JEE Chemistry Hydrogen MCQs Set 05

Answer: (a) \(\ce{CH3OH}\)

Explanation:
1. Industrial methanol is made by \(\ce{CO + 2H2 -> CH3OH}\) over a Cu/ZnO catalyst at high pressure.

Teacher's Note:
1. Water gas and syngas are mixtures of CO and H2.
2. Methanol synthesis is an important use of syngas.

Answer: (c) Loss of an electron from hydrogen leaves a nucleus of very small size which cannot exist free

Explanation:
1. H+ is a bare proton of radius about \(10^{-15}\) m, compared with about \(10^{-10}\) m for atoms.
2. Its very high charge density means it always attaches to another atom or molecule, e.g. \(\ce{H3O+}\).

Teacher's Note:
1. Hydronium ion is the form in water.
2. The tiny size gives enormous polarising power.

Answer: (c) \(\ce{CH4}\)

Explanation:
1. Electron-precise hydrides have exactly the electrons needed for normal bonds (group 14).
2. CH4 is electron precise; B2H6 is electron deficient; NH3 and H2O are electron rich.

Teacher's Note:
1. Group 13: electron deficient, group 14: electron precise, groups 15-17: electron rich.
2. Electron-precise hydrides are tetrahedral.

Answer: (b) \(\ce{2I- + 2H+ + H2O2 -> I2 + 2H2O}\)

Explanation:
1. As an oxidant, H2O2 is reduced (O from -1 to -2) and forms water.
2. In (c), iodide is oxidised to iodine while H2O2 becomes H2O.
3. In (a), (b) and (d), H2O2 is oxidised to O2, so it acts as a reductant.

Teacher's Note:
1. Look for O2 as a product (reducing role) or H2O (oxidising role).
2. Acidic H2O2 oxidises iodide, sulphides and iron(II).

Answer: (c) \(\ce{I2 + H2O2 + 2OH- -> 2I- + 2H2O + O2}\)

Explanation:
1. In (b), H2O2 is converted to O2 (oxygen goes from -1 to 0), so it is oxidised and acts as a reducing agent.
2. In the others H2O2 is reduced to water or hydroxide.

Teacher's Note:
1. Reducing role in basic medium; oxidising role in acidic medium for many cases.
2. Products with O2 mean H2O2 was the reductant.

Answer: (a) Groups 7, 8, 9

Explanation:
1. Metals of groups 7, 8 and 9 (in the d-block) do not form hydrides; this region is called the hydride gap.
2. The p-block elements form molecular hydrides.

Teacher's Note:
1. Hydride gap: groups 7-9.
2. Cr is an exception in group 6, forming CrH.

Answer: (a) Both A and R are true and R is the correct explanation of A

Explanation:
1. Washing soda precipitates \(\ce{Ca^{2+}}\) and \(\ce{Mg^{2+}}\) as insoluble carbonates, which removes permanent hardness.
2. Statement 2 gives the reason for statement 1.

Teacher's Note:
1. Permanent hardness is due to chlorides and sulphates of Ca and Mg.
2. Calgon and ion-exchange are other methods.

Answer: (a) Both A and R are true and R is the correct explanation of A

Explanation:
1. Pd and Pt form interstitial (non-stoichiometric) hydrides by absorbing hydrogen.
2. This property is why they can store hydrogen, so statement 2 explains statement 1.

Teacher's Note:
1. Palladium absorbs up to 900 times its volume of hydrogen.
2. Such hydrides are called metallic or interstitial hydrides.

Answer: (b) 5.6 V

Explanation:
1. Normality of H2O2 = volume strength / 5.6.
2. So 1 N H2O2 has volume strength 5.6 V.

Teacher's Note:
1. 11.2 V is 2 N.
2. Volume strength = 11.2 x molarity.

Answer: (c) electrolysis of water

Explanation:
1. Electrolysis of water (with traces of acid or alkali) gives hydrogen of very high purity.
2. Steam reforming gives a mixture containing CO and CO2.

Teacher's Note:
1. Electrolysis of warm aqueous barium hydroxide gives pure H2.
2. Industrial hydrogen comes mostly from steam reforming.

Answer: (a) I and II

Explanation:
1. D2O slows neutrons and is used as a moderator, and has stronger association owing to stronger D-bonds, so I and II are correct.
2. D2O is a less effective solvent for ionic compounds than H2O, so III is wrong.

Teacher's Note:
1. D2O has a higher boiling point and density.
2. It is less effective as solvent.

Answer: (b) Copper and aqueous HCl

Explanation:
1. Copper lies below hydrogen in the electrochemical series and does not displace H2 from dilute acids.
2. Iron with steam or dilute acid and sodium with alcohols all release hydrogen.

Teacher's Note:
1. Only metals above H in the series displace hydrogen.
2. Cu reacts with oxidising acids, not with HCl.

Answer: (a) Magnesium and chlorine

Explanation:
1. Hardness is due to dissolved salts of calcium and magnesium (Mg2+, Ca2+) as chlorides, sulphates or bicarbonates.
2. Sodium, potassium and ammonium salts do not cause hardness.

Teacher's Note:
1. Hard water does not lather with soap because of insoluble Ca/Mg soaps.
2. Temporary hardness is due to bicarbonates.

Answer: (a) ice crystals have a hollow hexagonal arrangement of H2O molecules

Explanation:
1. In ice each water molecule is tetrahedrally hydrogen bonded in an open hexagonal structure with empty spaces.
2. On melting, this structure partly collapses and the molecules pack closer, so water is denser.

Teacher's Note:
1. Water is densest at 4 degrees C.
2. Ice floats, protecting aquatic life.

Answer: (b) 40

Explanation:
1. 48 mg of MgSO4 = 48/120 = 0.4 mmol, equivalent to 0.4 mmol of CaCO3.
2. Mass of CaCO3 = 0.4 x 100 = 40 mg per kg, i.e. 40 ppm.

Teacher's Note:
1. Hardness is expressed in terms of CaCO3 equivalents.
2. 1 ppm = 1 mg per kg.

Answer: (d) 4

Explanation:
1. \(\ce{2MnO4- + 5H2O2 + 6H+ -> 2Mn^{2+} + 5O2 + 8H2O}\).
2. 2 mol of KMnO4 react with 5 mol of H2O2, so 10 mol of H2O2 need 4 mol of KMnO4.

Teacher's Note:
1. Use the balanced equation ratio 2 : 5.
2. Permanganate is reduced to Mn2+ in acid.

Answer: (c) 4

Explanation:
1. Clark's method removes only temporary hardness (using lime).
2. Ion exchange, synthetic resins, Calgon and washing soda all remove permanent hardness, so there are 4 methods.

Teacher's Note:
1. Boiling and Clark's method remove temporary hardness.
2. Permanent hardness needs removal of Ca/Mg chlorides and sulphates.