P-block elements practice · Mixed difficulty

Option B (Two moles of phosphine) is correct because calcium phosphide ($Ca_3P_2$) reacts with water to produce calcium hydroxide and phosphine ($PH_3$).

Option A (3) is correct because $IF_5$, $BrF_5$, and $XeOF_4$ have a square pyramidal structure.

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 C ([Ne]3s², 3p⁵) is correct because it represents a halogen, while the others belong to the same family (transition metals or group 2).

Option A (Oxygen is completely reacted) is correct because, based on the stoichiometry of the reaction, oxygen is the limiting reagent when one mole of each reactant is taken.

Option D (${N_2}{O_5}$) is correct because dinitrogen pentoxide is a solid at room temperature, while the other oxides of nitrogen listed are gases.

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 B is correct. Chlorine acts as a bleaching agent only in the presence of moisture because it forms hypochlorous acid (HOCl), which is the active bleaching agent.

Option C (7) is correct because iodine reacts with concentrated nitric acid to form iodic acid ($HIO_3$), where the oxidation state of iodine is +5. The question asks for the oxidation state of I in Y, where Y is $HIO_3$. The reaction is $I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O$. Thus, Y is $HIO_3$ and the oxidation state of I is +5.

Option C ($XeO_3$) is correct because $XeF_6$ undergoes complete hydrolysis to form $XeO_3$ and HF.

Option C ($L{i^ + }$) is correct because both Helium and $Li^+$ have 2 electrons and thus have similar electronic structures and spectra.

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 A ($SC{l_4}$) is correct because it has a see-saw shape due to the presence of a lone pair, while the others are tetrahedral.

Option A (It is less soluble in blood than Nitrogen under high pressure) is correct because helium's lower solubility in blood prevents nitrogen narcosis, a dangerous condition that can occur at high pressures underwater.

Option D is correct because the $P - O - P$ angle in ${P_4}{O_{10}}$ is not 180 degrees; it's approximately 120-140 degrees due to the tetrahedral arrangement around the oxygen atoms.

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 (AgI) is correct because silver halides, excluding AgF, are generally insoluble in water due to their high lattice enthalpy and low hydration enthalpy.

Option B (Radon) is correct because Radon (Rn) is the last member of the inert gas family (Group 18) in the periodic table.

Option D (${\mathrm{N}{\mathrm{H}_\mathrm{4}}\mathrm{OH}}$) is correct because the reaction of $\mathrm{NaN}{\mathrm{O}_\mathrm{2}}$ with $\mathrm{N}{\mathrm{H}_\mathrm{4}}\mathrm{OH}$ produces $N_2$, $NaOH$ and $H_2O$.

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 A is correct. Heating ammonium dichromate (NH4)2Cr2O7 produces nitrogen gas (N2) and water (H2O) along with chromium(III) oxide (Cr2O3).

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 A (2) is correct because the reaction of one mole of $Ca_3P_2$ with excess water produces two moles of $PH_3$.

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.

The correct answer is A ($M$) because the equivalent weight of $Cl_2$ in the given reaction is equal to its molecular weight due to the disproportionation reaction.

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 (${H_3}P{O_3} + HCl$) is correct because phosphorus trichloride reacts with water to form phosphorous acid and hydrochloric acid.

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 D (${H_2}Te$) is correct because reducing character increases down the group due to the decreasing bond dissociation energy of the H-E bond.

Option D ($N{a_2}{O_2}$) is correct because it is a peroxide, while the others are dioxides.