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 8.
Q75 (2026)
Consider the following reactions:
$NaCl + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} \rightarrow A + KHSO_{4} + NaHSO_{4} + H_{2}O$
$A + NaOH \rightarrow B + NaCl + H_{2}O$
$B + H_{2}SO_{4} + H_{2}O_{2} \rightarrow C + Na_{2}SO_{4} + H_{2}O$
In the product 'C', 'X' is the number of $O_{2}^{2-}$ units, 'Y' is the total number oxygen atoms present and 'Z' is the oxidation state of Cr. The value of $X + Y + Z$ is ____.
### Core Logic
The first reaction is the classical **Chromyl Chloride Test**:
$4\mathrm{NaCl} + \mathrm{K_2Cr_2O_7} + 6\mathrm{H_2SO_4} \rightarrow 2\mathrm{CrO_2Cl_2} (\text{A}) + 2\mathrm{KHSO_4} + 4\mathrm{NaHSO_4} + 3\mathrm{H_2O}$
Product A is Chromyl chloride ($\mathrm{CrO_2Cl_2}$), a red-orange gas.
When Chromyl chloride gas is passed into NaOH solution, it forms a yellow solution of sodium chromate (B):
$\mathrm{CrO_2Cl_2} (\text{A}) + 4\mathrm{NaOH} \rightarrow \mathrm{Na_2CrO_4} (\text{B}) + 2\mathrm{NaCl} + 2\mathrm{H_2O}$
Acidifying the sodium chromate solution with $H_2SO_4$ and adding $H_2O_2$ yields a deep blue solution of Chromium(VI) peroxide, $CrO_5$ (C):
$\mathrm{Na_2CrO_4} (\text{B}) + \mathrm{H_2SO_4} + 2\mathrm{H_2O_2} \rightarrow \mathrm{CrO_5} (\text{C}) + \mathrm{Na_2SO_4} + 3\mathrm{H_2O}$
Structure of $CrO_5$:
{{SOL_IMG1}}
- It has a butterfly structure.
- Number of peroxy units ($O_2^{2-}$), $X = 2$.
- Total number of oxygen atoms, $Y = 5$.
- Oxidation state of Cr, $Z = +6$.
Sum: $X + Y + Z = 2 + 5 + 6 = 13$.
### Step 1: Final Calculation
$X + Y + Z = 13$
### Pattern Recognition
Chromyl chloride test $\rightarrow$ $CrO_2Cl_2$ (red gas). Absorbed in NaOH $\rightarrow$ $Na_2CrO_4$ (yellow). Tested with $H_2O_2/H^+$ $\rightarrow$ $CrO_5$ (butterfly structure, blue, two peroxy links, Cr in +6).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Class 11 Chemistry: Redox Reactions
Q66 (2026)
Given below are some of the statements about $\text{Mn}$ and $\text{Mn}_2\text{O}_7$. Identify the correct statements:
A. Mn forms the oxide $\text{Mn}_2\text{O}_7$ 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. $\text{Mn}_2\text{O}_7$ is an ionic oxide.
D. The structure of $\text{Mn}_2\text{O}_7$ consists of one bridged oxygen.
Choose the correct answer from the options given below:
- $(1) \text{ A, B, C and D}$
- $(2) \text{ A, B and D Only}$
- $(3) \text{ A, C and D Only}$
- $(4) \text{ A, B and C Only}$
### Core Logic
- A is correct: $\text{Mn}_2\text{O}_7$ features Mn in +7 state (its highest oxidation state).
- B is correct: Oxygen stabilizes high oxidation states via multiple bonding.
- C is incorrect: $\text{Mn}_2\text{O}_7$ is a covalent green oil/oxide, not ionic.
- D is correct: Structure consists of two $\text{MnO}_4$ tetrahedra sharing one bridging oxygen atom ($\text{O}_3\text{Mn}-\text{O}-\text{MnO}_3$).
### 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 $\text{Mn}_2\text{O}_7$.
Trap: Assuming high oxidation state oxides of transition metals are ionic.
### Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
Q58 (2026)
A first row transition metal (M) does not liberate $H_{2}$ gas from dilute HCl. 1 mol of aqueous solution of $MSO_{4}$ is treated with excess of aqueous KCN and then $H_{2}S(g)$ is passed through the solution. The amount of MS (metal sulphide) formed from the above reaction is ____ mol.
- $\text{2}$
- $\text{1}$
- $\text{3}$
- $\text{0}$
### Related Formula
$$Cu^{2+} + 4CN^{-} \rightarrow [Cu(CN)_{4}]^{3-} \quad (\text{after redox with } CN^-)$$
### Core Logic
The first-row transition metal that does not liberate $H_{2}$ gas from dilute HCl is Copper (Cu), because its standard reduction potential is positive ($E^{\circ}_{Cu^{2+}/Cu} = +0.34\text{ V}$).
When $CuSO_{4}$ is treated with excess KCN, it forms a very stable soluble cyano complex:
$$CuSO_{4} + 2KCN \rightarrow Cu(CN)_{2} + K_{2}SO_{4}$$
$2Cu(CN)_{2} \rightarrow 2CuCN + (CN)_{2}$
$CuCN + 3KCN \rightarrow K_{3}[Cu(CN)_{4}]$
The complex ion $[Cu(CN)_{4}]^{3-}$ is highly stable (a perfect complex). When $H_{2}S$ is passed through this solution, it does not yield sufficient $Cu^{+}$ ions to exceed the solubility product ($K_{sp}$) of $Cu_{2}S$.
### Step 1: Final Conclusion
Since no copper sulphide precipitates, the amount of MS formed is 0 moles.
### Pattern Recognition
Cu and Cd separation: $Cu^{2+}$ forms a very stable cyanide complex that does not precipitate with $H_2S$, whereas $Cd^{2+}$ forms a less stable complex that does precipitate as $CdS$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d and f Block Elements
Class 12 Chemistry: Coordination Compounds
Q54 (2026)
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:
- Both Statement-I and Statement-II are false.
- Statement-I is true but Statement-II is false.
- Both Statement-I and Statement-II are true.
- Statement-I is false but Statement-II is true.
### Related Formula
$$\text{Electronic Configurations: } \text{Cr} = [\text{Ar}]3d^5 4s^1, \quad \text{Mn} = [\text{Ar}]3d^5 4s^2$$
### Core Logic
Step 1: Compare $IE_1$:
$$\text{Cr}: 4s^1 \implies \text{Removal of single } 4s \text{ electron requires less energy than removing } 4s^2 \text{ in Mn.}$$
Hence, $IE_1(\text{Cr}) < IE_1(\text{Mn})$ (Statement-I is TRUE).
Step 2: Compare $IE_2$ and $IE_3$:
$$\text{Cr}^+ = 3d^5 \implies \text{stable half-filled configuration } d^5, \text{ so } IE_2(\text{Cr}) > IE_2(\text{Mn})$$
$$\text{For } IE_3, \text{Mn}^{2+} = 3d^5 \implies \text{removing electron from stable } 3d^5 \text{ in Mn}^{2+} \text{ requires more energy than Cr}^{2+} (3d^4).$$
Hence, $IE_3(\text{Cr}) < IE_3(\text{Mn})$. Thus Statement-II is FALSE.
### Pattern Recognition
Sees: Ionization enthalpy comparison of Cr and Mn.
Shortcut: Stable $3d^5$ configuration in $\text{Cr}^+$ makes $IE_2$ very high, whereas $3d^5$ in $\text{Mn}^{2+}$ makes $IE_3$ of Mn higher than Cr.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d- and f-Block Elements
Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q73 (2026)
Among the following oxides of 3d elements, the number of mixed oxides are ____.
$$Ti_2O_3, V_2O_4, Cr_2O_3, Mn_3O_4, Fe_3O_4, Fe_2O_3, Co_3O_4$$
### Related Formula
$$\text{Mixed Oxide Formulation: } M_3O_4 \equiv MO \cdot M_2O_3$$
### Core Logic
Step 1: Check stoichiometric compositions of oxides:
- $Mn_3O_4 = MnO \cdot Mn_2O_3$ (Mixed oxide)
- $Fe_3O_4 = FeO \cdot Fe_2O_3$ (Mixed oxide)
- $Co_3O_4 = CoO \cdot Co_2O_3$ (Mixed oxide)
- $Ti_2O_3, V_2O_4, Cr_2O_3, Fe_2O_3$ are simple binary oxides.
Step 2: Total number of mixed oxides is 3.
### Pattern Recognition
Sees: Transition metal oxides list.
Shortcut: Oxides with $M_3O_4$ formula ($Mn_3O_4, Fe_3O_4, Co_3O_4$) are mixed oxides containing metal in both +2 and +3 oxidation states.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d- and f-Block Elements
