JEE Chemistry Hydrogen MCQs Set 10

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Multiple Choice Questions

Question 1: When electric current is passed through an ionic hydride in the molten state:
(a) hydrogen is liberated at the anode
(b) hydrogen is liberated at the cathode
(c) hydride ion migrates towards the cathode
(d) hydride ion remains in solution
Show Answer & Explanation

Answer: (a) hydrogen is liberated at the anode

Explanation:
1. Ionic (saline) hydrides contain the hydride ion, \( \ce{H-} \).
2. During electrolysis of the melt, \( \ce{H-} \) moves to the anode and is oxidised, so hydrogen gas is liberated at the anode.

Teacher's Note:
1. Anions move to the anode.
2. This is evidence for the H- ion in saline hydrides.

Question 2: The adsorption of hydrogen by metals is called:
(a) dehydrogenation
(b) hydrogenation
(c) occlusion
(d) adsorption
Show Answer & Explanation

Answer: (c) occlusion

Explanation:
1. Absorption of a gas, especially hydrogen, in the interior of a metal such as palladium is called occlusion.

Teacher's Note:
1. Palladium can occlude about 900 times its own volume of hydrogen.

Question 3: Hydrogen can be fused to form helium at:
(a) high temperature and high pressure
(b) high temperature and low pressure
(c) low temperature and high pressure
(d) low temperature and low pressure
Show Answer & Explanation

Answer: (a) high temperature and high pressure

Explanation:
1. Nuclear fusion needs the nuclei to overcome their repulsion.
2. This requires very high temperature and high pressure, as in the Sun.

Teacher's Note:
1. Stars fuse hydrogen at millions of kelvin.

Question 4: The oxidation states of the most electronegative element in the products of the reaction of \( \ce{BaO2} \) with dilute \( \ce{H2SO4} \) are:
(a) 0 and -1
(b) -1 and -2
(c) -2 and 0
(d) -2 and +1
Show Answer & Explanation

Answer: (b) -1 and -2

Explanation:
1. \( \ce{BaO2 + H2SO4 -> BaSO4 + H2O2} \).
2. The most electronegative element is oxygen, which is -1 in \( \ce{H2O2} \) and -2 in \( \ce{BaSO4} \).

Teacher's Note:
1. Barium peroxide contains the peroxide ion.
2. Oxygen is -1 in peroxides.

Question 5: \( \ce{H2O2 -> 2H+ + O2 + 2e-} \), E = -0.68 V. This equation represents which of the following behaviours of \( \ce{H2O2} \)?
(a) reducing
(b) oxidising
(c) acidic
(d) catalytic
Show Answer & Explanation

Answer: (a) reducing

Explanation:
1. Hydrogen peroxide loses electrons, and oxygen is oxidised from -1 to 0.
2. A species that loses electrons acts as a reducing agent.

Teacher's Note:
1. Loss of electrons means reduction of the other species.

Question 6: HCl is added to the following oxides. Which one would give \( \ce{H2O2} \)?
(a) \( \ce{MnO2} \)
(b) \( \ce{PbO2} \)
(c) \( \ce{BaO} \)
(d) none
Show Answer & Explanation

Answer: (d) none

Explanation:
1. \( \ce{MnO2} \) and \( \ce{PbO2} \) are oxidising oxides and give chlorine with HCl.
2. \( \ce{BaO} \) is a normal oxide and gives \( \ce{BaCl2} \) and water.
3. Only the peroxide \( \ce{BaO2} \) (not listed) gives \( \ce{H2O2} \).

Teacher's Note:
1. Only peroxides give H2O2 with a dilute acid.

Question 7: Chemical A is used for water softening to remove temporary hardness. A reacts with sodium carbonate to generate caustic soda. When \( \ce{CO2} \) is bubbled through a solution of A, it turns cloudy. What is the chemical formula of A?
(a) \( \ce{CaCO3} \)
(b) \( \ce{CaO} \)
(c) \( \ce{Ca(OH)2} \)
(d) \( \ce{Ca(HCO3)2} \)
Show Answer & Explanation

Answer: (c) \( \ce{Ca(OH)2} \)

Explanation:
1. Slaked lime \( \ce{Ca(OH)2} \) is used in Clarke's method to remove temporary hardness.
2. \( \ce{Ca(OH)2 + Na2CO3 -> CaCO3 + 2NaOH} \) gives caustic soda.
3. \( \ce{CO2} \) turns lime water cloudy by forming \( \ce{CaCO3} \).

Teacher's Note:
1. Clarke's process uses lime.
2. Lime water turning milky is the test for carbon dioxide.

Question 8: Which physical property of dihydrogen is wrong?
(a) odourless gas
(b) tasteless gas
(c) colourless gas
(d) non-inflammable gas
Show Answer & Explanation

Answer: (d) non-inflammable gas

Explanation:
1. Dihydrogen is colourless, odourless and tasteless.
2. It is highly inflammable and burns with a pale blue flame.

Teacher's Note:
1. Hydrogen is a fuel.

Question 9: The dielectric constant of \( \ce{H2O} \) is 80. The electrostatic force of attraction between \( \ce{Na+} \) and \( \ce{Cl-} \) will be:
(a) reduced to 1/40 in water than in air
(b) reduced to 1/80 in water than in air
(c) increased to 80 in water than in air
(d) will remain unchanged
Show Answer & Explanation

Answer: (b) reduced to 1/80 in water than in air

Explanation:
1. The force between charges is inversely proportional to the dielectric constant of the medium.
2. In water the force is 1/80 of its value in air.

Teacher's Note:
1. Coulomb force is F = q1q2/(4 pi eps r squared).

Question 10: \( \ce{H2O2} \) is always stored in dark (black) bottles because:
(a) it decomposes in the presence of light
(b) it reacts with the glass
(c) it is volatile
(d) it is hygroscopic
Show Answer & Explanation

Answer: (a) it decomposes in the presence of light

Explanation:
1. Hydrogen peroxide undergoes slow decomposition to water and oxygen, and light accelerates it.
2. Dark bottles cut off light and also contain stabilisers.

Teacher's Note:
1. Light speeds up decomposition.

Question 11: The purity of \( \ce{H2O2} \) in a given sample is 85%. Calculate the weight of impure sample of \( \ce{H2O2} \) which requires 10 mL of M/5 \( \ce{KMnO4} \) solution in titration in acidic medium.
(a) 2 g
(b) 0.2 g
(c) 0.17 g
(d) 0.15 g
Show Answer & Explanation

Answer: (b) 0.2 g

Explanation:
1. Milliequivalents of \( \ce{KMnO4} \) = 10 × (1/5) × 5 = 10.
2. Equivalents of pure \( \ce{H2O2} \) = 0.01, so mass = 0.01 × 17 = 0.17 g.
3. Mass of the impure sample = 0.17/0.85 = 0.2 g.

Teacher's Note:
1. Find the pure mass, then correct for purity.

Question 12: The numbers of protons, electrons and neutrons in a molecule of heavy water are respectively:
(a) 8, 10, 11
(b) 10, 10, 10
(c) 10, 11, 10
(d) 11, 10, 10
Show Answer & Explanation

Answer: (b) 10, 10, 10

Explanation:
1. \( \ce{D2O} \): each D has 1 proton and 1 neutron, and oxygen has 8 protons and 8 neutrons.
2. Protons = 2 + 8 = 10, electrons = 10 and neutrons = 2 + 8 = 10.

Teacher's Note:
1. Deuterium has one neutron.

Question 13: Which of the following terms is not correct for hydrogen?
(a) its molecule is diatomic
(b) it exists both as \( \ce{H+} \) and \( \ce{H-} \) in different chemical compounds
(c) it is the only species which has no neutrons in the nucleus
(d) heavy water is unstable because hydrogen is substituted by its isotope deuterium
Show Answer & Explanation

Answer: (d) heavy water is unstable because hydrogen is substituted by its isotope deuterium

Explanation:
1. Heavy water is stable and is used as a moderator in nuclear reactors.
2. The other statements are correct about hydrogen (protium has no neutron).

Teacher's Note:
1. Isotopic substitution does not make a compound unstable.

Question 14: Hard water is not fit for washing clothes because:
(a) it contains \( \ce{Na2SO4} \) and KCl
(b) it gives a precipitate with soap
(c) it contains impurities
(d) it is acidic in nature
Show Answer & Explanation

Answer: (b) it gives a precipitate with soap

Explanation:
1. Calcium and magnesium ions react with soap to give an insoluble scum of calcium and magnesium stearate.
2. This wastes soap and leaves deposits on clothes.

Teacher's Note:
1. Hardness ions precipitate soaps.

Question 15: The amount of \( \ce{H2O2} \) present in 1 litre of 1.5 N \( \ce{H2O2} \) solution is:
(a) 25.5 g
(b) 3.0 g
(c) 8.0 g
(d) 2.5 g
Show Answer & Explanation

Answer: (a) 25.5 g

Explanation:
1. The equivalent mass of \( \ce{H2O2} \) is 34/2 = 17.
2. Mass = normality × equivalent mass × volume = 1.5 × 17 × 1 = 25.5 g.

Teacher's Note:
1. Mass = N times equivalent mass times volume in litres.

Question 16: When the same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide solution, the ratio of the volumes of hydrogen evolved is:
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 9 : 4
Show Answer & Explanation

Answer: (a) 1 : 1

Explanation:
1. \( \ce{Zn + H2SO4 -> ZnSO4 + H2} \) and \( \ce{Zn + 2NaOH -> Na2ZnO2 + H2} \).
2. Each mole of zinc gives one mole of hydrogen in both cases, so the ratio is 1 : 1.

Teacher's Note:
1. Amphoteric zinc reacts with both acid and alkali.

Question 17: Hydrogen at room temperature contains:
(a) 25% ortho + 75% para
(b) 25% para + 75% ortho
(c) 50% ortho + 50% para
(d) 60% ortho + 34% para
Show Answer & Explanation

Answer: (b) 25% para + 75% ortho

Explanation:
1. At room temperature ordinary hydrogen is an equilibrium mixture of 75% ortho and 25% para hydrogen.

Teacher's Note:
1. At 0 K it is nearly pure para.
2. At higher temperatures the ratio is 3 : 1.

Question 18: \( \ce{D2O} \) is preferred to \( \ce{H2O} \) as a moderator in nuclear reactors because:
(a) \( \ce{D2O} \) slows down fast neutrons better
(b) \( \ce{D2O} \) has high specific heat
(c) \( \ce{D2O} \) is cheaper
(d) none of these
Show Answer & Explanation

Answer: (a) \( \ce{D2O} \) slows down fast neutrons better

Explanation:
1. A moderator must slow fast neutrons without absorbing them.
2. Deuterium absorbs far fewer neutrons than ordinary hydrogen, so heavy water is the better moderator.

Teacher's Note:
1. Hydrogen captures neutrons to form deuterium.

Question 19: Water is:
(a) more polar than \( \ce{H2S} \)
(b) more or less identical in polarity with \( \ce{H2S} \)
(c) less polar than \( \ce{H2S} \)
(d) none of these
Show Answer & Explanation

Answer: (a) more polar than \( \ce{H2S} \)

Explanation:
1. Oxygen is much more electronegative than sulphur, so the O-H bonds are more polar than the S-H bonds.
2. Water is therefore more polar than \( \ce{H2S} \).

Teacher's Note:
1. Electronegativity difference decides polarity.

Question 20: In which of the following reactions is \( \ce{H2O2} \) acting as a reducing agent?
(a) \( \ce{H2O2 + SO2 -> H2SO4} \)
(b) \( \ce{2KI + H2O2 -> 2KOH + I2} \)
(c) \( \ce{PbS + 4H2O2 -> PbSO4 + 4H2O} \)
(d) \( \ce{Ag2O + H2O2 -> 2Ag + H2O + O2} \)
Show Answer & Explanation

Answer: (d) \( \ce{Ag2O + H2O2 -> 2Ag + H2O + O2} \)

Explanation:
1. In the last reaction oxygen of \( \ce{H2O2} \) is oxidised to \( \ce{O2} \) (-1 to 0), so \( \ce{H2O2} \) is the reducing agent.
2. In the other three reactions \( \ce{H2O2} \) oxidises SO2, iodide and sulphide.

Teacher's Note:
1. Reducing agent is itself oxidised.
2. Look for O2 as a product.

Question 21: When a sample of hard water is passed through a layer of sodium zeolite, which of the following ions will not be present in the resulting sample of water obtained?
(a) \( \ce{Mg^2+} \) and \( \ce{Ca^2+} \)
(b) \( \ce{Ca^2+} \) and \( \ce{Na+} \)
(c) \( \ce{Mg^2+} \) and \( \ce{CO3^2-} \)
(d) \( \ce{CO3^2-} \) and \( \ce{Cl-} \)
Show Answer & Explanation

Answer: (a) \( \ce{Mg^2+} \) and \( \ce{Ca^2+} \)

Explanation:
1. Zeolite exchanges \( \ce{Ca^2+} \) and \( \ce{Mg^2+} \) ions of hard water for \( \ce{Na+} \) ions.
2. Hence calcium and magnesium ions are absent from the treated water.

Teacher's Note:
1. Permutit process exchanges cations.

Question 22: Hydrogen peroxide acts both as an oxidising and a reducing agent depending upon the nature of the reacting species. In which of the following cases does \( \ce{H2O2} \) act as a reducing agent in acid medium?
(a) \( \ce{MnO4-} \)
(b) \( \ce{SO3^2-} \)
(c) \( \ce{I-} \)
(d) \( \ce{Fe^2+} \)
Show Answer & Explanation

Answer: (a) \( \ce{MnO4-} \)

Explanation:
1. \( \ce{MnO4-} \) is a stronger oxidant than \( \ce{H2O2} \), so \( \ce{H2O2} \) reduces it to \( \ce{Mn^2+} \) and is itself oxidised to \( \ce{O2} \).
2. Sulphite, iodide and ferrous ions are oxidised by \( \ce{H2O2} \).

Teacher's Note:
1. H2O2 reduces strong oxidants and oxidises reductants.

Question 23: Pick out the correct statement:
(a) by decreasing the temperature pure para-hydrogen can be obtained
(b) by increasing the temperature pure ortho-hydrogen can be obtained
(c) by decreasing the temperature pure ortho-hydrogen can be obtained
(d) by increasing the temperature pure para-hydrogen can be obtained
Show Answer & Explanation

Answer: (a) by decreasing the temperature pure para-hydrogen can be obtained

Explanation:
1. Para hydrogen is the lower energy form.
2. At temperatures near 0 K the equilibrium mixture is almost entirely para hydrogen.

Teacher's Note:
1. Para is stable at low temperature.
2. Pure ortho cannot be obtained.

Question 24: An inorganic compound gives off \( \ce{O2} \) when heated, turns an acidic solution of KI violet and reduces acidified \( \ce{KMnO4} \). The compound is:
(a) \( \ce{SO3} \)
(b) \( \ce{KNO3} \)
(c) \( \ce{H2O2} \)
(d) all of these
Show Answer & Explanation

Answer: (c) \( \ce{H2O2} \)

Explanation:
1. \( \ce{H2O2} \) decomposes to give \( \ce{O2} \).
2. It oxidises iodide to iodine (violet in organic solvent) and, with the strong oxidant \( \ce{KMnO4} \), it acts as a reducing agent.

Teacher's Note:
1. Only H2O2 shows both oxidising and reducing behaviour.

Question 25: Saline hydrides react explosively with water, and such fires can be extinguished by:
(a) water
(b) carbon dioxide
(c) sand
(d) none of these
Show Answer & Explanation

Answer: (c) sand

Explanation:
1. Saline hydrides react with water to form hydrogen and alkali, which makes the fire worse.
2. They also react with \( \ce{CO2} \), so dry sand is used.

Teacher's Note:
1. Do not use water or CO2 on metal hydride fires.

Question 26: 30 volume hydrogen peroxide means:
(a) 30% \( \ce{H2O2} \) solution
(b) 30 cm3 solution contains 1 g of \( \ce{H2O2} \)
(c) 1 cm3 of the solution liberates 30 cm3 of \( \ce{O2} \) at STP
(d) 30 cm3 of solution contains 1 mole of \( \ce{H2O2} \)
Show Answer & Explanation

Answer: (c) 1 cm3 of the solution liberates 30 cm3 of \( \ce{O2} \) at STP

Explanation:
1. Volume strength is the volume of oxygen at STP liberated by 1 volume of the solution.
2. A 30 volume solution gives 30 cm3 of \( \ce{O2} \) per 1 cm3 of solution.

Teacher's Note:
1. Volume strength is a ratio of volumes.

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