P-block elements practice · Mixed difficulty

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 D is correct because in the reaction $2HN{O_3} + {P_2}{O_5} \to \mathrm{ }2HP{O_3} + {N_2}{O_5}$, $HN{O_3}$ acts as a dehydrating agent, not an oxidizing agent.

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 ($NO$) is correct because nitric oxide ($NO$) has an odd number of electrons, resulting in an unpaired electron and making it paramagnetic.

Option D (HF is a stronger acid than HCl) is not true because HF is a weak acid due to strong hydrogen bonding, while HCl is a strong acid.

Option B is correct because Hypophosphorous acid is a monoprotic acid, not a diprotic acid, due to the presence of two P-H bonds and one P-OH bond.

Option C (${N_3}H$) is correct because in hydrazoic acid ($N_3H$), the oxidation state of nitrogen is -1/3, which is higher than in the other options.

Option A (Much lesser than $\mathrm{N}{\mathrm{H}_3}$) is correct because the bond angle in $P{H_3}$ is closer to 90 degrees due to less s-character in the bonding orbitals compared to $N{H_3}$.

Option A is correct. Oxygen (O2) is a paramagnetic molecule due to the presence of unpaired electrons in its molecular orbital configuration.

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 (${}_{88}^{226}\mathrm{Ra}$) is correct because Radium-226 decays through alpha decay, producing Radon (an inert gas) and then further decays to other inert gases.

Option A ($CO_2$) is correct because it reacts with CaO (a basic oxide) to form $CaCO_3$, but does not react with $NaHCO_3$ (a weak base).

Option B ($Ca{C_2} + C{a_3}{P_2}$) is correct because the mixture of calcium carbide and calcium phosphide produces acetylene and phosphine, which ignite spontaneously and produce a bright light.

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 D (${\mathrm{I}_2} < \mathrm{B}{\mathrm{r}_2} < C{\mathrm{l}_2} < {\mathrm{F}_2}$) is correct because oxidizing power increases as you move up the halogen group due to increasing electronegativity and decreasing bond dissociation energy.

Option A (To remove arsenic impurity) is correct because $Fe(OH)_3$ is used to adsorb arsenic impurities in the Contact process for sulfuric acid production.

Option A (200 mL) is correct because oxygen and nitrogen are removed, leaving mostly noble gases and some residual gases.

Option B (${\mathrm{O}_2},\;{\mathrm{O}_3}$) is correct because alkaline pyrogallol absorbs oxygen (O2) and oil of turpentine absorbs ozone (O3).

Option D (Metaphosphoric acid) is correct because orthophosphoric acid ($H_3PO_4$) loses water upon heating to form metaphosphoric acid ($HPO_3$).

Option A (${\mathrm{I}_4}{\mathrm{O}_9}$) is the correct answer because iodine can exhibit a +9 oxidation state only in ionic compounds.

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 ($Ca_3P_2$) is correct because it reacts with water to produce phosphine ($PH_3$), which has a rotten fish smell and reacts with $CuSO_4$ to form $Cu_3P_2$.

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.

Option C (${H_2}PO < {H_2}Te < {H_2}Se < {H_2}S < {H_2}O$) is correct because thermal stability decreases down the group due to decreasing bond dissociation energy.

Option A (${N_2}O$) is correct because ammonium nitrate decomposes upon heating to produce nitrous oxide (laughing gas) and water.

Option A is correct because acidic nature of oxides increases with the electronegativity of the central atom and oxidation state.

Option C is correct. Chlorine is used in the extraction of both Platinum (Pt) and Gold (Au).

Option D ($POC{l_3}$) is correct because the reaction of white phosphorus with thionyl chloride ($SOCl_2$) produces $PC{l_3}$, $S{O_2}$, and ${S_2}C{l_2}$, but not $POC{l_3}$.