D And F Block Elements – JEE Main ChemistryPractice Questions & PYQs

Lanthanoid contraction is primarily driven by the diffuse geometry of $4f$ orbitals, which results in remarkably ineffective shielding of the outermost electrons from the positive nuclear charge. Since this poor shielding cannot adequately offset the rising nuclear charge as atomic number increases, the nucleus exerts a progressively stronger attraction on the valence electrons. This intense pulling force causes a systematic reduction in atomic and ionic radii across the lanthanide series.

According to Fajans' rule, the ionic or covalent nature of a chemical bond is determined by the polarizing capacity of the central metal cation. Because the higher oxidation state of $\mathrm{M}$ involves a greater positive charge and a smaller cationic radius compared to the lower +2 state, the electron cloud of the chloride ion becomes more strongly polarized in the presence of the +4 cation, shifting the character of the bond towards covalency. This makes $\mathrm{MCl}_4$ more covalent, which inherently implies that $\mathrm{MCl}_2$ maintains a higher degree of ionic character.

Lanthanide contraction fundamentally arises from the inadequate ability of $4 f$ electrons to screen the nuclear charge from outer electrons. Specifically, the complex and diffuse spatial distribution of the $4 f$ subshell means these electrons provide significantly poorer shielding for $5 d$ electrons than would otherwise be expected. Consequently, this deficient shielding results in a stronger net attractive force from the nucleus on the outer electron shells, leading to the characteristic reduction in atomic radii observed across the lanthanide series.

The reaction between calomel ($\mathrm{Hg}_2\mathrm{Cl}_2$) and ammonium hydroxide serves as a characteristic chemical test involving the displacement of ions within the mercury(I) compound. When these substances interact, the aqueous ammonia facilitates a chemical change that yields amino-mercury(II) chloride, $\mathrm{HgNH}_2\mathrm{Cl}$, along with ammonium chloride and water as byproducts. This specific transformation is described by the balanced equation $\mathrm{Hg}_2\mathrm{Cl}_2 + 2\mathrm{NH}_4\mathrm{OH} \longrightarrow \mathrm{HgNH}_2\mathrm{Cl} + \mathrm{NH}_4\mathrm{Cl} + 2\mathrm{H}_2\mathrm{O}$.

When acidified potassium dichromate reacts with potassium iodide, the dichromate ion acts as a strong oxidizing agent and undergoes reduction. In the resulting balanced redox reaction, $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 6\mathrm{I}^{-} + 14\mathrm{H}^{+} \longrightarrow 3\mathrm{I}_2 + 7\mathrm{H}_2\mathrm{O} + 2\mathrm{Cr}^{3+}$, the chromium species is reduced to the stable $\mathrm{Cr}^{3+}$ ion, which corresponds to an oxidation state of +3.

Heating a mixture of copper(I) oxide and copper(I) sulfide is a fundamental example of self-reduction, a key process in copper metallurgy where no additional reducing agent is necessary. In this reaction, copper(I) oxide acts as an oxidizing agent while copper(I) sulfide acts as a reducing agent, causing the sulfur to be liberated as sulfur dioxide gas while metallic copper is produced. This chemical transformation is represented by the equation:

$2\mathrm{Cu}_2\mathrm{O} + \mathrm{Cu}_2\mathrm{S} \rightarrow 6\mathrm{Cu} + \mathrm{SO}_2$

The lanthanide contraction occurs because 4f electrons offer poor shielding of the nuclear charge, which results in a stronger attraction on valence electrons that counteracts the expected size increase from adding a new shell. Because this effect offsets the expansion typically seen when moving from the 4d series to the 5d series in groups 4 through 12, the atomic radii remain nearly constant. Consequently, $\mathrm{Zr}$ and $\mathrm{Hf}$ possess essentially the same atomic radius.

The reaction between copper(II) sulfate and excess potassium iodide is a redox process where copper(II) acts as an oxidizing agent, converting iodide ions into iodine while being reduced itself to copper(I). The balanced equation for this transformation is $4\mathrm{KI} + 2\mathrm{CuSO}_4 \rightarrow \mathrm{I}_2 + \mathrm{Cu}_2\mathrm{I}_2 + 2\mathrm{K}_2\mathrm{SO}_4$, which clearly shows that the copper product is copper(I) iodide, $\mathrm{Cu}_2\mathrm{I}_2$, rather than the unstable copper(II) iodide, $\mathrm{CuI}_2$. Consequently, the assertion that $\mathrm{CuI}_2$ is formed is chemically incorrect. In the subsequent titration, the liberated iodine is reduced to iodide as it reacts with sodium thiosulfate according to $\mathrm{I}_2 + 2\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 \rightarrow \mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6 + 2\mathrm{NaI}$, confirming that the remaining options accurately describe the standard reaction pathway.

Cerium, which has an atomic number of 58, commonly exhibits both +3 and +4 oxidation states, with the +3 state being significantly more stable. Contrary to the suggestion that it cannot exist in solution, the +4 oxidation state is well-documented and stable in aqueous environments. Because the $Ce^{4+}$ ion has a strong tendency to gain an electron to revert to the more stable $Ce^{3+}$ form, it functions as a powerful oxidizing agent in chemical reactions.

The maximum oxidation state of transition metal atoms is determined by the total number of electrons present in the $(n-1) d$ and $n s$ subshells that are energetically accessible for removal. When evaluating the provided electronic configurations, the $(n-1) d^5 n s^2$ arrangement contains a sum of $5 + 2 = 7$ electrons. This total number of available valence electrons allows the element to achieve a maximum oxidation state of $+7$, which is higher than the potentials of the other listed configurations.

This phenomenon is known as lanthanide contraction, where the ionic radius decreases as the atomic number increases across the lanthanide series. Because the 4f electrons offer poor shielding against the increasing nuclear charge, the valence shell is pulled more strongly toward the nucleus as we move from $\mathrm{La}^{3+}$ (atomic number 57) to $\mathrm{Lu}^{3+}$ (atomic number 71). Since the ionic radius is inversely proportional to the atomic number, we can determine the unknown radius by solving the equation $\frac{r_{\mathrm{Lu}}}{1.06} = \frac{57}{71}$. Performing this calculation results in a radius of approximately 0.85 Å.

The thermal decomposition of metal nitrates is fundamentally governed by the position of the metal in the reactivity series. While most nitrates of moderately reactive metals decompose to form a stable metal oxide, nitrogen dioxide, and oxygen, the low affinity for oxygen exhibited by noble metals like silver prevents the formation of a stable oxide at high temperatures.

As a result, heating silver nitrate leads to the complete breakdown of the compound, leaving behind the pure metal residue rather than an oxide. This process is represented by the reaction $\mathrm{AgNO}_3 \rightarrow \mathrm{Ag} + \mathrm{NO}_2 + \frac{1}{2} \mathrm{O}_2$, which confirms that silver is the only element among the choices that remains in its metallic state following strong heating.

The interconversion between chromate and dichromate ions relies on the pH of the solution, with chromate ions existing in alkaline or neutral conditions and dichromate ions becoming stable in acidic media. Treating potassium chromate with excess dilute nitric acid provides the necessary acidic environment to shift the equilibrium, yielding potassium dichromate and water as described by the chemical equation $2\mathrm{K}_2\mathrm{CrO}_4 + 2\mathrm{HNO}_3 \rightarrow \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + 2\mathrm{KNO}_3 + \mathrm{H}_2\mathrm{O}$. Consequently, this reaction leads to the production of dichromate ions alongside water.

The red solid is mercuric iodide, $\mathrm{HgI}_2$, which remains insoluble in water but readily dissolves upon the addition of potassium iodide to form the soluble complex $\mathrm{K}_2[\mathrm{HgI}_4]$. When this compound is subjected to heat, it undergoes thermal decomposition, resulting in the liberation of violet iodine vapors and the appearance of mercury metal droplets on the cooler parts of the test tube.

Creating premium quality flat glass requires floating the molten mixture over a liquid metal surface to ensure perfect smoothness and consistency. Mercury is chosen for this purpose because it uniquely exists as a liquid at room temperature, providing a stable, molten foundation that does not solidify prematurely during the glass production process.

When determining the electronic configuration of transition metal ions, the fundamental rule is that electrons are always removed from the orbital with the highest principal quantum number first. Iron has an atomic number of 26 and a ground-state electronic configuration of $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2$. Because the $4s$ subshell has a higher principal quantum number than the $3d$ subshell, the two electrons required to form the $\mathrm{Fe}^{2+}$ ion are pulled from the $4s$ orbital. This leaves the configuration as $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^0$, which confirms that 6 electrons remain in the $d$-subshell.

The second ionization enthalpy depends on the stability of the electronic configuration of the singly charged ion, where an exceptionally stable state makes further ionization significantly more demanding. Among the elements vanadium (23), chromium (24), manganese (25), and iron (26), the chromium atom displays unique behavior because its singly charged ion $\text{Cr}^+$ achieves a stable half-filled $3d^5$ configuration. Because this specific arrangement provides extra stability to the ion, removing a second electron requires a notably higher amount of energy compared to the other options.

Qualitative separation between silver halides relies on the differences in their solubility product values, which determine whether a specific complexing agent can effectively draw the ions into solution. Because $\mathrm{AgCl}$ is significantly more soluble than $\mathrm{AgI}$, the addition of ammonia allows for the formation of the soluble complex $[\mathrm{Ag}(\mathrm{NH}_3)_2]^+$, whereas the extremely low solubility of $\mathrm{AgI}$ prevents this reaction from occurring. The remaining statements are chemically inaccurate: ferric ions react with potassium ferrocyanide to produce a Prussian blue precipitate rather than a green one, boiling a solution containing calcium and bicarbonate ions results in the formation of calcium carbonate precipitate instead of a double salt, and manganese salts exhibit their characteristic violet color only in an oxidizing flame, not the reducing flame.

The differentiation of group III cations relies on the common ion effect, where increasing the concentration of $\mathrm{NH}_4^+$ ions suppresses the dissociation of $\mathrm{NH}_4\mathrm{OH}$ and thereby limits the availability of $\mathrm{OH}^-$ ions. This precise control ensures that $\mathrm{Fe}^{3+}$ ions form a characteristic reddish-brown $\mathrm{Fe(OH)_3}$ precipitate, whereas $\mathrm{Cr}^{3+}$ ions generate a distinct bluish-green $\mathrm{Cr(OH)_3}$ precipitate. By modulating the $\mathrm{NH}_4^+$ concentration, the reaction environment is optimized to stabilize these specific hydroxide precipitates, allowing for their effective identification based on their unique visual appearances.

Cerium, situated at the beginning of the lanthanide series, demonstrates a distinct preference for +3 and +4 oxidation states due to the stability of its electronic structure. While the +3 state is common among all lanthanides, Cerium uniquely benefits from the +4 state because it allows the ion to achieve a highly stable, empty f-orbital state represented by the configuration $\text{[Xe]} 4f^0$, which mimics the inert noble gas structure of Xenon.