P-block elements practice · Mixed difficulty

Option D (Calcium phosphide) is correct because calcium phosphide reacts with water to produce phosphine gas, which spontaneously ignites and creates a smoke screen.

Option A is correct because interhalogen compounds can only be of the type XX^\'${XX}^{\'}$, XX^\'₃${XX}^{\'}_3$, XX^\'₅${XX}^{\'}_5$ and XX^\'₇${XX}^{\'}_7$.

Option B (16) is correct because the structure of $P_4O_{10}$ contains 16 sigma bonds.

Option A ($\mathrm{NH}_3$) is correct because ammonia gas reacts with Nessler's reagent to form a brown precipitate.

Option D is correct because chlorous acid and its salts can act as both oxidizing and bleaching agents due to the release of nascent oxygen.

The correct answer is A (decrease the concentration of ${OH}^{-}$) because $NH_4Cl$ suppresses the ionization of $NH_4OH$ by the common ion effect, decreasing the hydroxide ion concentration and preventing the precipitation of group IV hydroxides.

Option A is correct because acidic character of oxides increases across a period, and Cl2O7 is the most acidic among the given options.

Option D (All of the above) is correct because when excess KI is added to copper sulphate solution, cuprous iodide is formed, and iodine is liberated, indicating that potassium iodide is oxidized.

Option A (Reduction) is correct because sulfur dioxide bleaches by reduction, where it removes oxygen from the colored substance.

Option A (Both A and R are true and R is the correct explanation of A) is correct because N3H is a weak acid and NH3 is a weak base, and the proton-donor acceptor equilibrium explains their behavior.

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because acidity increases down the group due to decreasing bond strength of H-X bond.

Option B is correct because both the assertion and reason are true, but the reason doesn't explain why ozone is a better oxidizing agent than hydrogen peroxide.

Option B is correct because SO₂ is used as a food preservative due to its antioxidant properties, while NO₂ is not.

Option C (${O_3}$) is correct because ozone is used for dry bleaching processes.

Option D is correct because $HBrO_2$ is the least stable among the given oxyacids of halogens.

Option A ($Ca{F_2} + {H_2}S{O_4} \to CaS{O_4} + 2HF$) is correct because this reaction is not a redox reaction; the oxidation states of all elements remain unchanged.

Option B ($H{g_2}O$) is correct because ozone oxidizes mercury to form mercurous oxide, which causes tailing.

Option C (F₂) is correct because fluorine is a strong oxidizing agent and can oxidize water to liberate oxygen.

Option D (Goitre) is correct because iodine deficiency leads to the enlargement of the thyroid gland, causing goitre.

Option A is correct because XeF6 reacts with SiO2 in glass or quartz to form XeO3, which is highly explosive, making it unsafe to handle XeF6 in such apparatus.

Option A (Reducing agent) is correct because H2SO4 acts as an oxidizing agent due to sulfur being in its highest oxidation state (+6).

The correct answer is C (sp³, sp³d) because in solid state, BrF₃ exists as [BrF₂]⁺[BrF₄]⁻ and IF₅ exists as [IF₄]⁺[IF₆]⁻. The hybridizations of the cationic parts [BrF₂]⁺ and [IF₄]⁺ are sp³ and sp³d, respectively.

Option A (White P) is correct because white phosphorus is highly reactive and readily reacts with oxygen in the air at ordinary temperatures, often igniting spontaneously.

Option A is correct because ${H_4}{P_2}{O_7}$ (pyrophosphoric acid) has four ionizable hydrogen atoms, making it a tetrabasic acid.

Option B (Fluorspar $-\mathrm{BF}_3$) is incorrectly matched because Fluorspar is $\mathrm{CaF}_2$, not $\mathrm{BF}_3$.

Option A (Less) is correct because a larger bond angle implies more s-character in the bonding orbitals and less p-character in the lone pair.

Option D (If both Assertion & Reason are false) is correct because SO2 is used to make oleum by reacting with H2SO4, and SO2 does not form H2SO4 in air in the way described.

Option A (Both S-I and S-II are true) is correct because the number of lone pairs in ${XeOF}_4$ is indeed 15, and ${XeF}_2$ hydrolyzes to give ${Xe}, {HF}$, and ${O}_2$.

Option A is correct because halogens absorb visible light due to electronic transitions, which results in them being colored.

Option B ($+3,-\frac{1}{3}$ and -3) is correct because these are the calculated oxidation states of nitrogen in the respective compounds.