P-block elements practice · Mixed difficulty

Option D (${I_2}$) is correct because iodine (${I_2}$) is an essential trace element required for the proper functioning of the thyroid gland, and its deficiency leads to goiter.

The correct answer is D (If both Assertion & Reason are false) because as temperature increases, the equilibrium shifts towards the formation of NO2, increasing the intensity of the brown color, and the formation of NO2 is endothermic.

Option C (Heating with a cobalt salt) is correct because cobalt salts act as catalysts in the decomposition of bleaching powder to produce oxygen.

Option C is correct because the reaction of $NO$ with $N_2O_4$ at 250 K results in the formation of dinitrogen trioxide ($N_2O_3$).

Option D (S-I is false, S-II is true) is correct because water is less volatile and has the highest boiling point due to hydrogen bonding.

Option A (Both S-I and S-II are true) is correct because XeF2 does react with PF5 to form [XeF+][PF6]-, and Helium is indeed used in NMR spectrometers and MRI systems.

Option B ($\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3}$) is correct because ammonium nitrate decomposes upon heating to produce nitrous oxide (laughing gas) and water.

Option A (Crystalline structure) is correct because allotropes of sulfur differ in their arrangement of atoms in the solid state, leading to different crystal structures.

Option A (Statement I and Statement II are correct and Statement II is the correct explanation of Statement I) is correct because the inert pair effect, which is more pronounced in heavier p-block elements, causes the stability of the +5 oxidation state to decrease and the +3 state to increase down Group 15.

Option C is correct because hydrochloric acid forms a constant boiling azeotropic mixture with water at a concentration of 20.24%.

Option D (If both the assertion and reason are false) is correct because the correct IUPAC name is pent-1-en-4-yne, and the triple bond is not always given a lower locant.

Option C ($N{H_4}N{O_3} \to {N_2}O + 2{H_2}O$) is correct because it is not a disproportionation reaction as it involves two different nitrogen atoms with different oxidation states reacting to form a product.

Option A ($\mathrm{F}$) is correct because fluorine is the strongest oxidizing agent and cannot be further oxidized by $MnO_2$.

Option D (All of these) is correct because P, As, and Sb can all react with chlorine to form pentachlorides.

Option C (Both show +2, +4 and +6 oxidation states) is the incorrect statement because oxygen typically exhibits -2, -1, and +2 (in OF2) oxidation states, but not +4 or +6.

Option D ($\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_3+\mathrm{H}_2 \mathrm{SO}_4$ (conc.) $+\mathrm{KCl} \stackrel{\Delta}{\rightarrow}$) will not produce chlorine gas, instead, it produces chromyl chloride ($CrO_2Cl_2$).

Option D (If ${X}={Br}$ then ${Y}={F}, {Cl}$) is correct because bromine is a weaker oxidizing agent than fluorine and chlorine, so it cannot oxidize their ions.

Option C is correct because X is F and Y is Cl, which satisfy the given reactions.

Option D (36 N) is correct because the normality of pure sulfuric acid is approximately 36 N.

Option C is correct because carbon dioxide turns lime water milky, water turns cobalt chloride paper pink, and sulfur dioxide forms an acidic solution.

Option D ($A - {H_2}O,\,B - C{O_2},\,C - NAHC{O_3}$) is correct because it accurately represents the reactants involved in the Solvay process for the production of sodium bicarbonate.

Option D (Chlorine) is correct because chlorine always has an oxidation number of -1 in both $PCl_3$ and $PCl_5$.

Option D (${SO}_2$) is correct because sulfur dioxide is highly soluble in water, reacting to form sulfurous acid, and thus cannot be collected over water.

Option B (Phosphorus) is correct because phosphorus exists as a tetra-atomic molecule ($P_4$) rather than a stable diatomic molecule like iodine, nitrogen, and oxygen.

Option B (b and c) is correct because lead storage batteries use a 38% H2SO4 solution as an electrolyte and are commonly used in automobiles and inverters, but the anode is lead and the cathode is lead packed with PbO2.

The correct answer is B because both statements are correct, but statement II does not explain statement I. Triple phosphate is more soluble, and the calcium sulfate in superphosphate is an insoluble byproduct.

Option C (Both (1) and (2)) is correct because both lead(II) acetate and sodium nitroprusside can distinguish between S²⁻ and SO₃²⁻ based on the formation of distinct precipitates or colored complexes.

Option A (Iodine liberation) is correct because ozone oxidizes KI to liberate iodine, which turns starch paper blue.

Option C ($\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3>\mathrm{BiH}_3$) is correct because dipole moment decreases down the group due to decreasing electronegativity difference.

Option A (He) is correct because Helium is used in very low temperature thermometers due to its extremely low boiling point.