P Block Elements – JEE Main ChemistryPractice Questions & PYQs

Oxides are classified based on their behavior in water, where those that react to form acids are termed acidic oxides, while those that do not react in this manner are classified as neutral. Carbon dioxide ($CO_2$) is an acidic oxide because it undergoes a reaction with water to form carbonic acid, denoted as $CO_2 + H_2O \rightleftharpoons H_2CO_3$. Conversely, carbon monoxide ($CO$) is a neutral oxide because it is sparingly soluble in water and does not form an acid. Since the formation of an acid upon dissolution is the chemical property that distinguishes these oxides, the observation regarding the reactivity of $CO_2$ with water provides the fundamental explanation for its acidic nature as described in the assertion.

Substances classified as polar molecular solids are composed of individual molecules that feature a permanent dipole, leading to solid-state assembly primarily through dipole-dipole interactions. Because the molecular structures of $\mathrm{SO}_2$ and $\mathrm{NH}_3$ are inherently asymmetrical, they create a persistent separation of charge that results in polar behavior. Consequently, these specific molecules associate through the attraction of their respective dipoles to form stable polar molecular solids.

To determine the shift, first calculate the oxidation state of xenon in the reactant $\mathrm{XeF}_4$, where each of the four fluorine atoms maintains a state of -1, placing the xenon atom at +4. Complete hydrolysis of $\mathrm{XeF}_4$ results in the formation of $\mathrm{XeO}_3$ as the oxidized xenon product. Within the structure of $\mathrm{XeO}_3$, the three oxygen atoms each possess an oxidation state of -2, which dictates that the xenon atom must be in the +6 state. Comparing these values, the difference between the +6 oxidation state of the product and the +4 oxidation state of the reactant is 2.

The covalency of an atom corresponds to the total number of covalent bonds it establishes, so in the $[\mathrm{BF}_4]^{-}$ ion, where boron bonds with four fluorine atoms, the covalency is 4. The oxidation state is determined by setting the sum of the individual oxidation numbers equal to the total charge of the ion, such that $x + 4(-1) = -1$, where $x$ represents the oxidation state of boron. Solving this equation results in $x = +3$.

The borax bead test identifies metal cations through the production of characteristic colored beads during fusion with borax. During the heating process, copper(II) sulfate first undergoes thermal decomposition to release sulfur trioxide and form copper(II) oxide according to the equation $\mathrm{CuSO}_4 \xrightarrow{\Delta} \mathrm{CuO} + \mathrm{SO}_3$. This copper(II) oxide then combines with boric anhydride within the bead to generate copper(II) metaborate, represented by $\mathrm{CuO} + \mathrm{B}_2\mathrm{O}_3 \rightarrow \mathrm{Cu}(\mathrm{BO}_2)_2$, which is responsible for the characteristic blue-green color observed in an oxidizing flame.

According to Fajans' rule, the covalent character of a compound increases with the oxidation state of the central metal cation because a higher positive charge enhances the polarizing power on the associated anions. Following this principle, $\mathrm{SbCl}_5$ is more covalent than $\mathrm{SbCl}_3$ because the antimony atom possesses a higher $+5$ oxidation state compared to the $+3$ state in $\mathrm{SbCl}_3$. Furthermore, the higher oxides of halogens are generally more stable than their lower counterparts, primarily because these higher oxidation states are characteristically less reactive and often involve larger atomic configurations that contribute to greater overall stability.

The dichromate ion, $Cr_2O_7^{2-}$, contains chromium in the $+6$ oxidation state, calculated from the ionic charge where $2x + 7(-2) = -2$. In an acidic medium, this ion acts as a strong oxidizing agent by gaining six electrons to be reduced to the $Cr^{3+}$ ion, resulting in the chromium oxidation state shifting from $+6$ to $+3$ according to the reduction half-reaction $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$.

To determine the total number of $\pi$-bonds, we identify the sulfur-oxygen double bonds in the molecular structures of these acids, as each $S=O$ bond represents exactly one $\pi$-bond. Peroxodisulphuric acid ($H_2S_2O_8$) contains two sulfur atoms that each feature two terminal oxygen atoms bonded through double bonds, resulting in a total of 4 $\pi$-bonds. Similarly, pyrosulphuric acid ($H_2S_2O_7$) also contains two sulfur centers where each is connected to two oxygen atoms through double bonds, which contributes another 4 $\pi$-bonds. Combining these, the total number of $\pi$-bonds present in both acids is 8.

Electron gain enthalpy represents the energy change occurring when a neutral gaseous atom accepts an electron, where more exothermic processes correspond to lower, more negative numerical values. Assessing the experimental data for these elements, $\Delta_{eg} H(\text{Cl})$ is -345 kJ/mol and $\Delta_{eg} H(\text{F})$ is -328 kJ/mol, confirming that -345 is less than -328. Similarly, $\Delta_{eg} H(\text{I})$ at -295 kJ/mol is lower than $\Delta_{eg} H(\text{At})$ at -270 kJ/mol, and $\Delta_{eg} H(\text{Te})$ at -190 kJ/mol is lower than $\Delta_{eg} H(\text{Po})$ at -183 kJ/mol, upholding those comparisons. However, for the pair of sulfur and selenium, $\Delta_{eg} H(\text{Se})$ is -195 kJ/mol and $\Delta_{eg} H(\text{S})$ is -200 kJ/mol; because -195 is numerically greater than -200, the inequality $\Delta_{eg} H(\text{Se}) < \Delta_{eg} H(\text{S})$ is mathematically false.

The ability of a phosphorus oxoacid to act as a reducing agent and form a silver mirror with $\mathrm{AgNO}_{3}$ depends entirely on the presence of at least one $\mathrm{P-H}$ bond. This specific bond renders the phosphorus atom capable of donating electrons, which allows it to reduce silver ions into metallic silver. Among the options provided, $\mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{5}$ is the only compound that contains a $\mathrm{P-H}$ bond, making it the only species capable of carrying out this reduction.

The distinct physical properties of boron ($B$) arise primarily from its very small atomic size and its unique ability to form a complex, rigid crystalline lattice. Because the atoms are tightly packed and connected by exceptionally strong covalent bonds throughout this network, the element possesses a very high lattice energy, which directly accounts for its extreme hardness. This same structural stability makes it difficult to disrupt the network through thermal energy, causing boron to exhibit significantly higher melting and boiling points than any other element in its group.

Acidified potassium dichromate functions as a strong oxidizing agent, turning green when it reacts with sulfur dioxide due to the reduction of dichromate ions to chromium(III) ions. This redox reaction is described by the chemical equation $3\mathrm{SO}_2+\mathrm{Cr}_2\mathrm{O}_7^{2-}+2\mathrm{H}^{+}\rightarrow3\mathrm{SO}_4^{2-}+2\mathrm{Cr}^{+3}+\mathrm{H}_2\mathrm{O}$ where the sulfur dioxide is oxidized to sulfate ions. This visible color transition serves as a diagnostic test because the reduction process effectively converts the orange dichromate species into a green state.

The ratio of silica ($SiO_2$) to alumina ($Al_2O_3$) serves as a critical indicator of cement quality, directly influencing the setting time and the strength development of the material. Engineering standards dictate that this ratio must be maintained between 2.5 and 4 to ensure the cement possesses optimal hydraulic properties. Because 3 is the only value among the given choices that falls within this prescribed range, it represents the appropriate ratio for producing high-quality cement.

The Hell-Volhard-Zelinsky reaction facilitates the alpha-halogenation of carboxylic acids using $\text{Br}_2$ and red phosphorus followed by hydrolysis, whereas the Iodoform reaction identifies methyl-containing ketones or alcohols using $\text{NaOH}$ and $\text{I}_2$. The Etard reaction utilizes chromyl chloride (\text{CrO}_2\text{Cl}_2$) \in$\text{CS}_2$followed by$\text{H}_2\text{O}$to selectively oxidize toluene to benzaldehyde, and the Gatterman-Koch reaction introduces a formyl group to benzene using$\text{CO}$and$\text{HCl}$with anhydrous$\text{AlCl}_3$. These specific reagent-reaction combinations establish the pattern of A-III, B-I, C-II, and D-IV.

The water-gas shift reaction relies on the principle of oxidation-reduction, as illustrated by the chemical equation $\mathrm{CO} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{CO}_2 + \mathrm{H}_2$. In this process, the carbon atom within the carbon monoxide molecule transitions from an oxidation state of +2 to an oxidation state of +4 upon forming the carbon dioxide molecule. Since an increase in oxidation state signifies the loss of electrons, it is evident that carbon monoxide is oxidized to carbon dioxide during this transformation.

Borax is chemically structured as $\mathrm{Na}_2\mathrm{B}_4\mathrm{O}_5(\mathrm{OH})_4 \cdot 8 \mathrm{H}_2\mathrm{O}$, which clarifies the stoichiometry of its anhydrous, hydroxide, and hydrated components. By mapping the provided template $\mathrm{Na}_2\mathrm{B}_4\mathrm{O}_{x}(\mathrm{OH})_{y} \cdot z\mathrm{H}_2\mathrm{O}$ to this established formula, the specific values are identified as $x = 5$, $y = 4$, and $z = 8$. Calculating the requested sum yields $5 + 4 + 8 = 17$.

The spontaneous oxidation of halide ions by oxygen in an acidic environment depends on whether the reduction potential of the specific halogen is low enough to allow the reaction to proceed. Among the common halide ions, only iodide possesses a sufficiently low reduction potential for oxygen to act as an oxidizing agent under these conditions. Specifically, oxygen reacts with iodide in an acidic medium according to the equation $4\mathrm{I}^{-}(\mathrm{aq}) + 4\mathrm{H}^{+}(\mathrm{aq}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{I}_2(\mathrm{s}) + 2\mathrm{H}_2\mathrm{O}(\mathrm{l})$, while other halides like chloride and bromide do not undergo this transformation.

Bond enthalpy is primarily determined by the efficiency of atomic orbital overlap, which decreases significantly as the atomic radius increases down a periodic group. As we move from carbon down to tin in Group 14, the valence electrons occupy orbitals that are further from the nucleus, resulting in less effective electron sharing and weaker covalent bonds. This relationship between atomic size and bond strength ensures that the energy required to break these bonds follows the descending order $\mathrm{C}-\mathrm{C} > \mathrm{Si}-\mathrm{Si} > \mathrm{Ge}-\mathrm{Ge} > \mathrm{Sn}-\mathrm{Sn}$.

Identifying oxides with odd electrons requires summing the total number of valence electrons for each molecule. Nitrogen contributes 5 valence electrons and each oxygen atom contributes 6, leading to a total of $5 + 2(6) = 17$ valence electrons in $\mathrm{NO}_2$. Because this sum is an odd number, the molecule necessarily possesses an unpaired electron, whereas the other listed nitrogen oxides result in an even number of total valence electrons, allowing for complete pairing.

Each oxoacid of phosphorus is defined by a specific molecular formula, and we can directly compare these to find which one contains the most oxygen. Phosphoric acid is represented as $\mathrm{H}_3\mathrm{PO}_4$, whereas pyrophosphorous acid, hypophosphoric acid, and pyrophosphoric acid have the formulas $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_5$, $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_6$, and $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_7$, respectively. Among these options, pyrophosphoric acid contains the highest quantity of oxygen atoms with seven in its structure.