P-block elements practice · Easy

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

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 ($SC{l_4}$) 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 (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 (2) is correct because the reaction of one mole of $Ca_3P_2$ with excess water produces two moles of $PH_3$.

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

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

Option A ($N_3H$) is correct because the oxidation state of N in hydrazoic acid ($N_3H$) is -1/3, which is higher than in the other options.

Option A ($MCl$) is correct because the dihydrogen phosphate $M{H_2}P{O_4}$ indicates that the metal M has a +1 charge, thus forming $MCl$.

Option D ($T{i^{4 + }}$) is correct because after losing 4 electrons, Titanium achieves the electronic configuration of Argon, a noble gas.

Option A (3, 2 and 1) is correct because $XeF_2$ has 3 lone pairs, $XeF_4$ has 2 lone pairs, and $XeO_3$ has 1 lone pair on the Xe atom.

Option B ($\mathrm{HI} > HBr > HCl > HF$) is correct because acid strength increases down the group due to decreasing bond dissociation enthalpy.

Option A ($\mathrm{HOCl} > HOBr > HOI$) is correct because the acidic character of hypohalous acids decreases down the group due to decreasing electronegativity of the halogen.

Option A (${I_2} < B{r_2} < C{l_2} < {F_2}$) is correct because oxidizing power increases up the halogen group due to increasing electronegativity.

Option A ($SiC{l_4}$) is correct because silicon tetrachloride reacts with moisture in the air to produce HCl fumes and silica, creating a dense white smoke.

Option B (Reducing power) is correct because the reducing power increases down the group, so the order should be $HF < HCl < HBr < HI$.

Option A (Birkeland and Eyde’s process) is correct because it is the industrial process specifically used for the manufacture of nitric acid (HNO3).

Option B (${N_2}{O_5}$) is correct because when nitric acid reacts with phosphorus pentoxide, it forms metaphosphoric acid and dinitrogen pentoxide.

Option A (Dissolves more in ${\mathrm{C}_6}{\mathrm{H}_6}$) is correct because iodine is non-polar and dissolves better in non-polar solvents like benzene.

Option A ($PC{l_5}$) is correct because in $PCl_5$, the axial and equatorial $P-Cl$ bonds have different bond lengths and thus different bond energies.

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

Option C ($KMn{O_4}$) is correct because $SO_2$ reduces $KMn{O_4}$ by decolorizing it, acting as a reducing agent itself.

Option A (Helium is much less soluble in blood than nitrogen) is correct because the lower solubility of helium in blood at high pressures prevents nitrogen narcosis, a dangerous condition for deep-sea divers.

Option C ($\mathrm{A\;mixture\;of\;}{\mathrm{O}^{16}},\mathrm{\;}{\mathrm{O}^{17}}\mathrm{\;and\;}{\mathrm{O}^{18}}\mathrm{\;isotopes}$) is correct because ordinary oxygen exists as a mixture of its stable isotopes: O-16, O-17, and O-18.