P-block elements practice · Easy

Option C (HI > HBr > HCl) is correct because the acidity of halogen acids increases down the group due to decreasing bond strength.

Option D is correct. Potassium chlorate reacts with concentrated sulfuric acid to produce chlorine dioxide, perchloric acid, and potassium bisulfate.

Option B (Ore forming elements) is correct because chalcogens, which are group 16 elements (O, S, Se, Te, Po), are often found in ores, especially as sulfides and oxides.

Option D (All of the above) is correct because finely divided Pd and Pt, colloidal Pd, and coconut charcoal are all known to adsorb noble gases.

Option B (Polonium) is correct because Polonium is a radioactive element, specifically ${{\;}_{84}}Po$ is the only radioactive element of group 16.

Option D ($\mathrm{BrC}{\mathrm{l}_7}$) is correct because bromine cannot achieve a +7 oxidation state to bond with seven chlorine atoms due to its electronic configuration and size.

Option B (35%) is correct because the percentage of available chlorine in a commercial sample of bleaching powder is approximately 35%.

Option D is correct because the assertion is false (the mass ratio of S:O in SO2 is 1:1), but the reason is true (law of definite proportions).

Option C (III > I > II) is correct because the acidic nature of oxoacids increases with the electronegativity of the central atom.

Option D ($Kr{F_6}$) is correct because Krypton is less reactive than Xenon, and $KrF_6$ is not a known compound.

Option A ($2KBr + {H_2}S{O_4}\xrightarrow[{\mathrm{Conc}.}]{{}}{K_2}S{O_4} + 2HBr$) is correct because HBr is a strong reducing agent and gets oxidized by concentrated H2SO4, which is an oxidizing agent. Therefore, this reaction is not suitable for preparing HBr.

Option C ($He$) is correct because Helium (He) does not form clathrates due to its small size and weak intermolecular forces.

Option A ($KCl{O_3}$) is correct because $KCl{O_3}$ is the chemical formula for Berthelot's salt, also known as potassium chlorate.

Option D ($\mathrm{Ba}{\mathrm{O}_\mathrm{2}}$) is correct because barium peroxide contains the peroxide ion ($O_2^{2-}$).

Option A (N) is correct because nitrogen is the only element in the nitrogen family that does not exhibit allotropy.

Option A is correct because cooling nitrogen dioxide causes it to dimerize, pairing its odd electrons and reducing the brown color.

The correct answer is B (Pentagonal bipyramidal) because $IF_7$ has $sp^3d^3$ hybridization with 7 bond pairs and 0 lone pairs, resulting in a pentagonal bipyramidal shape.

Option C (Yellow lines) is correct because Helium exhibits a characteristic emission spectrum with prominent yellow lines.

Option A (${F_2}$ and $C{l_2}$) is correct because fluorine and chlorine react with starch solution differently than bromine and iodine.

Option A (Both Assertion and Reason are true and Reason is the correct explanation for Assertion) is correct because Group 15 elements exhibit multiple oxidation states due to having five valence electrons.

Option A ($I > II > III$) is correct because acid strength decreases as the electronegativity of the halogen decreases.

Option A (15) is correct because pnicogens are the elements belonging to Group 15 of the periodic table.

Option A (Mixture of helium and oxygen) is correct because helium is less soluble in blood than nitrogen, making it a life-saving mixture for asthma patients.

Option B (2) is correct because a 3p orbital has one angular node (l=1) and one radial node (n-l-1 = 3-1-1 = 1).

Option C (8) is correct because Argon (Ar) has an atomic number of 18, and its electronic configuration is 2, 8, 8, meaning it has 8 electrons in its outermost shell.

Option D (${I_2}$) is correct because iodine (${I_2}$) is an essential trace element required for the proper functioning of the thyroid gland, and its deficiency leads to goiter.

Option B ($\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3}$) is correct because ammonium nitrate decomposes upon heating to produce nitrous oxide (laughing gas) and water.

Option A (Crystalline structure) is correct because allotropes of sulfur differ in their arrangement of atoms in the solid state, leading to different crystal structures.

Option A (Statement I and Statement II are correct and Statement II is the correct explanation of Statement I) is correct because the inert pair effect, which is more pronounced in heavier p-block elements, causes the stability of the +5 oxidation state to decrease and the +3 state to increase down Group 15.

Option A ($\mathrm{F}$) is correct because fluorine is the strongest oxidizing agent and cannot be further oxidized by $MnO_2$.