correct statements from below: A. The chromate ion is square planar. B. Dichromates are generally prepared from chromates. C. The green manganate ion is diamagnetic. D. Dark green coloured

K₂MnO₄

disproportionates in a neutral or acidic medium to give permanganate. E. With increasing oxidation number of transition metal, ionic character of the oxides decreases. Choose the correct answer from the options given below:

A. $B, C, D only$ $\text{B, C, D only}$
B. $A, D, E only$ $\text{A, D, E only}$
C. $A, B, C only$ $\text{A, B, C only}$
D. $B, D, E only$ $\text{B, D, E only}$

Solution

Step 1: Statement A Analysis

$CrO₄²⁻$ $CrO_4^{2-}$ (chromate ion) is tetrahedral, not square planar. Statement A is incorrect.

Step 2: Statement B Analysis

$2Na₂CrO₄ + 2H^+ arrow Na₂Cr₂O₇ + 2Na^+ + H₂O$ . Dichromates are indeed prepared from chromates. Statement B is correct.

Step 3: Statement C Analysis

The green manganate ion ( $MnO₄²⁻$ $MnO_4^{2-}$) has manganese in the +6 oxidation state ( $3d¹$ $3d^1$). Thus, it contains 1 unpaired electron and is paramagnetic, not diamagnetic. Statement C is incorrect.

Step 4: Statement D Analysis

Dark green coloured $K₂MnO₄$ $K_2MnO_4$ undergoes disproportionation in neutral or acidic media to yield permanganate ( $MnO₄^-$ $MnO_4^-$) and manganese dioxide ( $MnO₂$ $MnO_2$). Statement D is correct.

Step 5: Statement E Analysis

Fajans' rule dictates that as the oxidation state increases, polarizing power increases, leading to a decrease in ionic character (increase in covalent character). Statement E is correct.

Final Conclusion

The correct statements are B, D, and E.

Chapter Mix

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

In chromyl chloride test for confirmation of Cl^-$Cl^-$ ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and 10%$10\%$ H₂O₂$H_2O_2$ turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is

Solution & Explanation

Core Logic

Step 1: Oxidation State Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More d and f Block Elements Previous-Year Questions — Page 9

Solution

Core Logic

Step 1: Checking configurations

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Match execution

Chapter Mix

Solution

Core Logic

Step 1: Final Selection

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Oxidation State Calculation

Chapter Mix

Solution

Step 1: Statement A Analysis

Step 2: Statement B Analysis

Step 3: Statement C Analysis

Step 4: Statement D Analysis

Step 5: Statement E Analysis

Final Conclusion

Chapter Mix

More d and f Block Elements Questions — jee_main_2024_29_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