P Block Elements – JEE Main ChemistryPractice Questions & PYQs

Sulphur exists in two primary crystalline allotropic forms, $\alpha$-sulphur (rhombic) and $\beta$-sulphur (monoclinic), which maintain a reversible equilibrium defined by a transition temperature of $369\,\text{K}$. Because heating rhombic sulphur above this threshold triggers a conversion to the monoclinic form and subsequent cooling reverses the change, the first statement is true. However, the second statement is false because $\alpha$-sulphur is the thermodynamically stable allotrope at room temperature, whereas the monoclinic structure only becomes stable above $369\,\text{K}$.

The stability and structural arrangement of phosphorus allotropes are fundamentally dependent on specific temperature and pressure conditions. Heating red phosphorus to 803 K within a sealed tube causes it to undergo a phase transformation, resulting in the formation of alpha-black phosphorus. This specific outcome occurs because the defined thermal environment facilitates the rearrangement of atoms into this more thermodynamically stable allotropic configuration.

The purification of bauxite ore relies on the amphoteric nature of aluminum oxide, which allows it to dissolve in a hot concentrated sodium hydroxide solution to form the soluble sodium tetrahydroxoaluminate complex, denoted as $X$ or $\mathrm{Na}[\mathrm{Al}(\mathrm{OH})_4]$. When this basic solution is treated by passing carbon dioxide gas, $Y$, through it, the environment is neutralized, facilitating the precipitation of aluminum in the form of hydrated aluminum oxide, $Z$, expressed as $\mathrm{Al}_2\mathrm{O}_3 \cdot x\mathrm{H}_2\mathrm{O}$. This transformation occurs via the following chemical process:
$\mathrm{Al}_2\mathrm{O}_3 + 2\mathrm{NaOH} + 3\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{Na}[\mathrm{Al}(\mathrm{OH})_4]$
The subsequent carbonation step with $\mathrm{CO}_2$ then yields the desired solid precipitate from the sodium aluminate solution.

Carbon monoxide is hazardous because it binds with haemoglobin to form carboxyhaemoglobin, which drastically reduces the oxygen-carrying capacity of the blood. Sulphur dioxide is known to cause the stiffness of flower buds in plants, leading to their eventual shedding, while polychlorinated biphenyls are persistent synthetic organic pollutants that are classified as carcinogenic. Oxides of nitrogen contribute to atmospheric pollution but can be absorbed and metabolized by certain types of vegetation, such as plants belonging to the Pyrus genus.

Interhalogen compounds typically demonstrate higher reactivity compared to pure halogen molecules because the bond between dissimilar atoms is inherently polar and weaker. Specifically, the reactivity follows the trend $\mathrm{F}_2 > \mathrm{ClF} > \mathrm{Cl}_2$, where $\mathrm{F}_2$ remains the most reactive due to the significant interelectronic repulsion experienced between its fluorine atoms. Because interhalogens are more reactive than their homonuclear counterparts, the assertion that $\mathrm{Cl}_2$ is more reactive than $\mathrm{ClF}$ is chemically inaccurate.

The structure of oxoacids is dictated by the central chlorine atom, which anchors a single hydroxyl group ($Cl-OH$) while using its remaining available electrons to form double bonds with other oxygen atoms. Chlorous acid ($HClO_2$) corresponds to a +3 oxidation state for chlorine, resulting in exactly one $Cl=O$ bond. Chloric acid ($HClO_3$) reflects a +5 oxidation state, which accommodates two $Cl=O$ bonds. Perchloric acid ($HClO_4$) features the chlorine atom in a +7 oxidation state, allowing it to support three $Cl=O$ bonds.

The stability of $X_2O$ oxides is governed by specific atomic properties that dictate the strength of the halogen-oxygen interaction. Iodine-oxygen bonds are the most stable due to their greater polarity, whereas chlorine-oxygen bonds derive stability from multiple bond formation facilitated by the d-orbitals of the chlorine atom. Bromine lacks both of these stabilizing features, which results in the lowest stability for its oxides and leads to an overall order of $I > Cl > Br$.

Buckminsterfullerene ($C_{60}$) is a carbon allotrope characterized by a soccer ball-like cage structure where every carbon atom undergoes $sp^2$-hybridization to form three sigma bonds with neighboring atoms. This geometric arrangement consists of exactly 20 six-membered rings and 12 five-membered rings, obeying the rule that five-membered rings fuse exclusively with six-membered rings, while six-membered rings can fuse with both types. The claim that the molecule contains 12 six-membered rings and 24 five-membered rings is therefore incorrect because it provides inaccurate values for the structural components of the cage.

The synthesis sequence described illustrates core stages of the Solvay process, where stoichiometry dictates the addition of specific reactants to form the desired products. Ammonia combines with carbon dioxide and water to generate ammonium carbonate, identifying A as $\mathrm{H}_2\mathrm{O}$. This intermediate then reacts with additional carbon dioxide and water to form ammonium bicarbonate, confirming B as $\mathrm{CO}_2$, and finally interacts with sodium chloride to yield ammonium chloride and sodium bicarbonate, revealing C to be $\mathrm{NaHCO}_3$.

The reducing power of group-15 hydrides is determined by the stability of the bond between the central atom and the hydrogen atom, which decreases as the atomic size of the central element increases moving down the group. Since Bismuth has the largest atomic radius in this group, the $\mathrm{Bi-H}$ bond in $\mathrm{BiH}_3$ possesses the lowest bond dissociation enthalpy, making it the most unstable and prone to breaking. This ease of bond cleavage allows $\mathrm{BiH}_3$ to release hydrogen atoms more readily than any of its group counterparts, thereby establishing it as the strongest reducing agent.

The complete hydrolysis of $\mathrm{PCl}_5$ results in the formation of orthophosphoric acid, which has the chemical formula $\mathrm{H}_3\mathrm{PO}_4$. In this molecular structure, each of the three hydrogen atoms is covalently bonded to an oxygen atom forming $\mathrm{P}-\mathrm{OH}$ groups, which are characteristically ionisable. Because the molecule lacks any direct $\mathrm{P}-\mathrm{H}$ linkages, there are zero non-ionisable hydrogen atoms present in the final product.

Determining the oxidation state of phosphorus in these oxo acids requires assigning standard values of +1 to hydrogen and -2 to oxygen, and then solving for the central atom by setting the algebraic sum of all atoms equal to zero. Hypophosphorus acid $\mathrm{H}_{3}\mathrm{PO}_{2}$ follows the calculation $3(1) + x + 2(-2) = 0$, which results in $x = +1$. Orthophosphoric acid $\mathrm{H}_{3}\mathrm{PO}_{4}$ is calculated via $3(1) + x + 4(-2) = 0$, giving $x = +5$. For Hypophosphoric acid $\mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{6}$, the balance $4(1) + 2x + 6(-2) = 0$ leads to $2x = 8$, providing $x = +4$. Finally, Orthophosphorus acid $\mathrm{H}_{3}\mathrm{PO}_{3}$ satisfies $3(1) + x + 3(-2) = 0$, which simplifies to $x = +3$.

The reaction between phosphorus trichloride ($\mathrm{PCl}_{3}$) and phosphonic acid ($\mathrm{H}_{3}\mathrm{PO}_{3}$) in the presence of water results in the production of $\mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{5}$. Within the structural framework of this molecule, only hydrogen atoms covalently bonded to oxygen atoms are capable of dissociating as protons, whereas any hydrogen atoms directly bonded to a phosphorus atom remain non-ionisable. By evaluating the structural configuration of this specific product, it is determined that there are exactly two such ionisable hydrogen atoms present.

Back donation in boron trihalides depends fundamentally on the efficiency of the orbital overlap between the filled p-orbital of the halogen and the vacant p-orbital of the central boron atom. As the size of the halide atom decreases, its p-orbital becomes smaller, allowing for a better size match and significantly more effective overlap with the p-orbital of boron. Among the given molecules, the fluorine atom in $\mathrm{BF}_3$ is the smallest, which facilitates the most efficient electron density transfer through back donation.

Diborane is synthesized through the chemical reaction between sodium borohydride and iodine, which is expressed by the balanced equation $2 \mathrm{NaBH}_4 + \mathrm{I}_2 \rightarrow \mathrm{B}_2\mathrm{H}_6 + 2 \mathrm{NaI} + \mathrm{H}_2$. Regarding its structural characteristics, the boron atoms are $sp^3$ hybridized, which forces the molecule into a non-planar geometry rather than a flat, planar configuration. Additionally, the structure of diborane contains two three-center-two-electron bonds that act as bridges between the boron atoms, which invalidates the claim that the molecule contains only one such bond.

The solubility of metal sulphides in nitric acid relies on the strong oxidizing potential of the acid, which converts sulphide ions to elemental sulphur and leads to the formation of soluble metal nitrates. When subjected to 50% $\mathrm{HNO}_3$, compounds such as $\mathrm{PbS}$, $\mathrm{CuS}$, $\mathrm{As}_2\mathrm{S}_3$, and $\mathrm{CdS}$ effectively react and dissolve.

In contrast, $\mathrm{HgS}$ and $\mathrm{Sb}_2\mathrm{S}_3$ display stability in this environment and do not dissolve, remaining as insoluble residues. Because 4 of the 6 sulphides listed are capable of undergoing this reaction, the total number of soluble sulphides is 4.

Haber's process serves as the industrial foundation for the synthesis of ammonia, $\mathrm{NH}_{3}$, whereas the Ostwald's process is the standard chemical route for the production of nitric acid, $\mathrm{HNO}_{3}$. In the realm of heavy chemical manufacturing, the Contact process is employed for the synthesis of sulfuric acid, $\mathrm{H}_{2}\mathrm{SO}_{4}$, by converting sulfur dioxide into sulfur trioxide. Finally, the Hall-Heroult process represents the electrolytic technique used for the extraction of aluminium from its oxide, $\mathrm{Al}_{2}\mathrm{O}_{3}$.

Diborane, $\mathrm{B}_2\mathrm{H}_6$, is an electron-deficient molecule where each boron atom undergoes $sp^3$ hybridization to form both terminal and bridging bonds. The terminal B-H bonds are conventional covalent bonds with higher $s$-character, which inherently results in less $p$-character compared to the unique 3-center-2-electron bridging bonds. The bridging B-H-B bond angle is 90 degrees rather than 120 degrees, and the two bridging bonds are equal in length rather than unequal. Additionally, because $\mathrm{BH}_3$ acts as an electron-deficient species, it functions as a Lewis acid and cannot be classified as a Lewis base.

Neutral oxides are chemical compounds that exhibit neither acidic nor basic properties, which means they do not react with either acids or bases. Among the various oxides of nitrogen, $N_2O$ and $NO$ are uniquely categorized as neutral because they remain chemically inert when exposed to acidic or alkaline conditions. In contrast, other nitrogen oxides such as $NO_2$ and $N_2O_3$ function as acidic oxides, leaving $N_2O$ and $NO$ as the only pair that fits the criteria.

The number of $S=O$ bonds in oxoacids of sulfur is determined by the distinct arrangement of oxygen atoms around each sulfur center, which reflects their specific molecular geometry and oxidation states. Sulphurous acid, $H_2SO_3$, features a central sulfur atom bonded to two hydroxyl groups and one oxygen atom through a double bond, resulting in exactly one $S=O$ bond. Conversely, both peroxodisulphuric acid, $H_2S_2O_8$, and pyrosulphuric acid, $H_2S_2O_7$, are dimerized structures where each of the two sulfur atoms is bonded to two oxygen atoms via double bonds, yielding a total of four $S=O$ bonds in each molecule.