Classification Of Elements – JEE Main ChemistryPractice Questions & PYQs

Atomic radius is fundamentally dictated by the principal energy level of an element and the effective nuclear charge attracting electrons toward the nucleus. Carbon and sulfur are non-metals with relatively smaller radii of 170 pm and 180 pm, which occupy the beginning of this sequence. Aluminum, as a metal in the third period, is slightly larger with an atomic radius of 184 pm. Because cesium resides in the sixth period, it possesses multiple occupied electron shells, leading to a much larger atomic radius of 300 pm. Comparing these physical quantities establishes the relationship $170 < 180 < 184 < 300$, which dictates the sequence $C < S < Al < Cs$.

As we descend a group in the periodic table, the addition of successive electron shells increases the distance between the nucleus and the valence electrons, which directly results in a larger atomic radius. This greater distance and increased shielding effect simultaneously weaken the nuclear attraction for shared electrons, leading to a consistent decrease in electronegativity.

Systematic IUPAC nomenclature assigns a temporary three-letter symbol to elements with an atomic number $Z > 100$ by identifying the root names associated with each individual digit. The number 119 comprises the digits 1, 1, and 9, which correspond to the roots "un" and "enn" respectively. By taking the first letter from each of these roots, the resulting symbol is constructed as uue.

Isoelectronic species possess the same number of electron shells, which isolates the nuclear charge as the primary variable influencing their size. Since all these species contain an equal number of electrons, a higher proton count exerts a stronger attractive force on the outer orbitals, pulling them closer to the center. Consequently, as the nuclear charge increases across the series from $Cl^{-}$ to Ar to $Ca^{2+}$, the electrostatic pull intensifies, resulting in a progressive decrease in the ionic radius.

The variation in oxidation states between aluminum and thallium is a classic demonstration of the inert pair effect, which becomes increasingly significant as we move down group 13 of the p-block. In heavier elements like thallium, the $ns^2$ valence electrons are shielded poorly by intervening $d$ and $f$ subshells, causing the nucleus to hold them so strongly that they become reluctant to participate in chemical bonding. Consequently, while aluminum consistently loses all three valence electrons to exhibit a +3 state, thallium often exhibits a +1 state because it effectively loses only the single $np^1$ electron, leaving the "inert" $ns^2$ pair behind.

The gap between successive ionization energies spikes dramatically when the removal of an electron disrupts a stable, fully filled noble gas shell. Potassium has a single valence electron, and its first ionization energy is relatively low because losing this electron allows the atom to attain the stable argon noble gas configuration. Since the second ionization of Potassium requires extracting an electron from this exceptionally stable, inner-shell noble gas core, the energy demand is significantly higher than that for Calcium, Barium, or Scandium, where subsequent electrons are removed from valence shells rather than complete cores.

Lanthanide ions exhibit colour due to f-f electronic transitions, which require partially filled 4f orbitals to occur. Ions possessing configurations such as $f^0$ in $La^{3+}$, $f^7$ in $Gd^{3+}$, and $f^{14}$ in $Lu^{3+}$ remain colourless because their f-orbitals are entirely empty, half-filled, or completely occupied, leaving no opportunity for these specific energy transitions. Unlike these cases, $Sm^{3+}$ contains a partially filled 4f subshell, allowing for light absorption and the manifestation of a visible colour.

The energy of an orbital in a multi-electron system is primarily dictated by the electrostatic attraction between the nucleus and the electrons, which strengthens as the number of protons increases. Because a higher proton count creates a stronger pull on the electrons, the energy of the $2s$ orbital decreases as the atomic number rises. Among the provided options, Potassium contains the highest number of protons, and consequently, it exerts the greatest attractive force on the $2s$ orbital, resulting in the lowest energy level.

The periodic table is organized such that elements with the same number of valence electrons exhibit similar chemical behaviors and fall into the same group. For the yet-undiscovered element with an atomic number of 120, the predicted valence shell electronic configuration is $[Noble gas] ns^2$. This specific arrangement of two electrons in the outermost s-orbital characterizes the alkaline earth metals, which occupy the second group of the periodic table.

Electronegativity generally increases as you move from left to right across a period due to the increasing effective nuclear charge that draws bonding electrons closer to the nucleus. This fundamental trend dictates that elements in Group 13, such as gallium, naturally possess lower electronegativity values than elements situated to their right in Group 14, such as germanium. Consequently, gallium is less electronegative than germanium, which aligns with the observed periodic behavior.

Beryllium has an atomic number of 4 while boron has an atomic number of 5, which results in beryllium possessing a lesser nuclear charge. Despite this difference, beryllium maintains a greater first ionization enthalpy because its valence electrons occupy a completely filled $2s$ orbital, creating a highly stable electronic configuration that requires more energy to remove an electron compared to the less stable $2p$ electron found in boron.

The similarity in electronegativity between aluminium and beryllium is a direct consequence of the diagonal relationship found between these two elements in the periodic table. Because of this shared periodic trend, both aluminium and beryllium exhibit an identical electronegativity value of 1.5.

The species $Na^{+}$, $Mg^{2+}$, $F^{-}$, and $O^{2-}$ are isoelectronic, as each contains 10 electrons in the $1s^{2} 2s^{2} 2p^{6}$ configuration. In such a series, ionic radius is inversely proportional to nuclear charge because a higher atomic number increases the electrostatic attraction between the nucleus and the electron cloud, drawing the electrons closer to the nucleus. Since the atomic numbers follow the order $Mg (12) > Na (11) > F (9) > O (8)$, the ionic radii follow the order $Mg^{2+} < Na^{+} < F^{-} < O^{2-}$.

Electron affinity is determined by the effective nuclear charge and the inter-electronic repulsion encountered by an incoming electron within a subshell. While $F$ has a higher effective nuclear charge than $Cl$, the incoming electron in $F$ experiences significant repulsion due to its compact $2p$ subshell, making the addition of an electron less exothermic than in the larger $3p$ subshell of $Cl$. Oxygen ($O$) possesses a lower effective nuclear charge than both halogens, resulting in a lower electron affinity compared to both $Cl$ and $F$.

The group of an element is identified by the sharp jump in successive ionization enthalpies, which occurs after the valence electrons are removed. For element A, the substantial increase between the second and third ionization enthalpies, from 1757 kJ/mol to 14847 kJ/mol, indicates it possesses two valence electrons and belongs to group 2. Similarly, element B exhibits a significant jump between its second and third ionization enthalpies, increasing from 1450 kJ/mol to 7731 kJ/mol, confirming it also possesses two valence electrons and belongs to group 2. Because ionization enthalpy decreases down a group due to the increase in atomic size and screening effect, and the first ionization enthalpy of element B is 737 kJ/mol, which is lower than that of element A at 899 kJ/mol, element B must be positioned below element A.

A normal oxide contains oxygen with an oxidation state of -2. Compounds such as $KO_2$ and $CsO_2$ are superoxides containing the $O_2^-$ ion where each oxygen atom has an oxidation state of -1/2, whereas $BaO_2$ is a peroxide containing the $O_2^{2-}$ ion where each oxygen atom has an oxidation state of -1. Silicon dioxide ($SiO_2$) is the only compound among the choices where oxygen exists with an oxidation state of -2, identifying it as a true oxide.

Ionization enthalpy is governed by the stability of electronic configurations and the effective nuclear charge, with half-filled subshells providing enhanced stability due to symmetrical charge distribution. While both $[Ne]3s^{2}3p^3$ and $[Ar]3d^{10}4s^{2}4p^3$ possess stable half-filled $p$-orbitals, the former involves electrons in the $n=3$ shell, which are closer to the nucleus and more strongly attracted than those in the $n=4$ shell. Since ionization energy decreases down a group due to the increase in atomic size and shielding effect, the configuration $[Ne]3s^{2}3p^3$ requires the highest energy to remove an electron.

The ionization enthalpy of elements is primarily governed by electronic stability, where Group 15 elements possess a stable half-filled $np^3$ configuration, requiring significantly more energy to remove an electron than the $np^4$ configuration found in Group 16. As a result, Group 16 elements exhibit lower ionization enthalpy values than their corresponding Group 15 elements within the same period. The other statements are false because Group 17 contains elements existing in gaseous, liquid, and solid states, Boron in Group 13 is classified as a metalloid, and the stability of the $+5$ oxidation state in Group 15 decreases down the group due to the inert pair effect.

These ions are isoelectronic, as each species contains 10 electrons despite having different nuclear charges. Because the atomic number increases from nitrogen to fluorine, the positive charge of the nucleus grows, creating a stronger attraction for the valence electrons and causing the ionic radius to contract. Consequently, the radius follows the decreasing order $N^{3-} > O^{2-} > F^{-}$, which corresponds to ionic radii of 1.71 Å, 1.40 Å, and 1.36 Å respectively.

Ionization enthalpy generally increases across a period due to stronger effective nuclear charge and decreases down a group as the valence electrons become more shielded and distant from the nucleus. Among these elements, argon represents the noble gas configuration with the highest energy requirement, while the group 16 non-metals sulfur and selenium have higher ionization values than the group 2 alkaline earth metals calcium and barium. Within these specific groups, ionization enthalpy drops as we move down the periodic table, which places barium below calcium and selenium below sulfur, resulting in the overall increasing trend of $Ba < Ca < Se < S < Ar$.