Classification Of Elements – JEE Main ChemistryPractice Questions & PYQs

The first ionization potential of an atom is defined as the energy required to remove an electron from the gaseous neutral atom to form a cation. For sodium, the reaction $\mathrm{Na} \rightarrow \mathrm{Na}^+ + \mathrm{e}^-$ involves an energy absorption of 5.1 eV. Electron gain enthalpy represents the energy change associated with the reverse process, where an electron is added back to the cation as $\mathrm{Na}^+ + \mathrm{e}^- \rightarrow \mathrm{Na}$. Since this reverse reaction is energetically symmetrical to the ionization process, the energy released is equal in magnitude but opposite in sign to the ionization energy, resulting in a value of -5.1 eV.

When comparing isoelectronic species, the ionic radius is inversely proportional to the atomic number because the number of electrons remains constant, making the effective nuclear charge dependent solely on the number of protons. As the atomic number increases, the nucleus exerts a stronger attractive force on the electron shell, which follows the relationship $Size \propto \frac{1}{Z}$. Given the atomic numbers for $S^{2-}$ (16), $Cl^{-}$ (17), $K^{+}$ (19), and $Ca^{2+}$ (20), the ionic radius decreases as the atomic number increases, resulting in the increasing order $Ca^{2+} < K^{+} < Cl^{-} < S^{2-}$.

The stability of a half-filled 4f subshell is the governing principle behind the unique electronic structure of Gadolinium. While lanthanide elements typically fill the 4f orbital sequentially, Gadolinium with an atomic number of 64 achieves a lower energy state by maintaining a stable, half-filled $4f^7$ configuration, which necessitates placing one electron into the 5d orbital. This specific electronic arrangement results in an outer configuration of $4f^7 5d^1 6s^2$.

The basic nature of metallic oxides is directly linked to the metallic character of the elements, which decreases across a period from left to right and increases down a group. For elements within the same period, the basicity of the oxides decreases from sodium to magnesium to aluminum, following the order $\mathrm{Al}_2\mathrm{O}_3 < \mathrm{MgO} < \mathrm{Na}_2\mathrm{O}$. Because potassium is located below sodium in the same group, its oxide exhibits a higher basic character, resulting in the final increasing sequence of $\mathrm{Al}_2\mathrm{O}_3 < \mathrm{MgO} < \mathrm{Na}_2\mathrm{O} < \mathrm{K}_2\mathrm{O}$.

Since these ions all possess the same number of electrons, they are considered isoelectronic, meaning their relative sizes are determined primarily by their nuclear charge. As the atomic number increases, the nucleus exerts a stronger electrostatic pull on the constant number of electrons, contracting the electron cloud and shrinking the radius. Therefore, species with a greater negative charge have lower effective nuclear attraction per electron and appear larger, whereas those with higher positive charge are more compact due to increased proton-electron attraction, establishing the decreasing sequence $\mathrm{O}^{2-} > \mathrm{F}^{-} > \mathrm{Na}^{+} > \mathrm{Mg}^{2+} > \mathrm{Al}^{3+}$.

The property listed in option C is incorrect because the hydrogen halides provided act as acids rather than bases, making their classification under basic strength inherently flawed. Furthermore, in the context of basicity for group 15 hydrides, the trend is governed by the availability of the lone pair of electrons for donation, which decreases as the atomic size of the central atom increases. Because nitrogen is the smallest atom in its group, its lone pair is concentrated into a smaller volume, making it more available for donation than the more diffuse lone pairs of phosphorus, arsenic, or antimony. This leads to a decreasing order of basic strength $\mathrm{NH}_3 > \mathrm{PH}_3 > \mathrm{AsH}_3 > \mathrm{SbH}_3$, which contradicts any sequence suggesting an increasing trend.

Ionic radius is primarily governed by the number of electron shells and the effective nuclear charge pulling on the valence electrons. When moving down a group in the periodic table, the addition of extra shells causes the ionic size to increase, ensuring that $\mathrm{Na}^{+}$ is larger than $\mathrm{Li}^{+}$, and $\mathrm{Mg}^{2+}$ is larger than $\mathrm{Be}^{2+}$. Concurrently, moving from left to right across a period increases the effective nuclear charge, which contracts the electron cloud and decreases the ionic radius. These combined periodic trends establish a clear size hierarchy among these ions, where the complete order is $\mathrm{Na}^{+} > \mathrm{Li}^{+} > \mathrm{Mg}^{2+} > \mathrm{Be}^{2+}$.

Alkali metals achieve chemical stability by losing their single valence electron, and as the number of electron shells increases down the group, the electrostatic attraction between the nucleus and this valence electron diminishes, making it progressively easier to remove and increasing chemical reactivity. In contrast, halogens react by acquiring an external electron, and this process becomes less efficient down the group because the increasing number of intervening electron shells shields the valence shell, weakening the nucleus's ability to attract that additional electron. This fundamental difference in how these groups achieve stability means that increasing the atomic number facilitates electron loss for alkali metals while simultaneously hindering electron gain for halogens.

Ionization enthalpy generally increases from left to right across a period because the effective nuclear charge increases, drawing electrons more strongly toward the nucleus. However, this trend is influenced by the stability of subshells, creating an exception for elements with half-filled orbitals like Phosphorus, which possesses a $3p^3$ configuration. This stable arrangement makes it harder to remove an electron from Phosphorus compared to the $3p^4$ configuration of Sulfur, resulting in a higher ionization enthalpy for the former. Consequently, when ordering these elements by their ionization energy, Boron occupies the lowest position due to its lower effective nuclear charge, while Fluorine sits at the highest, establishing the increasing order $B < S < P < F$.

An amphoteric oxide is defined by its ability to react with both acidic and basic solutions to yield salt and water. While $\mathrm{CaO}$ acts as a basic oxide and both $\mathrm{CO}_2$ and $\mathrm{SiO}_2$ behave as acidic oxides, tin is categorized among the elements such as $\mathrm{Zn}$, $\mathrm{Be}$, $\mathrm{Al}$, $\mathrm{B}$, $\mathrm{Cr}$, $\mathrm{Ga}$, and $\mathrm{Pb}$ that demonstrate amphoteric properties. Consequently, $\mathrm{SnO}_2$ is the only compound in the provided list that possesses this dual chemical character.

Ionization enthalpy generally follows an increasing trend across a period because the rise in effective nuclear charge makes it progressively harder to remove an electron from an atom. However, this trend is interrupted by the enhanced stability associated with half-filled and fully-filled subshells. Boron, Carbon, Nitrogen, and Oxygen have valence configurations of $2p^1$, $2p^2$, $2p^3$, and $2p^4$, respectively. Nitrogen possesses a stable, half-filled $2p^3$ subshell, which requires significantly more energy to ionize than the less stable $2p^4$ configuration of Oxygen. Therefore, the actual order of first ionization enthalpy is $\mathrm{B} < \mathrm{C} < \mathrm{O} < \mathrm{N}$, which contradicts the sequence provided in the question.

The other options accurately represent periodic trends. Atomic radii within a group increase from Lithium to Rubidium because each subsequent element adds an additional electron shell, increasing the distance from the nucleus. For electron gain enthalpy, Chlorine has a higher value than Fluorine; the extremely small size of the Fluorine atom creates strong inter-electronic repulsion when an electron is added to its $2p$ orbital, whereas the larger $3p$ orbital of Chlorine accommodates the incoming electron more readily. Finally, for isoelectronic species like $\mathrm{Al}^{3+}$, $\mathrm{Mg}^{2+}$, $\mathrm{Na}^{+}$, and $\mathrm{F}^{-}$, all containing 10 electrons, ionic size decreases as nuclear charge increases. Because Aluminum has the highest number of protons pulling on these electrons, it exhibits the smallest radius, making the given sequence correct.

The acid-base character of oxides is primarily determined by the position of the central atom within a period, as acidic strength increases from left to right due to the growing electronegativity and non-metallic nature of the elements. Aluminum, silicon, phosphorus, and sulfur are located in groups 13, 14, 15, and 16, respectively, which means the non-metallic character of these elements follows the progression $\mathrm{Al} < \mathrm{Si} < \mathrm{P} < \mathrm{S}$. As a result, the acidic character of their corresponding oxides increases steadily, yielding the sequence $\mathrm{Al}_2\mathrm{O}_3 < \mathrm{SiO}_2 < \mathrm{P}_2\mathrm{O}_3 < \mathrm{SO}_2$.

The formation of an oxide ion is fundamentally a battle between the stabilizing force of nuclear attraction and the destabilizing effect of inter-electronic repulsion. Adding the first electron to an oxygen atom is an exothermic process, releasing 142 kJ mol-1, because the neutral atom effectively attracts the incoming electron into its valence shell. However, the addition of a second electron to form $\mathrm{O}^{2-}$ is endothermic, requiring 844 kJ mol-1, because the incoming negative charge encounters severe electrostatic repulsion from the existing electron cloud surrounding the $\mathrm{O}^-$ ion. This repulsion outweighs the attraction from the nucleus, meaning the $\mathrm{O}^-$ ion resists the addition of another electron and requires an external energy input to force the reaction to proceed.

The ionic radius is primarily governed by the ratio of the atomic number ($Z$) to the number of electrons ($e$), as this ratio dictates the strength of the nucleus's attraction on the electron cloud. A lower $\frac{Z}{e}$ value implies a weaker pull from the nucleus on each electron, allowing the cloud to expand and thereby increasing the ionic radius.

Between the listed ions, $\mathrm{Li}^{+}$ and $\mathrm{B}^{3+}$ are significantly smaller because they have lost their outer shells and contain electrons only in the $1s$ shell, while $\mathrm{O}^{2-}$ and $\mathrm{F}^{-}$ retain electrons in $1s, 2s$, and $2p$ orbitals. Comparing these two anions, $\mathrm{O}^{2-}$ has a nuclear charge of $Z=8$ and $e=10$ electrons, resulting in a ratio of $\frac{Z}{e} = 0.8$, whereas $\mathrm{F}^{-}$ has $Z=9$ and $e=10$ electrons with a ratio of $\frac{Z}{e} = 0.9$. Since $\mathrm{O}^{2-}$ possesses the lowest ratio of nuclear charge to electron count, it exhibits the weakest effective attraction from the nucleus, resulting in the largest ionic radius.

The second ionization enthalpy is determined primarily by the stability of the electron configuration of the resulting cation and the specific orbital from which the second electron is removed. After the initial ionization, the Chromium ion achieves a stable, half-filled $3d^5$ configuration, while the corresponding ions of Vanadium, Manganese, and Iron still possess an electron in the $4s$ orbital. Because the $4s$ orbital is further from the nucleus and experiences less effective nuclear charge than the $3d$ orbital, removing an electron from it requires less energy. Consequently, the second ionization enthalpy for Chromium is significantly higher than that of the other elements because the electron must be pulled from a stable, inner $3d$ subshell rather than an outer $4s$ orbital.

Amphoteric oxides are characterized by their distinct ability to react with both acids and bases to form salts. While elements such as $\mathrm{Zn}, \mathrm{Be}, \mathrm{Al}, \mathrm{Pb},$ and $\mathrm{Sn}$ are recognized for forming these compounds, $\mathrm{ZnO}$ serves as a primary example of an oxide that exhibits this dual reactivity. In contrast, oxides like $\mathrm{Na}_2\mathrm{O}$ behave as basic oxides and $\mathrm{SO}_2$ functions as an acidic oxide, making $\mathrm{ZnO}$ the unique choice among the provided options that neutralizes both acidic and basic substances.

The physical and chemical characteristics of an element are determined by its internal atomic structure, specifically the number of protons contained within the nucleus. According to the modern periodic law, these properties exhibit a recurring, predictable pattern that serves as a periodic function of an element's atomic number. Consequently, organizing elements by their atomic number allows for the systematic alignment of those with similar behaviors.

The general trend where atomic size decreases as atomic number increases is driven by the growing effective nuclear charge, represented by $Z_{eff} = Z - S$, which pulls the electron cloud inward with greater force as more protons are added. This behavior is not restricted to the f-block, as the contraction effect is observable throughout the periodic table, nor is it unique to radioactive elements starting from atomic number 84, since the trend persists far earlier in the table. Because the pattern of decreasing size with rising atomic number is a broad feature that manifests as elements become heavier, it is characteristic of elements with high atomic masses.

The concept governing these ionic sizes is lanthanide contraction, which describes the steady decrease in ionic radii as the atomic number increases across the lanthanide series. In this trend, the poor shielding provided by the $4f$ electrons results in an increased effective nuclear charge, causing the electron cloud to be drawn more tightly toward the nucleus. Since $\mathrm{La}^{+3}$, with an atomic number of 57, sits at the beginning of the series, it possesses the largest ionic radius among the group. Moving across the series toward $\mathrm{Ce}^{+3}$, $\mathrm{Pm}^{+3}$, and finally $\mathrm{Yb}^{+3}$, the ionic size progressively diminishes, making the increasing order of their radii $\mathrm{Yb}^{+3} < \mathrm{Pm}^{+3} < \mathrm{Ce}^{+3} < \mathrm{La}^{+3}$.

The ionic radii of lanthanides systematically decrease across the series due to lanthanide contraction, a trend driven by the filling of the $4f$ subshell. Because $4f$ electrons provide poor shielding against the increasing nuclear charge as the atomic number rises from Cerium to Lutetium, the effective nuclear attraction strengthens and pulls the electron shell inward. Consequently, elements earlier in the series, such as Cerium, possess larger ionic radii compared to those that follow later in the sequence, such as Samarium, Terbium, and Lutetium.