Classification Of Elements – JEE Main ChemistryPractice Questions & PYQs

The ionization enthalpy of an atom is dictated primarily by the stability of its electronic configuration, as fully-filled and half-filled subshells require more energy to remove an electron compared to partially-filled subshells. Boron, represented by $1s^2 2s^2 2p^1$, possesses the lowest ionization enthalpy at 801 kJ/mol, while Beryllium ($1s^2 2s^2$) exhibits higher stability and an ionization enthalpy of 899 kJ/mol due to its fully-filled s-subshell. Nitrogen ($1s^2 2s^2 2p^3$) is exceptionally stable with a half-filled 2p subshell, resulting in the highest value of 1402 kJ/mol, followed by Oxygen ($1s^2 2s^2 2p^4$) with 1314 kJ/mol. Consequently, the matching associates $1s^2 2s^2$ with 899 kJ/mol, $1s^2 2s^2 2p^4$ with 1314 kJ/mol, $1s^2 2s^2 2p^3$ with 1402 kJ/mol, and $1s^2 2s^2 2p^1$ with 801 kJ/mol.

The size of isoelectronic ions is governed by the effective nuclear charge, which depends on the number of protons relative to a constant number of electrons. Since $\mathrm{N}^{3-}$, $\mathrm{O}^{2-}$, and $\mathrm{F}^{-}$ all contain 10 electrons, the species with the fewest protons experiences the weakest nuclear attraction, allowing the electron cloud to expand to a larger radius. Given that the ionic radius increases in the order $\mathrm{N}^{3-} > \mathrm{O}^{2-} > \mathrm{F}^{-}$, the nitride ion is necessarily larger than both the oxide ion at $1.4 \, \text{\AA}$ and the fluoride ion at $1.33 \, \text{\AA}$.

The IUPAC systematic nomenclature for elements with atomic numbers of 100 and above assigns specific roots to represent each digit in the atomic number. Deconstructing the name unnilennium reveals that the prefix 'un' stands for 1, 'nil' stands for 0, and 'ennium' stands for 9. Placing these digits in sequence results in an atomic number of 109.

The governing principle for trends across a period is the increase in effective nuclear charge ($Z_{\mathrm{eff}}$), which occurs as protons are added to the nucleus while electrons are filled into the same shell. This heightened electrostatic attraction pulls the electron cloud closer to the nucleus, causing the atomic radius to decrease from left to right. Conversely, this stronger nuclear hold makes it more difficult to remove or attract electrons, leading to an increase in ionization enthalpy, electron gain enthalpy, and electronegativity, thereby making atomic radius the only property among the group that exhibits a decreasing trend.

Atomic size follows a predictable periodic trend where the radius decreases moving up a group, confirming that Be is smaller than Mg, while Be exhibits a higher first ionization enthalpy than Al due to the stability provided by its fully filled $2s^2$ valence shell compared to the $3p^1$ configuration of aluminum. Beryllium and aluminum share a diagonal relationship characterized by high polarizing power, which leads both elements to form predominantly covalent compounds. Because the charge-to-radius ratio of Be is actually smaller than that of Al, the third statement is false, confirming that only the first, second, and fourth assertions are valid.

When considering electron gain enthalpy across the halogen group, we generally expect the energy released to decrease as the atomic size increases, due to the reduced effective nuclear attraction for the incoming electron. However, fluorine stands as a notable exception because its exceptionally small atomic size leads to significant inter-electronic repulsion, which makes its electron gain enthalpy less negative than that of chlorine. Consequently, the observed energy values are -349 kJ/mol for chlorine, -333 kJ/mol for fluorine, -325 kJ/mol for bromine, and -296 kJ/mol for iodine. Matching these values to the sequence of fluorine, chlorine, bromine, and iodine results in -333, -349, -325, and -296 kJ/mol respectively.

These ions are isoelectronic, meaning they possess the same electron configuration but differing nuclear charges. Because a greater nuclear charge exerts a stronger pull on the same number of electrons, the ionic radius decreases as the atomic number increases. This principle results in the ions being ordered by size as ${Al}^{3+} < {Mg}^{2+} < {Na}^{+} < {F}^{-} < {O}^{2-} < {N}^{3-}$.

Electron affinity generally decreases moving down a group because increasing atomic size diminishes the nuclear pull on an incoming electron, which explains why sulfur releases more energy than selenium, and lithium releases more energy than sodium. A distinct exception occurs among p-block elements, where the remarkably small size of second-period atoms leads to intense inter-electronic repulsion when an electron is added, causing them to release less energy than their larger third-period neighbors. Applying these principles, chlorine releases more energy than fluorine, identifying the elements with the higher energy release as chlorine, sulfur, and lithium.

Ionization enthalpy is determined by how effectively the nuclear charge binds the valence electrons, with orbital characteristics playing a crucial role. In Beryllium ($1s^{2} 2s^{2}$), the valence electrons reside in the $2s$ orbital, which possesses higher penetration power than the $2p$ orbital containing the valence electron in Boron ($1s^{2} 2s^{2} 2p^{1}$). Consequently, the $2p$ electron in Boron is more shielded from the nucleus by the inner electron core, making it easier to remove than the $2s$ electrons of Beryllium, even though Boron possesses a smaller atomic radius than Beryllium.

Ionization enthalpy reflects the energy required to remove an electron from an atom, a process heavily governed by the stability of the resulting electronic configuration. For Iron ($Z=26$), the removal of the third electron transitions the ion from a $[Ar] 3d^6$ configuration to a highly stable, half-filled $3d^5$ configuration. Because forming this stable state is energetically favorable, the third ionization enthalpy is lower for Iron compared to neighboring elements like Manganese, Cobalt, and Nickel, which do not benefit from reaching this specific half-filled electronic stabilization.

Ionization enthalpy trends offer a clear window into an element's electronic configuration. The metal exhibits a first ionization enthalpy of 496 kJ mol^-1, followed by a significantly higher second ionization enthalpy of 4560 kJ mol^-1. This substantial increase confirms the removal of the first electron from the valence shell and the second from a stable, completed inner shell, establishing the metal as a Group 1 element with a valency of +1. Therefore, the metal hydroxide is represented by the formula $\text{MOH}$.

Neutralization requirements follow from the stoichiometry of the respective reactions. The acid-base reaction with hydrochloric acid is represented by:
$\text{MOH} + \text{HCl} \rightarrow \text{MCl} + \text{H}_2\text{O}$
This equation indicates that 1 mole of $\text{MOH}$ consumes 1 mole of $\text{HCl}$. For sulfuric acid, the reaction is:
$2\text{MOH} + \text{H}_2\text{SO}_4 \rightarrow \text{M}_2\text{SO}_4 + 2\text{H}_2\text{O}$
Since 2 moles of the hydroxide react with 1 mole of sulfuric acid, it follows that 1 mole of $\text{MOH}$ reacts with exactly 0.5 moles of $\text{H}_2\text{SO}_4$.

Oxides are classified as acidic, basic, or amphoteric based on their reactivity, where non-metallic oxides typically form acidic solutions, group 1 and 2 metal oxides form basic solutions, and certain oxides like aluminum oxide exhibit amphoteric properties by reacting with both acids and bases. In the sequence $N_2O_3$, $Li_2O$, and $Al_2O_3$, nitrogen trioxide is an acidic oxide, lithium oxide is a basic oxide, and aluminum oxide is amphoteric, matching the required order.

These ions are isoelectronic, as each one possesses a total of 10 electrons, making their ionic radii dependent solely on the strength of nuclear attraction. As the atomic number $Z$ increases from 8 in $\mathrm{O}_2^{-}$ to 9 in $\mathrm{F}^{-}$, 11 in $\mathrm{Na}^{+}$, and 12 in $\mathrm{Mg}^{2+}$, the nucleus exerts a stronger pull on the fixed number of electrons, causing the ionic size to contract. This relationship is illustrated by the $\frac{Z}{e}$ ratio, which rises from 0.8 for $\mathrm{O}_2^{-}$ to 0.9, 1.1, and finally 1.2 for $\mathrm{Mg}^{2+}$. Because a higher ratio signifies a more compact ion, the size order is $\mathrm{O}_2^{-} > \mathrm{F}^{-} > \mathrm{Na}^{+} > \mathrm{Mg}^{2+}$.

Systematic IUPAC nomenclature constructs names for elements based on their atomic numbers by combining roots for each digit and appending -ium. In the case of Unnilunium, the roots correspond to 1 (un), 0 (nil), and 1 (un), directly indicating an atomic number of 101.

When forming metal cations, electrons are systematically removed from the outermost orbitals, typically beginning with the $6s$ shell before affecting the $4f$ subshell. Neutral Europium, with an atomic number of 63, has the electronic configuration $[\text{Xe}] 4f^7 5d^0 6s^2$, so bivalent $\text{Eu}^{2+}$ is formed by losing the two $6s$ electrons, leaving the configuration $[\text{Xe}] 4f^7$. Similarly, neutral Cerium with an atomic number of 58 has the configuration $[\text{Xe}] 4f^1 5d^1 6s^2$, and losing three electrons to form the $\text{Ce}^{3+}$ ion requires emptying both the $6s^2$ and $5d^1$ orbitals, which results in the remaining configuration $[\text{Xe}] 4f^1$.

Moving across a period in the periodic table from left to right, the nature of oxides predictably changes from basic to amphoteric and ultimately to acidic as non-metallic character increases. This systematic shift in chemical properties is directly linked to the rise in atomic number across the series. Because element X produces a basic oxide, Y produces an amphoteric oxide, and Z produces an acidic oxide, they must occupy positions in that specific order from left to right. Consequently, the atomic numbers follow the sequence $X < Y < Z$.

Ionization energy generally increases across a period as effective nuclear charge rises, though electronic structure deviations create specific exceptions. Sodium, with its $[Ne] 3s^1$ configuration, has the lowest ionization energy at 496 kJ/mol. Magnesium, having a stable, fully-filled $3s^2$ subshell, possesses a higher energy requirement of 737 kJ/mol, while Aluminum decreases to 577 kJ/mol because its $3p^1$ electron is more easily removed from a higher energy, more shielded orbital. Silicon continues the rising trend at 786 kJ/mol, as the increased nuclear pull outweighs the effects of subshell shielding.

Atomic radius is dictated by the increasing effective nuclear charge across a period and the addition of principal energy shells down a group. Within the second period, atomic size decreases from carbon to oxygen to fluorine as the nucleus exerts a stronger pull on valence electrons, creating the trend $F < O < C$. Because atomic radius increases down a column, chlorine in the third period and bromine in the fourth period are larger than the second-period elements, leading to the overall increasing order $F < O < C < Cl < Br$.

Ionization enthalpy represents the energy required to remove successive electrons from an atom, and a sudden, drastic increase in this energy level signals that the valence shell is empty and the next electron must be pulled from a stable, inner-shell noble gas configuration. Observing the given values of 800, 2427, 3658, 25024, and 32824 kJ mol⁻¹, the energy required to remove the first three electrons increases incrementally. A significant, sharp jump occurs between the third ionization enthalpy of 3658 kJ mol⁻¹ and the fourth ionization enthalpy of 25024 kJ mol⁻¹, which demonstrates that the removal of the fourth electron requires overcoming the stability of the inner shell of the $X^{+3}$ ion. Since the sudden jump follows the removal of three electrons, the element must have 3 valence electrons.

The addition of the first electron to a neutral atom is typically exothermic due to nuclear attraction, but adding a second electron to the resulting $O^{-}_{(g)}$ ion encounters strong electrostatic repulsion between the incoming electron and the negative charge already present. Overcoming this repulsive force requires an energy input, making the process $O^{-}_{(g)} + e^{-} \rightarrow O^{2-}_{(g)}$ endothermic. Consequently, the second electron gain enthalpy is always a positive value for every element.