P Block Elements – JEE Main ChemistryPractice Questions & PYQs

The Borax Bead Test involves the reaction of $\mathrm{CuSO}_4$ with $\mathrm{B}_2\mathrm{O}_3$ in an oxidizing non-luminous flame to yield copper(II) metaborate $\mathrm{Cu}(\mathrm{BO}_2)_2$ (Z), where copper exists in the $+2$ oxidation state. In the presence of a reducing luminous flame and carbon, this compound is reduced to copper(I) metaborate $\mathrm{CuBO}_2$ (Q), where copper exists in the $+1$ oxidation state. The oxidation states of copper in $\mathrm{Z}$ and $\mathrm{Q}$ are $+2$ and $+1$, respectively.

The first ionization enthalpy of group 14 elements is higher than that of group 13 elements because the increased effective nuclear charge ($Z_{eff}$) in group 14 elements exerts a stronger electrostatic pull on the valence electrons, making their removal more energy-intensive. Conversely, the melting and boiling points of group 14 elements are significantly higher than those of group 13 elements, as group 14 elements typically form robust, giant covalent network structures that require substantial energy to break, whereas group 13 elements predominantly exhibit weaker metallic bonding. Therefore, the assertion that group 13 elements possess higher melting and boiling points than group 14 elements is inaccurate.

The element in group 15 featuring the weakest $E-E$ bond is nitrogen, as the small atomic size leads to significant interelectronic repulsion between the lone pairs. Because nitrogen is restricted to the second period and lacks available $d$-orbitals in its valence shell to expand its octet, it is limited to a maximum covalency of 4, which is achieved when it donates its lone pair to form a coordinate bond, such as in the formation of the ammonium ion ($NH_{4}^{+}$).

The interconversion between chromate and dichromate ions is pH-dependent, with chromate ions existing in equilibrium as the stable form in basic media and dichromate ions being the stable form in acidic media. Potassium dichromate reacts with potassium hydroxide to yield potassium chromate, denoted as $[A]$, and the subsequent addition of sulfuric acid acidifies the medium to convert this chromate back into potassium dichromate, $[B]$, along with the formation of potassium sulfate. These transformations are represented by the following chemical equations:
$\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + 2\mathrm{KOH} \rightarrow 2\mathrm{K}_2\mathrm{CrO}_4 + \mathrm{H}_2\mathrm{O}$
$2\mathrm{K}_2\mathrm{CrO}_4 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + \mathrm{K}_2\mathrm{SO}_4 + \mathrm{H}_2\mathrm{O}$

The ability to form $d\pi - d\pi$ bonds with transition metals is a distinct characteristic of Group 15 elements that possess accessible vacant d-orbitals, which excludes Nitrogen from consideration. Phosphorus stands out within this group as it readily participates in such bonding and produces the hydride $\mathrm{PH}_3$, which acts as the strongest base among the hydrides of the other group members that share this specific bonding capability. This element is located in the periodic table with an atomic number of 15.

The significant disparity in the melting and boiling points of oxygen and sulfur is fundamentally dictated by their distinct atomicity. Oxygen exists as a small diatomic molecule, $O_2$, whereas sulfur molecules are composed of eight atoms linked together in a larger structure represented as $S_8$. Because of this difference in atomicity, sulfur molecules possess a much greater molecular mass and experience stronger van der Waals forces of attraction compared to oxygen, which requires more thermal energy to overcome during phase changes.

The $N - N$ single bond is weaker than the $P - P$ single bond due to significant repulsion between the lone pairs of the small nitrogen atoms, but it is actually shorter in length rather than longer, which renders the first statement false. Additionally, the capacity to undergo disproportionation reactions while in the +3 oxidation state is not a universal property shared by all elements within group 15, making the second statement false as well.

Nitrogen exhibits oxidation states ranging from +1 to +5 in its oxides primarily because its small atomic size and high electronegativity facilitate the formation of stable $p \pi - p \pi$ multiple bonds with oxygen. Furthermore, the absence of vacant d-orbitals in the valence shell of nitrogen restricts its maximum covalency to four, which renders the formation of pentahalides requiring a +5 oxidation state energetically and structurally impossible.

The atomic radius of group 13 elements typically increases down the group, but the relation $(Al < Ga)$ is incorrect because the presence of ten electrons in the inner $3d$ subshell of Gallium provides poor shielding, causing the atomic radius of Gallium to be smaller than that of Aluminium. For the ionic radius $(M^{3+})$, the value increases down the group such that $Al^{3+} < Ga^{3+}$, which identifies Gallium as the element $X$ with the higher ionic radius and an atomic number of 31.

In the molecule $\mathrm{SO}_2$, the central sulfur atom resides in an oxidation state of $+4$. Since this value falls between the extreme limits of $-2$ and $+6$ for sulfur, the molecule is inherently capable of participating in redox reactions in either direction. It can behave as an oxidizing agent by gaining electrons to drop to a lower oxidation state, or it can function as a reducing agent by losing electrons to increase its oxidation state toward $+6$. Therefore, the assertion that $\mathrm{SO}_2$ cannot act as a reducing agent is incorrect, as its intermediate oxidation state grants it the capacity to perform both roles depending on the reactants involved.

The inert pair effect explains why the lower oxidation state becomes increasingly stable relative to the higher one as we move down Group 13. Because $\mathrm{Al}^{3+}$ is highly stable and resistant to reduction, it differs from $\mathrm{Tl}^{3+}$, which is unstable in solution and acts as a powerful oxidizing agent by accepting electrons to reach the more stable $\mathrm{Tl}^{+}$ configuration. Given that $\mathrm{Al}^{3+}$ and $\mathrm{Tl}^{+}$ are the most thermodynamically favored forms, statements A, B, D, and E are accurate.

The chemical characteristics of Tellurium compounds are determined by the inert pair effect and periodic trends inherent to Group 16. As the stability of the higher oxidation state declines down the group, $\mathrm{TeO}_2$ acts as an oxidising agent, while the increase in acidic character due to the weakening of the bond strength leads to $\mathrm{TeH}_2$ exhibiting acidic properties.

Boron stands out among group 13 elements due to its exceptionally high melting point of 2453 K, which contrasts sharply with the much lower melting points observed in the rest of the group. This anomaly arises because solid boron possesses a uniquely robust and stable crystalline lattice that holds its atoms tightly together. The substantial thermal energy required to overcome these intense cohesive forces within the crystal structure directly explains why boron requires such extreme temperatures to transition into a liquid state.

Electronegativity is a periodic property that increases across a period, inherently making non-metals more electronegative than metals because they possess a greater ability to attract shared electrons. When these highly reactive non-metals interact with highly reactive metals, the substantial difference in electronegativity facilitates electron transfer, which is the defining characteristic for the formation of ionic compounds. Conversely, ionization enthalpy also follows an increasing trend across a period, meaning non-metals actually exhibit higher ionization energies than metals, and chemically, metal oxides typically act as bases while non-metal oxides function as acids.

Group 13 elements exhibit anomalous periodic trends due to the ineffective shielding provided by d-electrons. Specifically, the atomic radius does not simply increase down the group because the poor shielding in gallium results in a smaller atomic size than that of aluminum, leading to the actual decreasing order of $\mathrm{Tl} > \mathrm{In} > \mathrm{Al} > \mathrm{Ga} > \mathrm{B}$. Correspondingly, electronegativity also fluctuates rather than decreasing steadily, as the values show an irregular trend starting high at boron, decreasing at aluminum, and increasing again for heavier elements, which contradicts the premises in both statements A and B.

The stability of oxidation states in this group is dictated by the inert pair effect, which becomes increasingly pronounced for heavier members like thallium. While the +1 oxidation state is favored by heavier elements, light elements such as boron and aluminum are significantly more stable in the +3 oxidation state, meaning the claim that all group 13 elements share a highly stable +1 state is inaccurate.

Aluminum demonstrates distinct chemical behavior by reacting with dilute $\mathrm{HCl}$ to liberate $\mathrm{H}_2$, while concentrated $\mathrm{HNO}_3$ causes the formation of a persistent, protective oxide layer that passivates the metal, making the description in statement C correct. In the coordination sphere of the octahedral complex $[\mathrm{Al}(\mathrm{H}_2\mathrm{O})_6]^{3+}$, the central aluminum ion undergoes $\mathrm{sp}^3d^2$ hybridization to accommodate the ligands, validating statement E.

Noble gases consist of individual atoms with stable, fully occupied valence shells, rendering them chemically inert and unable to form strong interatomic bonds. Consequently, the only interactions existing between these atoms are London dispersion forces, which are significantly weaker than dipole-dipole or hydrogen bonding interactions. Because these attractive forces are so minimal, very little energy is required to transition the substance from a liquid to a gas, resulting in exceptionally low boiling points for all noble gases. Since both statements mischaracterize the strength of these intermolecular forces and the resulting thermal properties of the elements, they are both factually incorrect.

Disproportionation occurs when an element in a specific oxidation state simultaneously undergoes both oxidation and reduction, requiring that the species is not already at its maximum or minimum possible oxidation state. Solid sulfur, $\mathrm{S}_8$, exists in the zero oxidation state and acts as both an oxidizing and reducing agent in alkaline conditions to yield $\mathrm{S}^{2-}$ and $\mathrm{S}_2\mathrm{O}_3^{2-}$, making the first statement accurate. Conversely, the chlorine atom in the perchlorate ion, $\mathrm{ClO}_4^{-}$, already exists in its highest possible oxidation state of +7, rendering it unable to lose further electrons and thus incapable of undergoing disproportionation.

Oxygen is not restricted to a -2 oxidation state, as it actually exhibits a range of values from -2 to +2, which can be observed in compounds such as $OF_2$. Furthermore, the inert pair effect causes the stability of the +4 oxidation state to increase while the stability of the +6 oxidation state decreases as one moves down the group, making the second statement incorrect as well.

Semiconductors are materials characterized by electrical conductivity that falls between that of good conductors and insulators, typically governed by a specific energy band gap. While germanium, silicon, and copper oxide exhibit these essential semiconducting properties, graphite acts as a conductor due to its unique crystalline structure, which allows delocalized electrons to move freely across its layers. Because of this inherent ability to conduct electricity efficiently, graphite is classified as a conductor rather than a semiconductor.

Chemical stoichiometry rests on the law of conservation of mass, which demands that the total number of atoms for each element must remain constant across both sides of a reaction. For the oxidation of lead sulfide, $\mathrm{PbS}$, by ozone, $\mathrm{O}_3$, to yield lead sulfate, $\mathrm{PbSO}_4$, and oxygen gas, $\mathrm{O}_2$, we identify the stoichiometric coefficients by equating the atoms on both sides. The balanced reaction is expressed as $\mathrm{PbS} + 4\,\mathrm{O}_3 \longrightarrow \mathrm{PbSO}_4 + 4\,\mathrm{O}_2$, where the four oxygen atoms incorporated into the sulfate group plus the eight oxygen atoms released as gas balance exactly with the twelve oxygen atoms provided by the four ozone molecules. Given that the coefficient $X$ for ozone is 4 and the coefficient $Y$ for oxygen is 4, adding these values yields 8.