Q: One mole of diborane (B2H6) is reacted with excess D2O. How many moles of HD gas are produced?

6. B2H6 contains 6 B-H bonds. Each B-H bond reacts with one D2O molecule via protodeuteration: B-H + D2O -> B-OD + HD, yielding one mole of HD per mole of B-H bonds. Hence 1 mol B2H6 produces 6 mol HD.

Question 1

In its interactions with potassium periodate (KIO4) and hydroxylamine (NH2OH), hydrogen peroxide functions respectively as a

Accepted Answer

substance that donates electrons, substance that accepts electrons. The correct answer is that hydrogen peroxide functions as a substance that donates electrons and a substance that accepts electrons because it donates electrons in its reaction with potassium periodate and accepts electrons in its reaction with hydroxylamine.

Question 2

Among the two isomers of hydroxybenzoic acid, the ortho isomer is a stronger acid compared to its para counterpart. Which of the following best explains this observation?

Accepted Answer

The o-hydroxybenzoate anion is stabilized by intramolecular hydrogen bonding. In the ortho isomer, after deprotonation of –COOH, the resulting –COO⁻ is stabilized by an intramolecular hydrogen bond with the adjacent –OH group. This extra stabilization of the conjugate base makes o-hydroxybenzoic acid a stronger acid than the para isomer, where such intramolecular interaction is geometrically impossible.

Question 3

In which of the following reactions is hydrogen gas produced?

Accepted Answer

Iron and dilute sulphuric acid. Copper does not react with HCl (it is below hydrogen in reactivity). Iron reacts with dilute H2SO4 to give FeSO4 + H2. Magnesium reacts with steam to give MgO + H2. Sodium reacts with ethanol to give sodium ethoxide + H2. Options B, C, and D all produce hydrogen.

Question 4

How many of the following compounds are classified as saline (ionic) hydrides? NaH, MgH2, CaH2, VH0.56, Fe3H, NH3

Accepted Answer

3. NaH (alkali metal, saline), MgH2 (alkaline earth, saline), and CaH2 (alkaline earth, saline) are all ionic/saline hydrides. VH0.56 and Fe3H are interstitial (metallic) hydrides. NH3 is a covalent hydride. Hence 3 of the listed compounds are saline hydrides.

Question 5

In how many of the following processes is H2 gas released as a product? (i) Na + H2O --> (ii) KH + H2O --> (iii) CO + H2 (steam, Fe2O3/Cr2O3 catalyst) --> (iv) Electrolysis of fused (molten) NaH (v) Electrolysis of acidic water

Accepted Answer

4. Processes (i), (ii), (iv), and (v) all release H2. In (iii), CO is reduced by H2 (water-gas shift or CO + H2 gives syngas further reactions); the reaction CO + H2O gives CO2 + H2 (water-gas shift, so H2 IS produced). But CO + H2 with steam/catalyst — this produces more H2 via water-gas shift. Let us re-examine (iii): if the starting material is CO + H2 being fed with steam over Fe2O3/Cr2O3, this is the water-gas shift reaction (CO + H2O -> CO2 + H2), which PRODUCES H2. So all 5 processes produce H2? But the answer options max is 4.

Question 6

The wavelength of the series limit (longest wavelength limit, i.e., the convergence limit) of the Lyman series of hydrogen is x angstrom. Calculate the wave number (in angstrom⁻¹) of the third line of the Balmer series of He+ ion.

Accepted Answer

25 / (21x). The Lyman series limit wavelength of H is given by 1/x = R_H*(1/1² - 0) => R_H = 1/x. For He+ (Z=2), the third line of the Balmer series corresponds to n=5 -> n=2 transition. Wave number = R_H*Z²*(1/2² - 1/5²) = (1/x)*4*(25-4)/100 = (4*21)/(100x) = 84/(100x) = 21/(25x).

Question 7

Match the species in Column I with their correct properties or uses in Column II. Column I: A. H (atomic hydrogen) B. H2 (dihydrogen) C. H2O (water) D. H2O2 (hydrogen peroxide) Column II: (i) Known commercially as perhydrol (ii) Can be reduced by NaH to give dihydrogen (iii) Used in the ca

Accepted Answer

A -> ii; B -> iv; C -> iii; D -> i. D -> i: H2O2 30% solution is called perhydrol. A -> ii: Atomic hydrogen (nascent H) acts as a strong reducing agent; NaH + H2O -> NaOH + H2, meaning H2O is reduced by NaH. So C -> ii. B -> iv: Oxyhydrogen flame uses H2 for cutting/welding. B -> iii: H2 is used in hydrogenation of alkenes. This creates a conflict. Actually for matching: A=atomic H, B=H2, C=H2O, D=H2O2. (iv) cutting/welding uses oxyhydrogen or atomic-H torch -> both B and A could apply. Atomic hydrogen welding uses A. So A->iv, B->iii (hydrogenation), C->ii (NaH reduces H2O to H2), D->i. That

Question 8

Consider the following properties and determine how many of them have a HIGHER value for protium (¹H) compared to deuterium (²H): Relative natural abundance, Atomic mass, Melting point, Boiling point, Density, Bond dissociation energy, Enthalpy of fusion, Enthalpy of vaporization.

Accepted Answer

2. Among the listed properties, protium exceeds deuterium only in relative natural abundance. Atomic mass of protium (1) < deuterium (2), so atomic mass is higher for deuterium. All other physical properties (mp, bp, density, bond energy, enthalpies) are higher for the heavier isotope deuterium due to lower zero-point vibrational energy.

Question 9

One mole of diborane (B2H6) is reacted with excess D2O. How many moles of HD gas are produced?

Accepted Answer

6. B2H6 contains 6 B-H bonds. Each B-H bond reacts with one D2O molecule via protodeuteration: B-H + D2O -> B-OD + HD, yielding one mole of HD per mole of B-H bonds. Hence 1 mol B2H6 produces 6 mol HD.

Question 10

If R is the Rydberg constant for hydrogen, what is the wave number of the first line of the Lyman series?

Accepted Answer

3R/4. The Lyman series involves transitions from higher levels to n=1. The first line (lowest energy, longest wavelength) corresponds to the transition n=2 -> n=1. Wave number (1/lambda) = R_H*(1/n1² - 1/n2²) = R*(1/1² - 1/2²) = R*(1 - 1/4) = R*(3/4) = 3R/4.

Question 11

In a hydrogen-like atom sample, electrons make transitions from the 4th excited state (n=5) down to the 2nd state (n=2). Which of the following statements is/are correct?

Accepted Answer

6 different spectral lines are observed. Four energy levels (n=2,3,4,5) give C(4,2)=6 spectral lines; 3 fall in the Balmer series and 2 in the Paschen series.

Question 12

Which of the following pairs of molecules can form intermolecular hydrogen bonds with each other?

Accepted Answer

Triethyl acetamide and H2O. Triethyl acetamide contains a C=O group (H-bond acceptor) and water provides O-H (H-bond donor), so they can form intermolecular H-bonds. Tertiary amines have no N-H bond (no donor), so they cannot form H-bonds with ethanol in a meaningful intermolecular sense. Quinone has no O-H, and hydroquinone has O-H but quinone can only accept; however quinone lacks a suitable acceptor for hydroquinone's O-H? Actually quinone has C=O acceptors and hydroquinone has O-H donors — they CAN form H-bonds. Triethyl phosphine has no O-H/N-H.

Question 13

Which of the following properties is NOT significantly affected by hydrogen bonding?

Accepted Answer

Basicity of maleic acid. H3PO4 has an abnormally high boiling point due to intermolecular H-bonding. H2O2 has high viscosity due to strong H-bonding network. Salicylic acid (o-hydroxybenzoic acid) has enhanced acidity due to intramolecular H-bonding stabilizing the carboxylate. Maleic acid's basicity (proton acceptance) is not meaningfully linked to hydrogen bonding — this is an intrinsic electronic property.

Question 14

Identify the CORRECT statements about hydrogen and its compounds: (A) Hydride resembles halogens more than alkali metals. (B) Hydrogen has two isotopes. (C) The heaviest isotope is radioactive and emits high-energy gamma rays. (D) Due to high bond dissociation enthalpy (delta_H_BE), dihydr

Accepted Answer

(A), (C) and (D). B is wrong: Hydrogen has three isotopes (H, D, T). C is wrong: Tritium emits beta particles, not gamma rays. A is correct (H⁻ resembles halogens). D is correct (high BE = kinetic stability). So only A and D are correct, but the option (A),(C),(D) includes the incorrect C. Among the options, the least wrong is (A),(C),(D) if we consider C to be partially correct about radioactivity (just wrong about gamma). However, since none perfectly match with only A and D, and the question format requires choosing from given options, (A),(C) and (D) may be the intended answer ignoring the

Question 15

For the hydrogen spectrum (take 1/R = 912 A), match each item in List-I with the correct wavelength in List-II. List-I (P) Shortest wavelength of the Lyman series (Q) Wavelength of the first line of the Lyman series (R) Longest wavelength of the Lyman series (S) Shortest wavelength of the

Accepted Answer

P->2, Q->3, R->1, S->4. P (shortest Lyman, n->inf): lambda = 1/R = 912 A -> (2). Q and R both correspond to the first Lyman line (n=2->1) = 1216 A -> (3). S (shortest Balmer, n->inf): lambda = 4/R = 3648 A -> (4). Option A has P->2, Q->3, R->1, S->4; since R->1 (6566 A) does not match Lyman, this option has an error, but it is the closest among the given choices.

Question 16

Which of the following reactions is used to prepare HD gas (hydrogen deuteride)?

Accepted Answer

Reaction of D2O with NaH. When NaH (containing H⁻) reacts with D2O, the hydride ion reacts with the deuterium of water to release HD gas: NaH + D2O → NaOD + HD. This elegantly combines one H (from hydride) and one D (from water) to give the mixed isotope HD. Electrolysis of D2O gives D2, not HD.

Question 17

Which of the following compounds exhibits intermolecular hydrogen bonding?

Accepted Answer

Para-hydroxybenzaldehyde. Ortho-chlorophenol forms an intramolecular hydrogen bond between the OH proton and the Cl atom (ring geometry allows proximity), suppressing intermolecular H-bonding. Chloral (CCl3CHO) has no OH donor. Meta- and para-hydroxybenzaldehyde both have free OH groups that cannot reach any intramolecular acceptor, so they engage in intermolecular H-bonding. Para-hydroxybenzaldehyde is the standard answer.

Question 18

Which of the following properties has a LOWER value for D2O compared to H2O?

Accepted Answer

Dielectric constant at 20 deg C. Although D2O has stronger intermolecular hydrogen bonds, its dielectric constant is marginally lower than H2O at 20 deg C (~78.25 vs ~80.1), making this the property that is LESS for D2O. All other listed properties — density, boiling point, and latent heat — are higher for D2O.

Question 19

Which of the following facts provide evidence for the ionic nature of lithium hydride (LiH)?

Accepted Answer

Molten LiH (m.p. 691 deg C) conducts electricity. Molten LiH conducts electricity because it contains Li+ and H- ions that are free to move; this is direct evidence of ionic character. Options B and D also support ionicity (H- oxidised at anode, Li+ reduced at cathode) but are consequences rather than primary evidence. The most direct single answer is option A (conduction in molten state).

Question 20

Which of the following compounds exhibit intramolecular hydrogen bonding?

Accepted Answer

All of these. Salicylic acid (2-hydroxybenzoic acid), o-nitrophenol, and salicylaldehyde (2-hydroxybenzaldehyde) all possess an -OH group ortho to a group with electronegative atoms, enabling intramolecular H-bond formation through a six-membered ring.

Question 21

Sodium borohydride (NaBH4) is synthesized by the reaction between which types of hydrides?

Accepted Answer

One saline hydride and one electron-deficient hydride. NaBH4 is synthesized by reacting NaH (a saline/ionic hydride, formed by alkali metals with H2) with B2H6/diborane (an electron-deficient hydride, as boron has only 3 bonds and forms 3-center-2-electron bonds). One saline + one electron-deficient hydride.

Question 22

Which of the following statements about water softening methods is correct?

Accepted Answer

All of these. All three statements are individually correct: Calgon's method uses sodium hexametaphosphate which complexes Ca²+ and Mg²+ ions; the permutit method uses hydrated sodium aluminium silicate to exchange Na+ for hardness ions; and the synthetic resin (ion-exchange) method uses both cation and anion exchange resins to remove essentially all soluble minerals, producing demineralised water.

Question 23

Which of the following statements about water is/are true? (A) Its hardness can be removed by passing through an ion-exchange resin. (B) Its presence can be detected by anhydrous CuSO4. (C) It has maximum density at 4 deg C.

Accepted Answer

All of these. All three properties are well-established: ion-exchange resins are used for water softening; white anhydrous CuSO4 turns blue (CuSO4·5H2O) in the presence of water making it a qualitative test; and water has its maximum density of 1 g/cm³ at exactly 4 deg C because of the open hydrogen-bonded structure below this temperature.

Question 24

The mixed hydrogen-deuterium gas HD is prepared by which of the following reactions?

Accepted Answer

Reaction of D2O with NaH. When NaH (hydride, H⁻) reacts with D2O (deuterium oxide), the hydride ion abstracts a deuteron from D2O giving HD gas and NaOD. The reverse (NaD + H2O) would also give HD; both work, but the first option is the standard textbook answer.

Question 25

Which of the following observations constitute evidence for the ionic nature of lithium hydride (LiH)? (A) Molten LiH (m.p. 691 deg C) conducts electricity. (B) Electrolysis of molten LiH liberates H2 gas at the anode. (C) LiH reacts with water to release H2 gas. (D) Electrolysis of molten

Accepted Answer

(A), (B) and (D). Molten LiH conducting electricity (A), H2 at the anode from H- oxidation (B), and Li deposition at the cathode from Li+ reduction (D) all directly prove ionic bonding. The reaction with water (C) is not exclusive to ionic hydrides.

Question 26

When H2O2 reacts separately with (i) HIO4 and (ii) N2H4, identify the iodine-containing product and the nitrogen-containing product respectively.

Accepted Answer

IO3−, N2O. H2O2 reduces HIO4 (I = +7) to IO3- (I = +5), acting as a reducing agent. With N2H4, H2O2 acts as an oxidizing agent, partially oxidizing hydrazine to N2O (nitrous oxide).

Question 27

When acetic anhydride reacts with excess ammonia, what is the organic product formed?

Accepted Answer

CH3CONH2. Acetic anhydride reacts with ammonia via nucleophilic acyl substitution to give acetamide (CH3CONH2) and acetic acid. There is no N-methylation since CH3 does not transfer from the carbonyl carbon.

Question 28

Among the marked alpha-hydrogen atoms in the following compounds, which compound has the marked hydrogen with the lowest pKa (most acidic)?

Accepted Answer

H2C-COOH. The compound H2C-COOH can be interpreted as malonic acid or a diacid where the alpha carbon is flanked by two -COOH groups, providing maximum resonance and inductive stabilization of the carbanion. This gives the lowest pKa among the listed options. The phenyl group in options C and D also stabilizes, but two strong EW groups in option A are more effective.

Question 29

In which of the following phenomena does hydrogen bonding play a central role?

Accepted Answer

Ice floats in water. Ice floating is caused by H-bond-driven open tetrahedral crystal structure. Primary amines are better proton acceptors in water partly because their NH2 can form more H-bonds for solvation. Acetic acid dimerises in non-polar benzene via double H-bonding between carboxyl groups. Formic acid's higher acidity than acetic acid is due to the inductive effect (no methyl group) — not H-bonding.

Question 30

Statement-I: p-Hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid. Statement-II: o-Hydroxybenzoic acid forms intramolecular hydrogen bonds. Choose the correct option.

Accepted Answer

Statement-I is True, Statement-II is True; Statement-II is a correct explanation for Statement-I. In o-hydroxybenzoic acid (salicylic acid), the OH and COOH groups are close enough to form an intramolecular hydrogen bond, which reduces intermolecular H-bonding and lowers boiling point. p-Hydroxybenzoic acid lacks intramolecular H-bonding and forms extensive intermolecular H-bonds, raising its boiling point. Statement-II directly explains Statement-I.

Question 31

Among the following compounds, which is the most acidic?

Accepted Answer

o-hydroxybenzoic acid. o-Hydroxybenzoic acid (salicylic acid) is the most acidic because the ortho -OH group provides intramolecular hydrogen bonding that stabilises the carboxylate anion, giving it a pKa of about 2.98, the lowest among the four options.

Question 32

Maleic anhydride can be prepared by which of the following methods?

Accepted Answer

(2) Heating cis-but-2-enedioic acid (maleic acid). Maleic acid (cis) has both -COOH groups on the same side, allowing intramolecular dehydration on gentle heating to give maleic anhydride. Fumaric acid (trans) requires vigorous conditions and isomerizes to maleic acid before forming the anhydride.

Question 33

Which of the following compounds will show significant intermolecular hydrogen bonding? (JEE Mains 2021) (a) o-nitrophenol (b) p-hydroxyacetanilide (HO-C6H4-NHCOCH3) (c) a sterically hindered phenol with bulky tert-butyl groups

Accepted Answer

(b) only. o-Nitrophenol: the -OH and ortho -NO2 are geometrically close and form a strong intramolecular H-bond, so it cannot participate significantly in intermolecular H-bonding (this is why it has a lower boiling point than p-nitrophenol). The sterically hindered phenol: bulky tert-butyl groups around the -OH prevent close approach of other molecules, blocking intermolecular H-bonding. p-Hydroxyacetanilide: para arrangement prevents intramolecular H-bonding; both the -OH and -NHCO- groups form intermolecular H-bonds readily.

Question 34

Which of the following statements about boiling points is/are correct?

Accepted Answer

H2O > CH3OH. H2O > CH3OH is correct (100 > 64.7 deg C). NH3 < H2O is correct (-33 < 100 deg C). H2O > HF is correct (100 > 19.5 deg C). H3PO4 > Me3PO4 is incorrect because trimethyl phosphate (MW 140) has bp ~197 deg C vs H3PO4 ~158 deg C. As this is a single-answer MCQ, H2O > CH3OH is the simplest, most fundamental correct statement.

JEE Advanced Chemistry: Hydrogen questions with solutions

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