JEE D and F-Block Elements PYQs with Video Solutions PDF

Lanthanide contraction is the steady decrease in the size of lanthanide elements due to the poor shielding effect of the 4f electrons.

This poor shielding fails to counterbalance the increasing nuclear charge, causing the outer shells to be pulled in tightly.

As a result, the normal increase in size down a group from the 4d series (Zr) to the 5d series (Hf) is canceled out. Zr and Hf end up with almost identical atomic radii (∼160 pm).

Correct Option: (B)

1. Analyze the Disproportionation of Manganate Ion:
In a neutral, faintly alkaline, or acidic medium, the green manganate ion ($$\text{MnO}_4^{2-}$$) is thermodynamically unstable and undergoes self-oxidation and reduction (disproportionation). The balanced chemical equation for this process is:
$$3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}$$

2. Identify the Compounds based on Physical Properties:

• The purple color in the solution is due to the formation of the permanganate ion ($$\text{MnO}_4^-$$), which crystallizes out as potassium permanganate ($$\text{KMnO}_4$$). This is compound A .
• The dark brown precipitate is manganese dioxide ($$\text{MnO}_2$$). This is compound B .

3. Calculate the Changes in Oxidation States:
Let us calculate the oxidation state of Manganese ($$\text{Mn}$$) in each chemical species:

• In potassium manganate ($$\text{K}_2\text{MnO}_4$$): $$2(+1) + x + 4(-2) = 0 \implies x = +6$$
• In compound A ($$\text{KMnO}_4$$): $$1(+1) + x + 4(-2) = 0 \implies x = +7$$
• In compound B ($$\text{MnO}_2$$): $$x + 2(-2) = 0 \implies x = +4$$

The oxidation state changes from $$+6$$ in the starting reactant to $$+7$$ in compound A ($$\text{KMnO}_4$$). Therefore, the change in oxidation state is:

$$(+7) - (+6) = +1$$

Answer:
The correct formulas and corresponding oxidation state adjustment match Option B .

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The Reaction :
$$\underbrace{\mathrm{2Na}_{\mathrm{2}}\mathrm{CrO}_{\mathrm{4}}}_{\mathrm{(Yellow)}}\mathrm{+2CO}_{\mathrm{2}}\mathrm{+H}_{\mathrm{2}}\mathrm{O\longrightarrow }\underbrace{\mathrm{Na}_{\mathrm{2}}\mathrm{CrGrid}_{\mathrm{2}}\mathrm{O}_{\mathrm{7}}}_{\mathrm{(Orange)}}\mathrm{+2NaHCO}_{\mathrm{3}}$$
$$\text{}$$
Hence, Option A is correct.

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The primary reason why actinides show a much wider variety of oxidation states ( ranging from $$+2$$ up to $$+7$$, such as in Uranium, Neptunium, and Plutonium) compared to lanthanoids ( which mostly stick to $$+3$$) comes down to orbital energy gaps :

• In Lanthanoids: The energy gap between the $$4f$$ and $$5d$$ subshells is relatively large . Because the $$4f$$ electrons are deeply buried inside the atom, they are strongly held by the nucleus and do not easily participate in chemical bonding. Thus, bonding is mostly restricted to the outer $$6s$$ and $$5d$$ electrons, leading predominantly to a stable $$+3$$ oxidation state.
  $$\text{}$$
• In Actinides: The energy gap between the $$5f$$ and $$6d$$ subshells is significantly smaller . Additionally, $$5f$$ orbitals extend further out from the core into space compared to $$4f$$ orbitals. Because of this tiny energy difference, electrons from the $$5f$$, $$6d$$, and $$7s$$ subshells can all take part simultaneously in forming chemical bonds, allowing actinides to achieve much higher and more varied oxidation states.

Hence, Option B is correct

We need to determine the total number of equivalent $$\text{Cr-O}$$ bonds in the dichromate ion ($$\text{Cr}_2\text{O}_7^{2-}$$) by understanding its structural arrangement and bonding characteristics as per the JEE Main syllabus.

1. Structural Framework of $$\text{Cr}_2\text{O}_7^{2-}$$

The dichromate ion consists of two tetrahedral $$\text{CrO}_4$$ units sharing a single corner. This sets up a central, bent bridging bridge within the polyatomic ion:

$$\text{Cr - O}_{\text{bridge}} - \text{Cr}$$

Because of this arrangement, the total oxygen atoms are divided into two distinct structural categories:

• Bridging Oxygen Atom: $$1$$ central oxygen atom that links both chromium centers.
• Terminal Oxygen Atoms: $$6$$ peripheral oxygen atoms ($$3$$ attached to each individual chromium atom).

2. Valence Electron & Bonding Analysis

Let's look at how the valence electrons partition between these two structural regions:

• The Central Bridging Bonds:

  The central oxygen atom shares one electron pair with each of the two chromium atoms to form two localized, distinct single bonds ($$\text{Cr - O}_{\text{bridge}}$$). Because these electrons are locked in a fixed position between two specific nuclei, they do not participate in resonance with the rest of the molecule. This leaves these two bridging bonds longer than the others.

• The Terminal Bonds and Resonance Delocalization:

  Each chromium atom ($$\text{Cr}$$, Group 6) uses its remaining valence d-electrons to form bonds with its $$3$$ assigned terminal oxygen atoms. The negative charges ($$-2$$ total) along with the unshared d-orbital electron pairs are completely and symmetrically delocalized across these terminal positions via $$\pi$$-bonding ($$d\pi - p\pi$$ orbital overlap).

  Due to this continuous and symmetrical resonance, the double-bond character is evenly shared among all the terminal oxygen positions. This dynamic sharing averages out any single/double bond differences, making all the terminal positions structurally identical.

Conclusion:

While the $$2$$ internal bridging $$\text{Cr-O}$$ bonds are equal to each other, they are structurally different (longer) than the outer bonds. The $$6$$ terminal $$\text{Cr-O}$$ bonds, spread evenly across the two symmetrical halves of the ion ($$3 + 3 = 6$$), are completely identical to each other in bond length and bond energy due to resonance.

Therefore, the total number of exactly equivalent $$\text{Cr-O}$$ bonds is 6 .

Answer: 6

• Step 1: Write the electronic configuration for Trivalent Europium (Eu³⁺). Europium (Z=63) is $$[\text{Xe}]\,4f^7 6s^2$$. Removing 3 electrons gives $$\text{Eu}^{3+} = [\text{Xe}]\,4f^6$$.
• Step 2: Apply the spin-only calculation. A 4f⁶ system contains 6 unpaired electrons. According to Hund's rule, the ground state term should align total spins. However, when evaluating the total angular component, the vector summation for the lowest spin multiplet can approach a net angular state where J = 0.
• Step 3: For J = 0, the calculated magnetic moment at absolute zero temperature is exactly 0 BM.
• Step 4: At room temperature, the energy gap between this ground state (J=0) and the first excited states (J=1, J=2) is small enough that thermal energy populates those higher states. These excited states introduce a strong magnetic contribution, giving Eu³⁺ an experimental magnetic moment of roughly $$3.4\text{--}3.6\text{\ BM}$$.

Cerium has atomic number 58 and its ground-state electronic configuration is $$[Xe]\,4f^{1}5d^{1}6s^{2}$$.

Formation of $$Ce^{4+}$$ (Statement I)
To obtain $$Ce^{4+}$$ we remove four electrons: two from $$6s$$, one from $$5d$$ and one from $$4f$$, giving the configuration $$[Xe]$$ — a completely filled xenon core. Achieving a noble-gas configuration imparts extra stability, so the formation of $$Ce^{4+}$$ is indeed favoured. Hence Statement I is true.

Oxidising nature of $$Ce^{4+}$$ (Statement II)
Because the $$[Xe]$$ core is very stable, the $$Ce^{4+} \rightarrow Ce^{3+}$$ reduction (which adds one electron back into the empty 4f subshell) is highly favourable. The standard reduction potential is $$E^\circ = +1.61\;\text{V}$$, making $$Ce^{4+}$$ one of the strongest oxidants among the lanthanoids. Therefore Statement II is also true.

Since both statements are correct, the appropriate choice is:

Option B which is: Both Statement I and Statement II are true

The question asks for the pair that does NOT match the industrial process with the catalyst/reagent actually employed.

Option A (Photography - AgBr)
• Conventional black-and-white photographic films contain the light-sensitive salt $$AgBr$$ (often mixed with a little $$AgCl$$).
• When exposed to light, $$AgBr$$ decomposes to metallic silver, forming the latent image.
Hence this pair is correct.

Option B (Polythene preparation - $$TiCl_4,\;Al(CH_3)_3$$)
• High-density polyethylene (HDPE) is produced by Ziegler-Natta polymerisation of ethene.
• A typical Ziegler-Natta catalyst system consists of a titanium(IV) halide such as $$TiCl_4$$ and an organo-aluminium co-catalyst such as $$Al(C_2H_5)_3$$ or $$Al(CH_3)_3$$.
Therefore this pair is also correct.

Option C (Haber process - Iron)
• In the Haber process for ammonia synthesis, finely divided iron (promoted with $$K_2O$$ and $$Al_2O_3$$) is the standard heterogeneous catalyst.
So this pair is correct as well.

Option D (Wacker process - $$PtCl_2$$)
• The Wacker (or Wacker-Hoechst) process oxidises ethene to ethanal (acetaldehyde).
• The catalytic system employed is $$PdCl_2$$ with $$CuCl_2$$ in aqueous medium:
$$CH_2\!=\!CH_2 + PdCl_2 + H_2O \rightarrow CH_3CHO + Pd + 2HCl$$
• $$PtCl_2$$ is NOT used in the Wacker process.
Thus this is the mismatched (incorrect) pair.

Therefore, the incorrect pair is:
Option D which is: Wacker process - $$PtCl_2$$

1. Identify Compound X: The reaction of an acidic dichromate solution with $$\text{H}_2\text{O}_2$$ forms Chromium pentoxide, $$\text{CrO}_{5}$$ (also called chromium oxide peroxide). The ether layer stabilizes this blue-colored, unstable molecule.
2. Analyze Structure and Linkages: $$\text{CrO}_{5}$$ adopts a classic "butterfly" structure.
   • It contains one double-bonded oxo oxygen atom $$(\text{Cr}=\text{O})$$ [oxidation state = $$-2$$].
   • It contains four peroxide oxygen atoms forming two distinct peroxy rings $$(-\text{O}-\text{O}-)$$ [oxidation state of each peroxide oxygen = $$-1$$].
3. Calculate Oxidation State: Let $$x$$ be the oxidation state of Chromium:
   $$x+1(-2)+4(-1)=0\implies x=+6$$
   • Oxidation state of Cr = +6
   • Total peroxy linkages = 2

Correct Answer: A