P-block elements practice · Easy

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because Helium is used to avoid nitrogen narcosis, which is caused by nitrogen dissolving in the blood at high pressures.

Option A ($S{O_2}$) is correct because sulfurous anhydride is the chemical name for sulfur dioxide ($S{O_2}$).

Option C (As) is correct because Arsenic (As) is a metalloid, possessing properties of both metals and nonmetals.

Option D (${\left( {N{H_4}} \right)_2}C{O_3}$) is correct because ammonium carbonate decomposes to release ammonia gas, which has a pungent smell.

Option D (It contains impurity of $\;{P_2}{H_4}$) is correct because diphosphane ($P_2H_4$) is spontaneously flammable in air, causing the phosphine to ignite and produce smoky rings.

Option B (Ionization potential decreases) is correct because ionization potential generally decreases down a group as the atomic size increases and the outermost electrons are further from the nucleus.

Option B ($N{F_3}$) is correct because $NF_3$ is inert and does not readily hydrolyze due to the high electronegativity of fluorine and the absence of lone pairs on nitrogen to initiate hydrolysis.

Option C (A is true but R is false) is correct because fluorine always exhibits a -1 oxidation state, but chlorine, not fluorine, has the highest electron affinity.

Option B (White phosphorus) is correct because white phosphorus is the most reactive allotrope of phosphorus due to its tetrahedral structure and high angle strain.

Option C ((A) is true but (R) is false) is correct because fluorine is smaller than chlorine, but the F-F bond is weaker than the Cl-Cl bond due to interelectronic repulsions.

Option D (6) is correct because chlorine forms six different oxides: $Cl_2O$, $ClO$, $ClO_2$, $Cl_2O_6$, $Cl_2O_7$, and $ClO_4$.

Option A (Incombustible) is correct because helium is non-flammable (incombustible), making it safer than hydrogen for use in balloons.

Option B ($\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$) is correct because the acidic strength of hydrogen halides increases down the group due to decreasing bond strength.

Option C (Pnictogens) is correct because the elements of Group 15 (VA) are collectively known as pnictogens.

Option B (${I}_2 {O}_5$) is correct because iodine pentoxide is used to quantitatively determine carbon monoxide by oxidizing it to carbon dioxide.

Option C is correct because iodine can be separated from sand by sublimation, as it readily converts to a gas upon heating, leaving the sand behind.

The correct answer is D (All) because $\mathrm{ClO}^{-}$, $\mathrm{ClO}_2^-$, and $\mathrm{ClO}_3^{-}$ can all undergo disproportionation reactions.

Option D (All of these) is correct because ozone reacts with molecules containing C=C or C≡C bonds to form ozonides.

Option C (47) is correct because the percentage of nitrogen in urea (NH2CONH2) is approximately 47%.

Option A ($CaS{O_4}.2{H_2}O$) is correct because superphosphate of lime is a mixture of calcium dihydrogen phosphate and calcium sulfate dihydrate.

Option A is correct because HF is a liquid at room temperature due to strong hydrogen bonding between HF molecules, which significantly increases its boiling point.

The correct answer is B (Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.) because both statements are true, but electronegativity doesn't directly explain the electron gain enthalpy difference.

Option D (Nature of hydrides) is correct because oxygen forms a neutral hydride (${H_2}O$), while sulfur forms an acidic hydride (${H_2}S$).

Option A (${O^{16}}$) is correct because Oxygen-16 is the most abundant isotope of oxygen, making up the vast majority of naturally occurring oxygen.

Option A (${\mathrm{N}_2}$) is correct because nitrogen is relatively inert and does not readily react directly with fluorine under normal conditions.

Option D (Bleaching powder) is correct because Calcium hypochlorite ($Ca(OCl)_2$) and Calcium chloride ($CaCl_2$) are the main components of bleaching powder.

Option C (Helium) is correct because nuclear fusion involves the combination of light nuclei, such as deuterium, to form heavier nuclei, such as helium, releasing a large amount of energy.

Option B (Catalyst) is correct because $Mn{O_2}$ speeds up the decomposition of $KCl{O_3}$ without being consumed in the reaction.

Option D ($C{S_2}$) is correct because sulphur is soluble in carbon disulfide ($C{S_2}$) due to similar intermolecular forces.

Option A ($C{a_3}{P_2}$) is correct because it reacts with water to produce phosphine ($P{H_3}$), which has a rotten fish smell and reacts with $CuS{O_4}$ to form $C{u_3}{P_2}$.