JEE Chemistry Hydrogen MCQs Set 14

Answer: (c) at very high pressure

Explanation:
1. Under extremely high pressure the electrons of hydrogen become delocalised and the solid conducts like a metal.
2. So hydrogen behaves as a metal at very high pressure.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (c) Ortho hydrogen

Explanation:
1. Protium, deuterium and tritium are isotopes (they differ in the number of neutrons).
2. Ortho and para hydrogen are spin isomers (allotropic forms) of the same molecule H2, not isotopes.
3. So ortho hydrogen is not an isotope.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (a) 0, 1, 2

Explanation:
1. Mass numbers are 1, 2 and 3 and each has one proton.
2. Neutrons = mass number - 1 = 0, 1, 2.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (d) Both (a) and (b)

Explanation:
1. Zinc is amphoteric: \( \ce{Zn + 2NaOH -> Na2ZnO2 + H2} \).
2. Both hydrogen and sodium zincate are formed.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (c) Copper and dilute HCl

Explanation:
1. Iron liberates \( \ce{H2} \) from dilute acid and (as \( \ce{Fe3O4} \)) from steam, and sodium reacts with ethanol to give \( \ce{H2} \).
2. Copper lies below hydrogen in the electrochemical series and does not displace \( \ce{H2} \) from dilute HCl.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (a) cold water

Explanation:
1. Zinc displaces hydrogen from dilute acids and from hot NaOH solution (zincate formation).
2. Zinc does not react with cold water, so no hydrogen is obtained.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (b) a saline hydride such as NaH

Explanation:
1. \( \ce{NaH + H2O -> NaOH + H2} \) gives hydrogen free from other gases.
2. Steam reforming of methane or hydrocarbons gives CO + H2 mixtures, which need further purification.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (b) methanol

Explanation:
1. The reaction is \( \ce{CO + 2H2 ->[Co] CH3OH} \).
2. So methanol is formed.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (d) NaH

Explanation:
1. Ionic hydrides are formed by the s-block metals; NaH contains \( \ce{Na+} \) and \( \ce{H-} \).
2. \( \ce{H2S} \) and \( \ce{NH3} \) are covalent and \( \ce{TiH_{1.73}} \) is a non-stoichiometric metallic (interstitial) hydride.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (d) \( \ce{MgH2} \)

Explanation:
1. \( \ce{CaH2} \), NaH and \( \ce{BaH2} \) are ionic.
2. \( \ce{MgH2} \) has considerable covalent character and exists as a polymer.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

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

Explanation:
1. Oxygen is more electronegative than sulphur, so the O-H bond is more polar than the S-H bond.
2. Hence water is more polar than \( \ce{H2S} \).

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (d) 104.5°

Explanation:
1. Oxygen is \( sp^3 \) hybridised with two lone pairs, which compress the angle from 109.5° to about 104.5°.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (a) hydrogen-bonding interactions

Explanation:
1. In ice each water molecule is hydrogen bonded to four others giving an open cage structure with empty spaces.
2. Hence ice is less dense than water.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (b) Magnesium and chloride

Explanation:
1. Hardness is caused by the soluble salts of calcium and magnesium.
2. Magnesium chloride therefore causes (permanent) hardness.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (a) \( \ce{Na2[Na4(PO3)6]} \)

Explanation:
1. Calgon is sodium hexametaphosphate, written as \( \ce{Na2[Na4(PO3)6]} \).

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (a) slaked lime

Explanation:
1. Clark's process: \( \ce{Ca(HCO3)2 + Ca(OH)2 -> 2CaCO3 v + 2H2O} \).
2. Slaked lime precipitates the bicarbonates as carbonates.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (c) chlorides and sulphates of calcium and magnesium

Explanation:
1. Bicarbonates of Ca and Mg cause temporary hardness; their chlorides and sulphates cause permanent hardness.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (a) cold dilute \( \ce{H2SO4} \) + \( \ce{BaO2} \)

Explanation:
1. \( \ce{BaO2 + H2SO4 -> BaSO4 v + H2O2} \); the insoluble \( \ce{BaSO4} \) is filtered off.
2. Conc. acid or hot conditions would decompose \( \ce{H2O2} \).

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (a) 8.4

Explanation:
1. 1 N \( \ce{H2O2} \) corresponds to 5.6 volume strength.
2. Volume strength = 1.5 x 5.6 = 8.4.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

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

Explanation:
1. Volume strength is the volume of \( \ce{O2} \) at STP given by unit volume of the solution on decomposition.
2. So 1 cm³ gives 30 cm³ of \( \ce{O2} \).

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (d) 94.8°

Explanation:
1. \( \ce{H2O2} \) has an open-book, non-planar structure with O-O-H angle of about 94.8° (gas phase).

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (b) \( \ce{O2} \)

Explanation:
1. Oxygen is in the -1 state in \( \ce{H2O2} \). Oxidation raises it to 0, giving \( \ce{O2} \).

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (c) oxidising nature of \( \ce{H2O2} \)

Explanation:
1. Sulphur goes from -2 to 0 (oxidised), so \( \ce{H2O2} \) is the oxidising agent.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

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

Explanation:
1. \( \ce{CaC2 + 2D2O -> C2D2 + Ca(OD)2} \).

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.

Answer: (b) moderator

Explanation:
1. Heavy water slows down fast neutrons without absorbing them appreciably, so it is used as a moderator.

Teacher's Note:
1. Identify the key concept in the question first, then check each option against it.
2. Learn the exceptions to general trends, as they are the most common source of questions.