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

The Aufbau principle dictates that electrons fill atomic orbitals in order of increasing energy, with noble gas cores serving as a convenient shorthand for fully occupied inner shells. Since Neodymium has an atomic number of 60, we utilize the Xenon core to represent the first 54 electrons, leaving six electrons to be assigned to the $6s$ and $4f$ orbitals. These remaining electrons fill the $6s$ orbital with two electrons, while the other four enter the $4f$ orbital, resulting in the stable electronic configuration $\mathrm{[Xe]} \, 4f^{4} \, 6s^{2}$.

Transition metals typically exhibit variable oxidation states due to the participation of both $ns$ and $(n-1)d$ electrons in chemical bonding. Scandium, with its electronic configuration $[Ar]3d^14s^2$, loses its three valence electrons to attain a stable noble gas configuration, which limits it to a single $+3$ oxidation state. In contrast, Cobalt, Titanium, and Nickel are capable of showing multiple oxidation states, specifically $+2$, $+3$, and $+4$, because their atomic structures allow for the involvement of varying numbers of $d$-electrons.

The borax bead test relies on the formation of characteristic colored metaborates when metal salts are heated in specific zones of a Bunsen flame. Heating a salt at point B, which corresponds to the outer oxidizing flame, results in a green bead specifically for $\mathrm{Fe}^{+3}$ ions. Given that iron has an atomic number of 26, the $\mathrm{Fe}^{+3}$ ion possesses an electronic configuration of $[\mathrm{Ar}] 3d^5$, containing five unpaired electrons. Using the spin-only magnetic moment formula $\mu = \sqrt{n(n+2)}$ where $n$ represents the number of unpaired electrons, we calculate $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM. Rounding this value to the nearest integer yields 6.

A chemical species is diamagnetic only when all its electrons are paired, leaving no unpaired electrons to exhibit magnetic properties. Neutral Lanthanum possesses the configuration $[\mathrm{Xe}]\,4f^0\,5d^1\,6s^2$, so upon forming the $\mathrm{La}^{3+}$ ion, it loses these valence electrons to achieve the stable $[\mathrm{Xe}]$ core where all orbitals are fully paired. Similarly, Cerium, with its neutral configuration $[\mathrm{Xe}]\,4f^1\,5d^1\,6s^2$, becomes the diamagnetic $\mathrm{Ce}^{4+}$ ion by losing all four valence electrons, resulting in the same fully paired noble gas core.

Ions achieve a stable state when their electron count and arrangement mirror that of noble gases, characterized by a completely filled valence shell with the general electronic configuration $ns^{2} np^{6}$. By examining each species, we find that $\mathrm{Sr}^{2+}$ ($Z=38$) loses two electrons to reach a 36-electron count, matching the electronic structure of krypton, while $\mathrm{Cs}^{+}$ ($Z=55$) loses one electron to reach 54 electrons, which corresponds to the xenon configuration. Conversely, the remaining ions fail to achieve this specific stable state because $\mathrm{La}^{2+}$ ([\mathrm{Xe}] 5 d^{1}$) retains a$5d$electron,$\mathrm{Pb}^{2+}$([\mathrm{Xe}] 4 f^{14} 5 d^{10} 6 s^{2}$) keeps electrons in $6s$, $5d$, and $4f$ orbitals, $\mathrm{Yb}^{2+}$ ([\mathrm{Xe}] 4 f^{14}$) keeps a filled$4f$subshell, and$\mathrm{Fe}^{2+}$([\mathrm{Ar}] 3 d^{6}$) has partially filled $d$ orbitals, meaning only two of the listed ions successfully attain a noble gas configuration.

Unlike p-block elements, where the inert pair effect renders lower oxidation states more stable as you descend a group, transition elements exhibit increased stability of higher oxidation states when moving down the group, a trend attributed to an increase in effective nuclear charge. Regarding the reactivity of copper, the ability of a metal to displace hydrogen from an acid depends on its standard reduction potential relative to that of hydrogen, which is defined as $E_{\mathrm{H}^{+}/\mathrm{H}_2}^{\circ} = 0\ \mathrm{V}$. Because the standard reduction potential of copper is $E_{\mathrm{Cu}^{+2}/\mathrm{Cu}}^{\circ} = 0.34\ \mathrm{V}$, it acts as a weaker reducing agent than hydrogen and therefore lacks the electrochemical capability to liberate hydrogen gas from weak acids.

The oxidising ability of these oxoanions increases as the oxidation state of the central metal rises, following the order $\mathrm{VO}_2^+ < \mathrm{Cr}_2\mathrm{O}_7^{2-} < \mathrm{MnO}_4^-$, which identifies $\mathrm{VO}_2^+$ as the species with the least oxidising strength. Vanadium, with an atomic number of 23, exists in the $+5$ oxidation state within the $\mathrm{VO}_2^+$ ion, resulting in an electronic configuration of $[\mathrm{Ar}] 4s^0 3d^0$. Since there are no unpaired electrons present, the value for the spin-only magnetic moment $\mu = \sqrt{n(n+2)}$ is calculated as 0 BM.

Cerium, with the electronic configuration $[\text{Xe}] 4f^1 5d^1 6s^2$, achieves a stable noble gas configuration by losing four electrons to form $\text{Ce}^{+4}$, rendering its formation energetically favorable. This ion is highly reactive and acts as a powerful oxidizing agent as it eagerly accepts an electron to revert to the more stable $\text{Ce}^{+3}$ state. This strong oxidizing character is clearly demonstrated by the high reduction potential of $+1.74\ \text{V}$ for the $\text{Ce}^{4+}/\text{Ce}^{3+}$ couple.

When potassium permanganate is subjected to heat, it undergoes thermal decomposition to release oxygen gas alongside stable manganese-containing residues. At a temperature of 513 K, the compound breaks down specifically into potassium manganate and manganese dioxide, a process represented by the following chemical equation:

$2\mathrm{KMnO}_4 \xrightarrow{\Delta} \mathrm{K}_2 \mathrm{MnO}_4 + \mathrm{MnO}_2 + \mathrm{O}_2$

This specific reaction reflects the inherent thermal instability of the permanganate ion, which effectively serves as a reliable method for generating oxygen gas in a laboratory setting.

The chromate ion ($\mathrm{CrO}_4^{2-}$) adopts a tetrahedral geometry, while the green manganate ion is paramagnetic due to the presence of an unpaired electron, rendering statements A and C incorrect. Statement B is correct because dichromates are prepared by acidifying chromates, as demonstrated by the reaction $2\mathrm{Na}_2\mathrm{CrO}_4+2\mathrm{H}^{+}\rightarrow\mathrm{Na}_2\mathrm{Cr}_2\mathrm{O}_7+2\mathrm{Na}^{+}+\mathrm{H}_2\mathrm{O}$. Additionally, statements D and E are accurate, as dark green $\mathrm{K}_2\mathrm{MnO}_4$ undergoes disproportionation in neutral or acidic media to form permanganate ($\mathrm{MnO}_4^{-}$), and higher oxidation states of transition metals consistently lead to a decrease in the ionic character of their oxides.

The stability of a solution containing $\mathrm{FeSO}_4$ depends on maintaining the iron ions in their dissolved state rather than allowing them to react with water. If left untreated, $\mathrm{Fe}^{2+}$ ions undergo hydrolysis, a process that results in the formation of insoluble hydroxides and makes the solution appear murky or turbid. Adding dilute $\mathrm{H}_2\mathrm{SO}_4$ creates an acidic environment that effectively suppresses this hydrolysis, ensuring the ferrous sulphate remains fully dissolved so that pure crystals of Mohr's salt can form.

The intense colours observed in $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ and $\mathrm{KMnO}_4$ are not caused by $d \to d$ transitions because the central transition metal ions exist in high oxidation states that leave them completely devoid of $d$ electrons. Specifically, the chromium in $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ is present as $\mathrm{Cr}^{+6}$ and the manganese in $\mathrm{KMnO}_4$ is present as $\mathrm{Mn}^{7+}$, meaning neither species can support $d \to d$ electronic excitations. Instead, these striking orange and purple hues arise exclusively from charge transfer transitions, where an electron is momentarily transferred from the oxygen ligand to the central metal ion upon the absorption of light.

Battery technology relies on specific transition metals capable of sustaining reversible electrochemical reactions to store and release electrical energy efficiently. Among the listed elements, manganese, nickel, and cadmium are well-established materials frequently integrated into the cathode and anode components of various battery systems. These metals are selected for their distinct redox characteristics that make them suitable for industrial energy storage applications.

The colour of lanthanoid ions arises from f-f transitions, a phenomenon that necessitates the presence of unpaired electrons within the $4\mathrm{f}$ subshell. Ions with either an empty $4\mathrm{f}^{0}$ or a completely filled $4\mathrm{f}^{14}$ configuration lack such unpaired electrons, rendering them colourless. Among the species provided, $\mathrm{La}^{3+}$ with $[\mathrm{Xe}]\,4\mathrm{f}^{0}$ and $\mathrm{Lu}^{3+}$ with $[\mathrm{Xe}]\,4\mathrm{f}^{14}$ satisfy these criteria and are thus colourless. The remaining ions, specifically $\mathrm{Nd}^{3+}$ having $[\mathrm{Xe}]\,4\mathrm{f}^{3}$, $\mathrm{Sm}^{3+}$ with $[\mathrm{Xe}]\,4\mathrm{f}^{5}$, and $\mathrm{Eu}^{3+}$ possessing $[\mathrm{Xe}]\,4\mathrm{f}^{6}$, all contain unpaired electrons in their $4\mathrm{f}$ orbitals and exhibit colour.

Although the $3d^{10}$ configuration of $\mathrm{Cu(I)}$ suggests electronic stability, the overall stability of copper ions in aqueous solution is determined by the energy balance between ionization and hydration. The high energy required for the second ionization step, $\mathrm{IE}_2$, is substantial, but it is more than balanced by the much larger hydration energy released upon the formation of $\mathrm{Cu^{2+}}$ ions. Because this release of hydration energy effectively compensates for the energy input required to remove the second electron, $\mathrm{Cu(II)}$ is more stable than $\mathrm{Cu(I)}$.

The stability of a metal ion in an aqueous solution is determined by the balance between the ionization energy required to form the ion and the energy released during its hydration. Although forming $\mathrm{Cu}^{2+}$ from $\mathrm{Cu}^{+}$ requires overcoming a significant second ionization enthalpy, the much more negative enthalpy of hydration for $\mathrm{Cu}^{2+}$ (aq) compared to $\mathrm{Cu}^{+}$ (aq) more than compensates for this energy cost. Because the $\mathrm{Cu}^{2+}$ ion has a higher charge and smaller size, its enthalpy of hydration is significantly lower, which effectively stabilizes the divalent state relative to the monovalent state in water.

The basicity of metallic oxides is directly linked to the oxidation state of the metal and the proportion of oxygen; as the percentage of oxygen increases, the acidity of the oxide rises and its basic nature diminishes. In the case of vanadium oxides, moving from $\mathrm{V}_2\mathrm{O}_3$ to $\mathrm{V}_2\mathrm{O}_5$ involves an increasing percentage of oxygen, which progressively strengthens the acidic character and weakens the basic character. This pattern dictates that the oxide with the lowest oxygen content, $\mathrm{V}_2\mathrm{O}_3$, is the most basic, establishing the order $\mathrm{V}_2\mathrm{O}_3 > \mathrm{V}_2\mathrm{O}_4 > \mathrm{V}_2\mathrm{O}_5$.

When forming cations, electrons are systematically removed from the orbital with the highest principal quantum number because those electrons occupy the outermost shell and are most loosely bound to the nucleus. Neodymium, which has an atomic number of 60, possesses the ground state electronic configuration of $[\mathrm{Xe}] 4f^4 6s^2$. Since the $6s$ orbitals constitute the outermost energy level, ionizing the atom to a $2+$ state involves losing both $6s$ electrons, leaving the remaining $4f$ electrons unaffected. This results in the final configuration of $[\mathrm{Xe}] 4f^4$.

The chemical behavior of lanthanide ions is primarily governed by the stability of the $+3$ oxidation state. $\mathrm{Ce}^{4+}$ has a $[Xe] 4f^0$ electronic configuration but readily gains an electron to return to the preferred $+3$ state, making it a strong oxidizing agent. Conversely, $\mathrm{Eu}^{2+}$ possesses a stable half-filled $[Xe] 4f^7$ configuration, yet it acts as a strong reducing agent because it effectively loses an electron to achieve the standard $+3$ oxidation state.

Qualitative analysis relies on the redox potential of the sulfur dioxide gas evolved from the acidification of the sulfite ion ($\mathrm{SO}_3^{2-}$). This gas acts as a reducing agent against the dichromate ion ($\mathrm{Cr}_2\mathrm{O}_7^{2-}$), converting it into the chromium(III) ion ($\mathrm{Cr}^{3+}$). Consequently, the acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ solution shifts from its characteristic orange color to a distinctive green hue, indicating the successful reduction of the chromium.