P Block Elements – JEE Main ChemistryPractice Questions & PYQs

Hypophosphorous acid, denoted by the formula $\mathrm{H_3PO_2}$, possesses a molecular structure where the central phosphorus atom is arranged in a tetrahedral geometry to satisfy its valence requirements. This configuration consists of one oxygen atom connected by a double bond, one hydroxyl group, and two hydrogen atoms linked directly to the phosphorus center. Because these two hydrogen atoms are attached specifically to the phosphorus atom, they remain non-ionizable and do not contribute to the acidity of the compound.

The stability of hydrogen halides is primarily governed by the strength of the $HX$ bond, which consistently decreases as the size of the halogen atom increases moving down the group. This reduction in bond strength implies that while $HF$ and $HCl$ remain stable up to 1473 K, $HI$ dissociates readily at 713 K. This variation is further confirmed by bond dissociation energy values, which drop from 135 kcal mol-1 for $HF$ to 71 kcal mol-1 for $HI$, resulting in a stability order of $HF > HCl > HBr > HI$.

Silicon dioxide ($\mathrm{SiO}_2$) forms a robust, three-dimensional covalent network where every silicon atom is centrally positioned and bonded to four surrounding oxygen atoms in a tetrahedral arrangement. This specific structure relies on the fact that each oxygen atom acts as a bridge, shared between two adjacent $\mathrm{SiO}_4^{4+}$ tetrahedra. Because of this shared connectivity, every individual oxygen atom is effectively bonded to two different silicon atoms throughout the crystal lattice of quartz.

Diborane ($B_2H_6$) is an electron-deficient molecule that achieves stability through a bridged geometry rather than standard valence bonding. Four hydrogen atoms are attached to boron centers via typical 2c-2e bonds, where two electrons are shared between two atoms. The remaining two hydrogen atoms sit in bridge positions between the boron atoms, forming two 3c-2e bonds that allow a pair of electrons to be shared across three atomic centers.

Helium is actually heavier than hydrogen, contrary to the claim made in the first statement. While helium is indeed preferred for filling balloons because it is non-inflammable, the assertion that it is chosen because it is lighter is scientifically inaccurate. Understanding this specific density relationship is essential for correctly evaluating the properties of noble gases.

Aluminum chloride acts as a potent Lewis acid because its central aluminum ions possess high charge density and empty orbitals that readily accept electron lone pairs from surrounding water molecules. This hydration process stabilizes the metal center by forming the hexaaquaaluminum(III) complex, a transformation represented by the equilibrium reaction
$\mathrm{Al}_2\mathrm{Cl}_6 + 12 \mathrm{H}_2\mathrm{O} \rightleftharpoons 2 [\mathrm{Al}(\mathrm{H}_2\mathrm{O})_6]^{3+} + 6 \mathrm{Cl}^{-}$
Consequently, the species existing in an aqueous solution are the hydrated metal complexes alongside free chloride ions.

The deterioration of the buttons is a result of an allotropic phase transition, where the metal undergoes a spontaneous reorganization of its atomic lattice in response to extreme cold. Specifically, metallic white tin converts into a non-metallic, brittle grey allotropic form according to the equilibrium $\text{white tin} \rightleftharpoons \text{grey tin}$, causing the material to expand and crumble into a powdery substance. This characteristic structural shift, widely known as tin plaque or tin disease, explains why the buttons disintegrated during the freezing campaign.

Fluorine functions as the most powerful oxidizing agent among the halogens due to the distinct energetics involved in breaking its molecular bond. The bond dissociation energy of the $F-F$ bond is remarkably low, which significantly lowers the energy barrier required to generate reactive atomic fluorine. Since the resulting reactions involving these fluorine atoms are highly exothermic in nature, this favorable energy release promotes the overall oxidation process, making it superior to its counterparts.

Beryllium and aluminium exhibit a diagonal relationship, resulting in many similar properties such as the formation of amphoteric oxides, covalent halides, and polymeric hydrides. However, these elements fundamentally differ in their maximum bonding capacity due to their distinct electronic configurations. Beryllium is restricted to a maximum covalency of +2, while aluminium is able to reach a maximum covalency of +3.

When phosphine gas encounters chlorine gas, the chemical species react to form phosphorus pentachloride and hydrogen chloride through a specific stoichiometric process. This reaction follows the balanced equation $\mathrm{PH}_3 + 4\,\mathrm{Cl}_2 \longrightarrow \mathrm{PCl}_5 + 3\,\mathrm{HCl}$ As this transformation occurs, the system exhibits a cooling effect on the mixture.

Graphite's unique properties stem from its layered molecular architecture where carbon atoms undergo $sp^2$ hybridization to form expansive, tightly bonded hexagonal sheets. While the covalent bonds within these planar rings are incredibly strong, making the substance difficult to melt, the individual sheets are held together by only weak van der Waals forces. This structural arrangement allows the layers to easily slide past one another, directly conferring the material's characteristic softness and effectiveness as a solid lubricant.

Glass is best described as a super-cooled liquid because it lacks the rigid, ordered lattice structure that characterizes true crystalline solids. Its composition consists of amorphous silicates and borates represented by the general formula $R_2O \cdot MO \cdot 6SiO_2$, where $R$ is sodium or potassium, and $M$ is calcium, barium, zinc, or lead. Due to its disordered molecular arrangement, the material behaves like a liquid that has been cooled to an extremely high viscosity, allowing it to exist as a solid without forming crystals.

Concentrated hydrochloric acid naturally releases gaseous $\mathrm{HCl}$, which undergoes a chemical reaction upon encountering the oxygen present in the surrounding atmosphere. This process is governed by the stoichiometry $4\,\mathrm{HCl} + \mathrm{O}_2 \rightarrow 2\,\mathrm{Cl}_2 + 2\,\mathrm{H}_2\mathrm{O}$, where atmospheric oxygen interacts with the emitted acid gas. The resulting transformation produces chlorine gas and water, which manifest as the characteristic cloud of white fumes observed in open air.

Proton affinity represents the energy released when a molecule accepts a proton, a property that is intrinsically linked to the basicity of the substance. Within this specific set of compounds, $\mathrm{NH}_3$ stands out because the lone pair on its nitrogen atom is the most available to form a coordinate covalent bond with an incoming proton compared to the lone pairs on the other central atoms. Because $\mathrm{NH}_3$ acts as the strongest base among the options provided, it possesses the highest capacity for proton attachment.

The inherent instability of mercury(I) sulfide, $\mathrm{Hg}_2\mathrm{S}$, causes it to undergo a specific transformation when exposed to hydrogen sulfide, $\mathrm{H}_2\mathrm{S}$. This chemical interaction results in the formation of a mixture containing both $\mathrm{HgS} + \mathrm{Hg}_2\mathrm{S}$.

The number of lone pairs on the central xenon atom depends on its total of 8 valence electrons and the number of fluorine atoms it bonds with, calculated by taking the difference between these valence electrons and the bonded electrons, then dividing by two. In $\mathrm{XeF}_2$, where the central atom exhibits $sp^3 d$ hybridization, 2 electrons are used for bonding, leaving 6 electrons to form 3 lone pairs. For $\mathrm{XeF}_4$, the molecule uses $sp^3 d^2$ hybridization with 4 electrons dedicated to fluorine bonding, which leaves 4 electrons available to form 2 lone pairs. In $\mathrm{XeF}_6$, the central atom undergoes $sp^3 d^3$ hybridization and utilizes 6 electrons to bond with fluorine, resulting in 2 remaining electrons that constitute 1 lone pair.

The structure of $\mathrm{P}_4\mathrm{O}_{10}$ consists of a tetrahedral arrangement of four phosphorus atoms bridged by six oxygen atoms, with each phosphorus atom also possessing one terminal double-bonded oxygen atom. Within the cage framework, the six P-O-P bridging oxygen atoms form 12 P-O single bonds, each providing one sigma bond. Additionally, each of the four terminal P=O double bonds contains one sigma bond and one pi bond, contributing 4 further sigma bonds. Calculating the sum of these interactions, the total number of sigma bonds is $12 + 4 = 16$.

Bleaching powder, $\mathrm{CaOCl}_2$, is an ionic compound that dissociates into a calcium cation and two distinct types of anions: the hypochlorite ion $\mathrm{OCl}^{-}$ and the chloride ion $\mathrm{Cl}^{-}$. Within the hypochlorite ion, oxygen carries an oxidation state of -2, which requires the chlorine atom to balance the charge of -1 according to the equation $x + (-2) = -1$, resulting in an oxidation state of +1. Because the second chlorine atom exists as a chloride ion, it naturally possesses an oxidation state of -1, confirming that the compound contains chlorine in both +1 and -1 oxidation states.

The capacity of an element to form chemical bonds is fundamentally determined by its electronic configuration and the accessibility of specific orbitals. Nitrogen, represented by ${}_7\mathrm{N} = 1s^2 2s^2 2p^3$, only contains $s$ and $p$ orbitals in its valence shell, meaning it lacks the vacant $d$-orbitals required to accommodate the electrons needed for an expanded octet. Conversely, phosphorus has the configuration ${}_{15}\mathrm{P} = 1s^2 2s^2 2p^6 3s^2 3p^3$, which includes available $3d$-orbitals that can be utilized to expand its valence shell. This availability enables phosphorus to form higher-coordination compounds like $\mathrm{PCl}_5$, whereas nitrogen is restricted to forming trihalides such as $\mathrm{NCl}_3$.

Thermal decomposition of lead nitrate at 673 K yields nitrogen dioxide, lead monoxide, and oxygen, characterized by the reaction $2 \mathrm{Pb}(\mathrm{NO}_3)_2 \rightarrow 4 \mathrm{NO}_2 + 2 \mathrm{PbO} + \mathrm{O}_2$. The resulting nitrogen dioxide, identified as species A, dimerizes to produce dinitrogen tetroxide, which is compound B, written as $\mathrm{O}_2\mathrm{N}-\mathrm{NO}_2$. Since the structure of $\mathrm{N}_2\mathrm{O}_4$ features a direct nitrogen-nitrogen bond rather than a nitrogen-oxygen-nitrogen linkage, it contains no bridged oxygen atoms. Thus, the total number of bridged oxygen atoms in this molecule is 0.