Hydrogen practice · Mixed difficulty

Option B is correct. In reaction (A), H₂O₂ acts as an oxidizing agent, and in reaction (B), it acts as a reducing agent.

Option A (Passing $\mathrm{CO_2}$ into $\mathrm{BaO_2}$) is correct because hydrogen peroxide can be prepared in the laboratory by reacting barium peroxide with carbon dioxide and water.

Option C (spin of protons) is correct because para and ortho hydrogen differ in the relative spins of their protons.

Option D (All of these) is correct because H2(l), O2(l), and N2H4(l) are all used as rocket propellants, either as fuel or oxidizer.

Option D ($\mathrm{H}_2$) is correct because hydrogen is the component present in the greatest proportion in water gas.

Option D ($250 \mathrm{~mL}$) is correct because $20 \mathrm{~vol} \mathrm{~H}_2 \mathrm{O}_2$ means 1 L produces 20 L of $\mathrm{O}_2$ at STP, thus 0.25 L produces 5 L.

Option C (Reaction of zinc with aqueous alkali) is correct because it represents a common laboratory method for producing dihydrogen gas.

Option C ($H_2$) is correct because atomic hydrogen produced by the oxy-hydrogen flame can reach temperatures of 4000 K, suitable for welding metals with high melting points.

Option A (1) is correct because an 11.2 volume solution of H₂O₂ has a molarity of approximately 1M.

The correct answer is A ($\mathrm{H}_2 \mathrm{O}$) because the $1 \% \mathrm{H}_2 \mathrm{O}_2$ formed by auto-oxidation of 2-ethyl anthraquinol is extracted with water.

Option A (Basic solution) is correct because ionic hydrides react with water to form a metal hydroxide and hydrogen gas, resulting in a basic solution.

Option D (Silicol process) is correct because the Silicol process is the method by which hydrogen is obtained by the action of an alloy of silicon and iron with $\mathrm{NaOH}\mathrm{.}$

Option D (If Statement I is false but Statement II is correct) is correct because molarity can be less than, equal to, or greater than molality, and the ratio depends on the solvent's density.

Option A (Less stable than ortho hydrogen) is correct because para hydrogen is less stable at room temperature and tends to convert to the more stable ortho form.

Option C (It cannot be placed in first group of periodic table) is correct because Hydrogen can be placed in the first group due to its resemblance with alkali metals.

Option C (Blue flame) is correct because hydrogen gas burns with a pale blue, almost invisible flame in the presence of sufficient oxygen.

Option B is correct because potassium superoxide (KO2) reacts with water to produce potassium hydroxide, hydrogen peroxide, and oxygen gas.

Option D (Passing silen electric discharge through Hydrogen at low pressure) is correct because atomic hydrogen is produced by passing hydrogen gas through a silent electric discharge at low pressure, breaking the H-H bond.

Option C is correct because $H_2O_2$ does not decompose $Na_2CO_3$ to give $CO_2$ directly, and it is not a strong acid.

Option D is correct because the electronegativity difference between hydrogen and chlorine is less than 2.1, leading to a covalent bond in hydrogen chloride, whereas sodium chloride has a larger electronegativity difference, resulting in an ionic bond.

Option C (Both (a) and (b)) is correct because a 5.6 volume H₂O₂ solution is both 1.7% weight by volume and 1 N.

Option B is correct because water's high dielectric constant allows it to effectively reduce the electrostatic forces between ions, making it a good solvent for ionic compounds and therefore a universal solvent.

Option C (Carbon monoxide) is correct because in the water-gas shift reaction, carbon monoxide reacts with steam to produce hydrogen gas and carbon dioxide.

Option B (Na₂O₂) is correct because sodium peroxide reacts with water to produce hydrogen peroxide and sodium hydroxide.

Option B ($^2{H_2}{O^{16}}$) is correct because heavy water is deuterium oxide, where deuterium ($^2H$) is an isotope of hydrogen.

Option D (Fluorine) is correct because fluorine is a strong oxidizing agent and can oxidize water to oxygen.

Option C is correct because melts of ionic hydrides liberate dihydrogen gas at the anode, not the cathode.

Option C ($Mg{(HC{O_3})_2}$ is converted into $Mg{(OH)_2}$ and $Ca{(HC{O_3})_2}$ is converted into $CaC{O_3}$) is correct because boiling converts soluble bicarbonates of magnesium and calcium into insoluble magnesium hydroxide and calcium carbonate, respectively.

Option A (1) is correct because 11.2 volume H2O2 solution means 1 mL of the solution gives 11.2 mL of oxygen at NTP, which corresponds to a molarity of 1.

Option C (100) is correct because perhydrol is a 30% solution of $H_2O_2$, which corresponds to a volume strength of 100.