d and f Block Elements JEE Main previous year questions with solutions

5 solved JEE Main questions on d and f Block Elements, free to read — no sign-in needed. The full chapter has 259 questions; sign in to attempt the remaining 254 in the exam simulator.

1. Q1JEE Main 2026 (05 Apr, Shift 2)Compounds of Transition Metals
   A paper dipped in a dil. $H_2SO_4$ solution of '$X$' upon treatment with $SO_2$ gas turns into green. The compound '$X$' is :
   1. A.KI-starch
   2. B.$KMnO_4$
   3. C.$Pb(CH_3COO)_2$
   4. D.$K_2Cr_2O_7$
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   Answer: (D)

   $SO_2$ gas acts as a reducing agent. When a paper dipped in an acidified solution of potassium dichromate ($K_2Cr_2O_7$) is exposed to $SO_2$ gas, the orange-colored dichromate ion ($Cr_2O_7^{2-}$) is reduced to the green-colored chromium(III) ion ($Cr^{3+}$). The chemical reaction is: $K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$ The formation of chromium(III) sulfate, $Cr_2(SO_4)_3$, is responsible for the green color. Thus, the compound '$X$' is $K_2Cr_2O_7$. Answer: $K_2Cr_2O_7$
2. Q2JEE Main 2025 (04 Apr, Shift 2)F-Block elements
   The incorrect relationship in the following pairs in relation to ionisation enthalpies is :
   1. A.$\mathrm{Mn}^{+} \lt \mathrm{Cr}^{+}$
   2. B.$\mathrm{Mn}^{+} \lt \mathrm{Mn}^{2+}$
   3. C.$\mathrm{Fe}^{2+} \lt \mathrm{Fe}^{3+}$
   4. D.$\mathrm{Mn}^{2+} \lt \mathrm{Fe}^{2+}$
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   Answer: (D)

   $\mathrm{Mn}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^5$ Half filled stability More 1 E than $\mathrm{Fe}^{2+}$ $\mathrm{Fe}^{2+}:[\mathrm{Ar}] 3 \mathrm{~d}^6$
3. Q3JEE Main 2024 (09 Apr, Shift 2)Properties of D-block elements
   <p>A transition metal ' $\mathrm{M}$ ' among $\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}$ and $\mathrm{Fe}$ has the highest second ionisation enthalpy. The spin-only magnetic moment value of $\mathrm{M}^{+}$ion is $\underline{\phantom{0000}}. \mathrm{BM} (Near integer) (Given$ atomic number $\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26$)</p>
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   Answer: 6

   <p>Among given metals, $\mathrm{Cr}$ has maximum $\mathrm{IE}_2$ because Second electron is removed from stable configuration $3 d^5$ $\mathrm{Cr}^{+}:[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0$ $\therefore$ No of unpaired $\mathrm{e}^{-}$in $\mathrm{Cr}^{+}$is $5, \mathrm{n}=5$ So, Magnetic moment</p> <p>$\begin{aligned} & \mu = \sqrt{n(n+2)} \, \text{B.M.} \\ & = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \text{B.M.} \approx 6 \, \text{B.M.} \end{aligned}$</p>
4. Q4JEE Main 2026 (04 Apr, Shift 1)Compounds of Transition Metals
   Correct statements from the following are A. Potassium dichromate is an oxidising agent and it oxidises $FeSO_4$ to $Fe_2(SO_4)_3$ in acidic medium. B. Sodium dichromate can be used as primary standard in volumetric estimation. C. $CrO_4^{2-}$ and $Cr_2O_7^{2-}$ are interconvertible in aqueous solution by varying the $pH$ of the solution. D. $Cr$-$O$-$Cr$ bond angle in $Cr_2O_7^{2-}$ is $126^\circ$. Choose the correct answer from the options given below:
   1. A.A, B and C Only
   2. B.A, C and D Only
   3. C.A and C Only
   4. D.B and D Only
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   Answer: (B)

   Statement A is correct. Potassium dichromate acts as a strong oxidising agent in acidic medium and oxidises $Fe^{2+}$ to $Fe^{3+}$, converting $FeSO_4$ to $Fe_2(SO_4)_3$. Statement B is incorrect. Sodium dichromate is deliquescent (absorbs moisture from the atmosphere), which prevents it from being weighed accurately. Therefore, it cannot be used as a primary standard in volumetric estimation. Potassium dichromate is used instead. Statement C is correct. Chromate ($CrO_4^{2-}$) and dichromate ($Cr_2O_7^{2-}$) ions are interconvertible in aqueous solution depending on the $pH$ of the solution. In acidic medium, chromate changes to dichromate, and in basic medium, dichromate changes to chromate: $2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$. Statement D is correct. The structure of the dichromate ion consists of two tetrahedral $CrO_3$ groups joined by a bridging oxygen atom, and the $Cr-O-Cr$ bond angle is $126^\circ$. Therefore, statements A, C, and D are correct. Answer: A, C and D Only
5. Q5JEE Main 2026 (02 Apr, Shift 1)Compounds of Transition Metals
   Which of the following sets includes all the species that will change the orange colour of $K_2Cr_2O_7$ in acidic medium?
   1. A.$Fe^{2+}, Sn^{2+}, I^-, S^{2-}$
   2. B.$S^{2-}, Fe^{3+}, I^-, C_2O_4^{2-}$
   3. C.$Fe^{2+}, NO_2^-, SO_2, Sn^{4+}$
   4. D.$Fe^{3+}, SO_4^{2-}, S^{2-}, Sn^{4+}$
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   Answer: (A)

   $K_2Cr_2O_7$ in acidic medium acts as a strong oxidizing agent. It changes its colour from orange ($Cr_2O_7^{2-}$) to green ($Cr^{3+}$) when it reacts with a reducing agent. We need to identify the set in which all species can act as reducing agents (i.e., they can be oxidized to a higher oxidation state). In option (A), $Fe^{2+}$ oxidizes to $Fe^{3+}$, $Sn^{2+}$ oxidizes to $Sn^{4+}$, $I^-$ oxidizes to $I_2$, and $S^{2-}$ oxidizes to $S$. All of these are reducing agents. In option (B), $Fe^{3+}$ is already in its maximum stable oxidation state and cannot be oxidized further. In option (C), $Sn^{4+}$ cannot be oxidized further. In option (D), $Fe^{3+}$, $SO_4^{2-}$, and $Sn^{4+}$ cannot be oxidized further. Thus, only the species in option (A) will change the orange colour of $K_2Cr_2O_7$. Answer: $Fe^{2+}, Sn^{2+}, I^-, S^{2-}$