P Block Elements – JEE Main ChemistryPractice Questions & PYQs

These reactions demonstrate fundamental inorganic transformations of lead(II) ions, including precipitation, amphoteric complexation, and acetate-mediated dissolution. Reaction 1 yields the yellow precipitate lead(II) chromate ($\mathrm{PbCrO}_4$), which reacts with excess sodium hydroxide in Reaction 2 to form the soluble complex sodium tetrahydroxoplumbate(II) ($\mathrm{Na}_2[\mathrm{Pb}(\mathrm{OH})_4]$) alongside sodium chromate. Finally, the dissolution of insoluble lead(II) sulfate in ammonium acetate occurs via the formation of the coordination complex $(\mathrm{NH}_4)_2[\mathrm{Pb}(\mathrm{CH}_3 \mathrm{COO})_4]$, establishing that $\mathrm{A}$ is $\mathrm{PbCrO}_4$, $\mathrm{B}$ is $\mathrm{Na}_2[\mathrm{Pb}(\mathrm{OH})_4]$, and $\mathrm{X}$ is $(\mathrm{NH}_4)_2[\mathrm{Pb}(\mathrm{CH}_3 \mathrm{COO})_4]$.

Electronegativity in Group 15 decreases down the group following the order $N > P > As > Sb$, identifying $N$ as the most electronegative element ($X$) and $Sb$ as the least electronegative element ($Y$), with $N_2O_3$ being acidic and $Sb_2O_3$ being amphoteric, which confirms that Statement I is true. Although $BCl_3$ is covalent and undergoes hydrolysis in water to produce $[B(OH)_4]^{-}$, boron's high ionization energy prevents the formation of the $[B(H_2O)_6]^{3+}$ cation, which is a characteristic species of aluminum, thereby rendering Statement II false.

The lightest element in group 7 is manganese, which forms the manganate ion ($MnO_4^{2-}$) when in the +6 oxidation state. The corresponding potassium salt, potassium manganate ($K_2MnO_4$), exhibits a distinct green color due to the electronic transitions associated with the manganese(VI) center within the tetrahedral anion structure.

Electropositivity is intrinsically tied to an element's tendency to lose electrons, a trait clearly mirrored by its standard reduction potential. Aluminium demonstrates a significantly negative potential of $E^{\circ}_{\mathrm{Al}^{3+}/\mathrm{Al}} = -1.66\,\text{V}$, while Thallium displays a positive value of $E^{\circ}_{\mathrm{Tl}^{3+}/\mathrm{Tl}} = +1.26\,\text{V}$, making it evident that Aluminium is the more electropositive metal. The formation of a trivalent ion requires supplying enough energy to remove three valence electrons, a sum that is prohibitively high for Boron at $6887\,\text{kJ/mol}$. Consequently, while Boron is restricted to covalent bonding due to these high ionization demands, Aluminium can effectively overcome this energy barrier to form the $\mathrm{Al}^{3+}$ cation in ionic compounds.

The boiling point of hydrogen halides is determined by both hydrogen bonding and molecular mass, where $HF$ exhibits exceptionally strong hydrogen bonding leading to the highest boiling point, while the remaining halides show increasing boiling points with molecular mass due to stronger London dispersion forces, resulting in the order $HCl < HBr < HI < HF$. Melting points are governed by the packing efficiency of the molecules in the solid crystal lattice rather than boiling point trends, and because the hydrogen-bonded zig-zag chains of solid $HF$ pack less efficiently than the structure of solid $HI$, the melting point order is $HCl < HBr < HF < HI$.

The chemical nature of group 14 oxides, which trends from acidic for $\mathrm{CO}_2$ and $\mathrm{SiO}_2$ to amphoteric for $\mathrm{Sn}$ and $\mathrm{Pb}$ oxides, confirms that exactly two of the provided pairs, $[\mathrm{SnO}, \mathrm{SnO}_2]$ and $[\mathrm{PbO}, \mathrm{PbO}_2]$, consist of oxides that are both amphoteric. $\mathrm{BF}_3$ is an electron-deficient molecule with six valence electrons around the boron atom, which allows it to function as a Lewis acid by accepting a lone pair from $\mathrm{NH}_3$ to form an adduct, while its $sp^2$ hybridization maintains a trigonal planar geometry.

Oxygen typically exhibits a covalency of 2, yet it possesses the capacity to reach a covalency of 4 through specific bonding arrangements. When paired with less electronegative elements like sulfur in $\mathrm{SO}_2$, oxygen manifests an oxidation state of -2, while in its fluoride compound $\mathrm{OF}_2$, the dominance of fluorine's electronegativity results in an oxidation state of +2 for oxygen. The anomalous behavior of oxygen relative to the heavier elements of group 16 arises directly from its small atomic radius and high electronegativity, which set it apart from its group congeners.

Bond length in hydrogen halides $HX$ increases down the group as the atomic size of the halogen increases, implying that the longest bond corresponds to the element with the largest, not smallest, covalent radius, which invalidates the first statement. The second statement is false because while phosphine ($PH_3$) has the lowest boiling point among group 15 hydrides due to the absence of intermolecular hydrogen bonding, the element phosphorus has a maximum covalency of 6, not 4, owing to the availability of vacant $d$-orbitals.

The ionic radii of trivalent cations increase down group 13 due to the addition of successive principal quantum shells, while electronegativity does not uniformly decrease because of the anomalous effects caused by poor shielding of d-orbitals and f-orbitals in heavier elements. Boron exhibits the highest first ionization enthalpy among these elements because it possesses the smallest atomic radius, which results in the strongest electrostatic attraction between the nucleus and the valence electrons. Furthermore, the trichlorides and triiodides of group 13 elements are covalent in nature because the $M^{3+}$ ions provide sufficient polarizing power to distort the electron clouds of the relatively larger halide anions, facilitating covalent bond formation.

The isotope ${}^{13}C$ is a naturally occurring stable isotope of carbon, whereas ${}^{14}C$ is the radioactive isotope used in radiocarbon dating. Carbon is restricted to a maximum covalency of four because it lacks d-orbitals in its valence shell, while its ability to exhibit oxidation states of -4, +2, and +4 is well-established. Since the acidic nature of group 14 dioxides decreases down the group due to an increase in metallic character, $CO_2$ is correctly identified as the most acidic member of the series.

The stability of group 15 hydrides ($EH_3$) decreases down the group as the E-H bond dissociation enthalpy diminishes due to the increasing atomic size of the central element, while the basicity decreases because the lone pair on the larger central atom becomes more diffuse and less available for donation. The reducing character increases down the group as the weakening E-H bond facilitates the release of hydrogen, whereas the boiling point does not strictly increase due to the presence of intermolecular hydrogen bonding in $NH_3$, which makes its boiling point higher than that of $PH_3$. These trends confirm that statements A, B, and C are the correct descriptors for the properties of these hydrides.

The inert pair effect dictates that for Group 14 elements, the stability of the +2 oxidation state increases down the group, rendering $\mathrm{Pb}^{2+}$ more stable than $\mathrm{Pb}^{4+}$, while $\mathrm{Sn}^{4+}$ remains more stable than $\mathrm{Sn}^{2+}$. Reaction (1) involves the conversion of $\mathrm{Pb}^{4+}$ and $\mathrm{Pb}^{0}$ into the more stable $\mathrm{Pb}^{2+}$, making the process spontaneous such that $\Delta_{r} G^{\circ}(1) < 0$. In contrast, reaction (2) attempts to convert the more stable $\mathrm{Sn}^{4+}$ and $\mathrm{Sn}^{0}$ into the less stable $\mathrm{Sn}^{2+}$, which is a non-spontaneous process resulting in $\Delta_{r} G^{\circ}(2) > 0$.

The brown ring test serves as a reliable analytical technique for detecting nitrate ions, relying on the reduction of nitrate to nitric oxide gas. As nitric oxide gas reacts within a fresh ferrous sulfate solution, it coordinates with the iron(II) ions to form a distinct brown coordination complex. This interaction produces the compound $[Fe(NO)]SO_4$, establishing $NO$ as the gas $X$ and the resulting brown-ring complex as $Y$.

The reaction of $EH_3$ with Nessler's reagent ($K_2HgI_4$) in an alkaline medium to produce a precipitate of basic mercury(II) amido-iodide is the definitive analytical test for ammonia, which identifies the element $E$ as nitrogen. Nitrogen is positioned in Group 15 of the periodic table, and the provided ionization enthalpy values correspond to the first elements of Groups 13, 14, 15, and 16, which are Boron, Carbon, Nitrogen, and Oxygen, respectively. These values are $801 kJ mol^{-1}$ for B, $1086 kJ mol^{-1}$ for C, $1402 kJ mol^{-1}$ for N, and $1312 kJ mol^{-1}$ for O. Therefore, the first ionization enthalpy value for nitrogen is $1402 kJ mol^{-1}$.

Ionization enthalpy is governed by the effective nuclear charge and the extent of electron shielding provided by inner shells. When considering the second ionization enthalpy for Group 13 elements, the experimental values are 2427 kJ/mol for B, 1817 kJ/mol for Al, and 1979 kJ/mol for Ga. The observation that Gallium has a higher value than Aluminum is due to the poor shielding provided by its 3d electrons, which increases the effective nuclear charge experienced by the valence electrons. Consequently, the stated order B > Al > Ga is inaccurate, as the actual sequence is B > Ga > Al.

The first ionization enthalpy for Group 14 elements follows a trend influenced by atomic size and shielding effects, with experimental values of 786.5 kJ/mol for Si, 761-762 kJ/mol for Ge, 708 kJ/mol for Sn, and 715 kJ/mol for Pb. The assertion that these follow an ascending order of Si < Ge < Pb < Sn is refuted by the data, which demonstrates that the ionization enthalpy does not consistently increase across this sequence. Since neither statement aligns with these experimental values, both are false.

The atomic number of an element corresponds to the number of protons in its neutral state, which we determine by subtracting the charge of the monoatomic anion from its 36 electrons to find 35 protons. Summing these 35 protons with the 45 neutrons yields an atomic mass of $35 + 45 = 80$. This identification reveals the element is bromine, which is located in group 17 of the periodic table and exists in a liquid state at room temperature.

By matching the electronegativity value of 1.8 on the Pauling scale, we identify element E as Thallium, which is located in group 13. Due to the inert pair effect, which becomes increasingly pronounced in heavier group members, the +1 oxidation state of Thallium is significantly more stable than the +3 oxidation state. This energetic instability of $Tl^{3+}$ causes it to readily accept electrons, meaning it functions as a strong oxidizing agent rather than a reducing agent. This thermodynamic drive toward reduction is quantitatively confirmed by the standard electrode potential for the $Tl^{3+}/Tl$ couple, which is $+1.26 V$, confirming that the reaction potential is positive.

Nitrogen species in oxidation states ranging from +1 to +4 are chemically unstable in acidic solutions and inherently undergo disproportionation, a process illustrated by the following reaction:
$3\,\mathrm{HNO}_2 \longrightarrow \mathrm{HNO}_3+\mathrm{H}_2\mathrm{O}+2\,\mathrm{NO}$
Nitrogen is strictly limited to a maximum covalency of four because it possesses only four valence orbitals-one 2s and three 2p-and lacks the d-orbitals necessary to expand its coordination sphere. Other proposed properties are incorrect because nitrogen actually exhibits the lowest density among Group 15 elements, forms a weaker N-N single bond than the P-P single bond due to significant lone pair repulsion, and is unable to form $d\pi-p\pi$ bonds because it does not have available d-orbitals.

The properties of group 15 hydrides are primarily governed by the increasing atomic size and decreasing electronegativity of the central atom as one moves down the group. As the central atom grows larger, its valence orbitals become more diffuse, resulting in a less effective overlap with the hydrogen $1s$ orbital. This leads to an increase in $E-H$ bond length and a decrease in bond dissociation energy, which simultaneously causes the thermal stability of the hydrides to decline and their reducing character to increase from $\mathrm{NH}_3$ to $\mathrm{BiH}_3$.

Regarding the availability of the lone pair, the increasing size of the central atom causes the electron density of the lone pair to spread over a larger orbital volume, which decreases its availability for donation and results in a lower tendency to act as a Lewis base from $\mathrm{NH}_3$ to $\mathrm{BiH}_3$. Furthermore, as the electronegativity of the central atom drops, the bond pairs reside in orbitals that are increasingly pure $p$ in character rather than hybridized, causing the bond angle to contract towards $90^{\circ}$. This phenomenon explains why the $HEH$ bond angle consistently decreases as one progresses down the group.

Aluminium exhibits amphoteric behavior by reacting with concentrated sodium hydroxide to yield the soluble tetrahydroxoaluminate complex, written as $[\mathrm{Al}(\mathrm{OH})_4]^-$, which invalidates the assertion that a hexahydroxoaluminate species is produced. In terms of molecular shape, the $(\mathrm{ICl}_4)^-$ ion utilizes $\mathrm{sp}^3\mathrm{d}^2$ hybridization to form a square planar structure, the $(\mathrm{ClO}_3)^-$ ion is $\mathrm{sp}^3$ hybridized and pyramidal, and the $(\mathrm{IBr}_2)^-$ ion displays $\mathrm{sp}^3\mathrm{d}$ hybridization with a linear geometry, confirming the second statement is correct.