P Block Elements – JEE Main ChemistryPractice Questions & PYQs

The fundamental determinant of group 15 hydride properties is the strength of the $\mathrm{M}-\mathrm{H}$ bond, which weakens significantly as the central atom increases in size down the group. This decline in bond strength directly correlates to a decrease in thermal stability, following the sequence $\mathrm{NH}_3 > \mathrm{PH}_3 > \mathrm{AsH}_3 > \mathrm{SbH}_3 > \mathrm{BiH}_3$. Simultaneously, because these bonds become progressively easier to break, the molecules become increasingly effective at releasing hydrogen to act as reducing agents, which establishes the reducing ability trend of $\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 < \mathrm{SbH}_3 < \mathrm{BiH}_3$.

The acidic or basic nature of group 14 oxides is fundamentally governed by the metallic character of the central atom, which increases as one moves down the group. While oxides like $\mathrm{CO}$, $\mathrm{SiO}_2$, and $\mathrm{GeO}_2$ primarily exhibit acidic character, the elements tin and lead in their +4 oxidation state possess sufficient metallic character to form $\mathrm{SnO}_2$ and $\mathrm{PbO}_2$, which are amphoteric. This capacity to react with both acids and bases uniquely defines the chemical behavior of these specific dioxides among the given list.

The detection of phosphorus involves the oxidation of the compound to form phosphate ions, which subsequently react with ammonium molybdate in the presence of nitric acid. This reaction yields a characteristic canary yellow precipitate of ammonium phosphomolybdate, which has the chemical formula $(\mathrm{NH}_4)_3\mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3$.

Moving down Group 14, atomic size increases as new electron shells are filled, and electronegativity does not decrease gradually because of the poor shielding provided by filled inner d and f orbitals. Carbon's limited size and lack of vacant d orbitals restrict its covalence to 4, while heavier members utilize their d orbitals to expand their bonding capacity and fail to form $p \pi - p \pi$ bonds due to their inability to achieve sufficient lateral orbital overlap. These structural factors, combined with carbon's specific electronic properties, allow it to exhibit a broad range of oxidation states from $-4$ to $+4$ in various chemical environments.

The reducing character of Group 15 hydrides depends on the ease with which the central atom-hydrogen bond is broken, which is directly linked to the bond dissociation energy. Moving down the group, the atomic size of the central element increases, resulting in longer and weaker bonds between the central atom and hydrogen. Since $\mathrm{BiH}_3$ exhibits the lowest bond dissociation energy due to this trend, it is the most effective at releasing hydrogen and therefore functions as the strongest reducing agent in the list.

The identification of anions in qualitative salt analysis relies on distinguishing them by their reactivity with dilute mineral acids, which serves as a preliminary screening to assign them to a group rather than functioning as a confirmatory test. When a sulphide salt is warmed with dilute $\mathrm{H_2SO_4}$, it releases hydrogen sulphide gas that forms a black precipitate of lead sulphide, $\mathrm{PbS}$, upon contact with lead acetate paper according to the reaction $\mathrm{(CH_3COO)_2Pb} + \mathrm{H_2S} \longrightarrow \mathrm{PbS} + 2\,\mathrm{CH_3COOH}$. Because this reaction merely indicates the presence of a sulphide-containing group and the black precipitate is actually lead sulphide rather than lead sulphite, both provided statements are scientifically inaccurate.

The physical state of a substance at room temperature is fundamentally determined by the strength of the intermolecular forces holding its molecules together. Boric acid exists as a solid because its molecules are stabilized by an extensive network of strong hydrogen bonding, which holds them tightly in a rigid structure. In contrast, $\mathrm{BF}_3$ molecules lack such potent intermolecular interactions, meaning the molecules remain as discrete, independent particles that exist in the gaseous phase.

Chlorine interacts with oxygen to produce several unstable oxides, specifically $\mathrm{Cl}_2\mathrm{O}$, $\mathrm{ClO}_2$, $\mathrm{Cl}_2\mathrm{O}_6$, and $\mathrm{Cl}_2\mathrm{O}_7$, which are powerful oxidizing agents that possess an inherent tendency to explode. These reactions serve as fundamental examples of chemical behavior, aligning with the general principle that an element's reactivity is effectively determined by observing its propensity to bond with oxygen and halogens.

The Lewis acid strength of boron trihalides is fundamentally determined by the extent of back bonding between the filled $p$-orbital of the halogen and the vacant $p$-orbital of the boron atom. Because fluorine provides the most effective orbital overlap with boron ($\mathrm{2p-2p}$) compared to the increasingly larger orbitals of chlorine, bromine, and iodine ($\mathrm{2p-3p}$, $\mathrm{2p-4p}$, and $\mathrm{2p-5p}$ respectively), the back bonding effect is strongest in $\mathrm{BF}_3$ and weakest in $\mathrm{BI}_3$. Since efficient back bonding reduces the electron deficiency at the boron center, the compound with the least effective back bonding exhibits the highest Lewis acidity, establishing the order $\mathrm{BI}_3 > \mathrm{BBr}_3 > \mathrm{BCl}_3 > \mathrm{BF}_3$.

In the borax bead test, heating cupric sulphate within a non-luminous, oxidising flame produces blue or blue-green cupric metaborate, $\text{Cu}(\text{BO}_2)_2$. When exposed to a luminous, reducing flame instead, this compound undergoes reduction to either colourless cuprous metaborate, $\text{CuBO}_2$, or opaque red metallic copper, meaning that a green colour is not produced by the reducing flame. Because the luminous flame does not generate a green bead and the specified reduction product, $\text{CuBO}_2$, is colourless rather than green, both statements provided are incorrect.

The sequence of reactions described illustrates a crucial segment of the Solvay process, leveraging the chemical principle that strong bases displace ammonia from its salts. When calcium hydroxide, acting as compound A, reacts with ammonium chloride, it produces ammonia, identified as B, alongside calcium chloride and water according to the reaction $\mathrm{Ca}(\mathrm{OH})_2 + 2\,\mathrm{NH}_4\mathrm{Cl} \rightarrow 2\,\mathrm{NH}_3 + \mathrm{CaCl}_2 + 2\,\mathrm{H}_2\mathrm{O}$

This ammonia gas subsequently reacts with water and an excess of carbon dioxide to yield ammonium bicarbonate, which is compound C, through the reaction $\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \rightarrow \mathrm{NH}_4\mathrm{HCO}_3$ Finally, this ammonium bicarbonate reacts with sodium chloride to produce sodium hydrogen carbonate, establishing that the compounds A, B, and C are $\mathrm{Ca}(\mathrm{OH})_2$, $\mathrm{NH}_3$, and $\mathrm{NH}_4\mathrm{HCO}_3$, respectively.

Applying the law of conservation of mass ensures that the number of atoms for every element remains constant on both sides of the reaction between white phosphorus and thionyl chloride. Balancing the stoichiometry yields the complete equation $\mathrm{P}_4 + 8 \mathrm{SOCl}_2 \rightarrow 4 \mathrm{PCl}_3 + 2 \mathrm{S}_2\mathrm{Cl}_2 + 4 \mathrm{SO}_2$ By mapping this balanced representation to the reaction described \in the problem, it is clear that the 4 moles of product $A$ correspond to $\mathrm{PCl}_3$, the 2 moles of product $B$ correspond to $\mathrm{S}_2\mathrm{Cl}_2$, and the coefficient $x$ for the moles of $\mathrm{SO}_2$ equals 4.

Borazine, represented by the formula $\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6$, forms a cyclic structure that features electronic delocalization, allowing it to be effectively described as inorganic benzene. This molecule demonstrates reactivity with water, specifically following the reaction $\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6 + 9\mathrm{H}_2\mathrm{O} \rightarrow 3\mathrm{NH}_3 + 3\mathrm{H}_3\mathrm{BO}_3 + 3\mathrm{H}_2$, where it decomposes into ammonia, boric acid, and hydrogen gas. Because its bonding relies on conventional covalent and delocalized interactions rather than the electron-deficient 3-center 2-electron bridging structures known as banana bonds found in compounds like diborane, the statement claiming it contains such bonds is incorrect.

Ammonium salts in the atmosphere contribute significantly to the formation of fine particulate matter, which manifests as haze, while polychlorinated biphenyls are recognized for their historical utility as effective cleansing solvents. In contrast, atmospheric chlorine radicals are responsible for ozone depletion rather than its production, as they actively react with ozone to form oxygen via the process $\text{overset}{\cdot}{\mathrm{Cl}} + \mathrm{O}_3 \longrightarrow \mathrm{O}_2 + \text{overset}{\cdot}{\mathrm{ClO}}$. Furthermore, the condition known as 'blue baby' syndrome is specifically linked to the ingestion of water containing an excess of nitrate ions, rendering the claim regarding sulphate ions incorrect.

The structure of pyrophosphoric acid ($H_4P_2O_7$) is defined by two central phosphorus atoms joined by an oxygen bridge, which dictates the total number of covalent bonds present. Each phosphorus atom forms one double bond with a terminal oxygen atom, generating two $\pi$ bonds in the entire molecule. The remaining framework consists of four $P-OH$ groups and two $P-O$ bonds within the central bridge, accounting for a total of 12 $\sigma$ bonds. Comparing these quantities yields the final ratio:

$\frac{\sigma}{\pi} = \frac{12}{2} = 6$

The Ostwald process initiates with the catalytic air oxidation of $\mathrm{NH}_3$, which generates nitric oxide ($\mathrm{NO}$) as the primary product $A$. This gaseous product then undergoes further oxidation in the presence of air to yield nitrogen dioxide ($\mathrm{NO}_2$), identified as $B$. Upon hydration, $\mathrm{NO}_2$ disproportionates to produce nitric acid, which is an important oxoacid of nitrogen, while simultaneously evolving $\mathrm{NO}$ back into the system. Because nitric acid is a strong oxidizing agent and serves as the nitrate source for the brown ring test, this reaction sequence confirms that $A$ corresponds to $\mathrm{NO}$ and $B$ corresponds to $\mathrm{NO}_2$.

The compound $\mathrm{GaAlCl}_4$ is an ionic salt composed of a gallium cation and a tetrachloroaluminate anion, formally represented as $\mathrm{Ga}^{+}\mathrm{AlCl}_4^{-}$. In this structural arrangement, gallium exists in a +1 oxidation state rather than the +3 state commonly observed in other group 13 compounds, allowing it to function as the cationic part of the crystalline lattice. This configuration is characteristic of Gallous tetrachloro aluminate, where the electronegativity and electronic properties of gallium relative to aluminum facilitate its role as the stable, positively charged component of the salt.

Chemical reactions must obey the law of conservation of mass, where the identity of products is determined by balancing the atoms present in the reactants. In the first reaction, sodium oxide reacts with water to produce $\mathrm{NaOH}$, meaning $X$ contains 1 oxygen atom. For the second reaction, dichlorine heptoxide reacts with water to form $\mathrm{HClO}_4$, indicating that $Y$ contains 4 oxygen atoms. Combining the oxygen content from both products yields a total of 5 oxygen atoms.

In reactions involving noble gas fluorides, the tendency of the species to function as a fluoride donor or acceptor is governed by the relative Lewis acidity of the reactants. Because $\mathrm{SbF}_5$ acts as a powerful Lewis acid, it strips a fluoride ion from $\mathrm{XeF}_4$, resulting in the formation of the ionic species $[\mathrm{XeF}_3]^+ [\mathrm{SbF}_6]^-$. Comparing this chemical outcome to the general structure $[\mathrm{XeF}_m]^{n+} [\mathrm{SbF}_y]^{z-}$, we determine that $m=3$, $n=1$, $y=6$, and $z=1$. The sum of these values is calculated as $3 + 1 + 6 + 1 = 11$.

The chlorination of white phosphorus using thionyl chloride is a standard synthetic pathway that yields phosphorus trichloride as the primary halogenated product, as shown by the reaction
$\mathrm{P}_4 + 8 \mathrm{SOCl}_2 \rightarrow 4\mathrm{PCl}_3 + 4\mathrm{SO}_2 + 2\mathrm{S}_2\mathrm{Cl}_2$
This compound [A], which is $\mathrm{PCl}_3$, readily undergoes hydrolysis when exposed to water, following the chemical equation
$\mathrm{PCl}_3 + 3 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{H}_3\mathrm{PO}_3 + 3 \mathrm{HCl}$
The resulting product [B] is phosphorous acid ($\mathrm{H}_3\mathrm{PO}_3$), which is recognized as a dibasic acid because it possesses two hydroxyl groups bonded to the central phosphorus atom.