d and f Block Elements Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on the d- and f-block elements with step-by-step solutions covering oxidation states, lanthanoid and actinoid contraction, dichromate chemistry, magnetic moments, ionisation enthalpies and interstitial compounds.
A curated set of JEE Main 2026 previous-year questions on the d- and f-block elements, each solved step by step so you can check both the final answer and the reasoning.
Solutions are AI-generated and pending review.
Solution
Check each statement.
A. Going down a group, the higher oxidation state becomes more stable. So Mo(VI) and W(VI) are more stable than Cr(VI), not less. Incorrect.
B. $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$ tend to revert to the stable $+3$ state, so they act as oxidants; $\text{Eu}^{2+}$ and $\text{Yb}^{2+}$ tend to go to $+3$, so they act as reductants. Correct.
C. Configurations:
$$\text{Am}: [\text{Rn}]\,5f^{7}7s^{2}\ \Rightarrow\ 7\ \text{unpaired}$$$$\text{Cm}: [\text{Rn}]\,5f^{7}6d^{1}7s^{2}\ \Rightarrow\ 7+1 = 8\ \text{unpaired}$$Cm has 8 unpaired electrons, so “both have seven” is incorrect.
D. Because 5f electrons shield the nuclear charge even more poorly than 4f electrons, the contraction per element in the actinoids is greater than in the lanthanoids. Correct.
Correct statements: B and D.
Answer: C
Solution
A. In acidic medium $\text{Cr}_2\text{O}_7^{2-}$ oxidises $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$:
$$\text{Cr}_2\text{O}_7^{2-} + 6\,\text{Fe}^{2+} + 14\,\text{H}^{+} \rightarrow 2\,\text{Cr}^{3+} + 6\,\text{Fe}^{3+} + 7\,\text{H}_2\text{O}$$So $\text{FeSO}_4$ is oxidised to $\text{Fe}_2(\text{SO}_4)_3$. Correct.
B. Sodium dichromate is deliquescent/hygroscopic, so it cannot be a primary standard. The primary standard is potassium dichromate. Incorrect.
C. The equilibrium
$$2\,\text{CrO}_4^{2-} + 2\,\text{H}^{+} \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}$$shifts with pH (yellow chromate in base, orange dichromate in acid), so they are interconvertible. Correct.
D. In the $\text{Cr}_2\text{O}_7^{2-}$ ion the two tetrahedra share a corner oxygen with a Cr–O–Cr bridge angle of about $126^\circ$. Correct.
Correct statements: A, C and D.
Answer: B
Solution
A. The catalyst in the contact process is $\text{V}_2\text{O}_5$, which is acidic (amphoteric), not basic. Incorrect.
B. Ziegler–Natta uses a titanium halide ($\text{TiCl}_4/\text{TiCl}_3$). The spin-only moment
$$\mu = \sqrt{n(n+2)}\ \text{BM}$$gives $2.84$ BM only for $n = 2$ ($d^2$). Ti in $\text{TiCl}_4$ is $d^0$ ($\mu = 0$) and in $\text{TiCl}_3$ is $d^1$ ($\mu = 1.73$). Neither is $2.84$ BM. Incorrect.
C. The p-block co-catalyst is triethylaluminium $\text{Al}(\text{C}_2\text{H}_5)_3$, where Al is in the $+3$ state. Correct.
D. The Wacker process uses $\text{PdCl}_2$. For $\text{Pd}^{2+}$:
$$\text{Pd}^{2+}: [\text{Kr}]\,4d^{8}\ \Rightarrow\ 8\ d\text{-electrons}$$Correct.
Correct statements: C and D.
Answer: C
Solution
Write the 4f populations (ground-state configurations):
$$\text{Eu}(63): 4f^{7},\quad \text{Gd}(64): 4f^{7}5d^{1},\quad \text{Dy}(66): 4f^{10},\quad \text{Ho}(67): 4f^{11}$$$$\text{Tm}(69): 4f^{13},\quad \text{Yb}(70): 4f^{14},\quad \text{Lu}(71): 4f^{14}5d^{1},\quad \text{Hf}(72): 4f^{14}5d^{2}$$A. Eu $4f^{7}$, Gd $4f^{7}$ → same (7 each). Correct. B. Dy $4f^{10}$, Ho $4f^{11}$ → different. Incorrect. C. Yb $4f^{14}$, Hf $4f^{14}$ → same (14 each). Correct. D. Lu $4f^{14}$, Tm $4f^{13}$ → different. Incorrect.
Matching pairs: A and C.
Answer: D
Solution
Highest manganese fluoride is $\text{MnF}_4$ (Mn is $+4$); highest manganese oxide is $\text{Mn}_2\text{O}_7$ (Mn is $+7$).
$$|x| = |4 - 7| = 3$$Now find ions with exactly 3 unpaired electrons:
$$\text{Sc}^{3+}: d^{0} \to 0,\qquad \text{Zn}^{2+}: d^{10} \to 0$$$$\text{V}^{2+}: d^{3} \to 3\ \checkmark,\qquad \text{Fe}^{2+}: d^{6} \to 4,\qquad \text{Co}^{2+}: d^{7} \to 3\ \checkmark$$Ions with 3 unpaired electrons: C (V$^{2+}$) and E (Co$^{2+}$).
Answer: C
Solution
The borax bead test gives a coloured bead only for ions with unpaired $d$-electrons.
$$\text{Zn}^{2+}: 3d^{10}\ (\text{colourless, no bead colour})$$$$\text{Fe}^{2+}: 3d^{6},\quad \text{Fe}^{3+}: 3d^{5},\quad \text{Cr}^{2+}: 3d^{4}\ (\text{all coloured, positive test})$$So $\text{Zn}^{2+}$ is eliminated. Among the remaining coloured ions we need the highest ionisation enthalpy — i.e. the ion formed by removing the most electrons / from the most positive species. Forming $\text{Fe}^{3+}$ (removal of the third electron) requires much more energy than forming the divalent ions $\text{Fe}^{2+}$ or $\text{Cr}^{2+}$.
Positive bead test + highest ionisation enthalpy → $\text{Fe}^{3+}$.
Answer: D
Solution
An ion is paramagnetic if it has at least one unpaired electron.
$$\text{Mn}^{2+}: 3d^{5} \to 5\ \text{unpaired} \ \checkmark$$$$\text{Cu}^{2+}: 3d^{9} \to 1\ \text{unpaired} \ \checkmark$$$$\text{Zn}^{2+}: 3d^{10} \to 0 \quad(\text{diamagnetic})$$$$\text{Yb}^{2+}: 4f^{14} \to 0 \quad(\text{diamagnetic})$$$$\text{Sc}^{3+}: 3d^{0} \to 0,\quad \text{La}^{3+}: 4f^{0} \to 0 \quad(\text{diamagnetic})$$$$\text{Gd}^{3+}: 4f^{7} \to 7\ \text{unpaired} \ \checkmark$$$$\text{Lu}^{3+}: 4f^{14} \to 0,\quad \text{Ti}^{4+}: 3d^{0} \to 0,\quad \text{Ce}^{4+}: 4f^{0} \to 0 \quad(\text{diamagnetic})$$Paramagnetic ions: $\text{Mn}^{2+}$, $\text{Cu}^{2+}$, $\text{Gd}^{3+}$ → 3.
Answer: 3
Solution
Configurations: $\text{Cr} = [\text{Ar}]3d^{5}4s^{1}$, $\text{Mn} = [\text{Ar}]3d^{5}4s^{2}$.
First ionisation ($\Delta_iH_1$): one $4s$ electron is removed from each. Mn’s $4s^2$ pair experiences a higher effective nuclear charge, so
$$\Delta_iH_1:\ \text{Mn} > \text{Cr}\quad(717 > 653\ \text{kJ mol}^{-1})$$So statement C is correct (and A is wrong).
Second ionisation ($\Delta_iH_2$):
$$\text{Cr}^{+} = 3d^{5}\ (\text{stable half-filled — hard to remove}),\qquad \text{Mn}^{+} = 3d^{5}4s^{1}\ (\text{remove the loose }4s)$$Removing an electron from the extra-stable $3d^5$ of $\text{Cr}^{+}$ needs more energy:
$$\Delta_iH_2:\ \text{Cr} > \text{Mn}\quad(1592 > 1509\ \text{kJ mol}^{-1})$$So statement B is correct (and D is wrong).
Correct: B and C.
Answer: B
Solution
Interstitial compounds form when small atoms (H, C, N, B) occupy the interstitial holes of a transition-metal lattice. Their characteristic properties are:
- high melting points, often higher than the pure metal — a true property;
- they retain metallic conductivity — a true property;
- they are chemically inert and are usually non-stoichiometric — a true property;
- they are very hard and rigid (not soft) and are essentially metallic/covalent, not ionic.
Hence the statement “very soft and ionic in nature” is the one that is NOT correct.
Answer: B
Solution
Statement I. More unpaired $d$-electrons available for metallic bonding give stronger interatomic bonds, hence a higher enthalpy of atomisation. Correct.
Statement II. Match the geometry to the splitting pattern:
- $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$ is octahedral, where $t_{2g}\,(d_{xy}, d_{xz}, d_{yz})$ lie below $e_g\,(d_{x^2-y^2}, d_{z^2})$: $$d_{xy}=d_{xz}=d_{yz} < d_{x^2-y^2}=d_{z^2}\ \checkmark$$
- $[\text{NiCl}_4]^{2-}$ is tetrahedral, where $e\,(d_{x^2-y^2}, d_{z^2})$ lie below $t_2\,(d_{xy}, d_{xz}, d_{yz})$: $$d_{x^2-y^2}=d_{z^2} < d_{xy}=d_{xz}=d_{yz}\ \checkmark$$
Both splitting patterns are correctly assigned. Correct.
Both statements are correct.
Answer: A
Solution
Statement I. An ion is coloured if it has unpaired $d$-electrons ($d^1$–$d^9$).
$$\text{Ti}^{4+}: d^{0}\ (\text{colourless}),\ \ \text{V}^{2+}: d^{3},\ \ \text{Mn}^{2+}: d^{5},\ \ \text{Fe}^{3+}: d^{5},\ \ \text{Cr}^{2+}: d^{4}$$- $[\text{Ti}^{4+}, \text{V}^{2+}]$: Ti$^{4+}$ colourless → not both coloured.
- $[\text{V}^{2+}, \text{Mn}^{2+}]$: both coloured $\checkmark$
- $[\text{Mn}^{2+}, \text{Fe}^{3+}]$: both coloured $\checkmark$
- $[\text{V}^{2+}, \text{Cr}^{2+}]$: both coloured $\checkmark$
Count $= 3$. Correct.
Statement II. A species is diamagnetic if it has no unpaired electrons ($f^0$ or $f^{14}$ here).
$$\text{La}^{3+}: 4f^{0},\ \text{Yb}^{2+}: 4f^{14}\ \Rightarrow\ \text{both diamagnetic}\ \checkmark$$$$\text{Lu}^{3+}: 4f^{14},\ \text{Ce}^{4+}: 4f^{0}\ \Rightarrow\ \text{both diamagnetic}\ \checkmark$$$$\text{Ac}^{3+}: 5f^{0},\ \text{Lr}^{3+}: 5f^{14}\ \Rightarrow\ \text{both diamagnetic}\ \checkmark$$Count $= 3$. Correct.
Both statements are correct.
Answer: A
Solution
Statement I. Transition-metal catalysts use both their 3d and 4s electrons (their ability to adopt variable oxidation states and provide electrons) to form bonds with adsorbed reactants. Restricting this to “3d electrons only” is wrong. Incorrect.
Statement II. Adsorption does increase the concentration of reactants on the catalyst surface, but this weakens the bonds within the reacting molecules (lowering the activation energy), it does not strengthen them. Incorrect.
Both statements are incorrect.
Answer: B