Classification Of Elements – JEE Main ChemistryPractice Questions & PYQs

Electronegativity serves as a quantitative measure of an atom's ability to attract shared electrons toward itself within a chemical bond, allowing for a clear hierarchy based on specific values. Comparing the values of these elements, bromine has the highest electronegativity at 2.8, followed by carbon at 2.5, astatine at 2.2, and phosphorus at 2.1. By arranging these elements according to their respective values in descending order, the resulting relationship is $\mathrm{Br} > \mathrm{C} > \mathrm{At} > \mathrm{P}$.

An element remains liquid in boiling water only if it is chemically inert toward water and remains in its liquid phase at 100 °C. Alkali metals like $\mathrm{Li}$ and $\mathrm{Cs}$ react vigorously with water, while $\mathrm{Br}_2$ turns into a vapor because its boiling point is only 58 °C. In contrast, $\mathrm{Ga}$ is a suitable candidate because its melting point is 29 °C and its boiling point is 2400 °C, meaning it remains stable and liquid in boiling water and only reacts if the temperature exceeds 100 °C.

Noble gases possess remarkably stable octet or duplet configurations, which explains why their electron gain enthalpy values are positive, indicating that energy must be supplied to force an electron into an already occupied shell. Neon exhibits the highest resistance to accepting an electron at 116 kJ/mol due to the compact nature of its 2p orbitals, which creates significant inter-electronic repulsion. As we move down the group, the increasing atomic size eases the accommodation of an incoming electron, resulting in lower values for Krypton at 96 kJ/mol, Xenon at 77 kJ/mol, and finally Helium at 48 kJ/mol. This trend culminates in the ascending sequence $\mathrm{He} < \mathrm{Xe} < \mathrm{Kr} < \mathrm{Ne}$.

Electron gain enthalpy measures the energy involved when an atom gains an electron, and finding the maximum difference between two elements requires identifying the pair with the most extreme values. Neon represents the upper end of this spectrum because its stable electronic configuration makes electron addition highly unfavorable, resulting in the most positive $\Delta H_{eg}$ among all elements. Conversely, chlorine occupies the lower end, as it possesses the most negative $\Delta H_{eg}$ of all elements due to the significant energy released upon completing its octet. Pairing the most positive value with the most negative value naturally creates the greatest possible magnitude of difference, which occurs between neon and chlorine.

As you progress down Group 13, the metallic and electropositive character of the elements increases significantly, which fundamentally shifts the chemical nature of their oxides from acidic to basic. While $\mathrm{B}_2\mathrm{O}_3$ is acidic and both $\mathrm{Al}_2\mathrm{O}_3$ and $\mathrm{Ga}_2\mathrm{O}_3$ exhibit amphoteric properties, the basic character becomes more pronounced with the heavier elements. Therefore, $\mathrm{Tl}_2\mathrm{O}_3$ is the most basic oxide in the group because thallium possesses the highest electropositive character among the listed elements.

An f-subshell comprises seven orbitals, so the state of being half-filled corresponds to having exactly seven electrons in the $4f$ subshell, which is denoted as $4f^7$. Among the elements provided, Samarium with a $4f^6 6s^2$ configuration, Promethium with $4f^5 6s^2$, and Terbium with $4f^9 6s^2$ do not meet this requirement because their f-orbitals contain different numbers of electrons. In contrast, both Europium, which has an electronic configuration of $4f^7 6s^2$, and Gadolinium, which possesses $4f^7 5d^1 6s^2$, satisfy the condition of maintaining a stable, half-filled f-orbital arrangement in their ground state.

The assertion is false because the first ionization enthalpy of early 3d series elements is actually less than that of group 2 metals like magnesium. The provided data explicitly confirms that ionization energy values for elements from scandium through manganese are lower than the corresponding ionization energy of magnesium. Conversely, the reason statement is factually accurate, as the 3d series is defined by the progressive and successive filling of 3d orbitals with electrons as the atomic number increases.

Typically, electron gain enthalpy decreases as you descend a group because increasing atomic size reduces the electrostatic attraction for an incoming electron. Oxygen represents an important exception to this trend; its extremely compact electronic shell creates significant inter-electronic repulsion upon the arrival of a new electron, making the energy release less favorable compared to sulfur. Therefore, while the rest of the group follows the expected decline, sulfur possesses the highest electron gain enthalpy, resulting in the final order of $O < S > Se > Te$.

Ionization enthalpy generally increases across a period because the effective nuclear charge rises, pulling electrons closer to the nucleus. However, electronic configuration causes specific deviations, most notably between group 2 and group 13 elements. Magnesium holds a stable, fully filled $3s^2$ subshell, which makes removing its first electron more difficult than removing the single electron from the higher-energy $3p$ orbital of aluminum.

This difference in stability leads to an observed order of $\mathrm{Na} < \mathrm{Al} < \mathrm{Mg} < \mathrm{Si}$ for their first ionization enthalpies. Given that sodium is 496 kJ mol⁻¹, magnesium is 737 kJ mol⁻¹, and silicon is 786 kJ mol⁻¹, the value for aluminum must fall between 496 kJ mol⁻¹ and 737 kJ mol⁻¹. The figure 577 kJ mol⁻¹ is the only option that satisfies this condition, accurately reflecting the ionization energy trend dictated by subshell stability.

The electron gain enthalpy of an element depends on the effective nuclear charge and the size of the atom, influencing its affinity to accept an incoming electron. Rubidium and Cesium exhibit nearly identical electron gain enthalpy values of -46 kJ/mol. Similarly, noble gases like Argon and Krypton share positive electron gain enthalpy values of +96 kJ/mol because their stable electronic configurations resist the addition of an electron, resulting in a common $\Delta H_{\mathrm{eq}}$. These specific pairings are the ones where the values are either nearly identical or effectively the same.

Ionization enthalpy generally decreases down a group due to increased shielding and distance from the nucleus, while it increases from left to right across a period. This confirms that the ionization enthalpy of Potassium is lower than that of Sodium and Lithium, and that element 37 has a lower value than element 38. In Group 18, ionization energy also decreases as we move down the column, meaning Xenon is not the element with the lowest ionization energy. However, the claim that Gallium has a higher ionization enthalpy than Zinc, the element with atomic number 30, is incorrect because Zinc possesses a stable pseudo-inert gas configuration that results in a higher ionization energy than Gallium.

First ionization enthalpy follows the general periodic trend of increasing as we move left to right across the second period, largely because of the rising nuclear charge and shrinking atomic size. However, this pattern is interrupted by the enhanced stability of certain electronic configurations. Beryllium, with a completely filled $2s^2$ subshell, requires more energy to remove an electron than Boron, which holds a single electron in its $2p$ orbital. Similarly, Nitrogen possesses a stable, half-filled $2p^3$ configuration, which makes its ionization enthalpy significantly higher than that of Oxygen, where electron-electron repulsion in the $2p^4$ shell makes the removal of an electron easier. Integrating these specific stability factors into the periodic trend leads to the sequence $\mathrm{B} < \mathrm{Be} < \mathrm{O} < \mathrm{N}$.

Determining the identity of an element from its electron configuration requires summing the electrons provided by the noble gas core and the valence shell to calculate the atomic number $Z$. Since the [Rn] core accounts for 86 electrons and the outer shell configuration adds $14 + 1 + 2 = 17$ electrons, the total atomic number is 103. The IUPAC systematic naming convention assigns roots for each digit-specifically un for 1, nil for 0, and tri for 3-which, when combined with the suffix -ium, results in the name Unniltrium.

Metal oxides are typically basic in nature, which makes $\mathrm{CaO}$ a standard example of a basic oxide. Conversely, oxides of non-metals such as $\mathrm{SO}_3$ and $\mathrm{SiO}_2$ are classified as acidic, whereas $\mathrm{Al}_2\mathrm{O}_3$ is considered amphoteric because it reacts as both an acid and a base.

When comparing isoelectronic species, which possess the same number of electrons, ionic radius is primarily governed by the ability of the nucleus to attract that electron cloud. As the positive charge of an ion increases, the nucleus exerts a stronger relative grip on the electrons, resulting in a more compact size, whereas an increasing negative charge enhances electron-electron repulsion and diminishes the effective nuclear pull, leading to a larger radius. Following this relationship, the ionic size grows progressively from the cation with the highest positive charge to the anion with the highest negative charge in the order $\mathrm{Mg}^{2+} < \mathrm{Na}^{+} < \mathrm{F}^{-} < \mathrm{O}^{2-} < \mathrm{N}^{3-}$.

The chemical classification of oxides as acidic, basic, neutral, or amphoteric depends on their specific reactivity, which is determined by their metallic or non-metallic character. Evaluating the provided compounds, $\mathrm{Na}_2\mathrm{O}$ is a basic oxide, $\mathrm{Cl}_2\mathrm{O}_7$ is an acidic oxide, while $\mathrm{N}_2\mathrm{O}$ and $\mathrm{NO}$ are both classified as neutral. Since $\mathrm{As}_2\mathrm{O}_3$ is the only compound in the list that exhibits the dual ability to react with both acids and bases, there is exactly one amphoteric oxide present.

Ions that possess the same number of electrons are referred to as isoelectronic species, a classification that accurately describes both $\mathrm{O}^{2-}$ and $\mathrm{Mg}^{2+}$ since each contains exactly 10 electrons. Despite this commonality, their ionic radii are not identical because the size of an ion is heavily influenced by the effective nuclear charge; the nucleus of magnesium contains 12 protons, while oxygen contains only 8. This stronger electrostatic attraction in the magnesium ion draws the electron cloud inward more effectively than in the oxide ion, meaning the size of $\mathrm{O}^{2-}$ is actually greater than $\mathrm{Mg}^{2+}$, which makes the initial assertion false.

The electronic configuration of an atom plays a critical role in determining its ionization enthalpy, with half-filled subshells exhibiting enhanced stability due to their symmetric arrangement. Nitrogen achieves this stable configuration with a $2p^3$ state, which requires a significant amount of energy to disrupt, making its first ionization energy higher than that of oxygen, which possesses a $2p^4$ configuration. While it is accurate that the four electrons within the oxygen $2p$ orbitals experience greater electron-electron repulsion, this factor is secondary to the fundamental stability afforded by the half-filled $p$-orbital state in nitrogen. Consequently, both statements are factually sound, but the stability of the half-filled subshell is the primary reason for the trend, rather than the repulsive forces present in the oxygen atom.

Electron gain enthalpy represents the energy change associated with the addition of an electron to a neutral gaseous atom, where a greater magnitude of energy release indicates a stronger electron-accepting tendency. Halogens, such as $\mathrm{Cl}$ and $\mathrm{F}$, exhibit significantly higher exothermic values than chalcogens like $\mathrm{Te}$ and $\mathrm{Po}$ because of their high effective nuclear charge, and within these groups, the magnitude of the enthalpy decreases as one moves down the column due to increased atomic size and electron-electron repulsion. Given the values of $-349\ \mathrm{kJ}\ \mathrm{mol}^{-1}$ for $\mathrm{Cl}$, $-333\ \mathrm{kJ}\ \mathrm{mol}^{-1}$ for $\mathrm{F}$, $-190\ \mathrm{kJ}\ \mathrm{mol}^{-1}$ for $\mathrm{Te}$, and $-174\ \mathrm{kJ}\ \mathrm{mol}^{-1}$ for $\mathrm{Po}$, the correct order of these enthalpies in terms of increasing magnitude is $\mathrm{Po} < \mathrm{Te} < \mathrm{F} < \mathrm{Cl}$.

The covalent radius of a diatomic molecule like $Cl_2$ is defined by the relation $r_{covalent} = \frac{d}{2}$, where $d$ represents the internuclear distance between the bonded atoms, which contradicts the claim that the covalent radius is double the atomic radius. Conversely, the radius of an anion is always greater than that of its parent atom because adding an electron increases electron-electron repulsion and reduces the effective nuclear charge experienced by each electron, causing the electron cloud to expand.