P-block elements practice · Mixed difficulty

Option B (${H_3}P{O_2}$) is correct because it has the most P-H bonds, which are responsible for its stronger reducing property.

Option C is correct because sulfuric acid forms hydrates with water, releasing a large amount of heat, which explains its high affinity for water.

Option C is correct because Statement I is true, fluorine is produced electrolytically, but Statement II is false; water is not oxidized at a much higher potential than fluorine.

Option D ($1: 1: 1$) is correct because each of the given acids will neutralize an equal number of moles of NaOH, Ca(OH)2, and Al(OH)3 due to their respective n-factors.

Option B (Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is not the correct explanation of $\mathbf{A}$.) is correct because anhydrous $\mathrm{H}_2 \mathrm{SO}_4$ does undergo autoprotolysis, and it has high electrical conductivity, but the conductivity is not the direct reason for autoprotolysis.

Option D (Hypo phosphoric acid) is correct because its structure contains a direct P-P bond instead of a P-O-P bond.

Option D (Zn) is correct because Zinc (Zn) forms an amphoteric oxide (ZnO), which reacts with both acids and bases.

Option D ($PH_6^+$) is correct because it is not a known compound, so it cannot possess a stable P-H bond.

Option B (Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.) is correct because HBr is a stronger acid than HCl due to the weaker H-Br bond, and the reason correctly states that the H-X bond is weaker for HBr compared to HCl, but the reason is not the full explanation.

Option A (Xe) is correct because the ease of liquefaction increases with increasing atomic size and intermolecular forces.

Option A (${\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} {\mathrm{SO}_4}$) is correct because Schweitzer's reagent is the tetraamminecopper(II) hydroxide solution, which is often represented as the sulfate complex.

Option B (He) is correct because Helium is an inert gas and prevents magnesium from reacting with the atmosphere during welding.

Option B (If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I) is correct because ozone formation is non-spontaneous, and while atomic oxygen initiates ozone formation, it doesn't explain the non-spontaneity.

Option C (Strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke) is correct because HCl gas readily absorbs moisture from the air, forming small droplets of hydrochloric acid.

Option A ($Te$) is correct because $Te$ (Tellurium) has the electronic configuration [Kr] 4d¹⁰ 5s² 5p⁴, resulting in the same number of electrons in both the penultimate (n-1) and anti-penultimate (n-2) shells.

Option A is correct because the first real noble gas compound was based on the similar ionization energies of O2 and Xe, leading to the formation of O2+[PtF6]-.

Option B (The atmospheric air) is correct because the atmosphere is the primary source from which noble gases are extracted.

Option D (All of these) is correct because concentrated nitric acid is a strong oxidizing agent, it oxidizes moist iodine to iodic acid, and chromium does not dissolve in concentrated nitric acid due to the formation of a passive oxide film.

Option B is correct because the reaction sequence described is a test for bromide ions, where chlorine oxidizes bromide to bromine, which forms a brown layer in $CCl_4$.

Option B (Violet) is correct because ozone reacts with tetramethyl base paper to produce a violet color.

Option C ($Se{F_4}$) is correct because Selenium tetrafluoride ($Se{F_4}$) exists as a liquid at room temperature, while Sulfur tetrafluoride ($S{F_4}$) is a gas and Tellurium tetrafluoride ($Te{F_4}$) is a solid.

Option D (All of these) is correct because Freons are chlorofluorocarbons (CFCs), and $CC{l_2}{F_2}$, $CFC{l_3}$, and $CCl{F_3}$ are all examples of CFCs.

Option C (PH3) is correct because phosphorus in PH3 has a very small electronegativity difference with hydrogen, resulting in minimal hybridization.

Option B (Orange) is correct because the color of bromine in $C{S_2}$ is orange.

Option C ($IP{O_4}$) is correct because iodine phosphate has the formula $IP{O_4}$, indicating an ionic compound with $I^{3+}$ and $PO_4^{3-}$ ions.

Option C (Nelson cell) is correct because the Nelson cell is an electrolytic cell specifically designed for the electrolysis of brine solution to produce chlorine gas.

Option D is correct because $H_4P_2O_7$ reacts with water to form $H_3PO_4$, and upon strong heating, it dehydrates to form $P_2O_5$.

Option D (ICl) is correct because interhalogen compounds like ICl are more reactive than homonuclear diatomic halogens due to the polarity of the bond between the different halogen atoms.

Option B (${H_3}As{O_4}$) is correct because the chemical formula for Arsenic acid is ${H_3}As{O_4}$.

Option D (4) is correct because P4O10 has four P=O double bonds, which are shorter than the P-O single bonds.