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

The borax bead test provides a qualitative means to identify transition metal ions through distinct coloration, but $\mathrm{Zn}^{2+}$ is excluded as it does not show a positive borax bead test. Among the remaining ions $\mathrm{Fe}^{2+}$, $\mathrm{Fe}^{3+}$, and $\mathrm{Cr}^{2+}$, the ionization enthalpy is governed by effective nuclear charge and electronic stability. As a result, $\mathrm{Fe}^{3+}$ possesses the highest ionization enthalpy, fulfilling the required criteria.

The arrangement of electrons within the 4f subshell is determined by the element's specific electronic configuration, where orbital stability often dictates the population of these shells. By comparing these configurations, it is evident that Eu ($4f^7$) and Gd ($4f^7$) both contain 7 electrons in their 4f orbitals, while Yb ($4f^{14}$) and Hf ($4f^{14}$) share an identical count of 14 electrons. Since other combinations such as Dy ($4f^{10}$) paired with Ho ($4f^{11}$) or Lu ($4f^{14}$) paired with Tm ($4f^{13}$) feature unequal electron counts, only the pairs Eu-Gd and Yb-Hf demonstrate matching occupancy.

The reaction between potassium dichromate, sodium chloride, and concentrated sulfuric acid is known as the chromyl chloride test, which results in the evolution of reddish-brown vapors of chromyl chloride. The chemical process is represented by the following equation:

$K_2Cr_2O_7 + 4NaCl + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 2KHSO_4 + 4NaHSO_4 + 3H_2O$

In the resulting molecule $CrO_2Cl_2$, the oxidation state of the chromium atom is determined by assigning -2 to each oxygen atom and -1 to each chlorine atom, which leads to the calculation $x + 2(-2) + 2(-1) = 0$, yielding an oxidation state of +6.

Ionization enthalpy is governed by electronic stability, where completely filled or half-filled subshells offer greater resistance to electron removal. Chromium has a ground state configuration of $[Ar] 3d^5 4s^1$ while Manganese is $[Ar] 3d^5 4s^2$; consequently, the first electron in Manganese is harder to remove from its stable $4s^2$ orbital than the $4s^1$ electron in Chromium, resulting in a higher first ionization enthalpy for Manganese.

After the first ionization, the resulting Chromium ion $[Ar] 3d^5$ is exceptionally stable due to its half-filled $d$-subshell, requiring a much higher energy to remove a second electron. Conversely, the Manganese ion retains an $[Ar] 3d^5 4s^1$ configuration, making the removal of the second electron from the $4s$ orbital relatively easier, which confirms that the second ionization enthalpy of Chromium is higher than that of Manganese.

The reaction sequence begins with the formation of chromyl chloride as compound $\mathrm{A}$ ($\mathrm{CrO}_2\mathrm{Cl}_2$), which then reacts with $\mathrm{NaOH}$ to yield sodium chromate as compound $\mathrm{B}$ ($\mathrm{Na}_2\mathrm{CrO}_4$). The subsequent reaction of $\mathrm{Na}_2\mathrm{CrO}_4$ with $\mathrm{H}_2\mathrm{SO}_4$ and $\mathrm{H}_2\mathrm{O}_2$ produces chromium pentoxide as compound $\mathrm{C}$ ($\mathrm{CrO}_5$), which adopts a butterfly structure.

This structure contains one terminal oxo group and two peroxo ($\mathrm{O}_2^{2-}$) groups bonded to the central chromium atom. From this configuration, the number of peroxo units is $X = 2$, and the total number of oxygen atoms is $Y = 5$. The oxidation state of the chromium atom, $Z$, is determined by the balance of charges in $\mathrm{CrO}_5$ as $Cr + 1(-2) + 2(-2) = 0$, which results in $Z = 6$. The final calculation for the sum $X + Y + Z$ is $2 + 5 + 6 = 13$.

The first ionization enthalpy is determined by effective nuclear charge and orbital stability, where the chromium electron configuration $[Ar] 3d^5 4s^1$ loses its $4s$ electron more easily than the manganese configuration $[Ar] 3d^5 4s^2$ due to the greater nuclear charge and full $4s^2$ subshell of Mn, confirming that Statement I is true. Statement II is false because while the second ionization enthalpy of Cr is indeed higher than that of Mn due to the disruption of the stable $3d^5$ configuration, the third ionization enthalpy of Mn is higher than that of Cr; the removal of the third electron from Mn involves breaking the stable half-filled $3d^5$ subshell, whereas for Cr, the electron is removed from the less stable $3d^4$ configuration.

Both manganate ($MnO_4^{2-}$) and permanganate ($MnO_4^-$) ions exhibit tetrahedral geometry due to the $sp^3$ hybridization of the central manganese atom. Manganate, with manganese in the $+6$ oxidation state and a $3d^1$ configuration, contains one unpaired electron and is paramagnetic, whereas permanganate, with manganese in the $+7$ oxidation state and a $3d^0$ configuration, possesses no unpaired electrons and is diamagnetic. The other statements are incorrect because the oxidation states of manganese in these ions are reversed, the oxidation of $Mn^{2+}$ by peroxodisulphate yields the permanganate ion rather than manganate, and acidified permanganate acts as a strong oxidizing agent that oxidizes, rather than reduces, oxalate, nitrite, and iodide ions.

Potassium dichromate, $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$, functions as a potent oxidizing agent in an acidic medium, where the dichromate ion is responsible for the distinct orange hue. A change in this color indicates that a redox reaction has occurred, necessitating the presence of species capable of acting as reducing agents by donating electrons.

The set containing $\mathrm{Fe}^{2+}$, $\mathrm{Sn}^{2+}$, $\mathrm{I}^{-}$, and $\mathrm{S}^{2-}$ consists entirely of such reducing agents, each capable of being oxidized: $\mathrm{Fe}^{2+}$ converts to $\mathrm{Fe}^{3+}$, $\mathrm{Sn}^{2+}$ to $\mathrm{Sn}^{4+}$, $\mathrm{I}^{-}$ to $\mathrm{I}_2$, and $\mathrm{S}^{2-}$ to $\mathrm{S}$. Because these species can drive the reduction of the chromium species, they trigger the associated change in color, unlike the other options which include ions that are either already fully oxidized or are not reactive in these specific conditions.

The manufacture of sulfuric acid utilizes vanadium pentoxide, but this compound is amphoteric rather than basic, which makes the first statement incorrect. Similarly, the transition metal halide in the Ziegler-Natta catalyst, titanium tetrachloride ($\mathrm{TiCl}_{4}$), features titanium in the $+4$ oxidation state with a $3d^0$ configuration, resulting in a spin-only magnetic moment of 0 BM rather than 2.84 BM. In the same polymerization system, the p-block organometallic component involves aluminum in the $+3$ oxidation state, while the Wacker process utilizes palladium(II) chloride ($\mathrm{PdCl}_{2}$) where the palladium ion exhibits a $4d^8$ electronic configuration, confirming that the outermost d-orbital contains 8 electrons.

Paramagnetism occurs when a metal ion contains one or more unpaired electrons in its atomic orbitals, whereas species with entirely paired electrons exhibit diamagnetism. Upon examining the electronic configurations of the given ions, we find that $\mathrm{Mn}^{2+}$ ($3d^5$), $\mathrm{Cu}^{2+}$ ($3d^9$), and $\mathrm{Gd}^{3+}$ ($4f^7$) each contain unpaired electrons, which qualifies them as paramagnetic. In contrast, $\mathrm{Zn}^{2+}$, $\mathrm{Yb}^{2+}$, $\mathrm{Sc}^{3+}$, $\mathrm{La}^{3+}$, $\mathrm{Lu}^{3+}$, $\mathrm{Ti}^{4+}$, and $\mathrm{Ce}^{4+}$ all possess completely filled electron shells or lack unpaired electrons, resulting in a diamagnetic character. Therefore, exactly 3 ions from the provided list are paramagnetic.

A mixed oxide is defined as a compound that contains a metal in two different oxidation states, effectively acting as a combination of two distinct simple oxides, frequently exemplified by the spinel structure $\mathrm{M}_3\mathrm{O}_4$. Within the provided list, $\mathrm{Mn}_3 \mathrm{O}_4$ functions as a combination of $\mathrm{MnO}$ and $\mathrm{Mn}_2\mathrm{O}_3$, $\mathrm{Fe}_3 \mathrm{O}_4$ as a combination of $\mathrm{FeO}$ and $\mathrm{Fe}_2\mathrm{O}_3$, and $\mathrm{Co}_3 \mathrm{O}_4$ as a combination of $\mathrm{CoO}$ and $\mathrm{Co}_2\mathrm{O}_3$. Conversely, $\mathrm{Ti}_2 \mathrm{O}_3$, $\mathrm{V}_2 \mathrm{O}_4$, $\mathrm{Cr}_2 \mathrm{O}_3$, and $\mathrm{Fe}_2 \mathrm{O}_3$ are classified as simple oxides because the metal atoms in these structures exist in only a single oxidation state.

Transition metal ions exhibit colour due to $d-d$ electronic transitions facilitated by partially filled $d$-orbitals, whereas diamagnetism occurs when all electrons are paired in completely filled or empty subshells. In the first group, $\mathrm{Ti}^{4+}$ has a $d^0$ configuration, rendering it colourless, while $\mathrm{V}^{2+}$, $\mathrm{Mn}^{2+}$, $\mathrm{Fe}^{3+}$, and $\mathrm{Cr}^{2+}$ all possess unpaired $d$-electrons, making them coloured; thus, the pairs $[\mathrm{V}^{2+}, \mathrm{Mn}^{2+}]$, $[\mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}]$, and $[\mathrm{V}^{2+}, \mathrm{Cr}^{2+}]$ are the three pairs that each contain two coloured ions.

Regarding the second group, checking the electronic configurations confirms that $\mathrm{La}^{3+}$, $\mathrm{Ce}^{4+}$, and $\mathrm{Ac}^{3+}$ reach noble gas cores, while $\mathrm{Yb}^{2+}$, $\mathrm{Lu}^{3+}$, and $\mathrm{Lr}^{3+}$ feature completely filled $f^{14}$ subshells, resulting in no unpaired electrons. Since every ion in the pairs $[\mathrm{La}^{3+}, \mathrm{Yb}^{2+}]$, $[\mathrm{Lu}^{3+}, \mathrm{Ce}^{4+}]$, and $[\mathrm{Ac}^{3+}, \mathrm{Lr}^{3+}]$ is diamagnetic, there are indeed three such pairs.

Disproportionation occurs when a species in an intermediate oxidation state is simultaneously oxidized and reduced to form two different products. In the manganate ion ($MnO_4^{2-}$), manganese exists in the +6 oxidation state, which is unstable in an acidic medium and disproportionates into the more stable +7 oxidation state found in the permanganate ion ($MnO_4^-$) and the +4 oxidation state found in manganese dioxide ($MnO_2$). The corresponding chemical reaction is $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$.

Sulfur dioxide functions as a strong reducing agent, a characteristic often tested by its ability to induce specific color changes in chemical reagents. When $\mathrm{SO}_2$ gas is passed over paper soaked in acidified potassium dichromate, it performs a redox reaction that reduces orange chromium(VI) to green chromium(III) ions. This transformation is described by the equation $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + \mathrm{SO}_2 \xrightarrow{\mathrm{H}_2\mathrm{SO}_4} \mathrm{Cr}_2(\mathrm{SO}_4)_3 + \mathrm{SO}_4^{2-}$, where the production of the green compound $\mathrm{Cr}_2(\mathrm{SO}_4)_3$ identifies 'X' as $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$.

Catalytic activity in transition metals is fundamentally driven by their ability to provide a surface where reactant molecules adsorb using both $3d$ and $4s$ electrons to form chemical bonds, which makes the assertion that only $3d$ electrons are involved incorrect. Although adsorption does indeed increase the local concentration of reactants at the catalyst surface, this interaction serves to weaken the existing internal bonds of the reactant molecules rather than strengthen them. By weakening these bonds, the catalyst effectively lowers the activation energy required to form the activated complex, which is essential for accelerating the overall reaction rate.

Manganese achieves its maximum oxidation state of $+7$ in $Mn_2O_7$ by utilizing all its valence electrons ($3d^5 4s^2$), although this compound is a covalent, acidic molecule rather than an ionic solid. High electronegativity and the small size of the oxygen atom facilitate the formation of $p\pi-d\pi$ multiple bonds, which stabilize the transition metal in such high oxidation states. The structural arrangement of $Mn_2O_7$ consists of two $MnO_4$ tetrahedra linked by a single oxygen atom in an $O_3Mn-O-MnO_3$ bridging configuration.

Ions with partially filled d-orbitals exhibit color due to d-d transitions, whereas ions with $d^0$ or $d^{10}$ configurations are colorless because they lack the necessary electrons or available energy levels for such transitions. Analyzing the provided pairs, $\mathrm{Sc}^{3+}$ ($3d^0$), $\mathrm{Zn}^{2+}$ ($3d^{10}$), and $\mathrm{Ti}^{4+}$ ($3d^0$) are colorless, which means the pairs $[\mathrm{Sc}^{3+}, \mathrm{Ti}^{3+}]$, $[\mathrm{Cu}^{2+}, \mathrm{Zn}^{2+}]$, and $[\mathrm{Ni}^{2+}, \mathrm{Ti}^{4+}]$ each contain at least one colorless ion, leaving only $[\mathrm{Mn}^{2+}, \mathrm{Cr}^{2+}]$ as a pair where both ions are colored. Since there is only one such pair instead of three, Statement I is false.

A species acts as a strong reducing agent if it readily loses electrons to undergo oxidation. Among the given ions, $\mathrm{Eu}^{2+}$ is a strong reducing agent because it easily oxidizes to $\mathrm{Eu}^{3+}$, which possesses a stable half-filled $4f^7$ configuration, whereas $\mathrm{Th}^{4+}$ exists in a highly stable oxidation state and does not function as a strong reducing agent. Thus, characterizing $\mathrm{Th}^{4+}$ as the strongest reducing agent is chemically incorrect, making Statement II false.

Interstitial compounds form when small atoms like hydrogen, carbon, or nitrogen occupy the voids within a transition metal lattice, resulting in materials that are exceptionally hard and possess higher melting points than the parent metals themselves. These substances are defined by their retained metallic conductivity and chemical inertness, and they frequently exhibit non-stoichiometric compositions. Because these compounds rely on metallic and covalent-like bonding to achieve their significant hardness, the description of them as soft and ionic is fundamentally incorrect.

Potassium dichromate functions as an effective oxidizing agent that converts $\mathrm{FeSO}_4$ to $\mathrm{Fe}_2(\mathrm{SO}_4)_3$ in acidic conditions, while sodium dichromate is unsuitable for use as a primary standard in volumetric estimation because it is deliquescent. The chromate ($\mathrm{CrO}_4^{2-}$) and dichromate ($\mathrm{Cr}_2\mathrm{O}_7^{2-}$) ions exist in an aqueous equilibrium that allows them to interconvert based on changes in solution pH. Furthermore, structural analysis of the dichromate ion confirms a $\mathrm{Cr}-\mathrm{O}-\mathrm{Cr}$ bond angle of $126^{\circ}$.

Determining the oxidation states of manganese in its highest oxidation-state compounds is the first step toward finding the target number of electrons. Manganese reaches an oxidation state of +7 in $\mathrm{Mn}_2\mathrm{O}_7$ and a maximum oxidation state of +4 in its fluoride compound, $\mathrm{MnF}_4$. The difference between these oxidation states is $|7 - 4| = 3$, meaning we must identify which of the provided ions possess exactly 3 unpaired electrons.

Evaluating the electronic configurations reveals the number of unpaired electrons for each ion. The $\mathrm{Sc}^{3+}$ ion, with an $[\mathrm{Ar}]$ configuration, has 0 unpaired electrons, and $\mathrm{Zn}^{2+}$, which has a $3d^{10}$ configuration, also has 0 unpaired electrons. In contrast, $\mathrm{V}^{2+}$ adopts an $[\mathrm{Ar}] 3d^3$ configuration, providing 3 unpaired electrons, while $\mathrm{Fe}^{2+}$ with its $[\mathrm{Ar}] 3d^6$ configuration results in 4 unpaired electrons. Lastly, $\mathrm{Co}^{2+}$ holds an $[\mathrm{Ar}] 3d^7$ configuration, which results in 3 unpaired electrons. Consequently, $\mathrm{V}^{2+}$ and $\mathrm{Co}^{2+}$ are the ions matching the calculated difference.