d- and f-Block Elements: JEE Main 2026 PYQs

Answer: Option (1)

Key idea: Redox reactions with dichromate ion; identifying reducing agents that can reduce Cr(VI) to Cr(III)

1. $K_2Cr_2O_7$ is orange due to $Cr_2O_7^{2-}$ with Cr in +6 oxidation state. In acidic medium, it acts as a strong oxidizing agent.
2. When reduced, $Cr^{6+} \to Cr^{3+}$, changing color from orange to green. We need species that can act as reducing agents.
3. Analyze each option: Reducing agents include $Fe^{2+}$, $Sn^{2+}$, $I^-$, $S^{2-}$, $NO_2^-$, $SO_2$, $C_2O_4^{2-}$ (can be oxidized).
4. Non-reducing agents: $Fe^{3+}$, $Sn^{4+}$, $SO_4^{2-}$ (already in high/highest oxidation states).
5. Option (1): $Fe^{2+}, Sn^{2+}, I^-, S^{2-}$ - All are reducing agents that can reduce dichromate.
6. Options (2), (3), (4) contain species like $Fe^{3+}$, $Sn^{4+}$, or $SO_4^{2-}$ which cannot reduce dichromate.

Answer: 3

Key idea: Paramagnetism and electronic configuration of d-block and f-block ions

1. Determine the electronic configuration of each metal atom using atomic numbers
2. Remove appropriate electrons to get the ion configuration (remove $s$ electrons first, then $d$ or $f$)
3. For d-block: $\text{Mn}^{2+}$ has $3d^5$ (5 unpaired), $\text{Cu}^{2+}$ has $3d^9$ (1 unpaired), $\text{Zn}^{2+}$ has $3d^{10}$ (0 unpaired)
4. For f-block: $\text{Gd}^{3+}$ has $4f^7$ (7 unpaired), $\text{Yb}^{2+}$ has $4f^{14}$ (0 unpaired)
5. Ions with completely filled or empty subshells: $\text{Sc}^{3+}$, $\text{La}^{3+}$, $\text{Lu}^{3+}$, $\text{Ti}^{4+}$, $\text{Ce}^{4+}$ are diamagnetic
6. Count paramagnetic ions (those with unpaired electrons): $\text{Mn}^{2+}$, $\text{Cu}^{2+}$, $\text{Gd}^{3+}$ = 3

Answer: Option (3)

Key idea: Stability of oxidation states in d-block and f-block elements, electronic configurations, and periodic trends

1. **Statement A**: Mo(VI) and W(VI) are actually MORE stable than Cr(VI) as stability of +6 state increases down group 6. Cr(VI) is a strong oxidant. Statement A is **incorrect**.
2. **Statement B**: $Ce^{4+}$ has [Xe]$4f^0$ and $Tb^{4+}$ has [Xe]$4f^7$ (stable configs) so they are oxidants. $Eu^{2+}$ has [Xe]$4f^7$ and $Yb^{2+}$ has [Xe]$4f^{14}$ (stable configs) so they are reductants. Statement B is **correct**.
3. **Statement C**: $Cm^{3+}$ has [Rn]$5f^7$ (7 unpaired e⁻) but $Am^{3+}$ has [Rn]$5f^6$ (6 unpaired e⁻). Statement C is **incorrect**.
4. **Statement D**: Actinoid contraction is greater than lanthanoid contraction due to poorer shielding of 5f electrons compared to 4f electrons. Statement D is **correct**.
5. Correct statements are **B and D only**. Answer is option **(3)**.

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Key idea: Properties of chromium compounds - oxidizing nature of dichromate, pH-dependent chromate-dichromate equilibrium, and structural features

Answer: Option (3)

Key idea: Oxidation states of transition metals, electronic configuration of ions, and unpaired electrons

1. Find oxidation state of Mn in highest fluoride MnF₄: Mn is $+4$
2. Find oxidation state of Mn in highest oxide Mn₂O₇: Mn is $+7$
3. Calculate $|x| = |4-7| = 3$, so we need ions with 3 unpaired electrons
4. Check each ion: V²⁺ has configuration [Ar]3d³ with 3 unpaired electrons
5. Fe²⁺ has [Ar]3d⁶ with 4 unpaired electrons, Co²⁺ has [Ar]3d⁷ with 3 unpaired electrons
6. Zn²⁺ has [Ar]3d¹⁰ with 0 unpaired electrons, Se³⁺ has [Ar]3d¹⁰4p¹ with 1 unpaired electron
7. Only C (V²⁺) and E (Co²⁺) have 3 unpaired electrons

Answer: Option (4)

Key idea: Qualitative Analysis - Group III Cations and Oxidation States of d-block elements

1. Identify the group reagent: $H_2S$ in presence of $NH_4OH$ is the Group III reagent in qualitative analysis
2. Analyze precipitation behavior: Among $Fe^{3+}$, $Pb^{2+}$, $Cu^{2+}$, and $Mn^{2+}$, the Group III cations are $Fe^{3+}$ and $Mn^{2+}$
3. $Pb^{2+}$ and $Cu^{2+}$ precipitate in Group II (acidic medium), not in basic medium with $NH_4OH$
4. When $H_2S$ is passed in presence of $NH_4OH$: $Mn^{2+}$ forms $MnS$ precipitate (flesh-colored sulfide)
5. Determine highest oxidation state of Mn: Manganese exhibits oxidation states from $+2$ to $+7$
6. The highest oxidation state of Mn is $+7$ (found in permanganate ion $MnO_4^-$)

Answer: Option (4)

Key idea: Redox reactions of transition metal compounds; Reduction of dichromate by SO₂

1. Identify that $SO_2$ acts as a reducing agent in acidic medium
2. Analyze that the green color indicates formation of $Cr^{3+}$ ions
3. Write the reduction reaction: $Cr_2O_7^{2-}$ (orange) $+ 3SO_2 + 2H^+ \rightarrow 2Cr^{3+}$ (green) $+ 3SO_4^{2-} + H_2O$
4. Recognize this as a characteristic test for potassium dichromate
5. Therefore, compound X is $K_2Cr_2O_7$

Answer: Option (3)

Key idea: Industrial catalysts and electronic configurations of transition metals

1. Statement A: V₂O₅ (acidic oxide) is used in H₂SO₄ manufacture, not basic vanadium oxide. **FALSE**
2. Statement B: Ziegler-Natta uses TiCl₄ where Ti⁴⁺ has configuration [Ar]3d⁰, so $\mu = \sqrt{n(n+2)} = 0$ BM, not 2.84 BM. **FALSE**
3. Statement C: Ziegler-Natta uses Al(C₂H₅)₃ where Al (p-block) is in +3 oxidation state. **TRUE**
4. Statement D: Wacker process uses PdCl₂ where Pd²⁺ has configuration [Kr]4d⁸, giving 8 d-electrons. **TRUE**
5. Correct statements are C and D only

Answer: Option (2)

Key idea: Ionization enthalpy trends in d-block elements based on electronic configuration stability

1. Write electronic configurations: Cr = [Ar] $3d^5 4s^1$, Mn = [Ar] $3d^5 4s^2$
2. For first ionization: Mn has higher nuclear charge (Z=25) than Cr (Z=24), so $\Delta_i H_1$: Mn > Cr (Statement C is correct)
3. After first ionization: Cr$^+$ = [Ar] $3d^5$ (half-filled, very stable), Mn$^+$ = [Ar] $3d^5 4s^1$
4. For second ionization: Cr$^+$ must lose electron from stable $d^5$ (difficult), while Mn$^+$ loses from $4s$ to achieve stable $d^5$ (easier)
5. Therefore $\Delta_i H_2$: Cr > Mn (Statement B is correct)
6. Statements B and C are correct, so answer is option (2)

Answer: Option (2)

Key idea: Catalytic properties of transition metals - electron involvement and bond weakening mechanism

1. Analyze Statement I: First row transition metals use 3d electrons only for catalysis
2. Statement I is INCORRECT - transition metals use both $3d$ and $4s$ electrons (not only 3d) for bond formation in catalysis
3. Analyze Statement II: Adsorption increases reactant concentration and strengthens bonds
4. Statement II is INCORRECT - while adsorption does increase concentration, it WEAKENS bonds in reactant molecules (not strengthens)
5. Bond weakening is essential for catalysis as it lowers activation energy and facilitates reaction
6. Since both statements are incorrect, the answer is option (2)