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

The stability of group 6 transition metals increases down the group, meaning Cr(VI) is less stable than Mo(VI) and W(VI), which renders the first statement incorrect. Lanthanoids in the +4 oxidation state function as oxidizing agents because they easily gain electrons, whereas those in the +2 state serve as reducing agents by readily donating electrons, confirming the second statement. Although Americium has seven unpaired electrons, Curium possesses eight due to its $5f^{7}6d^{1}$ configuration, making the third statement false. Because $5f$ electrons provide poorer shielding than $4f$ electrons, the effective nuclear charge increases more sharply across the actinoids, resulting in a more pronounced contraction compared to the lanthanoid series, establishing the final statement as true.

The spin-only magnetic moment of transition metal ions depends directly on the number of unpaired electrons $n$, as calculated by the formula $\mu = \sqrt{n(n+2)} \text{ BM}$. Evaluating the electronic configurations, $\mathrm{Cr}^{2+}$ ($3d^4$) possesses four unpaired electrons, $\mathrm{Cr}^{3+}$ ($3d^3$) has three, $\mathrm{Cu}^{2+}$ ($3d^9$) contains one, and $\mathrm{Cu}^{+}$ ($3d^{10}$) has zero, which results in the decreasing magnetic moment order of $\mathrm{Cr}^{2+} > \mathrm{Cr}^{3+} > \mathrm{Cu}^{2+} > \mathrm{Cu}^{+}$.

The number of unpaired electrons in a transition metal ion is determined by its $d$-orbital electronic configuration, which utilizes Hund's rule of maximum multiplicity to fill orbitals singly before pairing occurs. The $\mathrm{V}^{2+}$ ion has a $[Ar] 3d^3$ configuration resulting in 3 unpaired electrons, while the $\mathrm{Co}^{2+}$ ion has a $[Ar] 3d^7$ configuration that also leaves 3 unpaired electrons after pairing occurs in the subshell. Because both ions contain an identical count of unpaired electrons, this pair correctly satisfies the condition.

Transition metals with a $d^8$ electronic configuration are unique because their orbital occupancy remains fixed as $t_{2g}^6 e_g^2$ within an octahedral crystal field, regardless of the nature of the ligand. This consistent electron distribution is responsible for the distinct violet color exhibited by the $\mathrm{Ni}^{2+}$ ion in a non-luminous flame under hot conditions during the borax bead test.

Matching biological structures with their primary functions requires identifying the specific physiological role each organelle plays within the cell. Mitochondria serve as the powerhouse of the cell by generating energy in the form of adenosine triphosphate, while Ribosomes function as the primary sites for protein synthesis. The Nucleus acts as the central repository for genetic material and regulates cell activity, and Lysosomes provide digestive enzymes to break down waste materials within the cell.

Among the oxides of vanadium provided, $\mathrm{V}_2\mathrm{O}_5$ is the amphoteric species that undergoes a reaction with alkali to produce the $\mathrm{VO}_4^{3-}$ anion. Because this process is an acid-base neutralization rather than a redox reaction, the oxidation state of vanadium remains unchanged. Calculating the oxidation state using the equation $x + 4(-2) = -3$ reveals that the vanadium in this anion exists in the +5 state.

The oxidative fusion of chromite ore $\mathrm{FeCr}_2\mathrm{O}_4$ with sodium carbonate $\mathrm{Na}_2\mathrm{CO}_3$ and atmospheric $\mathrm{O}_2$ converts the chromium content into a water-soluble chromate salt, while the iron component is converted into a distinct, water-insoluble oxide product, $\mathrm{Fe}_2\mathrm{O}_3$. To determine the molar mass of this insoluble substance, we calculate the sum of the atomic masses for its constituent elements, iron and oxygen.

Using the standard atomic masses of 56 for iron and 16 for oxygen, the calculation for $\mathrm{Fe}_2\mathrm{O}_3$ accounts for two iron atoms and three oxygen atoms:

$(2 \times 56) + (3 \times 16) = 112 + 48 = 160$

The molar mass of the insoluble product is 160 g mol^-1.

The melting points of transition metals are fundamentally linked to the strength of interatomic metallic bonding, which is influenced by electronic configuration and crystal structure. For the sets ($\mathrm{Mn}, \mathrm{Fe}$), ($\mathrm{Tc}, \mathrm{Ru}$), and ($\mathrm{Re}, \mathrm{Os}$), these bonding characteristics result in melting point relationships of $\mathrm{Mn} < \mathrm{Fe}$, $\mathrm{Tc} < \mathrm{Ru}$, and $\mathrm{Os} < \mathrm{Re}$.

The electronic behavior of transition metals is determined by the relative energies of orbitals and the extra stability associated with certain subshell arrangements, often requiring adjustments to the standard order of filling. Niobium possesses a ground state configuration of $[\mathrm{Kr}] 4d^4 5s^1$, which means there are 4 electrons residing in the $4d$ orbital, making $x = 4$.

Ruthenium follows a similar pattern in its stable ground state of $[\mathrm{Kr}] 4d^7 5s^1$, where 7 electrons occupy the $4d$ orbital, making $y = 7$. Adding the electron counts from these specific subshells results in $x + y = 11$.

The electronic stability of lanthanoid ions is largely governed by the presence of a half-filled $f$-subshell, which corresponds to a $4f^7$ configuration. Europium ($\mathrm{Eu}$, atomic number 63) possesses a ground-state configuration of $[Xe] 4f^7 6s^2$, so upon forming the $\mathrm{Eu}^{2+}$ ion, it loses its two $6s$ electrons to leave the stable $4f^7$ configuration intact. In the same way, Gadolinium ($\mathrm{Gd}$, atomic number 64) has a configuration of $[Xe] 4f^7 5d^1 6s^2$, meaning the $\mathrm{Gd}^{3+}$ ion, formed by the loss of two $6s$ electrons and one $5d$ electron, also retains the $4f^7$ structure. Because the other listed ions possess different configurations, only $\mathrm{Eu}^{2+}$ and $\mathrm{Gd}^{3+}$ represent the ions containing exactly seven electrons in the $4f$ orbital.

Alloys are metallic substances formed by combining two or more elements to achieve specific material properties, where each mixture corresponds to a distinct, standardized composition. Bronze consists of $\mathrm{Cu}$ and $\mathrm{Sn}$, whereas brass is made up of $\mathrm{Cu}$ and $\mathrm{Zn}$. Similarly, the UK silver coin is produced using $\mathrm{Cu}$ and $\mathrm{Ni}$, and stainless steel is engineered with a mixture of $\mathrm{Fe}$, $\mathrm{Cr}$, $\mathrm{Ni}$, and $\mathrm{C}$.

Oxidizing agents are substances that readily gain electrons to undergo reduction, typically to reach a more stable electronic configuration. In the lanthanide series, the +3 oxidation state is characteristically the most thermodynamically stable, so ions existing in higher oxidation states are driven to accept electrons to return to this preferred condition. Consequently, $\mathrm{Tb}^{4+}$ acts as the strongest oxidizing agent in this group because it readily gains an electron to reduce to the highly stable $\mathrm{Tb}^{3+}$ ion.

The color of transition metal ions in aqueous solution is primarily dictated by $d-d$ electronic transitions caused by the splitting of $d$-orbitals in a crystal field. Among the provided options, the set comprising $\mathrm{V}^{2+}$, $\mathrm{Cr}^{3+}$, and $\mathrm{Mn}^{3+}$ is unique because every ion in this group displays a consistent violet color in aqueous media. This characteristic uniformity in spectral behavior sets them apart from the other listed sets, which contain ions with varying or distinct color properties.

The magnetic moment of a transition metal ion is determined by the number of unpaired electrons in its $d$-orbitals, calculated using the spin-only formula $\mu = \sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons and $\mu$ is the magnetic moment in Bohr Magnetons.

The $\mathrm{Ti}^{3+}$ ion, with a $3d^1$ configuration, contains a single unpaired electron that corresponds to a magnetic moment of 1.73 B.M. Similarly, $\mathrm{V}^{2+}$ features a $3d^3$ configuration with three unpaired electrons, resulting in a value of 3.87 B.M. Following the same principle, $\mathrm{Ni}^{2+}$ has a $3d^8$ configuration with two unpaired electrons, producing a magnetic moment of 2.84 B.M. Finally, $\mathrm{Sc}^{3+}$ maintains a $3d^0$ configuration with zero unpaired electrons, yielding a magnetic moment of 0.00 B.M.

Vanadium ($V$) exhibits the highest enthalpy of atomisation in the first transition series because its electronic configuration facilitates strong metallic bonding. Its aquated ion $V^{3+}$ is characterized by a green colour, and the oxide it forms, vanadium(III) oxide ($V_2O_3$), is basic, consistent with the general trend where transition metal oxides in lower oxidation states exhibit basic properties.

The spin-only magnetic moment for a transition metal ion is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ B.M.}$, where $n$ represents the number of unpaired electrons. For the magnetic moment to equal 4.9 B.M., the value of $n$ must be 4, as $\sqrt{4(4+2)} = \sqrt{24} \approx 4.9$. Evaluating the electronic configurations shows that $\mathrm{Cr}^{2+}$ is a $d^4$ system, $\mathrm{Fe}^{2+}$ is a $d^6$ system that results in four unpaired electrons, and $\mathrm{Mn}^{3+}$ is a $d^4$ system, all of which contain exactly four unpaired electrons. In contrast, $\mathrm{Fe}^{3+}$ has five unpaired electrons and $\mathrm{Co}^{2+}$ contains three, confirming that only $\mathrm{Cr}^{2+}$, $\mathrm{Fe}^{2+}$, and $\mathrm{Mn}^{3+}$ produce the specified magnetic moment.

The enthalpy of atomisation, which represents the energy required to break the metallic bonds within a crystal lattice to produce isolated gaseous atoms, is primarily governed by the strength of interatomic metallic bonding. In the 3d transition series, metals with a greater number of unpaired electrons available for delocalisation typically exhibit higher atomisation enthalpies. Comparing the provided values, Cobalt possesses the highest enthalpy of atomisation at 425 kJ/mol, while Scandium, Manganese, and Copper have lower values of 326 kJ/mol, 281 kJ/mol, and 338 kJ/mol, respectively.

To determine the magnetic properties of the Co$^{2+}$ ion, we first establish its electronic configuration. The neutral Cobalt atom has the configuration [Ar] 3d$^7$ 4s$^2$; therefore, the Co$^{2+}$ ion is formed by losing the two 4s electrons, resulting in a [Ar] 3d$^7$ configuration. Within the 3d subshell, seven electrons populate the five orbitals such that 3 electrons remain unpaired. Substituting $n = 3$ into the spin-only magnetic moment formula $\mu = \sqrt{n(n+2)}$ BM gives $\sqrt{3(3+2)} = \sqrt{15}$, which equals approximately 3.87 BM. Rounding this value to the nearest integer yields 4.

The oxidizing power of group 6 oxides is directly linked to the thermodynamic stability of the metal's highest oxidation state, where lower stability typically corresponds to stronger oxidizing behavior. As we move down the group from chromium to molybdenum, the stability of the hexavalent state increases, meaning $\mathrm{Mo(VI)}$ is more stable than $\mathrm{Cr(VI)}$. This trend dictates that $\mathrm{CrO}_3$ functions as a stronger oxidizing agent than $\mathrm{MoO}_3$, which confirms that Statement I is true and Statement II is false.

The chemical sequence provided illustrates the chromyl chloride test, where the interconversion of chromium species is driven by changes in pH. The reaction of chromyl chloride vapour with an alkali produces the yellow chromate ion, $\mathrm{CrO}_4^{2-}$, which corresponds to compound $B$. Upon subsequent acidification of this species, dimerization occurs to form the dichromate ion, $\mathrm{Cr}_2\mathrm{O}_7^{2-}$, which is compound $C$.

Structurally, the dichromate ion is composed of two $\mathrm{CrO}_4$ tetrahedra that share a single oxygen vertex, establishing a $\mathrm{Cr}-\mathrm{O}-\mathrm{Cr}$ bridge. In this arrangement, each chromium atom is bonded to three oxygen atoms that are distinct from the shared bridge, meaning these three oxygens are considered terminal. Since there are two such chromium centers in the dichromate structure, the total count of terminal oxygen atoms is calculated by summing those attached to each center, resulting in 6.

The industrial synthesis of potassium permanganate relies on a two-stage process beginning with the oxidative fusion of manganese dioxide. When $\mathrm{MnO}_2$ is heated with potassium hydroxide ($\mathrm{KOH}$) in the presence of an oxidizing agent like potassium nitrate ($\mathrm{KNO}_3$), the manganese undergoes oxidation to reach a +6 oxidation state, resulting in the formation of dark green potassium manganate, denoted as $\mathrm{K}_2\mathrm{MnO}_4$. This intermediate compound is subsequently oxidized electrolytically in an alkaline solution to produce the final potassium permanganate.