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…

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More d and f Block Elements Previous-Year Questions — Page 2

Q73 jee_main_2026_22_january_evening Mixed Oxides Identification

Among the following oxides of 3d elements, the number of mixed oxides are ____.

Ti₂O₃, V₂O₄, Cr₂O₃, Mn₃O₄, Fe₃O₄, Fe₂O₃, Co₃O₄

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

Mixed Oxide Formulation: M₃O₄ \equiv MO \cdot M₂O₃

Core Logic

Step 1: Check stoichiometric compositions of oxides:

$Mn₃O₄ = MnO \cdot Mn₂O₃$ $Mn_3O_4 = MnO \cdot Mn_2O_3$ (Mixed oxide)
$Fe₃O₄ = FeO \cdot Fe₂O₃$ $Fe_3O_4 = FeO \cdot Fe_2O_3$ (Mixed oxide)
$Co₃O₄ = CoO \cdot Co₂O₃$ $Co_3O_4 = CoO \cdot Co_2O_3$ (Mixed oxide)
$Ti₂O₃, V₂O₄, Cr₂O₃, Fe₂O₃$ $Ti_2O_3, V_2O_4, Cr_2O_3, Fe_2O_3$ are simple binary oxides.

Step 2: Total number of mixed oxides is 3.

Pattern Recognition

Sees: Transition metal oxides list. Shortcut: Oxides with $M₃O₄$ $M_3O_4$ formula ( $Mn₃O₄, Fe₃O₄, Co₃O₄$ $Mn_3O_4, Fe_3O_4, Co_3O_4$) are mixed oxides containing metal in both +2 and +3 oxidation states.

Chapter Mix

Class 12 Chemistry: d- and f-Block Elements

Q67 jee_main_2026_23_january_evening Potassium Dichromate

The oxidation state of chromium in the final product formed in the reaction between KI and acidified

K₂Cr₂O₇

solution is:

A. $+4$ $+4$
B. $+3$ $+3$
C. $+2$ $+2$
D. $+6$ $+6$

Solution

Related Formula

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

2I^- arrow I₂ + 2e^-

Core Logic

Acidified potassium dichromate ( $K₂Cr₂O₇$ $K_2Cr_2O_7$) acts as a strong oxidizing agent. In acidic medium, the dichromate ion ( $Cr₂O₇²⁻$ $Cr_2O_7^{2-}$, where Cr is in the $+6$ $+6$ oxidation state) is reduced to the Chromium(III) ion ( $Cr³⁺$ $Cr^{3+}$).

Concurrently, iodide ions ( $I^-$ $I^-$ from KI) are oxidized to elemental iodine ( $I₂$ $I_2$). The full ionic equation is:

Cr₂O₇²⁻ + 6I^- + 14H^+ arrow 2Cr³⁺ + 3I₂ + 7H₂O

The final product containing chromium is the $Cr³⁺$ $Cr^{3+}$ ion, meaning its oxidation state is $+3$ $+3$.

Pattern Recognition

Always remember that in acidic media, $Cr₂O₇²⁻$ $Cr_2O_7^{2-}$ (orange) universally reduces to $Cr³⁺$ $Cr^{3+}$ (green). The oxidation state invariably goes from $+6 arrow +3$ $+6 \rightarrow +3$.

Chapter Mix

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

Q69 jee_main_2026_24_january_evening Compounds of Transition Elements

"X" is an oxoanion of the lightest element of group 7 (in the periodic table). The metal is in +6 oxidation state in "X". The color of the potassium salt of X is

A. green
B. purple
C. yellow
D. orange

Solution

Core Logic

The lightest element of Group 7 in the periodic table is Manganese ( $Mn$ $Mn$). The oxoanion of $Mn$ $Mn$ where it resides in the +6 oxidation state is the manganate ion ( $MnO₄²⁻$ $MnO_4^{2-}$). The potassium salt of this oxoanion is potassium manganate ( $K₂MnO₄$ $K_2MnO_4$).

Step 1: Deduce Color

$K₂MnO₄$ $K_2MnO_4$ (containing the $MnO₄²⁻$ $MnO_4^{2-}$ ion) is known to have a green color. (Contrast with the +7 state in $KMnO₄$ $KMnO_4$, which is intensely purple).

Pattern Recognition

Manganese oxoanions have signature colors: $MnO₄⁻$ $MnO_4^{-}$ (Permanganate, +7) is Purple/Pink, while $MnO₄²⁻$ $MnO_4^{2-}$ (Manganate, +6) is Green.

Chapter Mix

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

Q68 jee_main_2026_28_january_morning Color and Oxidation States of Transition Metals

Given below are two statements: Statement I: The number of pairs, from the following, in which both the ions are coloured in aqueous solution is 3.

[Sc³⁺, Ti³⁺], [Mn²⁺, Cr²⁺], [Cu²⁺, Zn²⁺]

and

[Ni²⁺, Ti⁴⁺]

Statement II:

Th⁴⁺

is the strongest reducing agent among

Th⁴⁺, Ce⁴⁺, Gd³⁺

and

Eu²⁺

. In the light of the above statements, choose the correct answer from the options given below

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

Solution

Step 1: Evaluate Statement I

Color in transition metal ions arises from d-d transitions, which require unpaired d-electrons ( $d¹$ $d^1$ to $d⁹$ $d^9$).\n$\bullet$ $[Sc³⁺, Ti³⁺]$ $[Sc^{3+}, Ti^{3+}]$: $Sc³⁺$ $Sc^{3+}$ is $3d⁰$ $3d^0$ (Colourless). Pair invalid.\n$\bullet$ $[Mn²⁺, Cr²⁺]$ $[Mn^{2+}, Cr^{2+}]$: $Mn²⁺$ $Mn^{2+}$ is $3d⁵$ $3d^5$ (Coloured), $Cr²⁺$ $Cr^{2+}$ is $3d⁴$ $3d^4$ (Coloured). Pair valid.\n$\bullet$ $[Cu²⁺, Zn²⁺]$ $[Cu^{2+}, Zn^{2+}]$: $Zn²⁺$ $Zn^{2+}$ is $3d¹⁰$ $3d^{10}$ (Colourless). Pair invalid.\n$\bullet$ $[Ni²⁺, Ti⁴⁺]$ $[Ni^{2+}, Ti^{4+}]$: $Ti⁴⁺$ $Ti^{4+}$ is $3d⁰$ $3d^0$ (Colourless). Pair invalid.\nOnly ONE pair contains both colored ions. Statement I is false.

Step 2: Evaluate Statement II

Thorium ( $Th$ $Th$) exhibits a stable $+4$ $+4$ oxidation state. $Th⁴⁺$ $Th^{4+}$ has an empty shell ( $5f⁰ 6d⁰ 7s⁰$ $5f^0 6d^0 7s^0$) and cannot lose more electrons to act as a reducing agent (which requires getting oxidized further). Statement II is false.

Final Conclusion

Both Statement I and Statement II are false.

Pattern Recognition

$d⁰$ $d^0$ and $d¹⁰$ $d^{10}$ configurations never absorb visible light for d-d transitions, rendering them strictly colourless. Maximum group oxidation states cannot act as reducing agents.

Chapter Mix

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

Q57 jee_main_2026_28_january_evening Properties Of Permanganate And Manganate Ions

Consider the following statements about manganate and permanganate ions. Identify the correct statements: (A) The geometry of both manganate and permanganate ions is tetrahedral. (B) The oxidation states of Mn in manganate and permanganate are +7 and +6, respectively. (C) Oxidation of Mn(II) salt by peroxodisulphate gives manganate ion as the final product. (D) Manganate ion is paramagnetic and permanganate ions is diamagnetic. (E) Acidified permanganate ion reduces oxalate, nitrite and iodide ions. Choose the correct answer from the options given below:

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

Solution

Core Logic

(A) Both $MnO₄²⁻$ $MnO_{4}^{2-}$ (Manganate) and $MnO₄⁻$ $MnO_{4}^{-}$ (Permanganate) have tetrahedral geometry utilizing $d³s$ $d^3s$ hybridization. (Correct)

(B) The oxidation state of Mn in manganate ( $MnO₄²⁻$ $MnO_{4}^{2-}$) is +6 and in permanganate ( $MnO₄⁻$ $MnO_{4}^{-}$) is +7. The statement swaps these. (Incorrect)

(D) $MnO₄⁻$ $MnO_{4}^{-}$ (Mn in +7, $d⁰$ $d^0$) is diamagnetic. $MnO₄²⁻$ $MnO_{4}^{2-}$ (Mn in +6, $d¹$ $d^1$) is paramagnetic. (Correct)

(E) Acidified permanganate ion is an oxidizing agent, meaning it OXIDIZES oxalate, nitrite, and iodide ions; it does not reduce them. (Incorrect)

Step 1: Final Conclusion

Statements A and D are correct.

Pattern Recognition

Recall $MnO₄^-$ $MnO_4^-$ is purple, diamagnetic, +7 state, powerful oxidizing agent. $MnO₄²⁻$ $MnO_4^{2-}$ is green, paramagnetic, +6 state. Oxidizing agent means it reduces itself, thus oxidizes other substrates.

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 2

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Deduce Color

Pattern Recognition

Chapter Mix

Solution

Step 1: Evaluate Statement I

Step 2: Evaluate Statement II

Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

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