JEE Chemistry Hydrogen MCQs Set 06

Answer: (b) \( \ce{H-_{(aq)} + H2O_{(l)} -> OH-_{(aq)} + H2_{(g)}} \)

Explanation:
1. Since \( \ce{H-} \) is a stronger base than \( \ce{OH-} \), it takes a proton from water.
2. The proton taken from water combines with \( \ce{H-} \) to give \( \ce{H2} \), and water is left as \( \ce{OH-} \).
3. So the reaction is \( \ce{H- + H2O -> OH- + H2} \).

Teacher's Note:
1. A stronger base always removes a proton from a weaker acid, and water is the weaker acid here.
2. Ionic hydrides such as NaH and CaH2 therefore release hydrogen gas with water.

Answer: (d) Lead sulphide

Explanation:
1. A substance that reduces \( \ce{H2O2} \) must itself be oxidised, so \( \ce{H2O2} \) must act as the oxidising agent.
2. Ozone and acidic \( \ce{KMnO4} \) are oxidants, and these are oxidised or reduced by \( \ce{H2O2} \), not the other way round. \( \ce{BaO2} \) is only a peroxide used to prepare \( \ce{H2O2} \).
3. \( \ce{PbS} \) is oxidised to \( \ce{PbSO4} \) by \( \ce{H2O2} \): \( \ce{PbS + 4H2O2 -> PbSO4 + 4H2O} \), so \( \ce{PbS} \) reduces \( \ce{H2O2} \).

Teacher's Note:
1. Decide first whether the question asks for an oxidising or a reducing role of H2O2.
2. H2O2 oxidises black PbS to white PbSO4, a classic restoration reaction for old paintings.

Answer: (d) Calcium hydroxide (slaked lime)

Explanation:
1. Clark's process removes temporary hardness by adding a calculated amount of slaked lime, \( \ce{Ca(OH)2} \).
2. \( \ce{Ca(HCO3)2 + Ca(OH)2 -> 2CaCO3 v + 2H2O} \); the insoluble \( \ce{CaCO3} \) is filtered off.
3. Calcium bicarbonate is the cause of the hardness, and sodium bicarbonate and alum do not remove it.

Teacher's Note:
1. Clark's process works only for temporary hardness.
2. Excess lime must be avoided because it would keep calcium in solution as bicarbonate after reacting with CO2.

Answer: (c) \( \ce{^3_1H} \)

Explanation:
1. Protium and deuterium are stable isotopes.
2. Tritium, \( \ce{^3_1H} \), has two neutrons and is unstable; it decays by beta emission with a half-life of about 12.3 years.
3. So only tritium is radioactive.

Teacher's Note:
1. Remember: H-1 and H-2 are stable, H-3 is radioactive.
2. Tritium is a weak beta emitter and occurs in traces in the upper atmosphere.

Answer: (b) \( \ce{Mg + NaOH ->} \)

Explanation:
1. Hydrogen is liberated from alkalis only by amphoteric metals such as Zn, Al, Be, Sn and Pb.
2. \( \ce{Zn + 2NaOH -> Na2ZnO2 + H2} \), \( \ce{2Al + 2NaOH + 2H2O -> 2NaAlO2 + 3H2} \), \( \ce{Be + 2NaOH -> Na2BeO2 + H2} \).
3. Magnesium is not amphoteric and does not react with NaOH, so no hydrogen is formed.

Teacher's Note:
1. Amphoteric metals dissolve in both acids and strong alkalis.
2. Mg is a basic metal, so it reacts with acids but not with alkalis.

Answer: (d) \( \ce{O3} \)

Explanation:
1. \( \ce{H2O2} \) oxidises \( \ce{Na2SO3} \) to \( \ce{Na2SO4} \), \( \ce{PbS} \) to \( \ce{PbSO4} \) and \( \ce{KI} \) to \( \ce{I2} \) in acidic medium.
2. Ozone is a stronger oxidising agent than \( \ce{H2O2} \); instead, \( \ce{H2O2} \) reduces it: \( \ce{H2O2 + O3 -> H2O + 2O2} \).
3. So ozone cannot be oxidised by H2O2.

Teacher's Note:
1. Compare the oxidising strength of the two species before deciding who oxidises whom.
2. H2O2 acts as a reducing agent towards strong oxidants such as O3, KMnO4 and Cl2.

Answer: (d) None of these

Explanation:
1. When \( \ce{H2O2} \) acts as an oxidising agent, its oxygen goes from -1 to -2 and no \( \ce{O2} \) is formed from it.
2. In (a), (b) and (c) oxygen gas is released from \( \ce{H2O2} \), meaning oxygen goes from -1 to 0; \( \ce{H2O2} \) is oxidised, so it acts as a reducing agent.
3. So none of the three reactions shows the oxidising property.

Teacher's Note:
1. Look at the oxygen product: O2 formed from H2O2 means H2O2 is the reductant.
2. H2O2 reduces strong oxidants such as MnO4-, hexacyanoferrate(III) in alkali and PbO2.

Answer: (d) \( \ce{MgCO3} \)

Explanation:
1. Hard water contains dissolved calcium and magnesium salts: bicarbonates (temporary hardness) and chlorides or sulphates (permanent hardness).
2. \( \ce{MgCO3} \) is insoluble in water, so it cannot be present in dissolved form in clear water.

Teacher's Note:
1. Clear hard water has only soluble Ca and Mg salts.
2. Insoluble carbonates settle out and do not cause hardness.

Answer: (a) \( \ce{C2D2} \)

Explanation:
1. Calcium carbide gives acetylene with water: \( \ce{CaC2 + 2H2O -> Ca(OH)2 + C2H2} \).
2. With heavy water the same reaction occurs, giving deuteroacetylene: \( \ce{CaC2 + 2D2O -> Ca(OD)2 + C2D2} \).

Teacher's Note:
1. Replace H by D in the usual reaction.
2. Deuterated acetylene is used as a labelled compound in research.

Answer: (d) Ca

Explanation:
1. Hydrolith is the common name of calcium hydride, \( \ce{CaH2} \).
2. It is formed by heating calcium in hydrogen: \( \ce{Ca + H2 -> CaH2} \).

Teacher's Note:
1. Hydrolith means calcium hydride.
2. It is used to prepare H2 in the laboratory: CaH2 + 2H2O gives Ca(OH)2 + 2H2.

Answer: (c) Bosch's process

Explanation:
1. Water gas (\( \ce{CO + H2} \)) is mixed with more steam over \( \ce{Fe2O3} \) and \( \ce{Cr2O3} \) at about 673 K.
2. \( \ce{CO + H2O -> CO2 + H2} \) (water gas shift reaction); this is Bosch's process.

Teacher's Note:
1. Bosch's process gives most of the commercial hydrogen.
2. CO2 is then removed by washing with water under pressure.

Answer: (a) Magnesium bicarbonate

Explanation:
1. Temporary hardness is caused by bicarbonates of calcium and magnesium; it is removed by boiling.
2. Chloride and sulphate salts cause permanent hardness, and \( \ce{CaCO3} \) is insoluble, so only magnesium bicarbonate fits.

Teacher's Note:
1. Bicarbonates give temporary hardness.
2. Chlorides and sulphates give permanent hardness.

Answer: (b) \( \ce{LiH < NaH < KH < RbH < CsH} \)

Explanation:
1. Ionic character increases as the electropositive character of the metal increases.
2. Electropositive character rises down the group from Li to Cs, so ionic character is LiH < NaH < KH < RbH < CsH.

Teacher's Note:
1. Larger electropositivity means more ionic bonding.
2. LiH is the least ionic and most covalent alkali metal hydride.

Answer: (b) \( \ce{BaO2.8H2O} \)

Explanation:
1. Hydrated barium peroxide reacts with dilute sulphuric acid: \( \ce{BaO2.8H2O + H2SO4 -> BaSO4 + H2O2 + 8H2O} \).
2. \( \ce{PbO2} \), \( \ce{MnO2} \) and \( \ce{TiO2} \) are oxides (not peroxides) and do not release \( \ce{H2O2} \).

Teacher's Note:
1. Only a true peroxide gives H2O2 with a dilute acid.
2. BaSO4 is insoluble and is filtered off, leaving the H2O2 solution.

Answer: (b) \( \ce{Na2CO3} \)

Explanation:
1. Washing soda precipitates calcium and magnesium ions as carbonates: \( \ce{CaCl2 + Na2CO3 -> CaCO3 v + 2NaCl} \).
2. \( \ce{Cl2} \) and bleaching powder only disinfect water, and \( \ce{KNO3} \) gives no precipitate.

Teacher's Note:
1. Washing soda removes both temporary and permanent hardness.
2. Calgon and ion exchangers also work.