Classification Of Elements – JEE Main ChemistryPractice Questions & PYQs

The ionization energy of atoms within the same period is primarily determined by the effective nuclear charge and the relative stability of the electron subshells. Substituting $n=2$ into the given list identifies the elements as Beryllium ($2s^2$), Boron ($2s^2 2p^1$), Nitrogen ($2s^2 2p^3$), and Neon ($2s^2 2p^6$). Boron, containing a single electron in the $2p$ subshell, has the lowest ionization energy of 800 kJ mol^-1 because its removal is easier compared to the more stable, fully filled $2s^2$ subshell of Beryllium, which requires 899 kJ mol^-1. Nitrogen requires 1402 kJ mol^-1 due to the extra stability provided by a half-filled $2p$ subshell, while Neon displays the highest ionization energy of 2080 kJ mol^-1, characteristic of its stable noble gas configuration. Consequently, the configuration $ns^2$ maps to 899 kJ mol^-1, $ns^2 np^1$ maps to 800 kJ mol^-1, $ns^2 np^3$ maps to 1402 kJ mol^-1, and $ns^2 np^6$ maps to 2080 kJ mol^-1.

Electronegativity increases across a period as effective nuclear charge rises, validating the sequence $F > O > N$ for these elements. Determining oxidation states relies on assigning negative values to the more electronegative atom in a compound. In $\mathrm{OF}_2$, fluorine's higher electronegativity leads to oxygen assuming a state of +2, satisfying the equation $x + 2(-1) = 0$. Conversely, in $\mathrm{Na}_2\mathrm{O}$, oxygen's higher electronegativity compared to sodium results in a state of -2, derived from the equation $2(+1) + x = 0$.

First ionization energy increases across a period and decreases down a group, with anomalies like Phosphorus having higher ionization energy than Sulfur due to its stable half-filled $3p^3$ configuration, making the order $\mathrm{F} > \mathrm{P} > \mathrm{S} > \mathrm{B}$ correct. Electron affinity generally increases across a period, but Chlorine exceeds Fluorine due to the smaller size and increased electron-electron repulsion in the latter, while Sulfur is higher than Phosphorus because the additional electron in Phosphorus disrupts its stable half-filled subshell, confirming $\mathrm{Cl} > \mathrm{F} > \mathrm{S} > \mathrm{P}$ is the correct trend. Metallic character decreases across a period, so Magnesium is more metallic than Aluminum, which invalidates the proposed trend $\mathrm{K} > \mathrm{Al} > \mathrm{Mg} > \mathrm{B}$, whereas the basic character of oxides consistently decreases from left to right and increases down a group, establishing $\mathrm{K}_2\mathrm{O} > \mathrm{Na}_2\mathrm{O} > \mathrm{MgO} > \mathrm{Al}_2\mathrm{O}_3$ as correct and identifying A, B, and D as the accurate statements.

Electronegativity values decrease down group 15, with nitrogen having a significantly higher value compared to phosphorus, arsenic, and subsequent elements. Using the given condition $|\chi_X - \chi_P| > |\chi_P - \chi_Y|$, assigning nitrogen to $X$ and arsenic to $Y$ results in the equation $|3.0 - 2.1| > |2.1 - 2.0|$, which simplifies to the inequality $0.9 > 0.1$. Thus, $X$ and $Y$ are identified as nitrogen and arsenic, respectively.

The second ionization potential depends on the electronic configuration of the singly ionized species and the increasing effective nuclear charge across the period. Since the $C^+$, $N^+$, $O^+$, and $F^+$ ions have outer configurations of $2p^1$, $2p^2$, $2p^3$, and $2p^4$ respectively, the $O^+$ ion exhibits enhanced stability due to its half-filled $2p^3$ subshell, making its second ionization energy higher than that of $F^+$. This electronic stability anomaly overrides the general trend of increasing ionization energy with atomic number, resulting in the correct order $C < N < F < O$.

Atomic radius decreases across a period from left to right as the increasing effective nuclear charge exerts a stronger attraction on the valence electrons, while the number of electron shells remains constant. Consequently, Potassium ($K$) in Group 1 exhibits the largest atomic radius in Period 4, whereas Bromine ($Br$) in Group 17 exhibits the smallest atomic radius because it possesses the highest effective nuclear charge among the main-group elements in that period.

Metallic character increases down a group and decreases across a period from left to right. Phosphorus, located in group 15, period 3, exhibits the most metallic character among the listed elements and has 5 valence electrons. Fluorine exhibits the least metallic character due to its position at the top right of group 17, period 2, and it contains 7 valence electrons.

The process of transforming gaseous lithium into a dipositive ion requires the successive removal of two electrons, which necessitates summing the first and second ionization energies. Combining the first ionization energy of 520 kJ/mol with the second ionization energy of 7297 kJ/mol reveals that $520 + 7297 = 7817\ \mathrm{kJ\ mol^{-1}}$ of energy is needed to ionize one mole of lithium. Since the sample mass is 3.5 mg, which is equal to 0.0035 g, the number of moles is determined by the calculation $\frac{3.5 \times 10^{-3}}{7} = 5 \times 10^{-4}\ \mathrm{mol}$. Multiplying the total molar energy requirement by the actual number of moles results in $7817 \times 5 \times 10^{-4} = 3.9085\ \mathrm{kJ}$, which rounds to 4 kJ.

The second ionization enthalpy of $Na$ is significantly higher than that of $Mg$ because removing the second electron from $Na^+$ involves disrupting a stable, noble gas configuration ($1s^2 2s^2 2p^6$), whereas for $Mg^+$, the second electron is removed from a $3s^1$ orbital, which is energetically more accessible. For the second statement, $O^{2-}$ and $F^-$ are isoelectronic species containing 10 electrons; since the ionic radius decreases as the nuclear charge increases in isoelectronic ions, the smaller nuclear charge of oxygen ($Z=8$) relative to fluorine ($Z=9$) results in a larger ionic radius for $O^{2-}$.

Across any period, elements situated at the extreme left belong to Group 1 and possess the largest atomic size, meaning their valence electrons are loosely held and require the least energy to remove, thereby resulting in the lowest first ionization enthalpy. In contrast, the elements at the extreme right, excluding the chemically inert noble gases, belong to Group 17 and feature high effective nuclear charge combined with a nearly complete valence shell, which creates a strong affinity for an additional electron and releases the greatest amount of energy, represented by the highest, most negative $\Delta H_{\text{eg}}$.

The classification of oxides is dictated by the electronic nature and oxidation state of the central element, where metals generally produce basic oxides and non-metals produce acidic oxides, with neutral and amphoteric types serving as intermediate states. In the first statement, the pairs $\mathrm{Al}_2 \mathrm{O}_3$ and $\mathrm{Cr}_2 \mathrm{O}_3$ are both amphoteric, $\mathrm{Cl}_2 \mathrm{O}_7$ and $\mathrm{Mn}_2 \mathrm{O}_7$ are both acidic, $\mathrm{Na}_2 \mathrm{O}$ and $\mathrm{V}_2 \mathrm{O}_3$ are both basic, and $\mathrm{CO}$ and $\mathrm{N}_2 \mathrm{O}$ are both neutral; as all four pairs contain oxides of matching nature, the first statement is true.

Regarding the second statement, $\mathrm{Na}_2 \mathrm{O}$ features an alkali metal leading to strongly basic character, whereas $\mathrm{Cl}_2 \mathrm{O}_7$ involves a non-metal in a high oxidation state resulting in strong acidity, making these the most basic and acidic oxides in the provided list, respectively. Consequently, both provided statements accurately describe the chemical properties of these compounds.

Removing an electron from a neutral atom requires an input of energy, meaning ionization enthalpy is inherently an endothermic process with a positive value. After the first electron is removed to form a cation, the remaining electrons are held more tightly by the nucleus due to increased effective nuclear charge and decreased electron-electron repulsion, which causes the second ionization enthalpy to be significantly higher than the first, represented by the relationship $\mathrm{IE}_2 > \mathrm{IE}_1$. Since the first ionization enthalpy for magnesium is +737 kJ/mol, the second ionization energy must be a positive value greater than this amount. Among the options provided, +1450 kJ/mol is the only value that correctly reflects an increase in the energy required to remove the subsequent electron.

Atomic radius decreases across a period from left to right due to increased effective nuclear charge, resulting in the order $Mg > Al$, while for isoelectronic species, the ion with the higher nuclear charge, $Al^{3+}$, possesses a smaller radius than $Mg^{2+}$, thereby invalidating the order given in Statement I. The magnitude of electron gain enthalpy generally increases across a period but is influenced by electron-electron repulsions in smaller atoms, resulting in the experimentally determined trend of $Cl > Br > S > O$, which confirms that Statement II is true.

Metallic character decreases from left to right across a period and increases down a group, confirming that the order $K > Mg > Al > B$ accurately reflects decreasing metallic character, thus making Statement I true. Conversely, Statement II is false because while cations have smaller radii than their neutral parent atoms due to the loss of electrons and increased effective nuclear charge, anions possess larger radii than their parent atoms due to increased inter-electronic repulsion and reduced effective nuclear charge per electron, meaning the atomic radius is not always greater than the ionic radius.

Chemical oxides are classified by their chemical behavior: basic oxides typically react with acids to form salts, acidic oxides react with bases, amphoteric oxides possess the dual capability to react with both, and neutral oxides show no acidic or basic character. To satisfy the requirement of the question, a set must include one distinct representative from each of these four categories. Within the provided options, $K_2O$ serves as the basic oxide, $Cl_2O_7$ acts as the acidic oxide, $As_2O_3$ exhibits amphoteric properties, and $NO$ remains neutral. This specific combination is the only one that successfully covers all four distinct types of oxides.

Ionization enthalpy generally increases across a period; however, the anomalous stability of nitrogen's half-filled $2p^3$ configuration makes its ionization enthalpy greater than that of oxygen, which establishes the correct trend $C < O < N < F$. The magnitude of electron gain enthalpy is highest for sulfur because oxygen's compact atomic size creates significant inter-electronic repulsion that hinders the addition of an electron, resulting in the valid decreasing order $|S| > |Se| > |Te| > |Po| > |O|$. As both of these established trends accurately reflect the electronic factors influencing these periodic properties, both statements are true.

Periodic trends dictate that first ionization enthalpy generally increases across a period due to rising effective nuclear charge, meaning the actual order for these elements is $\mathrm{Na} < \mathrm{Mg} < \mathrm{Cl} < \mathrm{Ar}$ rather than the reverse sequence provided. The second statement is accurate because calcium achieves a stable noble gas configuration after the loss of two electrons; consequently, removing a third electron necessitates disrupting this highly stable core, which requires 4910 kJ/mol-a value significantly higher than the third ionization energies of $\mathrm{Al}$, $\mathrm{Fe}$, and $\mathrm{B}$.

Ionization enthalpy generally increases across a period due to the rising effective nuclear charge, which strengthens the electrostatic attraction between the nucleus and the valence electrons. Because the half-filled $3s^{2}3p^{3}$ orbital of phosphorus is more stable than the $3s^{2}3p^{4}$ configuration of sulfur, the energy required to remove an electron from phosphorus is higher than that required for sulfur. These factors, combined with the overall periodic trend, result in the sequence $\mathrm{Al} < \mathrm{Si} < \mathrm{S} < \mathrm{P} < \mathrm{Cl}$.

Electronegativity generally increases across a period from left to right as effective nuclear charge rises, which strengthens an atom's ability to attract shared electrons. Within the third period, aluminum resides to the right of magnesium and therefore possesses a higher electronegativity value, expressed as $\mathrm{Al} > \mathrm{Mg}$. Consequently, any ordering suggesting that aluminum is less electronegative than magnesium is inaccurate.

Periodic trends in ionization enthalpy dictate that Germanium exhibits the highest first ionization enthalpy among these elements, which correlates to the stability of the +4 oxidation state common to group 14. Conversely, Indium has the lowest first ionization enthalpy, and as a member of group 13, its most stable oxidation state is +3. The respective stability of the oxidation states for the elements with the highest and lowest ionization enthalpies therefore corresponds to the +4 and +3 states.