Classification Of Elements – JEE Main ChemistryPractice Questions & PYQs

Oxidation states are assigned based on the electronegativity differences between atoms in a molecule, reflecting the hypothetical charge each atom would carry if shared electrons were allocated to the more electronegative partner, which contradicts the claim that electron gain enthalpy dictates these values. On the other hand, $p \pi - p \pi$ bond formation is indeed more prevalent among second-period elements because their small atomic radii allow for highly effective side-on orbital overlap, a phenomenon that becomes significantly less favorable as atomic size increases in heavier elements.

Chemical reactivity across a period generally decreases rather than increases because the tendency to lose electrons diminishes as metallic character fades from the highly electropositive alkali metals on the left toward the inert noble gases on the right, confirming that the first statement is false. Conversely, the second statement is correct because the periodic table exhibits a clear transition from metallic to non-metallic character, causing Group 1 alkali metals to produce basic oxides while the non-metal halogens in Group 17 form acidic oxides.

The third ionization enthalpy is primarily dictated by the stability of the electronic configuration of the ion formed after the removal of two electrons, as breaking into a stable, half-filled $d$-subshell requires significant energy. Manganese, with a ground state configuration of $3d^5 4s^2$, forms an $Mn^{2+}$ ion possessing a highly stable half-filled $3d^5$ configuration. Because removing the third electron from this specific ion necessitates disrupting the symmetry of the half-filled subshell, the energy requirement is notably higher than that for vanadium ($3d^3 4s^2$), chromium ($3d^5 4s^1$), or iron ($3d^6 4s^2$), which do not result in such stable configurations upon losing two electrons.

Ionization enthalpy trends in group 13 are primarily governed by the poor shielding effects of d and f subshells, which significantly increase the effective nuclear charge felt by valence electrons. Due to scandide contraction, where the intervening d-electrons provide ineffective screening, $\mathrm{Ga}$ exhibits a higher first ionization enthalpy than $\mathrm{Al}$. Furthermore, the presence of f-electrons leads to lanthanide contraction, which provides even poorer shielding and causes $\mathrm{Tl}$ to have an ionization enthalpy that is greater than both $\mathrm{Ga}$ and $\mathrm{Al}$.

The systematic IUPAC naming convention uses specific numerical roots where the prefix "un" corresponds to the digit 1, which identifies Unununnium as having an atomic number of 111. Locating this element within the periodic table confirms that it resides in group 11, positioned alongside other heavy transition metals. Accordingly, the value of $n=11$.

Electron gain enthalpy represents the energy change involved in adding an electron to a gaseous atom, where a positive value signifies an endothermic process and a negative value indicates energy release. Argon, being a noble gas with a stable octet, resists electron addition, resulting in a positive value of 96 kJ/mol, which makes it the highest in the sequence. In contrast, fluorine, bromine, and sulfur undergo exothermic electron gain because they seek stability by completing their valence shells. Among these, sulfur releases 200 kJ/mol, while bromine and fluorine are more strongly exothermic with values of 325 kJ/mol and 333 kJ/mol, respectively. Comparing these values algebraically, the order follows $96 > -200 > -325 > -333$, leading to the arrangement of Ar, S, Br, and F, which corresponds to $A > D > B > C$.

The inert pair effect explains why the $+1$ oxidation state becomes increasingly stable down Group 13; as we descend, the poor shielding provided by $d$ and $f$ electrons results in the outer $s$ electrons being held more strongly by the nucleus, effectively discouraging them from participating in chemical bonding. Contrary to the second statement, the atomic size of gallium is actually smaller than that of aluminium because the contraction in atomic radius caused by these poorly shielding $d$ electrons outweighs the increase in size typically expected from the addition of a new electron shell.

Electronegativity trends within a group typically depend on the interplay between effective nuclear charge and the shielding effect of inner electrons, which often leads to a decline down the group. However, for the Group 14 elements from silicon to lead, the electronegativity values are nearly constant at $1.8$, $1.8$, $1.8$, and $1.9$ respectively, indicating that there is no gradual decrease across this series. Regarding their chemical classification, Group 14 is distinctly heterogeneous, containing non-metallic carbon, the metalloids silicon and germanium, as well as the metallic elements tin and lead, which confirms that the second statement is correct.

The periodic table is structured to ensure that elements with similar properties are grouped vertically, and inserting the $4f$ and $5f$ series into the main body would disrupt this arrangement, thereby undermining the core principle of periodicity. By placing these series separately at the bottom, the organization of the main group elements remains consistent and logical. Regarding the nature of s-block elements, their extremely low ionization energy makes them highly reactive, meaning they readily participate in chemical reactions to form bonds. Consequently, these elements are never found in their pure, isolated form in nature, but rather exist exclusively in a combined state as parts of stable compounds.

Ionization enthalpy increases across a period due to increasing effective nuclear charge and decreases down a group due to increased atomic size, resulting in the correct order of $\text{Na} < \text{Li} < \text{Cl} < \text{F}$. For electron gain enthalpy, halogens possess much higher negative values than alkali metals because they readily accept an electron to achieve a noble gas configuration, while chlorine has a higher negative value than fluorine because fluorine's compact $2p$ orbitals create inter-electronic repulsion that opposes the incoming electron, establishing the order $\text{Na} < \text{Li} < \text{F} < \text{Cl}$. Both statements correctly identify these periodic trends.

When determining the electronic configuration of transition metal ions, always remove electrons from the orbitals with the highest principal quantum number first, which means emptying the $4s$ subshell before the $3d$ subshell. For chromium, which starts as $3d^5 4s^1$, forming $\mathrm{Cr}^{+2}$ requires losing the $4s$ electron and one from the $3d$ orbital to reach $3d^4$. Manganese has a neutral configuration of $3d^5 4s^2$, so removing one electron results in $\mathrm{Mn}^{+}$ with the configuration $3d^5 4s^1$. Similarly, nickel's $3d^8 4s^2$ arrangement loses its two $4s$ electrons to become $3d^8$ in $\mathrm{Ni}^{+2}$, while vanadium, initially $3d^3 4s^2$, loses one $4s$ electron to yield $\mathrm{V}^{+}$ with the configuration $3d^3 4s^1$.

Electron affinity represents the energy change associated with adding an electron to a gaseous atom, where a negative value signifies an endothermic process requiring energy input. Beryllium, with its stable fully filled $2s^2$ configuration, and Nitrogen, possessing a stable half-filled $2p^3$ configuration, both resist electron addition, resulting in a negative electron affinity for the processes $\mathrm{Be} \rightarrow \mathrm{Be}^{-}$ and $\mathrm{N} \rightarrow \mathrm{N}^{-}$. Similarly, the second step of oxide formation, $\mathrm{O}^{-} + \mathrm{e}^{-} \rightarrow \mathrm{O}^{2-}$, is highly endothermic due to the strong electrostatic repulsion between the incoming electron and the negatively charged ion, giving a negative electron affinity for the overall conversion $\mathrm{O} \rightarrow \mathrm{O}^{2-}$. Conversely, processes such as $\mathrm{Na} \rightarrow \mathrm{Na}^{-}$ and $\mathrm{Al} \rightarrow \mathrm{Al}^{-}$ are exothermic and therefore possess positive electron affinity values.

Ionization enthalpy follows a consistent periodic trend, decreasing as one moves down a group, which confirms that statement B is a correct observation. In contrast, the electron gain enthalpy of $\mathrm{F}$ is less negative than that of $\mathrm{Cl}$ because the small size of the $\mathrm{F}$ atom creates significant inter-electronic repulsion, rendering statement A incorrect. Similarly, while $\mathrm{Al}_2\mathrm{O}_3$ is an amphoteric oxide, $\mathrm{NO}$ is classified as a neutral oxide, which means statement D is also incorrect. Since statement B is a correct fact, any answer choice containing it must be invalid, which leaves the combination of A, C, and D as the only possible selection by process of elimination.

For isoelectronic species, which contain the same number of electrons, the ionic radius is determined by the magnitude of the nuclear charge, represented by the atomic number $Z$. Because these species share an identical electron configuration, the size of the ion is inversely proportional to this nuclear charge, expressed by the relationship $\text{size} \propto \frac{1}{Z}$. As the nuclear charge increases, the nucleus exerts a stronger electrostatic force of attraction on the electron cloud, pulling the shells tighter and resulting in a smaller ionic radius.

Comparing the provided species, we observe their respective atomic numbers: $\mathrm{Ca}^{2+}$ has $Z=20$, $\mathrm{K}^{+}$ has $Z=19$, $\mathrm{Cl}^{-}$ has $Z=17$, and $\mathrm{S}^{2-}$ has $Z=18$. Applying the inverse relationship, the ion with the highest atomic number must have the smallest radius, while the ion with the lowest atomic number possesses the largest radius. Following this trend, the increasing order of ionic radii is $\mathrm{Ca}^{2+} < \mathrm{K}^{+} < \mathrm{Cl}^{-} < \mathrm{S}^{2-}$.

The energy required to ionize an atom is dictated by the electrostatic attraction between the positively charged nucleus and its orbiting electrons. To form $Mg^+$, one must overcome the first ionization energy, $IE_1$, whereas forming $Mg^{2+}$ requires the total energy sum of $IE_1$ and $IE_2$ because the second electron is being removed from a pre-existing cation. Since $Mg^{2+}$ is smaller and carries a greater positive charge than $Mg^+$, the remaining electrons are held much more tightly by the nucleus, which explains why the second ionization process is significantly more demanding.

Ionization enthalpy generally increases across a period as effective nuclear charge rises, yet local deviations occur because of the added stability inherent in fully filled or half-filled subshells. Specifically, the fully filled $2s^2$ orbital of beryllium makes it harder to remove an electron than from the partially filled $2p^1$ orbital of boron, while the stable half-filled $2p^3$ configuration of nitrogen requires more energy to disrupt than the $2p^4$ configuration of oxygen. Arranging these values, which range from 520 kJ/mol for lithium up to 1681 kJ/mol for fluorine, results in the sequence $\text{Li} < \text{B} < \text{Be} < \text{C} < \text{O} < \text{N} < \text{F}$.

The classification of elements into blocks depends on the subshell being filled by the final electron, which is determined by the order of orbital energies defined by the Aufbau principle. An atomic number of 37 corresponds to an element with a valence configuration ending in $5s^1$, which places it in the s-block, whereas an atomic number of 78 indicates a transition metal with the last electron entering the $5d$ subshell, defining it as a d-block element. An atomic number of 52 represents a main group element with the final electron filling the $5p$ subshell, categorizing it as a p-block element, and an atomic number of 65 refers to a lanthanoid element with electrons filling the $4f$ subshell, which classifies it as an f-block member. These assignments result in the mapping where 37 pairs with the s-block, 78 with the d-block, 52 with the p-block, and 65 with the f-block.

The bond dissociation energy of $E-H$ bonds in these hydrides is primarily governed by bond length, which increases as the atomic radius of the central element grows down the group. Since a larger atomic size results in a longer, weaker bond that is easier to break, the energy required to dissociate the molecules decreases progressively from oxygen to tellurium. This relationship establishes the trend $H_2O > H_2S > H_2Se > H_2Te$, confirming that smaller central atoms facilitate the strongest bonds.

Acidic oxides are typically non-metallic oxides that react with water to form acidic solutions or interact with bases to form salts. Among the listed compounds, $\mathrm{N}_2\mathrm{O}_3$, $\mathrm{NO}_2$, $\mathrm{Cl}_2\mathrm{O}_7$, and $\mathrm{SO}_2$ fulfill these criteria as acidic oxides, while $\mathrm{N}_2\mathrm{O}$, $\mathrm{CO}$, and $\mathrm{NO}$ are categorized as neutral oxides, and $\mathrm{CaO}$ and $\mathrm{Na}_2\mathrm{O}$ are basic oxides. This evaluation identifies exactly four acidic oxides in the provided group.

Ionization enthalpy generally decreases as you move down a group because atomic size increases, effectively placing the outermost electrons further from the nucleus. However, the specific ionization enthalpy values for group 13 elements show that Boron has an ionization enthalpy of 801 kJ/mol, which drops significantly to 577 kJ/mol for Aluminum. In contrast, the ionization enthalpy for Gallium is 579 kJ/mol, indicating that the decrease between Boron and Aluminum is substantial, while the change between Aluminum and Gallium is negligible.

This anomalous trend in Gallium occurs because of the electronic configuration, which is $[\mathrm{Ar}]\,3d^{10}\,4s^{2}\,4p^{1}$. Because the $3d$ subshell contains 10 electrons, it is completely filled, providing poor shielding for the valence electrons against the increasing nuclear charge. This ineffective screening prevents the ionization enthalpy from decreasing as expected, confirming that both the observed magnitude of change between the elements and the description of the filled $d$ orbitals in Gallium are accurate.