P-block elements practice · Mixed difficulty

Option B (Ammonium nitrate) is correct because when heated, it decomposes to nitrous oxide ($\mathrm{N}_2\mathrm{O}$) and water, and nitrous oxide does not support combustion.

Option C (O > S > Se > Te - Electron gain enthalpy (magnitude)) is correct because the electron gain enthalpy of oxygen is less negative than sulfur due to its small size and high electron density.

Option B (${SO}_2$ is both oxidising as well reducing but ${SO}_3$ is only oxidising) is correct because sulfur in SO₂ has an intermediate oxidation state (+4) allowing it to both gain and lose electrons, while in SO₃ sulfur has its maximum oxidation state (+6) and can only gain electrons.

Option A (White phosphorous is soluble in CS₂ but red phosphorous is insoluble in CS₂) is correct because white phosphorus exists as discrete P₄ molecules and is soluble in CS₂, while red phosphorus is a polymer and insoluble.

Option B is correct because both the assertion and reason are true, but the reason (sublimation) does not explain why XeF4 is a good fluorinating agent.

Option C ($XePt{F_6}$) is correct because $XePtF_6$ was the first noble gas compound synthesized, paving the way for further noble gas chemistry.

Option D (${H}_2 {~S}$) is correct because as we go down the group, the acidity increases due to the decrease in bond dissociation energy.

Option D (Dissociation energy of its molecule is fairly high) is correct because nitrogen has a triple bond (N≡N), which requires a large amount of energy to break, making it relatively inert.

Option C ($Ar$) is correct because Argon (Ar) is an inert gas commonly used in electric bulbs and tubes to prevent the filament from oxidizing and prolong its life.

Option B (5) is correct because one molecule of the compound dissociates into 5 ions: Fe2+, 2 SO42-, and 2 NH4+.

Option D (Zero) is correct because the zero group (noble gases) lies between highly electronegative and electropositive elements, acting as a buffer.

Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because both the shape similarity and the larger covalent radius of Se are true, but the radius isn't the direct cause of the shape.

The correct answer is A (Both (A) and (R) are true and (R) is the correct explanation of (A)) because ${XeO}_3 {F}_2$ indeed has a trigonal bipyramidal geometry due to $sp^3d$ hybridization with the described arrangement.

Option B (Both (A) and (R) are true and (R) is not the correct explanation of (A)) is correct because both the weakness of the F-F bond and the small size of the fluorine atom are true, but the small size isn't the primary reason for the weak bond.

Option D (An oxidizing agent) is correct because concentrated sulphuric acid acts as an oxidizing agent and oxidizes HBr, preventing its formation.

Option B (Moisture) is correct because chlorine acts as a bleaching agent due to the formation of nascent oxygen in the presence of moisture.

Option B (Cancer) is correct because radon's radioactivity is used in cancer therapy.

Option A (8) is correct because rhombic sulfur exists as S8 molecules, thus having an atomicity of 8.

The correct answer is A ($\mathrm{NH}_3$) because ammonia is the strongest base among the given hydrides of group 15 elements due to nitrogen's small size and high electronegativity.

Option A (1.27 ${A^ \circ }$) is correct because the O-O bond length in ozone is approximately 1.27 Angstroms.

Option B (Ionisation potential) is correct because the oxidizing power of fluorine is primarily due to its low bond dissociation enthalpy, high hydration enthalpy, and high electron affinity, but not its ionization potential.

Option B ($N{H_3}$ molecule is triangular planar) is correct because ammonia ($N{H_3}$) has a pyramidal shape due to the presence of a lone pair of electrons on the nitrogen atom.

Option D (A mixed salt) is correct because bleaching powder (calcium hypochlorite) contains more than one type of anion.

Option A (${\mathrm{H}_2}\mathrm{S}{\mathrm{O}_3}$) is correct because it is the oxoacid formed when sulfur dioxide ($\mathrm{S}{\mathrm{O}_2}$) reacts with water.

Option C (The most abundant noble gas in the atmosphere is He) is false because Argon (Ar) is the most abundant noble gas in the atmosphere.

Option D ($2HI + {H_2}S{O_4} \to {I_2} + S{O_2} + 2{H_2}O$) represents the oxidizing behavior of ${H_2}S{O_4}$ because sulfur is reduced from +6 to +4, while iodide is oxidized to iodine.

Option B (Both (A) & (R) are true but the (R) is not the correct explanation of the (A)) is correct because the reaction of $\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3$ with $\mathrm{I}_2$ does indeed produce $\mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6$, and the reaction involves both color and electronic changes, but the color/electronic change isn't the direct *reason* for the reaction itself.

Option D (Transition temperature) is correct because the transition temperature is the temperature at which two allotropic forms of a substance are in equilibrium.

Option B is correct because both the assertion and the reason are true, but the reason does not correctly explain the assertion. The instability of HI in the presence of conc. H2SO4 is due to its reducing nature, not its bond strength.

Option A is correct because interhalogen compounds are better oxidizing agents due to weaker $X-X^{\prime}$ bonds compared to $X-X$ bonds, making them more reactive.