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 7.
Q76 (2024)
Match List-I with List-II.
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">List-I (Species)</th><th style="border: 1px solid #888; padding: 8px;">List-II (Electronic distribution)</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">(A) $Cr^{+2}$</td><td style="border: 1px solid #888; padding: 8px;">(I) $3d^8$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(B) $Mn^+$</td><td style="border: 1px solid #888; padding: 8px;">(II) $3d^54s^1$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(C) $Ni^{+2}$</td><td style="border: 1px solid #888; padding: 8px;">(III) $3d^4$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(D) $V^+$</td><td style="border: 1px solid #888; padding: 8px;">(IV) $3d^34s^1$</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- $\text{(A)-(I), (B)-(II), (C)-(III), (D)-(IV)}$
- $\text{(A)-(III), (B)-(II), (C)-(I), (D)-(IV)}$
- $\text{(A)-(IV), (B)-(III), (C)-(I), (D)-(II)}$
- $\text{(A)-(II), (B)-(I), (C)-(IV), (D)-(III)}$
### 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 $4s$ orbital.
(A) $Cr$ (Z=24): $[Ar] 3d^5 4s^1$ $\rightarrow$ $Cr^{2+}$: $[Ar] 3d^4$
(B) $Mn$ (Z=25): $[Ar] 3d^5 4s^2$ $\rightarrow$ $Mn^+$: $[Ar] 3d^5 4s^1$
(C) $Ni$ (Z=28): $[Ar] 3d^8 4s^2$ $\rightarrow$ $Ni^{2+}$: $[Ar] 3d^8$
(D) $V$ (Z=23): $[Ar] 3d^3 4s^2$ $\rightarrow$ $V^+$: $[Ar] 3d^3 4s^1$
### Step 1: Match execution
A $\rightarrow$ III
B $\rightarrow$ II
C $\rightarrow$ I
D $\rightarrow$ IV
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
Q1938 (2024)
Choose the correct statements from the following
A. $Mn_2O_7$ is an oil at room temperature
B. $V_2O_4$ reacts with acid to give $VO_2^{2+}$
C. $CrO$ is a basic oxide
D. $V_2O_5$ does not react with acid
Choose the correct answer from the options given below:
- $\text{A, B and D only}$
- $\text{A and C only}$
- $\text{A, B and C only}$
- $\text{B and C only}$
### Core Logic
(A) $Mn_2O_7$ is a covalent oxide and exists as a green oil at room temperature. (Correct)
(B) $V_2O_4$ dissolves in acids to give $VO^{2+}$ (vanadyl) salts, not $VO_2^{2+}$. (Incorrect)
(C) $CrO$ has chromium in the $+2$ oxidation state. Lower oxidation state metal oxides are typically basic in nature. (Correct)
(D) $V_2O_5$ 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).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
Q86 (2024)
In the reaction of potassium dichromate, potassium chloride and sulfuric acid (conc.), the oxidation state of the chromium in the product is (+) ________
### Related Formula
$$K_2Cr_2O_7(s) + 4KCl(s) + 6H_2SO_4(conc.) \rightarrow 2CrO_2Cl_2(g) + 6KHSO_4 + 3H_2O$$
### 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 ($CrO_2Cl_2$) are evolved.
### Step 1: Oxidation State Calculation
In chromyl chloride ($CrO_2Cl_2$):
Let the oxidation state of Chromium be $x$.
Oxygen is typically $-2$ and Chlorine is $-1$.
$$x + 2(-2) + 2(-1) = 0$$
$x - 4 - 2 = 0$
$x = +6$
Thus, the oxidation state of Chromium in the product is $6$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
Class 11 Chemistry: Practical Chemistry
Q71 (2024)
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 $K_2MnO_4$ 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:
- $\text{B, C, D only}$
- $\text{A, D, E only}$
- $\text{A, B, C only}$
- $\text{B, D, E only}$
### Step 1: Statement A Analysis
$CrO_4^{2-}$ (chromate ion) is tetrahedral, not square planar. Statement A is incorrect.
### Step 2: Statement B Analysis
$2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O$. Dichromates are indeed prepared from chromates. Statement B is correct.
### Step 3: Statement C Analysis
The green manganate ion ($MnO_4^{2-}$) has manganese in the +6 oxidation state ($3d^1$). Thus, it contains 1 unpaired electron and is paramagnetic, not diamagnetic. Statement C is incorrect.
### Step 4: Statement D Analysis
Dark green coloured $K_2MnO_4$ undergoes disproportionation in neutral or acidic media to yield permanganate ($MnO_4^-$) and manganese dioxide ($MnO_2$). 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.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: The d- and f-Block Elements
Q63 (2026)
$MnO_{4}^{2-}$, in acidic medium, disproportionates to :
- $Mn_{2}O_{7}\text{ and }MnO_{2}$
- $\mathrm{MnO}_4^-\text{ and }MnO$
- $\mathrm{MnO}_4^-\text{ and }\mathrm{MnO}_2$
- $\mathrm{Mn}_{2}\mathrm{O}_{7}\text{ and }MnO$
### Related Formula
$$3\mathrm{MnO}_4^{2-} + 4\mathrm{H}^+ \rightarrow 2\mathrm{MnO}_4^- + \mathrm{MnO}_2 + 2\mathrm{H}_2\mathrm{O}$$
### Core Logic
Manganate ion ($\mathrm{MnO}_4^{2-}$), where Mn is in +6 oxidation state, is unstable in acidic medium and undergoes disproportionation.
It oxidizes to Permanganate ($\mathrm{MnO}_4^-$, +7 state) and reduces to Manganese dioxide ($\mathrm{MnO}_2$, +4 state).
### Pattern Recognition
Manganate (green, +6) disproportionates in acid to Permanganate (purple, +7) and $MnO_2$ (brown/black precipitate, +4).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: d and f Block Elements
