P-block elements practice · Mixed difficulty

Option B (Square pyramidal) is correct because $XeF_5^+$ has five bonding pairs and one lone pair, resulting in a square pyramidal geometry.

Option D (${H_2}PO < {H_2}Te < {H_2}Se < {H_2}S < {H_2}O$) is correct because thermal stability decreases down the group due to decreasing bond dissociation energy, with water being an exception due to strong hydrogen bonding.

Option B (O2) is correct because oxygen supports combustion and can relight a glowing splinter due to its oxidizing properties.

Option C (3, 2 and 1) is correct because $XeF_2$ has 3 lone pairs, $SeF_4$ has 1 lone pair, and $XeF_6$ has 1 lone pair.

Option A (Solution state) is correct because manganese heptoxide is formed as a dark green oily liquid in solution when solid potassium permanganate reacts with concentrated sulfuric acid.

Option B (${I_2}{O_5}$) is correct because iodine pentoxide is a stable covalent oxide of iodine.

Option A is correct. Uranium isotopes are usually separated by using compounds of Fluorine (F2) due to the volatility of UF6.

Option B (4) is correct because in the reduction of $HN{O_3}$ to ${N_2}O$, the oxidation state of nitrogen changes from +5 to +1, involving 4 electrons.

Option D (40) is correct because the volume of ozone formed is calculated based on the stoichiometry of the reaction and the given condition that the volumes of oxygen and ozone become equal.

Option D ($\mathbf{A}$ is false but $\mathbf{R}$ is true.) is correct because the mass ratio of S:O in SO2 is not 1:2, but the law of definite proportions (Reason R) is always true.

Option D (0.125 mole) is correct because the amount of arsenic pentasulphide formed is calculated based on the stoichiometry of the reaction between arsenic acid and hydrogen sulphide.

Option C (${N_2}{O_5}$) is correct because, in general, for oxides of elements in the same group, acidity increases with increasing oxidation state and electronegativity. Nitrogen is more electronegative than As, P, and Sb, and $N_2O_5$ has a high oxidation state for N.

Option B (26.7 ml) is correct because the volume of $KMnO_4$ required is calculated using stoichiometry and the balanced redox reaction.

Option B (+1) is correct because the oxidation state of phosphorus in hypophosphorous acid ($H_3PO_2$) is +1.

Option C (Of the oxidation of ${H_2}S$ gas by some acid radicals) is correct because the yellow turbidity is due to the oxidation of ${H_2}S$ to elemental sulfur by oxidizing agents present in the solution.

Option A (${S_4}{O_6}^{2-} + I^−$) is correct because thiosulfate ion is oxidized to tetrathionate ion by iodine in acidic solution.

Option B is not correctly formulated. With very dilute $HN{O_3}$, Zinc ($Zn$) produces $N{H_4}N{O_3}$ instead of $NO$.

Option A ($Ca{F_2} + {H_2}S{O_4} \to CaS{O_4} + 2HF$) is correct because there is no change in oxidation state of any element in this reaction, hence no oxidation or reduction.

Option A ($90\% \mathrm{ Ca}{\mathrm{C}_{\mathrm{2}}} + 10\% \,\mathrm{CaC}{\mathrm{l}_{\mathrm{2}}}$) is correct because this mixture is used in the Fisher-Ringe method to absorb nitrogen from air, allowing noble gases to be separated.

Option A is correct. The reaction of nitrogen with calcium carbide (CaC2) produces calcium cyanamide, which has the formula Ca(CN)2.

Option D is correct because heating a mixture of $N{H_4}Cl$ and $NaN{O_2}$ produces pure nitrogen gas.

Option B ($s p^3 d$ and tetrahedral) is incorrect. The correct answer is $s p^3 d$ and trigonal bipyramidal because group 16 tetrafluorides have $s p^3 d$ hybridization and a see-saw or distorted tetrahedral structure, which is derived from a trigonal bipyramidal arrangement.

Option C is correct because liquid oxygen is pale blue in color.

Option C is correct because the magnetic moment depends on the number of unpaired electrons, which differs between $\mathrm{F}{\mathrm{e}^{\mathrm{II}}}$ and $F{e^{III}}$.

Option A is correct. Fluorine reacts with cold and dilute sodium hydroxide to produce sodium fluoride (NaF) and oxygen difluoride (OF2).

Option B (Bent T-shaped, square pyramidal) is correct because the geometries of $\mathrm{XeF}_3^+$ and $\mathrm{XeF}_5^+$ are bent T-shaped and square pyramidal, respectively, based on VSEPR theory.

Option C (All but fluorine show positive oxidation states) is the correct answer because fluorine is the most electronegative element and always exhibits a -1 oxidation state.

Option B (Intermolecular forces among molecules of chlorine are the weakest and those in iodine are the strongest) is correct because the physical state of halogens at room temperature depends on the strength of intermolecular forces, which increase with molecular size.

Option C (Neon) is correct because Neon (Ne) exists as a monoatomic gas, unlike the diatomic molecules of oxygen, fluorine, and nitrogen.

Option A ($\mathrm{Ca}{\mathrm{C}_2} + \mathrm{C}{\mathrm{a}_3}{\mathrm{P}_2}$) 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 on ships.