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

The electronic configurations associated with different oxidation states of transition metals dictate the wavelengths of light they absorb, which gives rise to their characteristic colors in aqueous solutions. The $\mathrm{Fe}^{2+}$ ion consistently exhibits a green color, while the $\mathrm{Fe}^{3+}$ ion presents as yellow. Therefore, in the reaction provided, the $\mathrm{Fe}^{2+}$ species contributes the green color to the reactant side, and the $\mathrm{Fe}^{3+}$ species contributes the yellow color to the product side.

Ruby gemstones achieve their striking red color strictly due to the incorporation of $Cr^{3+}$ ions into the $Al_2O_3$ crystal structure, rendering the claim about $Co^{3+}$ factually false. Because the statements describing $VOSO_4$ as a reducing agent, $Cr_2O_3$ as an amphoteric oxide, and $RuO_4$ as an oxidizing agent are all chemically accurate, the assertion concerning the cobalt ion remains the only incorrect statement in the group.

The magnetic quantum number determines the spatial orientation of an orbital and is strictly defined by the azimuthal quantum number of the electron's subshell. For the $\mathrm{Zn}^{+}$ ion, the full electronic configuration is $1\mathrm{s}^{2} 2\mathrm{s}^{2} 2\mathrm{p}^{6} 3\mathrm{s}^{2} 3\mathrm{p}^{6} 3\mathrm{d}^{10} 4\mathrm{s}^{1}$, which places the outermost electron in an s-orbital. Since the azimuthal quantum number $l=0$ for any s-orbital, the magnetic quantum number $m$, which is defined by the range $-l \leq m \leq +l$, must be $m=0$.

The electronic configuration of a transition metal ion in a specific oxidation state is determined by removing electrons from the outermost orbitals of the neutral atom. Since we are looking for an element that results in an $[Ar] 3d^5$ configuration after losing three electrons, we identify iron ($Fe$) as the candidate. The neutral iron atom has the configuration $Fe = [Ar] 3d^6 4s^2$, and when it loses three electrons-two from the $4s$ subshell and one from the $3d$ subshell-it results in the final configuration $Fe^{3+} = [Ar] 3d^5$.

In mild alkaline conditions, the permanganate ion functions as an oxidizing agent that drives the conversion of the thiosulphate ion into the sulphate ion, designated here as A. To find the oxidation state of sulphur in the sulphate ion, $\mathrm{SO}_4^{2-}$, we equate the sum of the oxidation numbers of one sulphur atom and four oxygen atoms to the overall ionic charge of $-2$. Applying the standard oxidation state of $-2$ for each oxygen, the calculation $x + 4(-2) = -2$ determines the oxidation state of sulphur to be $+6$.

Thermal decomposition serves as a reliable method to prepare potassium manganate, as heating potassium permanganate to 573 K successfully yields potassium manganate alongside manganese dioxide and oxygen. Although both species adopt a tetrahedral geometry, their magnetic properties differ fundamentally. In potassium permanganate, the manganese atom exists in the +7 oxidation state with a $3d^0$ configuration, resulting in a diamagnetic character. Conversely, potassium manganate features manganese in the +6 oxidation state with a single unpaired electron in its $3d^1$ configuration, rendering it paramagnetic; therefore, the statement that both are paramagnetic is incorrect.

The element with atomic number 24 is chromium, which possesses the ground state electronic configuration $[Ar] 4s^1 3d^5$. Because it contains six valence electrons distributed across its outer subshells, the atom is capable of losing varying numbers of these electrons to form stable ions. Reflecting this versatility in electron loss, chromium exhibits a range of common positive oxidation states extending from $+2$ to $+6$.

Determining the total number of f-electrons in an atom requires identifying all occupied f-subshells within its ground state electronic configuration, which is governed by the Aufbau principle. For Neptunium (Z=93), the electronic structure is $\mathrm{1s}^{2} \mathrm{2s}^{2} \mathrm{2p}^{6} \mathrm{3s}^{2} \mathrm{3p}^{6} \mathrm{3d}^{10} \mathrm{4s}^{2} \mathrm{4p}^{6} \mathrm{4d}^{10} \mathrm{5s}^{2} \mathrm{5p}^{6} \mathrm{4f}^{14} \mathrm{5d}^{10} \mathrm{6s}^{2} \mathrm{6p}^{6} \mathrm{5f}^{4} \mathrm{6d}^{1} \mathrm{7s}^{2}$. By observing the configuration, we find that the f-electrons are located within the $\mathrm{4f}$ and $\mathrm{5f}$ orbitals. Summing the electrons from these specific subshells, where the $\mathrm{4f}^{14}$ orbital holds 14 electrons and the $\mathrm{5f}^{4}$ orbital holds 4 electrons, provides a cumulative total of 18 f-electrons.

When dilute sodium hydroxide is introduced to a solution containing chromium(III) ions, the chemical environment facilitates the formation of an insoluble solid rather than keeping the metal in a dissolved state. This precipitation occurs as the chromium ions aggregate into a hydrated oxide structure, represented by the formula $\mathrm{Cr}_{2}\mathrm{O}_{3}(\mathrm{H}_{2}\mathrm{O})_{n}$, resulting directly from the reaction between the metal ions and the hydroxide source.

Across the actinide series, which includes Neptunium (atomic number 93) and Berkelium (atomic number 97), the size of ${M}^{3+}$ ions undergoes a consistent decrease as the atomic number increases. This phenomenon, known as actinoid contraction, results from the poor shielding of ${5f}$ electrons, which allows the nuclear charge to draw the electron cloud closer to the nucleus. Because Berkelium follows Neptunium in the series, the ${Bk}^{3+}$ ion is indeed smaller than the ${Np}^{3+}$ ion, but attributing this trend specifically to lanthanoid contraction is scientifically incorrect.

The electronic state of an ion is determined by the Aufbau principle and Hund's rule, which dictate the specific order of electron removal from the valence shells of a neutral atom. Neutral gadolinium has an atomic number of 64 and an electronic configuration of $[Xe]4f^{7}5d^{1}6s^{2}$; forming the $Gd^{3+}$ ion requires stripping the two electrons from the $6s$ orbital and the single electron from the $5d$ orbital. This leaves an electronic configuration of $[Xe]4f^{7}$, which results in 7 unpaired electrons within the $f$-subshell. Calculating the spin-only magnetic moment in Bohr Magnetons (B.M.) involves the formula $\mu = \sqrt{n(n+2)}$, where $n$ is the number of unpaired electrons. Substituting 7 for $n$ yields:

$\mu = \sqrt{7(7+2)} = \sqrt{63} \approx 7.9 \text{ B.M.}$

The magnetic properties of manganate and permanganate ions are determined by the oxidation state of the central manganese atom and the resulting presence of unpaired electrons. In the green manganate ion ($\mathrm{MnO}_{4}^{2-}$), the manganese atom has one unpaired electron, making it paramagnetic, while the purple permanganate ion ($\mathrm{MnO}_{4}^{-}$) features a $d^0$ configuration with no unpaired electrons, which renders it diamagnetic. Because both ions are tetrahedral and involve $\pi$-bonding through the overlap of oxygen $p$-orbitals and manganese $d$-orbitals, the assertion that both species are paramagnetic is factually incorrect.

The stability of specific oxidation states among lanthanoids is dictated by the energy required to reach stable electronic configurations, such as empty ($f^0$), half-filled ($f^7$), or fully filled ($f^{14}$) subshells. Europium, which possesses a $[Xe] 4f^7 6s^2$ configuration, achieves a highly stable half-filled $f$-orbital state by losing two electrons to form $Eu^{2+}$, meaning it does not exhibit a $+4$ oxidation state. In contrast, Cerium, Terbium, and Dysprosium can readily adopt a $+4$ oxidation state because the removal of electrons allows them to attain or approach configurations that are energetically favorable.

The atomic size typically increases as one descends a group in the periodic table; however, this trend is significantly altered when moving from the second to the third transition series. The poor shielding effect provided by the filling 4f orbitals causes the valence electrons to be held more strongly by the nucleus, a phenomenon known as the lanthanide contraction. This contraction effectively offsets the expected increase in size for elements in the third transition series, resulting in silver and gold having nearly identical atomic radii.

Electrons populate orbitals based on the $(n+l)$ rule, which dictates that orbitals with lower energy levels, determined by the sum of the principal quantum number $n$ and the azimuthal quantum number $l$, are filled first. In the sixth period, calculating these values shows that 6s ($n=6, l=0$) comes first with a sum of 6, followed by 4f ($n=4, l=3$), 5d ($n=5, l=2$), and 6p ($n=6, l=1$), which all share a sum of 7 but occupy energy levels in increasing order of $n$. This process dictates the filling sequence of 6s, 4f, 5d, and 6p for this period.

The chromate ion ($CrO_4^{2-}$) consists of a central chromium atom bonded to four oxygen atoms, where two oxygen atoms are linked by double bonds, each containing one $\sigma$ and one $\pi$ bond, and two are connected by single bonds, each containing one $\sigma$ bond. Summing these individual components yields a total of $2(2) + 2(1) = 6$ bonds. The dichromate ion ($Cr_2O_7^{2-}$) is formed by two $CrO_3$ units joined by a central bridging oxygen atom, where each chromium atom maintains two double bonds to terminal oxygen atoms, one single bond to a terminal oxygen atom, and one single bond to the bridging oxygen. Calculating the bonds for each chromium center as $2(2) + 1 + 1 = 6$ results in a combined total of $12$ bonds for the dichromate structure. Combining the bond counts from both species, the sum of all chromium-oxygen bonds in chromate and dichromate ions is $6 + 12 = 18$.

Transition elements are defined as elements possessing incompletely filled d-orbitals in their ground state or any common oxidation state, typically occupying groups 3 through 11. In the set 21, 25, 42, and 72, which correspond to Scandium ([Ar] 3d^1 4s^2$), Manganese ([Ar] 3d^5 4s^2$), Molybdenum ([Kr] 4d^5 5s^1$), and Hafnium ([Xe] 4f^{14} 5d^2 6s^2$) respectively, all elements satisfy the electronic configuration requirements for transition metals, whereas the other options include s-block, p-block, or f-block elements that do not qualify.

The underlying principle here is stoichiometry, where balanced chemical equations establish the fixed molar relationship between reactants and products. Given that 5 g of zinc reacts with an excess of reagents, we can represent the processes as:

$\mathrm{Zn} + 2\mathrm{HCl} \rightarrow \mathrm{ZnCl}_2 + \mathrm{H}_2$
$\mathrm{Zn} + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{ZnO}_2 + \mathrm{H}_2$

In each balanced reaction, one mole of zinc generates one mole of hydrogen gas, so using the same initial mass of zinc ensures that the molar amount of hydrogen produced is identical in both cases, leading to a volume ratio of 1:1.

The reaction sequence begins with the chromyl chloride test, where sodium chloride, potassium dichromate, and concentrated sulfuric acid produce the volatile red vapor chromyl chloride, $CrO_2Cl_2$, which represents (A). Reacting (A) with sodium hydroxide facilitates its conversion into sodium chromate, $Na_2CrO_4$, identified as (B), which subsequently reacts with dilute sulfuric acid and hydrogen peroxide to produce the characteristic blue complex chromium pentoxide, $CrO_5$, representing (C). Analyzing the atomic composition of each molecule reveals that $CrO_2Cl_2$ consists of 5 atoms, $Na_2CrO_4$ consists of 7 atoms, and $CrO_5$ consists of 6 atoms. Summing these values yields the final result of $5 + 7 + 6 = 18$.

Chromium(VI) compounds function as powerful oxidizing agents unlike the highly stable tungsten(VI) states, and lanthanoid oxides are well-established for their role in phosphor applications, confirming the validity of statements (a) and (c). In contrast, permanganate titrations are unsuitable in the presence of $HCl$ because $KMnO_4$ oxidizes the acid to release $Cl_2$, which interferes with the experimental accuracy. Thus, the claim that these titrations yield satisfactory results with $HCl$ is the only incorrect statement.