P Block Elements – JEE Main ChemistryPractice Questions & PYQs

When performing Lassaigne's test to detect halogens, any nitrogen or sulphur present in the original organic compound is converted into $\mathrm{NaCN}$ and $\mathrm{Na}_2\mathrm{S}$ during the sodium fusion process. Because these species would interfere with the identification of halides by forming their own precipitates upon the addition of silver nitrate, the extract is boiled with $\mathrm{HNO}_3$ to decompose them into volatile $\mathrm{HCN}$ and $\mathrm{H}_2\mathrm{S}$. This critical pretreatment removes the interfering ions from the solution, ensuring that the subsequent chemical testing is specific only to the presence of halogen ions.

Amphoteric oxides represent a specific class of compounds that display dual reactivity, allowing them to act as both an acid and a base depending on the environment. To determine the number of amphoteric oxides, we must evaluate the chemical nature of each compound provided in the series. Among the given substances, $\mathrm{Cl}_2\mathrm{O}_7$, $\mathrm{SiO}_2$, and $\mathrm{N}_2\mathrm{O}_5$ are classified as acidic oxides because they react with bases. Meanwhile, $\mathrm{CO}$, $\mathrm{NO}$, and $\mathrm{N}_2\mathrm{O}$ are recognized as neutral oxides, as they do not show any significant reaction with either acids or bases. This distinguishes $\mathrm{Al}_2\mathrm{O}_3$, $\mathrm{SnO}_2$, and $\mathrm{PbO}_2$ as the only amphoteric oxides in the collection, as these are the specific compounds capable of interacting with both acidic and basic reagents to form salts and water. Counting these identifies exactly 3 amphoteric oxides in the specified list.

To determine the oxidation state of sulfur in each compound, assign standard oxidation values to the surrounding atoms-typically -2 for oxygen, +1 for hydrogen, -1 for halogens like chlorine and fluorine, and +2 for barium-and ensure the total sum equals zero for these neutral molecules. Through this calculation, we find that $\mathrm{H_2SO_3}$, $\mathrm{SOCl}_2$, and $\mathrm{SF}_4$ all contain sulfur in a +4 oxidation state. Since the remaining compounds, $\mathrm{SO}_3$, $\mathrm{BaSO}_4$, and $\mathrm{H_2S_2O_7}$, feature sulfur in a +6 oxidation state, there are exactly 3 compounds from the list that meet the criteria.

In group 14, the oxides $\mathrm{SiO}_2$ and $\mathrm{GeO}_2$ exhibit acidic character, whereas the oxides $\mathrm{SnO}$ and $\mathrm{PbO}$ are amphoteric, which confirms that the first statement is true. Carbon's ability to exist in multiple allotropic forms arises from its unique property of catenation and its capacity for $p \pi-p \pi$ bond formation, rather than $p \pi-d \pi$ bond formation, as carbon lacks d-orbitals entirely, rendering the second statement false.

Fuming sulphuric acid, also commonly referred to as oleum, acts as a mixture of sulphur trioxide and concentrated sulphuric acid. This combination is chemically represented by the formula $\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7$. By observing the subscript associated with the oxygen atoms in this specific structure, one can determine that there are 7 oxygen atoms present in total.

Identifying the color of inorganic salts often involves distinguishing between substances that lack electronic transitions in the visible region, which typically appear white, and those that absorb light due to charge transfer or specific electronic configurations. Among the provided compounds, $\mathrm{SrSO}_4$, $\mathrm{Mg}(\mathrm{NH}_4)\mathrm{PO}_4$, $\mathrm{Mn}(\mathrm{OH})_2$, $\mathrm{PbSO}_4$, and $\mathrm{CaC}_2\mathrm{O}_4$ possess a white color. The remaining substances, specifically $\mathrm{BaCrO}_4$, $\mathrm{PbCrO}_4$, $\mathrm{AgBr}$, $\mathrm{PbI}_2$, and $[\mathrm{Fe}(\mathrm{OH})_2(\mathrm{CH}_3\mathrm{COO})]$, exhibit yellow or brown-red shades, confirming that there are exactly five white salts in this set.

Chromyl chloride behaves as an acid chloride of chromic acid, facilitating its conversion into a soluble chromate salt when treated with a strong base. Through the reaction $\mathrm{CrO}_2\mathrm{Cl}_2 + 4\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{CrO}_4 + 2\mathrm{NaCl} + 2\mathrm{H}_2\mathrm{O}$, we identify $A$ as sodium chromate, $\mathrm{Na}_2\mathrm{CrO}_4$, because the chromium remains in the same oxidation state during this neutralization process.

When sodium chromate encounters acidified hydrogen peroxide, it undergoes a transformation into a peroxo complex, which serves as a classic qualitative test for detecting chromium in the $+6$ oxidation state. This interaction proceeds as $\mathrm{Na}_2\mathrm{CrO}_4 + 2\mathrm{HCl} + 2\mathrm{H}_2\mathrm{O}_2 \rightarrow \mathrm{CrO}_5 + 2\mathrm{NaCl} + 3\mathrm{H}_2\mathrm{O}$, identifying $B$ as the deep blue chromium pentoxide, $\mathrm{CrO}_5$.

Nessler's reagent is a specific chemical indicator used primarily for the detection of ammonia gas. When ammonia, denoted as $\mathrm{NH_3}$, is passed through this reagent, it triggers a reaction that produces a characteristic brown precipitate known as the iodide of Millon's base, which is expressed by the formula $\mathrm{HgO\cdot Hg(NH_2)I}$. This distinctive color change serves as a reliable, qualitative identification test for the presence of $\mathrm{NH_3}$ because none of the other gases listed would react with this compound to form the same brown-colored precipitate.

Boron is predominantly found as the isotope $^{11}\text{B}$, which has an atomic number of 5. Subtracting this atomic number from the mass number 11 reveals that the number of neutrons is $11 - 5 = 6$, so $x = 6$.

Upon heating in air, amorphous boron reacts with oxygen to produce boron trioxide according to the equation $4\text{B} + 3\text{O}_2 \rightarrow 2\text{B}_2\text{O}_3$. In this compound, oxygen carries an oxidation state of -2, which requires the boron atoms to balance the charge at +3, giving $y = 3$. Combining these values yields $x + y = 6 + 3 = 9$.

Minerals are naturally occurring inorganic substances defined by specific chemical compositions, and accurately recognizing these formulas is fundamental to chemistry. While cryolite, fluoroapatite, and carnallite are correctly paired with their respective chemical structures, the term fluorspar refers to calcium fluoride, which has the formula $\mathrm{CaF}_2$, rather than $\mathrm{BF}_3$. Because $\mathrm{BF}_3$ represents boron trifluoride, the initial pairing is factually incorrect.

The inert pair effect dictates that the lower oxidation state, +3, becomes progressively more stable than the +5 state as one moves down Group 15, rendering the claim that the stability of +5 increases down the group false. Additionally, nitrogen belongs to the second period and lacks the vacant d-orbitals necessary to expand its octet, meaning its maximum covalency is limited to 4 rather than 6. Consequently, because statements regarding the inert nature of dinitrogen, the common oxidation states of -3, +3, and +5, and the unique $p\pi - p\pi$ multiple bonding of nitrogen are scientifically accurate, only (D) and (E) contain incorrect information.

Fluorine remains unique among the halogens because it is the most electronegative element in the periodic table, consistently maintaining a -1 oxidation state throughout its chemistry. Unlike chlorine, bromine, and iodine, which possess vacant $d$-orbitals allowing for expanded valence shells and multiple positive oxidation states, fluorine is strictly limited by its small atomic radius and lack of available $d$-orbitals.

When lead sulphide reacts with dilute nitric acid, the acid acts as an oxidizing agent, resulting in the formation of lead nitrate, sulphur, water, and nitric oxide as the chemical products. The reaction proceeds according to the equation $\mathrm{PbS} + \mathrm{HNO}_3 \rightarrow \mathrm{Pb}(\mathrm{NO}_3)_2 + \mathrm{NO} + \mathrm{S} + \mathrm{H}_2\mathrm{O}$ Because the chemical process consistently yields nitric oxide ($\mathrm{NO}$) instead of nitrous oxide ($\mathrm{N}_2\mathrm{O}$), the latter is not formed.

A disproportionation reaction requires the element to be capable of existing in both a higher and lower oxidation state than its initial elemental state. Because fluorine is the most electronegative element, it cannot exhibit a positive oxidation state, effectively prohibiting $\mathrm{F}_2$ from undergoing this type of reaction. However, chlorine, bromine, and iodine are less electronegative and possess the ability to adopt positive oxidation states, which allows $\mathrm{Cl}_2$, $\mathrm{Br}_2$, and $\mathrm{I}_2$ to undergo disproportionation.

Gallium is utilized in the manufacturing of thermometers primarily because of its exceptionally high boiling point, which allows it to remain in the liquid state across a wide range of elevated temperatures. However, its relatively high melting point makes it unsuitable for detecting low temperatures, such as the $256\ \mathrm{K}$ freezing point of a brine solution. Since gallium would solidify long before reaching this freezing point, it cannot function as a reliable indicator in such low-temperature environments, making the first statement correct while identifying the second as false.

When determining the color of inorganic compounds, identifying their distinct physical characteristics is essential. Lead iodide ($PbI_2$) is characterized by its vibrant, bright yellow appearance, and ammonium arsinomolybdate is similarly recognized as a yellow substance. Meanwhile, ammonium sulphide is distinguished primarily by its solubility in water, leaving lead sulphate ($PbSO_4$) as the only compound in the group that naturally exists as a white solid.

Group 13 trivalent halides exhibit significant covalent character, which makes them highly susceptible to hydrolysis when exposed to water. The central metal atoms in these compounds possess empty orbitals that readily accept lone pairs from water molecules, facilitating the reaction and confirming the validity of the first statement. In the specific case of aluminum chloride, the hydrolysis process occurring in an acidified aqueous solution involves the displacement of chloride ions by water ligands, resulting in the formation of the stable octahedral complex $[Al(H_2O)_6]^{3+}$. Because this chemical behavior describes the coordination environment of aluminum accurately, both statements are correct.

The inert pair effect explains why the $ns^2$ electrons in heavier elements become increasingly reluctant to participate in chemical bonding due to the poor shielding provided by inner $d$ and $f$ orbitals. This reluctance to lose the $s$-electrons results in a greater stability for the lower oxidation state as one descends the group. Consequently, the stability of the $+1$ oxidation state increases from $\mathrm{Ga}$ to $\mathrm{In}$ to $\mathrm{T}\mathrm{l}$, making the stated order $\mathrm{Ga}^{+1} < \mathrm{In}^{+1} < \mathrm{T}\mathrm{l}^{+1}$ a direct consequence of this electronic phenomenon.

The fundamental difference in the molecular structures of oxygen and sulphur arises from their atomic size and their ability to undergo effective orbital overlap. Oxygen atoms are small enough to facilitate the efficient side-to-side overlap of $2p$ orbitals, allowing them to form stable $2p\pi-2p\pi$ multiple bonds with themselves, which leads to the formation of diatomic $O_2$ molecules. Conversely, sulphur atoms possess a larger atomic radius, which prevents the effective overlap of $3p$ orbitals required for stable homonuclear multiple bonding. As a result, sulphur prefers to form stable single bonds, arranging itself into puckered $S_8$ rings in both its rhombic and monoclinic allotropic forms.

While the assertion correctly describes these structural differences, the provided reason is factually inaccurate. Although it is true that sulphur struggles to form stable $3p\pi-3p\pi$ homonuclear multiple bonds, the claim that it cannot form multiple bonds with other elements is incorrect. Sulphur successfully engages in multiple bonding with small, highly electronegative atoms like carbon and nitrogen, as evidenced by compounds such as carbon disulphide ($S=C=S$) and thiocyanate species. Therefore, because the reasoning incorrectly generalizes sulphur's limitation to all elements, it fails to provide a correct explanation for the assertion.

Group 16 elements readily produce oxides of the form $\mathrm{EO}_2$ and $\mathrm{EO}_3$ for $\mathrm{E} = \mathrm{S}, \mathrm{Se}, \mathrm{Te},$ and $\mathrm{Po}$, with both types displaying acidic behavior. Chemical trends show that $\mathrm{SO}_2$ acts as a reducing agent, whereas $\mathrm{TeO}_2$ behaves as an oxidizing agent because the stability of the higher oxidation states decreases down the group. Regarding the incorrect claims, the reducing strength of hydrides such as $\mathrm{H_2S}$ and $\mathrm{H_2Te}$ intensifies down the group due to falling bond dissociation enthalpy, and an ozone molecule actually contains six lone pairs of electrons rather than five.