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 2.
Q44 (2025)
Which of the following oxidation reactions are carried out by both $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ and $\mathrm{KMnO}_4$ in acidic medium?
A. $\mathrm{I}^- \rightarrow \mathrm{I}_2$
B. $\mathrm{S}^{2-} \rightarrow \mathrm{S}$
C. $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}$
D. $\mathrm{I}^{-} \rightarrow \mathrm{IO}_{3}^{-}$
E. $\mathrm{S}_2\mathrm{O}_3^{2-} \rightarrow \mathrm{SO}_4^{2-}$
Choose the correct answer from the options given below:
- $\text{B, C and D only}$
- $\text{A, D and E only}$
- $\text{A, B and C only}$
- $\text{C, D and E only}$
### Core Logic
In an acidic medium, both $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ and $\mathrm{KMnO}_4$ act as strong oxidizing agents and carry out the following transformations:
- **A:** Oxidize iodide to iodine: $\mathrm{I}^- \rightarrow \mathrm{I}_2$
- **B:** Oxidize sulfide to elemental sulfur: $\mathrm{S}^{2-} \rightarrow \mathrm{S}$
- **C:** Oxidize ferrous ions to ferric ions: $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}$
For reactions D and E:
- Iodide is oxidized to iodate ($\mathrm{IO}_3^-$) by $\mathrm{KMnO}_4$ primarily in a neutral or faintly alkaline medium, not acidic.
- Thiosulfate ($\mathrm{S}_2\mathrm{O}_3^{2-}$) undergoes a disproportionation/precipitation reaction in acid rather than clean oxidation to sulfate by dichromate.
Thus, statements A, B, and C are valid for both under acidic conditions.
### Pattern Recognition
Sees: Shared oxidation products in an acidic environment.
Shortcut: Remember that $\mathrm{I}^- \rightarrow \mathrm{IO}_3^-$ is the signature reaction for alkaline permanganate, allowing quick elimination of choices containing D.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d-and f-Block Elements
Q800 (2025)
Consider the following reactions:
$$A + \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{CrO}_2\text{Cl}_2 + \text{Side Products}$$
$$\text{CrO}_2\text{Cl}_2(\text{vapour}) + \text{NaOH} \rightarrow B + \text{NaCl} + \text{H}_2\text{O}$$
$$B + \text{H}^+ \rightarrow C + \text{H}_2\text{O}$$
The number of terminal 'O' present in the compound 'C' is ______
### Core Logic
Let us identify the sequential chemical components via the chromyl chloride test pathway:
1. Reactant **A** represents a dichromate salt such as $\text{K}_2\text{Cr}_2\text{O}_7$. Heating it with a metal chloride and concentrated sulfuric acid generates deep red chromyl chloride vapors ($\text{CrO}_2\text{Cl}_2$).
2. Passing these vapors into sodium hydroxide dissolves them, producing yellow sodium chromate compound **B** ($\text{Na}_2\text{CrO}_4$).
3. Acidifying the chromate solution dimerizes it into orange sodium dichromate compound **C** ($\text{Na}_2\text{Cr}_2\text{O}_7$).
### Step 1: Structural Analysis of Dichromate
The dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$) consists of two tetrahedral chromium units sharing a single bridging oxygen atom ($\text{Cr}-\text{O}-\text{Cr}$). Each chromium atom retains 3 localized terminal oxygen atoms:
$$\text{Total terminal 'O' atoms} = 7 - 1 = 6$$
Thus, the total count of terminal oxygen atoms present in compound $C$ is **6**. {{SOL_IMG1}}
### Pattern Recognition
Shortcut: The chromyl chloride sequence moves from dichromate $\rightarrow$ chromate $\rightarrow$ dichromate. In the dichromate ion ($\text{Cr}_2\text{O}_7^{2-}$), out of the 7 oxygen atoms, exactly 1 is bridging, leaving $7 - 1 = 6$ terminal oxygen atoms.
### Evaluation Rubric / Model Answer
6
### Chapter Mix
Class 12 Chemistry: d- and f-Block Elements
Q27 (2025)
The incorrect relationship in the following pairs in relation to ionisation enthalpies is :
- $Mn^{+} < Cr^{+}$
- $Mn^{+} < Mn^{2+}$
- $Fe^{2+} < Fe^{3+}$
- $Mn^{2+} < Fe^{2+}$
### Related Formula
$$\text{IE} \propto \frac{1}{\text{Stability of electronic configuration}}$$
### Core Logic
Let's examine the configurations:
- For $Mn^{2+}$, the electronic configuration is $[Ar]3d^5$, which features a highly stable, symmetric half-filled d-subshell.
- For $Fe^{2+}$, the configuration is $[Ar]3d^6$.
Because of the extra exchange energy and stability of the half-filled $3d^5$ state, it is harder to remove an electron from $Mn^{2+}$ than from $Fe^{2+}$. Therefore, the ionisation enthalpy of $Mn^{2+}$ is greater than that of $Fe^{2+}$:
$$\text{IE}(Mn^{2+}) > \text{IE}(Fe^{2+})$$
Hence, the expression $Mn^{2+} < Fe^{2+}$ is incorrect.
### Pattern Recognition
Whenever you see manganese ($Mn$) in the $+2$ oxidation state, remember its exceptionally stable $d^5$ config. This creates anomalous spikes in successive ionisation energies compared to neighboring iron ($Fe$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d and f Block Elements
Q44 (2025)
Pair of transition metal ions having the same number of unpaired electrons is:
- $V^{2+}, Co^{2+}$
- $Ti^{2+}, Co^{2+}$
- $Fe^{3+}, Cr^{2+}$
- $Ti^{3+}, Mn^{2+}$
### Core Logic
Let's map the electronic configurations and count the unpaired d-orbital electrons for each option:
* For pair (1):
$$V^{2+} \implies [Ar] 3d^3 4s^0 \implies 3 \text{ unpaired electrons}$$
$$Co^{2+} \implies [Ar] 3d^7 4s^0 \implies t_{2g}^5 e_g^2 \implies 3 \text{ unpaired electrons}$$
Both ions contain exactly 3 unpaired electrons.
* For other ions:
$$Ti^{2+} \implies [Ar] 3d^2 \implies 2 \text{ unpaired e-}, \quad Fe^{3+} \implies [Ar] 3d^5 \implies 5 \text{ unpaired e-}$$
$$Cr^{2+} \implies [Ar] 3d^4 \implies 4 \text{ unpaired e-}, \quad Ti^{3+} \implies [Ar] 3d^1 \implies 1 \text{ unpaired e-}$$
$$Mn^{2+} \implies [Ar] 3d^5 \implies 5 \text{ unpaired e-}$$
### Pattern Recognition
D-orbital counts follow a predictable symmetry: a $3d^n$ system contains the same number of unpaired electrons as a $3d^{10-n}$ system under high-spin conditions. This explains why $3d^3$ ($V^{2+}$) and $3d^7$ ($Co^{2+}$) match perfectly with 3 unpaired electrons each.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
Q47 (2025)
$KMnO_4$ acts as an oxidising agent in acidic medium. 'X' is the difference between the oxidation states of Mn in reactant and product. 'Y' is the number of 'd' electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of $X + Y$ is ______.
### Core Logic
Let's resolve both components step by step:
1. **Finding X:** In an acidic medium, the permanganate ion ($KMnO_4$, where $Mn$ is in the $+7$ state) is reduced to the divalent manganese cation ($Mn^{2+}$, state $+2$):
$X = 7 - 2 = 5$
2. **Finding Y:** During qualitative salt analysis, the acetate ion reacts with neutral ferric chloride to produce a characteristic blood-red coordination solution. Boiling this solution throws down a **brown-red precipitate** of basic ferric acetate, $[Fe(OH)_2(CH_3COO)]$. In this complex, Iron retains its $+3$ oxidation state:
$$Fe^{3+} \implies [Ar] 3d^5 4s^0 \implies \text{Number of d-electrons } (Y) = 5$$
Summing the values yields:
$$X + Y = 5 + 5 = 10$$
### Pattern Recognition
This problem elegantly links standard redox transitions with qualitative inorganic salt tests. Remember that throughout the basic ferric acetate precipitation test, Iron remains steadily in its ferric $+3$ ($d^5$) core configuration.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
Inorganic Qualitative Analysis
