d-and f-Block Elements Previous Year Questions — JEE Main Chemistry
44 past-year d-and f-Block Elements questions from JEE Main (Chemistry) — page 9.
Q67 (2026)
The oxidation state of chromium in the final product formed in the reaction between KI and acidified $K_{2}Cr_{2}O_{7}$ solution is:
- $+4$
- $+3$
- $+2$
- $+6$
### Related Formula
$$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$
$$2I^- \rightarrow I_2 + 2e^-$$
### Core Logic
Acidified potassium dichromate ($K_2Cr_2O_7$) acts as a strong oxidizing agent. In acidic medium, the dichromate ion ($Cr_2O_7^{2-}$, where Cr is in the $+6$ oxidation state) is reduced to the Chromium(III) ion ($Cr^{3+}$).
Concurrently, iodide ions ($I^-$ from KI) are oxidized to elemental iodine ($I_2$).
The full ionic equation is:
$$Cr_2O_7^{2-} + 6I^- + 14H^+ \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$$
The final product containing chromium is the $Cr^{3+}$ ion, meaning its oxidation state is $+3$.
### Pattern Recognition
Always remember that in acidic media, $Cr_2O_7^{2-}$ (orange) universally reduces to $Cr^{3+}$ (green). The oxidation state invariably goes from $+6 \rightarrow +3$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Class 11 Chemistry: Redox Reactions
Q69 (2026)
"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
- green
- purple
- yellow
- orange
### Core Logic
The lightest element of Group 7 in the periodic table is Manganese ($Mn$).
The oxoanion of $Mn$ where it resides in the +6 oxidation state is the manganate ion ($MnO_4^{2-}$).
The potassium salt of this oxoanion is potassium manganate ($K_2MnO_4$).
### Step 1: Deduce Color
$K_2MnO_4$ (containing the $MnO_4^{2-}$ ion) is known to have a green color.
(Contrast with the +7 state in $KMnO_4$, which is intensely purple).
### Pattern Recognition
Manganese oxoanions have signature colors: $MnO_4^{-}$ (Permanganate, +7) is Purple/Pink, while $MnO_4^{2-}$ (Manganate, +6) is Green.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
Q68 (2026)
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.
$[Sc^{3+}, Ti^{3+}], [Mn^{2+}, Cr^{2+}], [Cu^{2+}, Zn^{2+}]$ and $[Ni^{2+}, Ti^{4+}]$
Statement II: $Th^{4+}$ is the strongest reducing agent among $Th^{4+}, Ce^{4+}, Gd^{3+}$ and $Eu^{2+}$.
In the light of the above statements, choose the correct answer from the options given below
- Statement I is true but Statement II is false
- Statement I is false but Statement II is true
- Both Statement I and Statement II are false
- Both Statement I and Statement II are true
### Step 1: Evaluate Statement I
Color in transition metal ions arises from d-d transitions, which require unpaired d-electrons ($d^1$ to $d^9$).\n$\bullet$ $[Sc^{3+}, Ti^{3+}]$: $Sc^{3+}$ is $3d^0$ (Colourless). Pair invalid.\n$\bullet$ $[Mn^{2+}, Cr^{2+}]$: $Mn^{2+}$ is $3d^5$ (Coloured), $Cr^{2+}$ is $3d^4$ (Coloured). Pair valid.\n$\bullet$ $[Cu^{2+}, Zn^{2+}]$: $Zn^{2+}$ is $3d^{10}$ (Colourless). Pair invalid.\n$\bullet$ $[Ni^{2+}, Ti^{4+}]$: $Ti^{4+}$ is $3d^0$ (Colourless). Pair invalid.\nOnly ONE pair contains both colored ions. Statement I is false.
### Step 2: Evaluate Statement II
Thorium ($Th$) exhibits a stable $+4$ oxidation state. $Th^{4+}$ has an empty shell ($5f^0 6d^0 7s^0$) 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
$d^0$ and $d^{10}$ configurations never absorb visible light for d-d transitions, rendering them strictly colourless. Maximum group oxidation states cannot act as reducing agents.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
Q57 (2026)
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:
- $(1)\text{ A, C and D Only}$
- $(2)\text{ A, B and C Only}$
- $(3)\text{ A, D and E Only}$
- $(4)\text{ A and D Only}$
### Core Logic
(A) Both $MnO_{4}^{2-}$ (Manganate) and $MnO_{4}^{-}$ (Permanganate) have tetrahedral geometry utilizing $d^3s$ hybridization. (Correct)
(B) The oxidation state of Mn in manganate ($MnO_{4}^{2-}$) is +6 and in permanganate ($MnO_{4}^{-}$) is +7. The statement swaps these. (Incorrect)
(C) $Mn^{2+} + S_{2}O_{8}^{2-} \rightarrow MnO_{4}^{-}$ (Permanganate ion), not manganate. (Incorrect)
(D) $MnO_{4}^{-}$ (Mn in +7, $d^0$) is diamagnetic. $MnO_{4}^{2-}$ (Mn in +6, $d^1$) 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 $MnO_4^-$ is purple, diamagnetic, +7 state, powerful oxidizing agent. $MnO_4^{2-}$ is green, paramagnetic, +6 state. Oxidizing agent means it reduces itself, thus oxidizes other substrates.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
