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

The redox behavior of potassium permanganate is significantly dictated by the pH of the solution, which determines the final oxidation state of the iodine product. In an acidic environment, the permanganate ion efficiently oxidizes iodide ions ($\mathrm{I}^{-}$) to produce elemental iodine ($\mathrm{I}_2$). However, under neutral or faintly alkaline conditions, the reduction process follows a different pathway involving manganese dioxide, which provides a strong enough oxidizing environment to convert the iodide ions ($\mathrm{I}^{-}$) into the iodate ion ($\mathrm{IO}_3^{-}$).

In the metal carbonyls $\mathrm{Ni}(\mathrm{CO})_4$ and $\mathrm{Fe}(\mathrm{CO})_5$, the carbon monoxide ligand is a neutral molecule, meaning the oxidation state of the central metal atom is mathematically determined to be zero, which confirms that Assertion A is correct. Reason R is incorrect because CO is classified as a $\pi$-acceptor ligand, which stabilizes low metal oxidation states via metal-to-ligand back-bonding ($d_\pi \rightarrow p_\pi^*$) rather than by acting as a $\pi$-donor to the metal.

Nessler's reagent is chemically represented by the formula $\mathrm{K}_2 [ \mathrm{HgI}_4 ]$, which identifies it as a solution containing potassium tetraiodomercurate. Evaluating this chemical structure reveals that the compound is strictly composed of potassium ($\mathrm{K}$), mercury ($\mathrm{Hg}$), and iodine ($\mathrm{I}$). Consequently, nitrogen ($\mathrm{N}$) is the only element among the choices that is not found within the composition of this reagent.

Nickel metal serves as a crucial catalyst for the hydrogenation of unsaturated vegetable oils into solid, edible fats. In contrast, silicon hydrides, such as $SiH_4$, are classified as electron-precise because they possess the exact number of electrons required to form standard covalent bonds, which directly contradicts the claim that they are electron-rich or electron-deficient. Consequently, the first statement aligns with established chemical principles regarding catalysis while the second statement incorrectly characterizes the nature of silicon hydrides.

The third ionization energy refers to the energy required to remove an electron from a $+2$ ion, making the inherent stability of that divalent electronic configuration the primary determining factor. In the lanthanide series, Europium and Ytterbium possess notably high third ionization energies because their divalent states, $\mathrm{Eu}^{+2}$ and $\mathrm{Yb}^{+2}$, correspond to the highly stable half-filled $4f^7$ and fully-filled $4f^{14}$ configurations, respectively. Because these subshells are already energetically favored, removing an additional electron requires significantly more energy compared to other lanthanides that do not possess such stable electron arrangements.

Dimethylglyoxime is a highly selective chelating agent used to identify nickel ions within a basic medium. When mixed with an ammoniacal solution of $\mathrm{Ni}^{+2}$, it coordinates with the metal to produce an insoluble, brilliant red precipitate. This chemical transformation is represented by:

$\mathrm{Ni}^{+2} + 2\,\mathrm{DMG} \xrightarrow{\mathrm{NH}_3(\mathrm{aq})} [\mathrm{Ni}(\mathrm{DMG})_2]$

The qualitative detection of chloride ions relies on the precipitation of $\mathrm{AgCl}$, which manifests as a curdy white solid $[A]$ upon adding silver nitrate to the acidified solution. When treated with ammonium hydroxide, this insoluble salt transforms into the soluble coordination complex $[\mathrm{Ag}(\mathrm{NH}_3)_2]\mathrm{Cl}$, generating the clear solution $[B]$. This reaction demonstrates the ability of silver chloride to form a stable ammine complex in the presence of ammonia, confirming that $[A]$ is $\mathrm{AgCl}$ and $[B]$ is $[\mathrm{Ag}(\mathrm{NH}_3)_2]\mathrm{Cl}$.

The stability of ferrous ions ($Fe^{2+}$) is a critical factor during the synthesis of ferrous ammonium sulphate, as they are inherently susceptible to chemical transformation when exposed to heat. Prolonged heating increases the reaction rate between these ions and dissolved oxygen, facilitating the undesired oxidation of ferrous ions into ferric ions ($Fe^{3+}$). By limiting the duration the mixture spends at elevated temperatures, the procedure effectively prevents this oxidation and ensures the final product maintains its intended chemical composition.

$\mathrm{Mn}_2\mathrm{O}_7$ adopts a covalent structure consisting of two $\mathrm{MnO}_4$ tetrahedra sharing a common oxygen atom, which inherently forms an Mn-O-Mn bridge. In this geometry, each manganese atom is tetrahedrally coordinated by four oxygen atoms, and there is no direct Mn-Mn bond present.

This specific redox transformation relies on the fact that the copper(II) ion acts as a mild oxidizing agent, capable of extracting electrons only from the iodide ion. Because iodide is a sufficiently potent reducing agent, it readily gives up electrons to copper(II), leading to the formation of the insoluble precipitate copper(I) iodide and the release of diatomic iodine, as represented by $2\mathrm{Cu}^{2+}+4\mathrm{I}^{-}\rightarrow\mathrm{Cu}_2\mathrm{I}_2+\mathrm{I}_2$. In contrast, chloride and bromide ions lack the necessary reducing strength to undergo this reaction with copper(II), making iodine the only halogen that facilitates this process.

In an acidic medium, potassium permanganate acts as a strong oxidizing agent, where the manganese atom is reduced from the +7 oxidation state in $\mathrm{KMnO}_4$ to the +2 oxidation state in the product $\mathrm{Mn}^{2+}$. The oxidation state of manganese in the permanganate ion, $\mathrm{MnO}_4^-$, is calculated by $x + 4(-2) = -1$, which yields $x = +7$. The reduction process during the reaction with potassium iodide is represented by the following half-reaction:

$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5\mathrm{e}^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$

The total change in the oxidation state of manganese is the absolute difference between the final and initial states, calculated as $|2 - 7| = 5$.

Copper matte is primarily a molten mixture of metal sulfides formed during the extraction process, characterized specifically by the presence of $\mathrm{FeS}$ and $\mathrm{Cu}_2\mathrm{S}$. Within the provided list of chemical species, only $\mathrm{Cu}_2\mathrm{S}$ is a constituent of this mixture, as $\mathrm{CuCO}_3$, $\mathrm{Cu}_2\mathrm{O}$, and $\mathrm{FeO}$ are not components of the copper matte phase.

The spatial orientation of atomic orbitals dictates their ability to interact during chemical reactions, with 4f orbitals being deeply buried within the atom and consequently less available for engagement. Because 5f orbitals possess greater radial extension from the nucleus, they are not as deeply buried as 4f orbitals, which significantly increases their accessibility. This structural difference allows 5f electrons to overlap more effectively with surrounding orbitals, facilitating much broader participation in chemical bonding for actinoids compared to their 4f counterparts.

The reduction of the permanganate ion, where manganese shifts from an oxidation state of +7 to +4, corresponds to a change of 3 units and results in the formation of manganese dioxide as represented by $MnO_4^- \rightarrow MnO_2$. This specific shift in oxidation state is characteristic of reactions occurring in neutral or weakly alkaline conditions. Given that the reaction between permanganate and thiosulphate involves this precise 3-unit change, it is favored in an aqueous neutral medium.

Nessler's reagent is an analytical solution frequently used in the detection of ammonia, and its chemical identity is given by the formula $\mathrm{K}_2 [\mathrm{HgI}_4]$. By observing this composition, it is apparent that the reagent consists of potassium, mercury, and iodine atoms bonded together. Since oxygen does not appear anywhere in this specific chemical formula, it is the element that is not present.

Transition metals exhibit variable oxidation states due to the participation of d-electrons, where Manganese achieves a +7 state in $\mathrm{Mn}_2\mathrm{O}_7$ and Ruthenium and Osmium reach the maximum +8 state in their oxides $\mathrm{RuO}_4$ and $\mathrm{OsO}_4$. Chromium in its +6 state acts as a potent oxidizing agent, contrasting with Scandium, which is limited to the +3 oxidation state and does not form a +4 state. These chemical properties confirm the validity of the first, second, and fourth statements while indicating the inaccuracy of the third.

The number of electrons involved in a reduction reaction is determined by the net change in the oxidation state of the central atom. The manganese in the permanganate ion ($\mathrm{MnO}_4^{-}$) possesses an oxidation state of $+7$, which decreases to $+4$ in manganese dioxide ($\mathrm{MnO}_2$). As the oxidation state shifts by 3 units, the balanced reduction half-reaction $\mathrm{MnO}_4^{-} + 4 \mathrm{H}^{+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{MnO}_2 + 2 \mathrm{H}_2\mathrm{O}$ confirms that 3 electrons must be consumed.

The oxidation state of a central metal atom in a compound is determined by equating the algebraic sum of the oxidation numbers of all constituent atoms to the net charge of the molecule, where oxygen is typically assigned -2 and chlorine is assigned -1. For chromyl chloride ($CrO_2Cl_2$), assigning $x$ as the oxidation state of chromium, the calculation is represented by the equation $x + 2(-2) + 2(-1) = 0$. Simplifying this expression leads to $x - 6 = 0$, which results in $x = +6$.

Copper(II) ions act as oxidizing agents in the presence of iodide ions, which function as reducing agents to convert $\mathrm{Cu}^{2+}$ into $\mathrm{Cu}^{+}$. The initial reaction product, $\mathrm{CuI}_2$, is unstable and quickly decomposes into a white precipitate of cuprous iodide, $\mathrm{Cu}_2\mathrm{I}_2$, along with the release of iodine. This liberated iodine is then titrated against sodium thiosulphate, resulting in the reduction of iodine and the simultaneous oxidation of thiosulphate ions into tetrathionate ions, represented by the chemical formula $\mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6$. Therefore, the white precipitate $X$ corresponds to $\mathrm{Cu}_2\mathrm{I}_2$, while the compound $Y$ formed during titration is $\mathrm{Na}_2\mathrm{S}_4\mathrm{O}_6$.

The ability of oxide ions to stabilize higher oxidation states in transition metals is a fundamental characteristic of d-block chemistry, confirming the validity of statement (B). In comparing electrochemical behaviors, the standard reduction potential for manganese, $E^0_{\mathrm{Mn}^{3+}/\mathrm{Mn}^{2+}} = +1.57$ V, is substantially greater than that of iron, $E^0_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = +0.77$ V, which contradicts the assertion in statement (A).

Because the electrode potential for the $\mathrm{Cr}^{3+}/\mathrm{Cr}^{2+}$ couple is negative at $-0.26$ V, the $\mathrm{Cr}^{2+}$ ion acts as a strong reducing agent capable of liberating hydrogen gas from dilute acids, making statement (C) true. Regarding magnetic properties, the $\mathrm{V}^{2+}$ ion contains 3 unpaired electrons, and applying the spin-only formula $\mu = \sqrt{n(n+2)}$ results in $\sqrt{3(5)} = 3.87$ B.M., which invalidates the range cited in statement (D).