P-block elements practice · Hard

Option D is correct. The assertion is false because SO2 reacts with SO3 to form oleum, not H2SO4. The reason is also false because SO2 forms H2SO3, not H2SO4, in the presence of water.

Option C (400) is correct because ppm (parts per million) is calculated as (mass of solute / mass of solution) * 10⁶, which in this case is (0.2 g / 500 g) * 10⁶ = 400 ppm.

Option A (Anhydrous $\mathrm{Ba}{\left( {\mathrm{Cl}{\mathrm{O}_4}} \right)_2}$) is correct because it is a known desiccant due to its strong affinity for water.

Option C ($N{O_2}$) is correct because the reaction of concentrated sulfuric acid with dry potassium nitrate produces nitrogen dioxide gas, which is brown in color.

Option A (Heating a mixture of phosphorite mineral with sand and coke in electric furnace) is correct because this is the modern industrial method for producing phosphorus.

Option B is correct because both the Assertion and Reason are True, but the Reason is not a correct explanation of the Assertion. $\mathrm{H}_3 \mathrm{PO}_2$ is a strong reducing agent and gives disproportionation on heating, and it is a monobasic bronsted acid due to one $-\mathrm{OH}$ group and two $\mathrm{P}-\mathrm{H}$ groups, but the reducing property is not *because* it's monobasic.

Option B ($[PF_6]^−$) is correct because $XeF_2$ reacts with $PF_5$ to form $[XeF]^+[PF_6]^−$. The anionic species is $[PF_6]^−$.

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because $XeO_3F_2$ has a trigonal bipyramidal geometry with Xe undergoing $sp^3d$ hybridization, with the O atoms in the equatorial positions and F atoms in the axial positions.

The correct answer is C ($N{H_4}I + {N_2} + {I_2}$) because iodine reacts with ammonia to form nitrogen triiodide, which is unstable and decomposes explosively into ammonium iodide, nitrogen, and iodine.

Option A (1.5 litre) is correct because it is the volume of chlorine calculated from the percentage of available chlorine in the bleaching powder sample.

Option A is correct because both statements are correct. Hypophosphoric acid contains two ${P}={O}$ bonds, four ${P}-{OH}$ bonds, and one $P-P$ bond, and hypophosphorous acid reduces ${AgNO}_3$ to metallic silver.

Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because both the assertion and reason are true, but the reason does not explain the assertion.

Option A ($\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{S}{\mathrm{O}_4}$) is correct because Schweitzer’s reagent is a solution of tetraamminecopper(II) hydroxide, which is typically prepared by dissolving copper(II) sulfate in ammonia solution.

Option C ($Ca{S_5} + {H_2}$) is correct because the reaction of $CaS$ with ${H_2}S$ produces calcium pentasulfide ($CaS_5$) and hydrogen gas (${H_2}$).

Option D ($Xe{O_3} + 6HF \to Xe{F_6} + 3{H_2}O$) is correct because $XeO_3$ is explosive and cannot be directly converted to $XeF_6$ by reaction with HF.

Option C (Both (1) and (2) in different stages of rocket propulsion) is correct because PSLV uses a combination of composite solid propellant in some stages and biliquid propellants in other stages.

Option B (${\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}{\mathrm{S}_\mathrm{x}}$) is correct because yellow ammonium sulfide contains polysulfide ions ($S_x^{2-}$), giving it a yellow color.

Option B is correct. SO2 has a bent shape at room temperature, and SeO2 also has a bent shape in the vapor phase at higher temperatures. The covalent radius difference is true but doesn't directly explain the shape similarity.

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because Bartlett's discovery of $O_2^+[PtF_6]^-$ led him to realize that Xenon, with a similar ionization energy, could also react with $PtF_6$.

Option D ($Ca{\left( {CN} \right)_2}$) is correct because calcium cyanide ($Ca(CN)_2$) does not react with water to produce ammonia ($NH_3$).

Option B is correct. Fluorine does oxidize H2O to O2, and chlorine does not under similar conditions. However, the reason given about electronegativity is not the direct explanation for this difference in oxidizing power.

Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because both the assertion and reason are true, but the reason does not explain the assertion.

Option B (CAN is hygroscopic) is correct because Calcium Ammonium Nitrate (CAN) absorbs moisture from the air, and the calcium silicate coating prevents this.

Option A ($N{a_2}S{O_3}$) is correct because it reacts with hot dilute ${H_2}S{O_4}$ to produce $S{O_2}$ and also decolourises $B{r_2}$ water due to its reducing nature.

Option B ($O{F_2}$) is correct because $OF_2$ reacts with water to produce $O_2$ and $HF$, not an oxyacid.

Option A ($BaCl_2$) is correct because barium chloride does not react with sulfide ions to produce a characteristic precipitate or gas that confirms the presence of sulfide ions.

Option A is correct because ${H_4}{P_2}{O_6}$ (Hypophosphoric acid) contains a P-P bond in its structure.

Option C ($\mathrm{Po}{\mathrm{O}_2}$) is correct because as metallic character increases down the group, the oxides become less soluble in water. Polonium is the most metallic in the group.

Option C (Bromine) is correct because concentrated sulfuric acid oxidizes bromide ions to bromine, which is a reddish-brown gas.

Option C (Iodine) is correct because sulfur dioxide reduces iodates to iodine in acidic conditions.