JEE Chemistry Hydrogen MCQs Set 09

Answer: (b) the mass of deuterium and tritium is two to three times that of protium

Explanation:
1. The isotopic effect arises from a difference in mass.
2. For hydrogen the mass doubles or triples on going to \( D \) or \( T \), while for isotopes such as \( C-12/C-14 \) the change is only a small fraction of the mass.
3. Answer: (b)

Teacher's Note:
1. Large relative mass change gives a large effect.
2. Mainly physical properties change.

Answer: (a) (i), because the C–H bond is weaker than the C–D bond

Explanation:
1. The heavier isotope has a lower zero-point energy, so the C–D bond has a higher bond dissociation energy than the C–H bond.
2. Breaking the C–H bond needs less energy, so reaction \( (i) \) is faster (kinetic isotope effect).
3. Answer: (a)

Teacher's Note:
1. Heavier isotope gives stronger bond.
2. Stronger bond reacts slower.

Answer: (b) increases from H₂ to T₂

Explanation:
1. The bond energies are about 436.0 (H₂), 443.3 (D₂) and 446.9 (T₂) kJ mol⁻¹.
2. The bond length is the same (74 pm), but the heavier isotopes have lower zero-point energy and so stronger effective bonds.
3. Answer: (b)

Teacher's Note:
1. Heavier isotope means stronger bond.
2. Bond length is unaffected.

Answer: (b) the anode, because the hydride ion is the anion

Explanation:
1. Molten NaH contains Na⁺ and H⁻.
2. H⁻ migrates to the anode and is oxidised: 2H⁻ → H₂ + 2e⁻; sodium is deposited at the cathode.
3. Answer: (b)

Teacher's Note:
1. Anions move to the anode.
2. This is proof of the hydride ion.

Answer: (b) the impurities form tiny galvanic cells with zinc, which speeds up the reaction

Explanation:
1. Pure zinc has a smooth, unbroken surface and reacts very slowly because bubbles of hydrogen coat the surface.
2. Impurities such as copper form local galvanic cells with zinc, so hydrogen evolves faster.
3. Answer: (b)

Teacher's Note:
1. Local cells accelerate corrosion of zinc.
2. A few drops of CuSO₄ do the same.

Answer: (d) Fe

Explanation:
1. Amphoteric metals such as Zn, Al, Sn, Pb and Be form oxo-anions and liberate hydrogen with hot alkali.
2. Iron is not amphoteric and does not react with NaOH.
3. Answer: (d)

Teacher's Note:
1. Be, Zn, Al, Sn, Pb are amphoteric.
2. Fe is basic in nature.

Answer: (b) Electrolysis of water made slightly alkaline or acidic with inert electrodes

Explanation:
1. Electrolysis of dilute acid or alkali solution gives hydrogen at the cathode and oxygen at the anode, and the hydrogen is free from gaseous impurities.
2. The other methods give hydrogen contaminated with H₂S, AsH₃, CO or CO₂.
3. Answer: (b)

Teacher's Note:
1. Electrolysis of water gives very pure H₂.
2. Acid or alkali only makes water conducting.

Answer: (b) CO + H₂O → CO₂ + H₂

Explanation:
1. Steam passed over red-hot coke gives water gas, CO + H₂.
2. In the Bosch step (water-gas shift reaction) CO is converted to CO₂ by steam over the iron–chromium catalyst, giving extra hydrogen: CO + H₂O → CO₂ + H₂.
3. Answer: (b)

Teacher's Note:
1. Water-gas shift enriches hydrogen.
2. CO₂ is removed by scrubbing with water under pressure.

Answer: (b) 0.896 L

Explanation:
1. Moles of Fe = \( 1.68/56 = 0.03 \).
2. From the equation, 3 mol Fe give 4 mol H₂, so 0.03 mol gives 0.04 mol.
3. Volume = \( 0.04 \times 22.4 = 0.896 L \).
4. Answer: (b)

Teacher's Note:
1. Use mole ratio 3 : 4.
2. Iron is regenerated from Fe₃O₄ by water gas.

Answer: (b) Na with cold water; Mg with hot water; Fe with steam

Explanation:
1. Highly reactive metals (Na, \( K \), Ca) react with cold water, moderately reactive metals (Mg, Al, Zn) with hot water, and less reactive metals (Fe, Ni, Sn) only with steam.
2. Answer: (b)

Teacher's Note:
1. Reactivity decides the temperature needed.
2. Alkali metals react violently with cold water.

Answer: (c) TiH_1.8

Explanation:
1. Interstitial hydrides are formed by \( d- \) and f-block metals, in which hydrogen occupies holes in the metal lattice and the composition is non-stoichiometric.
2. TiH_1.8 fits this; NaH is ionic, CH₄ is covalent and BeH₂ is a polymeric covalent hydride.
3. Answer: (c)

Teacher's Note:
1. Non-stoichiometric: metallic hydride.
2. Hydrogen occupies interstitial sites.

Answer: (b) 7, 8 and 9

Explanation:
1. Metals of groups 7, 8 and 9 do not form hydrides.
2. This region of the d-block is therefore called the hydride gap.
3. Answer: (b)

Teacher's Note:
1. Group 6 (Cr) forms CrH, but groups 7–9 form none.
2. Hydride gap is a d-block feature.

Answer: (a) SiH₄

Explanation:
1. Electron-precise hydrides are formed by group 14 elements, where all valence electrons are used in bonding with hydrogen and there are no lone pairs.
2. SiH₄ is electron precise.
3. Answer: (a)

Teacher's Note:
1. Group 14: tetrahedral hydrides.
2. No lone pairs on central atom.

Answer: (c) MgH₂

Explanation:
1. The hydrides of Be and Mg are covalent and polymeric, while those of other s-block elements are ionic (saline).
2. MgH₂ is therefore the correct choice.
3. Answer: (c)

Teacher's Note:
1. BeH₂ and MgH₂ are exceptions.
2. Heavier s-block hydrides are ionic.

Answer: (a) sodium hexametaphosphate, Na₂[Na₄(PO₃)₆]

Explanation:
1. Calgon is sodium hexametaphosphate, Na₂[Na₄(PO₃)₆].
2. It forms a soluble complex with Ca²⁺ and Mg²⁺: 2Ca²⁺ + \( \ce{[Na4P6O18]^{2-}} \) → \( \ce{[Na2Ca2P6O18]^{2-}} \) + 2Na⁺, preventing them from forming scum with soap.
3. Answer: (a)

Teacher's Note:
1. Calgon sequesters Ca²⁺ and Mg²⁺.
2. Works for permanent hardness.

Answer: (b) 7.4 g

Explanation:
1. Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃ + 2H₂O.
2. One mole of lime is needed per mole of bicarbonate, so 0.1 mol of Ca(OH)₂ = \( 0.1 \times 74 = 7.4 g \).
3. Answer: (b)

Teacher's Note:
1. Clarke's process uses lime.
2. Molar mass of Ca(OH)₂ is \( 74 g \).

Answer: (a) dilute acid

Explanation:
1. A cation exchanger (R–COOH or R–SO₃H) exchanges H⁺ for Ca²⁺ and Mg²⁺.
2. When it is exhausted, it is regenerated by treating with dilute acid, which replaces the metal ions with H⁺; the anion exchanger is regenerated with dilute NaOH.
3. Answer: (a)

Teacher's Note:
1. Cation resin: dilute acid.
2. Anion resin: dilute NaOH.

Answer: (a) D₂O has a higher mass, so D⁺ and OD⁻ are discharged more slowly than H⁺ and OH⁻

Explanation:
1. In the electrolysis of water H⁺ is discharged at the cathode faster than D⁺ because of the lower mass and higher mobility of H⁺.
2. D₂O therefore concentrates in the residual solution.
3. Answer: (a)

Teacher's Note:
1. Lighter isotope reacts faster.
2. Heavy water is enriched in the residue.

Answer: (b) H₂O₂ > D₂O > H₂O

Explanation:
1. Hydrogen peroxide has more extensive hydrogen bonding and a higher molar mass: \( b.p \). about \( 150 ^{\circ}C \).
2. D₂O boils at \( 101.4 ^{\circ}C \) and H₂O at \( 100 ^{\circ}C \).
3. Answer: (b)

Teacher's Note:
1. Stronger intermolecular forces mean higher boiling point.
2. Isotopic mass raises boiling point slightly.

Answer: (b) anhydrous barium peroxide forms an insoluble coating of BaSO₄ that stops the reaction

Explanation:
1. BaO₂·8H₂O + H₂SO₄ → BaSO₄ + H₂O₂ + 8H₂O.
2. Anhydrous BaO₂ becomes coated with insoluble BaSO₄, which prevents further action of the acid, so the hydrated salt is used.
3. Answer: (b)

Teacher's Note:
1. Insoluble BaSO₄ coats the solid.
2. H₃PO₄ may be used in place of H₂SO₄.

Answer: (a) auto-oxidation of 2-alkylanthraquinol

Explanation:
1. In the modern industrial process 2-ethylanthraquinol is auto-oxidised by air to 2-ethylanthraquinone and H₂O₂, and the quinone is reduced back with H₂ over Pd.
2. The process is cyclic.
3. Answer: (a)

Teacher's Note:
1. Cyclic process with a quinone.
2. Hydrogen peroxide is extracted with water.

Answer: (c) non-planar, like a half-open book with an O–O single bond

Explanation:
1. H₂O₂ has an O–O single bond, and the two O–H bonds lie in different planes, like pages of a half-open book.
2. This is why it has a dipole moment of about \( 2.1 D \).
3. Answer: (c)

Teacher's Note:
1. Dihedral angle about 111° in gas phase.
2. Non-zero dipole shows it is non-planar.

Answer: (b) 5.6 L

Explanation:
1. 2 mol KMnO₄ give 5 mol O₂, so 0.1 mol gives 0.25 mol.
2. Volume = \( 0.25 \times 22.4 = 5.6 L \).
3. Answer: (b)

Teacher's Note:
1. Use the ratio 2 : 5.
2. All oxygen comes from H₂O₂.

Answer: (b) It is stored in dark, plastic or wax-lined bottles with a stabiliser such as acetanilide or H₃PO₄

Explanation:
1. Light, rough surfaces, alkali from glass and traces of metals catalyse the decomposition of H₂O₂.
2. Hence it is stored in dark, coated or plastic containers with inhibitors such as acetanilide or phosphoric acid.
3. Answer: (b)

Teacher's Note:
1. Catalysts for decomposition: MnO₂, metals, light.
2. Inhibitors slow the decomposition.

Answer: (b) 1 M

Explanation:
1. 11.2 volume means 1 mL gives 11.2 mL of O₂ at STP, i.e. \( 1 L \) gives \( 11.2 L = 0.5 \) mol of O₂.
2. 2H₂O₂ → 2H₂O + O₂, so 0.5 mol of O₂ comes from 1 mol of H₂O₂.
3. The solution is \( 1 M \).
4. Answer: (b)

Teacher's Note:
1. Volume strength × (1/11.2) = molarity.
2. Two moles of H₂O₂ give one mole of O₂.