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

The $d^{10}$ electronic configuration represents a highly stable, fully filled subshell, which occurs in transition metals that often exhibit variations from the expected orbital filling order. Both Copper and Silver attain this stability by shifting an electron into the $d$ orbital, resulting in the configurations $[\text{Ar}] 4s^1 3d^{10}$ and $[\text{Kr}] 5s^1 4d^{10}$ respectively. Similarly, Zinc and Cadmium satisfy this criterion with the electronic structures $[\text{Ar}] 4s^2 3d^{10}$ and $[\text{Kr}] 5s^2 4d^{10}$, ensuring that every element in this set possesses the fully saturated $d$ subshell.

To determine the correct IUPAC name for $\mathrm{K}_2\mathrm{MnO}_4$, calculate the oxidation state of the central manganese atom by assuming the oxidation states of potassium are +1 and oxygen are -2. By applying the equation $2(+1) + x + 4(-2) = 0$, we solve to find that $x = +6$, confirming the oxidation state of manganese is +6. The standard nomenclature involves naming the cation first, followed by the anion, which includes the prefix for four oxygen atoms as tetraoxido, the root of the central metal as manganate, and the oxidation state in parentheses, yielding the name potassium tetraoxidomanganate (VI).

The characteristic stability of the +3 oxidation state across the lanthanoid series is the driving force behind the redox behavior of these elements. A reducing agent is a species that readily donates electrons, and $\mathrm{Eu}^{2+}$ is unique among these options because it easily oxidizes to the more stable $\mathrm{Eu}^{3+}$ state by releasing an electron. This transformation, expressed as $\mathrm{Eu}^{2+} ([\mathrm{Xe}] 4f^7 6s^0) \rightarrow \mathrm{Eu}^{3+} ([\mathrm{Xe}] 4f^6 6s^0) + 1 e^-$, effectively characterizes its role as a reducing agent. In contrast, $\mathrm{Ce}^{4+}$ acts as an oxidizing agent by accepting an electron to return to the +3 state, whereas $\mathrm{Lu}^{3+}$ and $\mathrm{Gd}^{3+}$ exist in their most stable configurations and do not readily undergo further oxidation.

The magnetic behavior of transition metal oxides is governed by the number of unpaired electrons in the metal $d$-orbitals, which is quantified by the spin-only magnetic moment formula $\mu = \sqrt{n(n+2)}$ BM. Among the listed compounds, $\mathrm{CrO}$ functions as a basic oxide while $\mathrm{Cr}_2\mathrm{O}_3$ behaves as an amphoteric oxide, but $\mathrm{CrO}_3$ is acidic and is therefore disregarded in the requested sum.

Regarding $\mathrm{CrO}$, the metal resides in the $+2$ oxidation state with a $3d^4$ configuration, leading to four unpaired electrons and a magnetic moment of 4.90 BM. Within the amphoteric $\mathrm{Cr}_2\mathrm{O}_3$, the chromium center holds a $+3$ oxidation state and a $3d^3$ configuration, which results in three unpaired electrons and a magnetic moment of 3.87 BM. Adding these values yields a total of 8.77 BM, equivalent to $877 \times 10^{-2}$ BM, which identifies the final integer value as 877.

Lanthanoids comprise the inner transition elements with atomic numbers ranging from 57 to 71. Among the provided elements, Europium, Erbium, Terbium, Ytterbium, and Lutetium all belong to this series. Since Curium is classified as an actinide, it stands as the only element in the given list that does not fall within the lanthanoid group.

The reaction described is the chromyl chloride test, which is a qualitative method for detecting chloride ions by generating the volatile compound chromyl chloride ($\mathrm{CrO}_2\mathrm{Cl}_2$). When sodium chloride reacts with concentrated sulfuric acid and potassium dichromate, the reddish-brown vapors released are identified as $\mathrm{CrO}_2\mathrm{Cl}_2$, which represents compound (B). Upon further reaction with sodium hydroxide, these vapors produce sodium chromate ($\mathrm{Na}_2\mathrm{CrO}_4$), which results in a characteristic yellow solution, corresponding to compound (C).

The second ionization enthalpy of a transition metal peaks significantly when the electron is removed from a stable, half-filled $d$-subshell, a condition met by chromium because the formation of its $Cr^+$ ion results in a stable $3d^5$ configuration. With an electronic configuration of $[Ar] 3d^5 4s^0$, the $Cr^+$ ion contains 5 unpaired electrons. Applying the spin-only magnetic moment equation $\mu = \sqrt{n(n+2)}$ BM with $n = 5$ leads to $\mu = \sqrt{35}$ BM, which is approximately 5.92 BM, rounding to 6.

Manganese distinguishes itself among the 3d transition series due to its electronic configuration of $3d^5 4s^2$, which makes seven valence electrons available for participation in chemical bonding. This configuration enables the metal to reach a maximum oxidation state of $\mathrm{Mn}^{+7}$. Among the choices provided, manganese is the element that achieves this highest oxidation state, which is the maximum value found in the 3d series.

The number of unpaired electrons in transition metals depends on their ground-state electronic configurations and the distribution of electrons according to Hund's rule. Following this principle, $\mathrm{Sc}: [\mathrm{Ar}] 4s^2 3d^1$ has 1 unpaired electron, $\mathrm{Ti}: [\mathrm{Ar}] 4s^2 3d^2$ has 2, $\mathrm{V}: [\mathrm{Ar}] 4s^2 3d^3$ has 3, $\mathrm{Mn}: [\mathrm{Ar}] 4s^2 3d^5$ has 5, and $\mathrm{Cr}: [\mathrm{Ar}] 4s^1 3d^5$ contains 6 unpaired electrons. Sorting these elements by their unpaired electron counts leads to the increasing order $(A) < (D) < (C) < (E) < (B)$.

Transition metal oxides exhibit diverse chemical characteristics depending on their oxidation state and bonding nature. $\mathrm{Mn}_2\mathrm{O}_7$ is a covalent oxide that exists as a green oil at room temperature, which confirms that statement A is correct. While $\mathrm{V}_2\mathrm{O}_4$ reacts with acids to produce the $\mathrm{VO}^{2+}$ ion rather than $\mathrm{VO}_2^{2+}$, $\mathrm{CrO}$ functions as a basic oxide, making statement C correct. Finally, $\mathrm{V}_2\mathrm{O}_5$ is amphoteric and interacts with both acids and bases, which invalidates the claim in statement D.

The chemical behavior of transition metal ions in aqueous solutions is fundamentally governed by the relative stability of their electronic configurations. For $\mathrm{Cr}^{2+}$, the oxidation to $\mathrm{Cr}^{3+}$ involves a transition from a $d^4$ configuration to a $d^3$ configuration, which is particularly stable because the $t_{2g}$ level becomes half-filled in an octahedral crystal field. Conversely, $\mathrm{Mn}^{3+}$ possesses a $d^4$ configuration and readily gains an electron to form $\mathrm{Mn}^{2+}$, resulting in a $d^5$ configuration, which is exceptionally stable due to the half-filled nature of the d-subshell. Because the energetic drive to achieve these stable, half-filled electronic states dictates whether a species will lose or gain electrons, the general principle of extra stability for half-filled configurations directly accounts for why $\mathrm{Cr}^{2+}$ acts as a reducing agent while $\mathrm{Mn}^{3+}$ functions as an oxidizing agent.

The flame test is an analytical procedure that identifies metal ions based on the characteristic emission of light caused by the excitation of electrons when heated in a non-luminous Bunsen burner flame. Only certain metal ions exhibit the specific electronic transitions required to produce distinct, observable colors, while others do not possess this property under standard testing conditions. Among the provided list, $\mathrm{Sr}^{2+}$, $\mathrm{Ba}^{2+}$, $\mathrm{Ca}^{2+}$, and $\mathrm{Cu}^{2+}$ are the only ions that produce a characteristic flame coloration, whereas $\mathrm{Zn}^{2+}$, $\mathrm{Co}^{2+}$, and $\mathrm{Fe}^{2+}$ do not, resulting in a total of 4 metal ions.

The capacity of a metal ion to liberate hydrogen gas from a dilute acid is fundamentally dictated by its strength as a reducing agent. For such a reaction to occur, the metal ion must be capable of donating electrons to hydrogen ions, effectively reducing them into hydrogen gas. The ions $\mathrm{Ti}^{2+}$, $\mathrm{V}^{2+}$, and $\mathrm{Cr}^{2+}$ are all recognized as strong reducing agents, meaning they possess a high tendency to lose electrons and become oxidized to a higher oxidation state. Because each of these three transition metal ions is strong enough to facilitate the reduction of $\mathrm{H}^+$ to $\mathrm{H}_2$, all three of them will successfully produce hydrogen gas when placed in a dilute acid solution.

The oxidative fusion of chromite ore with sodium carbonate occurs under atmospheric conditions to produce sodium chromate and ferric oxide as the primary solid products. This process follows the chemical equation:
$4\,\mathrm{FeCr}_2\mathrm{O}_4 + 8\,\mathrm{Na}_2\mathrm{CO}_3 + 7\,\mathrm{O}_2 \rightarrow 8\,\mathrm{Na}_2\mathrm{CrO}_4 + 2\,\mathrm{Fe}_2\mathrm{O}_3 + 8\,\mathrm{CO}_2$
In the resulting sodium chromate ($\mathrm{Na}_2\mathrm{CrO}_4$), chromium is in the +6 oxidation state with a $d^0$ electronic configuration, meaning it has no unpaired electrons and consequently a spin-only magnetic moment of 0 B.M. In ferric oxide ($\mathrm{Fe}_2\mathrm{O}_3$), iron exists in the +3 oxidation state with a $d^5$ electronic configuration; this state possesses five unpaired electrons, which corresponds to a magnetic moment of 5.9 B.M. Adding the magnetic moments of these two products yields a total of 5.9 B.M., which rounds to 6 when expressed as the nearest integer.

The reaction mechanism is driven by a homogeneous redox cycle where the iron(III) catalyst facilitates electron transfer between iodide and persulphate by cycling through its oxidation states. Initially, $\mathrm{Fe}^{3+}$ functions as an oxidizing agent, pulling electrons from the iodide ions to generate iodine while becoming reduced to $\mathrm{Fe}^{2+}$ as represented by $2\mathrm{Fe}^{3+}+2\mathrm{I}^-\rightarrow2\mathrm{Fe}^{2+}+\mathrm{I}_2$. In the subsequent step, these generated $\mathrm{Fe}^{2+}$ ions act as a reducing agent for the persulphate ions, converting them into sulphate ions and effectively regenerating the original $\mathrm{Fe}^{3+}$ catalyst according to $2\mathrm{Fe}^{2+}+\mathrm{S}_2\mathrm{O}_8^{2-}\rightarrow2\mathrm{Fe}^{3+}+2\mathrm{SO}_4^{2-}$, which confirms that iron(III) oxidizes iodide while the resulting iron(II) reduces the persulphate.

The reaction produces potassium manganate, which contains manganese in the $+6$ oxidation state and is inherently unstable in acidic or neutral solutions. This instability leads to a disproportionation reaction yielding permanganate, where manganese is in the $+7$ state, and manganese dioxide, where it is in the $+4$ state. For the permanganate ion, the electronic configuration is $3d^0$, resulting in zero unpaired electrons and a magnetic moment of 0 BM. In contrast, the manganese ion in manganese dioxide exhibits a $3d^3$ configuration with 3 unpaired electrons, providing a magnetic moment of $\sqrt{3(3+2)} \approx 3.87$ BM, which results in a total sum that rounds to 4.

The magnetic moment of an electronic configuration is determined by the number of unpaired electrons, $n$, according to the formula $\mu = \sqrt{n(n+2)}$. By applying Hund's rule to fill the five $d$ orbitals, we find that the $[\mathrm{Ar}] 3d^7$ and $[\mathrm{Ar}] 3d^3$ configurations each contain 3 unpaired electrons, resulting in a magnetic moment of $\sqrt{15}$ BM. The $[\mathrm{Ar}] 3d^8$ configuration contains 2 unpaired electrons with a magnetic moment of $\sqrt{8}$ BM, while the $[\mathrm{Ar}] 3d^6$ configuration contains 4 unpaired electrons, yielding the highest magnetic moment of $\sqrt{24}$ BM.

Vanadium possesses the highest enthalpy of atomisation among first-row transition elements at 515 kJ/mol. The amphoteric oxide $\mathrm{V}_2\mathrm{O}_5$ features vanadium in the +5 oxidation state, which corresponds to the electronic configuration $[\mathrm{Ar}]3d^0$. With zero unpaired electrons, the spin-only magnetic moment calculated using the formula $\sqrt{n(n+2)}$ is 0 BM.

Elements in the actinide series are characterized by the gradual filling of the $5f$ subshell after the $7s$ orbitals are completed. Since Einsteinium possesses an atomic number of 99, we subtract the 86 electrons covered by the Radon core to determine the remaining 13 electrons. These electrons populate the $7s$ and $5f$ orbitals accordingly, resulting in the electronic configuration $[\mathrm{Rn}]\,5\,f^{11}\,6\,d^{0}\,7\,s^{2}$.

The group 12 elements, zinc, cadmium, and mercury, possess a fully filled $(n-1)d^{10} ns^2$ electronic configuration that minimizes metallic bonding, resulting in notably low enthalpies of atomization and characteristic softness compared to other transition metals. While zinc and cadmium are restricted to a $+II$ oxidation state, mercury uniquely displays both $+I$ and $+II$ states, with the $+I$ state typically found in the $Hg_2^{2+}$ dimer. Furthermore, because these elements contain no unpaired electrons in their completely filled shells, all their compounds are diamagnetic rather than paramagnetic.