P Block Elements – JEE Main ChemistryPractice Questions & PYQs

The nitrogen molecule ($\mathrm{N}_2$) features an exceptionally strong triple bond, rendering it chemically inert under standard atmospheric conditions. Furthermore, the combination of nitrogen and oxygen to form nitric oxide is a strongly endothermic process that requires significant activation energy to initiate. Consequently, the reaction $\mathrm{N}_2 + \mathrm{O}_2 \rightleftharpoons 2\mathrm{NO}$ remains impossible at ambient temperatures and only becomes feasible when subjected to the intense thermal environment of 1483-2000 K.

The borax bead test works because heating borax causes it to dehydrate and decompose into sodium metaborate and boric anhydride, with the latter acting as a glassy flux capable of dissolving metallic oxides. Specifically, the boric anhydride ($\mathrm{B}_2\mathrm{O}_3$) reacts with cobalt(II) oxide ($\mathrm{CoO}$) to produce cobalt metaborate ($\mathrm{Co}(\mathrm{BO}_2)_2$), which is the stable, blue-colored compound that forms on the platinum loop.

The formation of product 'B' is governed by the Lewis acid-base interaction between the electron-deficient diborane, $\mathrm{B_2H_6}$, and the trimethylamine ligand, which leads to the symmetrical cleavage of the diborane bridge to form the stable adduct $\mathrm{BH}_3 \leftarrow \mathrm{NMe}_3$. Within this adduct, the boron atom is bonded to three hydrogen atoms and one nitrogen atom, totaling four electron pairs around the central atom. This arrangement of four sigma bonds necessitates $sp^3$ hybridization of the boron, which fixes the geometry around the boron atom as tetrahedral.

Boron resides in the second period, meaning its valence electrons occupy only the $n=2$ shell, which lacks the empty $d$-orbitals necessary to expand its coordination number. Consequently, Boron is restricted to a maximum covalency of 4, rendering the formation of $\mathrm{BF}_6^{3-}$ impossible. While the statement regarding Boron's very small atomic size is factually true, this physical attribute is not the fundamental chemical cause for the limitation on its covalency, making the provided reason independent of the specific explanation for why the ion cannot be formed.

Stannane, denoted as $\text{SnH}_4$, is classified as a molecular hydride due to the covalent nature of the bond between tin and hydrogen. Because the central tin atom undergoes $sp^3$ hybridization, the molecule maintains a tetrahedral geometry in three-dimensional space, which confirms that it is a non-planar structure.

Caesium is highly valued for photoelectric cells because its low ionization energy allows for efficient electron emission when exposed to light. Gallium is ideal for high-temperature thermometers since it maintains a liquid state across a vast temperature range due to its remarkably high boiling point. Boron fibers are exceptionally strong, making them the preferred material for constructing bullet-proof vests. Meanwhile, silicones are widely used in water-repellent sprays because they are lined with non-polar alkyl groups that create a hydrophobic surface.

The adsorption of gases onto activated charcoal is driven by their ease of liquefaction, which correlates directly with higher critical temperatures and stronger intermolecular forces. Because $\mathrm{SO}_2$ exhibits a higher critical temperature than $\mathrm{CH}_4$, it is more readily adsorbed, validating the assertion. However, the reason is incorrect because it falsely suggests that lower critical temperatures promote adsorption, whereas higher values are actually required for the process to occur efficiently.

Concentrated nitric acid serves as a vigorous oxidizing agent that forces elemental iodine to undergo oxidation into its stable oxoacid, iodic acid. In this redox transformation, the iodine is converted to $\mathrm{HIO}_3$, while the nitric acid is concurrently reduced to nitrogen dioxide, generating water molecules in the process. The reaction stoichiometry is balanced as follows:
$\mathrm{I}_2 + 10 \mathrm{HNO}_3 \rightarrow 2 \mathrm{HIO}_3 + 10 \mathrm{NO}_2 + 4 \mathrm{H}_2 \mathrm{O}$
These products consist exclusively of $\mathrm{HIO}_3$, $\mathrm{NO}_2$, and $\mathrm{H}_2 \mathrm{O}$.

Melting points of group 16 hydrides generally increase down the group because the rising molecular mass enhances van der Waals intermolecular forces. Following this trend, the values increase from $\mathrm{H}_2\mathrm{S}$ at 188K to $\mathrm{H}_2\mathrm{Se}$ at 208K and finally $\mathrm{H}_2\mathrm{Te}$ at 222K. However, $\mathrm{H}_2\mathrm{O}$ stands apart because extensive hydrogen bonding creates significantly higher intermolecular attraction, resulting in a much higher melting point of 273K.

To classify oxides based on their chemical behavior, we distinguish between neutral oxides, which do not react with acids or bases, and acidic oxides, which typically react with bases to form salts. Among the list provided, $\mathrm{NO}$, $\mathrm{N}_2\mathrm{O}$, and $\mathrm{CO}$ are recognized as neutral oxides. The remaining substances, which are $\mathrm{B}_2\mathrm{O}_3$, $\mathrm{N}_2\mathrm{O}_5$, $\mathrm{SO}_3$, and $\mathrm{P}_4\mathrm{O}_{10}$, are acidic oxides, meaning there are 4 such compounds in total.

The thermal decomposition of metal azides provides a highly effective route to generating pure nitrogen gas because the azide anion readily breaks down at high temperatures without producing competing side products like nitrogen oxides. Subjecting barium azide to heat initiates this process, resulting in the formation of solid barium metal alongside the release of nitrogen gas, as shown by the reaction $\mathrm{Ba}(\mathrm{N}_3)_2 \xrightarrow{\Delta} \mathrm{Ba} + 3\,\mathrm{N}_2$. In contrast, common nitrogen-containing alternatives such as nitrates and nitrites typically release oxygen or various oxides of nitrogen during thermal decomposition, which confirms that the azide is the required reagent for this specific transformation.

Borazine, which is often termed inorganic benzene due to its cyclic structure, is prepared through the reaction of diborane and ammonia. This process requires a specific stoichiometric balance involving three equivalents of diborane ($\mathrm{B}_2\mathrm{H}_6$) and six equivalents of ammonia ($\mathrm{NH}_3$), which combine to yield two molecules of borazine ($\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6$). Consequently, this exact pairing provides the necessary components to assemble the $\mathrm{B}_3\mathrm{N}_3$ ring system, confirming that diborane and ammonia are the required reactants.

Liquation is a metallurgical technique specifically designed to purify metals that possess significantly lower melting points than the impurities they contain. When applied to tin ($Sn$), the impure solid is carefully heated on a tilted surface, causing the metal to melt and flow downward while the less fusible impurities remain behind as a solid residue. This process relies on physical separation by exploiting the temperature difference required to liquefy the desired metal compared to the unwanted contaminants.

White phosphorus acts as a strong reducing agent when it interacts with the chlorinating agent $SOCl_2$. During this transformation, $P_4$ undergoes oxidation and chlorination to produce $PCl_3$, while the thionyl chloride decomposes into $SO_2$ gas and $S_2Cl_2$. The balanced stoichiometric equation for this specific chemical process is $P_4 + 8 SOCl_2 \rightarrow 4 PCl_3 + 4 SO_2 + 2 S_2Cl_2$.

When an aqueous solution of ammonium chloride is mixed with sodium nitrite, the two compounds undergo a metathesis reaction to form ammonium nitrite, which is inherently unstable. This intermediate species immediately breaks down into nitrogen gas, water, and sodium chloride as shown in the reaction $\mathrm{NH}_4\mathrm{Cl} + \mathrm{NaNO}_2 \rightarrow \mathrm{NaCl} + \mathrm{N}_2(\mathrm{g}) + 2 \mathrm{H}_2\mathrm{O}$. As a result, the primary gaseous product evolved from this chemical interaction is $\mathrm{N}_2$.

When borax dissolves in water, it undergoes hydrolysis according to the reaction $\mathrm{Na}_2 \mathrm{B}_4 \mathrm{O}_7 + 7 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_3 \mathrm{BO}_3 + 2 \mathrm{Na} [\mathrm{B}(\mathrm{OH})_4]$. This process establishes an aqueous buffer system with a $\mathrm{pH} \simeq 9$, which imparts a stronger basic character to the solution compared to the other provided boron compounds.

Industrial chemical processes rely heavily on heterogeneous catalysis, where the physical state of the catalyst differs from that of the reactants. By providing an alternative reaction pathway with a lower activation energy, these catalysts significantly enhance the efficiency and yield of large-scale manufacturing operations without being consumed in the process.

For the oxidation of sulfur dioxide, the reaction $2 \mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{SO}_3(\mathrm{g})$ is catalyzed by vanadium pentoxide, $\mathrm{V}_2\mathrm{O}_5$. The production of nitrogen monoxide through the reaction $4 \mathrm{NH}_3(\mathrm{g}) + 5 \mathrm{O}_2(\mathrm{g}) \rightarrow 4 \mathrm{NO}(\mathrm{g}) + 6 \mathrm{H}_2\mathrm{O}(\mathrm{g})$ employs a platinum-rhodium alloy, $\mathrm{Pt}(\mathrm{s})-\mathrm{Rh}(\mathrm{s})$, to drive the process forward efficiently. In the Haber process, represented by the equation $\mathrm{N}_2(\mathrm{g}) + 3 \mathrm{H}_2(\mathrm{g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{g})$, finely divided iron, $\mathrm{Fe}(\mathrm{s})$, is the standard catalyst used. Lastly, the conversion of unsaturated vegetable oil into solid vegetable ghee via hydrogenation, which involves the addition of hydrogen to the oil, utilizes nickel, $\mathrm{Ni}(\mathrm{s})$, as the active catalyst. Mapping these processes to their respective catalysts yields the pairings of (A) with (III), (B) with (II), (C) with (I), and (D) with (IV).

The structural determination of peroxo ($\mathrm{O}-\mathrm{O}$) bonds in sulphur oxoacids identifies whether an acid contains a peroxide linkage rather than standard oxo linkages. Among the listed compounds, $\mathrm{H}_2\mathrm{SO}_3$, $\mathrm{H}_2\mathrm{SO}_4$, and $\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7$ contain only $\mathrm{S}=\mathrm{O}$, $\mathrm{S}-\mathrm{OH}$, and $\mathrm{S}-\mathrm{O}-\mathrm{S}$ bonds, whereas peroxodisulphuric acid ($\mathrm{H}_2\mathrm{S}_2\mathrm{O}_8$) uniquely possesses an $\mathrm{S}-\mathrm{O}-\mathrm{O}-\mathrm{S}$ structure. Therefore, only one of the provided oxoacids contains a peroxo bond.

The reaction of ethanol with $\mathrm{PCl}_3$ yields ethyl chloride and phosphorous acid, $\mathrm{H}_3\mathrm{PO}_3$, which is identified as product A. In the subsequent reaction, product A reacts with $\mathrm{PCl}_3$ to form pyrophosphorous acid, $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_5$, which is product B. Protons that are non-ionisable in these phosphorus-based oxyacids are specifically those bonded directly to the phosphorus atom, as opposed to acidic protons bonded to oxygen atoms. Within the molecular structure of pyrophosphorous acid, there are two phosphorus atoms, and each phosphorus atom retains a direct P-H bond. Consequently, the presence of these two P-H bonds indicates that there are 2 non-ionisable protons in the molecule.

The formation of stable hypervalent molecules like $\mathrm{PCl}_5$ relies on the central atom's ability to accommodate more than eight electrons in its valence shell, a phenomenon known as octet expansion. Nitrogen, being a second-period element, is strictly limited to its valence orbitals and lacks the vacant $d$ orbitals required to house additional bonding pairs. In contrast, phosphorus belongs to the third period and possesses vacant $3d$ orbitals, which allow it to promote electrons and successfully expand its valence octet to form five covalent bonds with chlorine atoms.