P-block elements practice · Medium

Option A (${HF}$) is correct because it forms intermolecular hydrogen bonds, leading to a higher boiling point compared to other hydrogen halides.

Option B is correct because ClO2 is paramagnetic due to having an odd number of electrons, making it not diamagnetic.

Option D ($O$) is correct because oxygen exists as a diatomic molecule ($O_2$), which has the lowest atomicity among the given elements.

Option B (${CCl}_3{NO}_2$) is correct because ${CCl}_3{NO}_2$, also known as chloropicrin, is a well-known tear gas.

Option B ($\mathrm{A}{\mathrm{s}_2}{\mathrm{O}_3}$) is correct because arsenic trioxide ($\mathrm{A}{\mathrm{s}_2}{\mathrm{O}_3}$) is a known poison.

Option A is correct. XeF4 and [PtCl4]2- have square planar geometry due to their electronic configurations and hybridization.

Option A (Statement I and II are correct) is correct because both statements accurately describe chemical reactions involving phosphorus compounds.

Option B ($I{\left( {C{H_3}COO} \right)_3}$) is correct because iodine acetate is an ionic compound with iodine having a +3 oxidation state and acetate having a -1 charge.

Option B ($BrO_4^ -$) is correct because the oxidizing power of perhalate ions increases down the group: $BrO_4^ - > ClO_4^ - > IO_4^ -$.

Option C (P and NaOH) is correct because phosphine is produced by the reaction of phosphorus with sodium hydroxide and water.

Option A (Statement I and Statement II are correct and Statement II is the correct explanation of Statement I) is correct because both statements about ammonia's properties and its freezing/boiling points are accurate, and the latter explains the former due to hydrogen bonding.

Option A ($\mathrm{N}_2 \mathrm{O}_5$) is correct because it balances the chemical equation with the given reactants.

The correct answer is D because iodine is non-polar and thus more soluble in carbon tetrachloride (a non-polar solvent) than in water (a polar solvent).

Option C is correct because Se in $SeCl_4$ has one lone pair and seesaw shape, making statement I correct and statement II false.

Option A ($\mathrm{NO}, \mathrm{NO}_{2}$) is correct because ammonia oxidation yields NO (A), which further oxidizes to NO2 (B). NO2 then reacts with water and oxygen to form HNO3, which gives NO and a positive brown ring test.

Option A (Both $\mathbf{A}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$.) is correct because the high viscosity of anhydrous sulfuric acid is due to extensive hydrogen bonding between molecules.

Option B (${N_2}O < \mathrm{ }NO < \mathrm{ }{N_2}{O_3} < \mathrm{ }N{O_2} < {N_2}{O_5}$) is correct because the acidic nature of oxides increases with the oxidation number of the central atom.

Option B ($\mathrm{BrF}$) is correct because as the electronegativity difference between the halogens decreases, the reactivity decreases. $\mathrm{BrF}$ has the smallest difference among the options.

The correct answer is B ($C{l^ - }\mathrm{ }\,\mathrm{and}\,\mathrm{ }ClO_3^ -$) because chlorine reacts with hot and concentrated sodium hydroxide to form chloride and chlorate ions.

The correct answer is A (${\mathrm{I}_4}{\mathrm{O}_9}$) because it exists as $\mathrm{I}{\left( {\mathrm{I}{\mathrm{O}_3}} \right)_3}$, which is an ionic compound composed of $I^+$ and $IO_3^-$ ions.

Option B is correct because pyrophosphorous acid (H4P2O5) contains a P-O-P bond, while hypophosphoric acid (H4P2O6) does not. The structure of H4P2O5 has a direct P-P bond, and the structure of H4P2O6 has a P-O-P bond.

Option B (Nitrogen) is correct because the largest industrial use of hydrogen is in the Haber-Bosch process to produce ammonia (NH₃) from nitrogen and hydrogen.

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 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 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.