Hydrogen JEE Advanced Previous Year Questions with Solutions

Hydrogen JEE Advanced Previous Year Questions with Solutions has been specially prepared and provided here to help JEE aspirants get familiar with the actual entrance examination questions. We have listed down a set of questions from this chapter and students can go through them to understand the type of questions that can appear in the exam, the difficulty level and the format of the question paper. After revising and solving these Hydrogen JEE Advanced Previous Year Questions with Solutions students will further be able to figure out their preparation level and plan things accordingly.

Additionally, the solutions that have been provided will help students get better clarity while solving the papers. As aspirants get more familiar with the question types, it will become very easy for them to enhance their study hours and study productively. Students can access the Hydrogen JEE Advanced Previous Year Questions with Solutions PDF below.

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JEE Advanced Previous Year Questions on Hydrogen

Question 1. Hydrogen peroxide in its reaction with KIO₄ and NH₂OH, respectively is acting as a:

A. Reducing agent, oxidising agent

B. Reducing agent, reducing agent

C. Oxidising agent, oxidising agent

D. Oxidising agent, reducing agent

Solution: (A)

KIO₄ + H₂O₂ → KIO₃ + H₂O + O₂

NH₂OH + 3H₂O₂ → HNO₃ + 4H₂O.

If we look at the given two reactions, we can easily figure out that hydrogen peroxide acts as a reducing agent in its reaction with KlO₄ and s an oxidising agent in its reaction with NH₂OH.

Question 2. Polyphosphates are used as water softening agents because they;

A. Form soluble complexes with anionic species

B. Precipitate anionic species

C. Form soluble complexes with cationic species

D. Precipitate cationic species

Solution: (C)

Polyphosphates are often used as water softening agents because they tend to form soluble complexes with cationic species of hard water.

2CaSO₄ + Na₂[Na₄(PO₃)₆] → 2NaSO₄ + Na₂[Ca₂(PO₃)₆]

soluble complex

Question 3. Heavy water is

A. H₂O₁₈

B. Water obtained by repeated distillation

C. D₂O

D. Water at 4°C

Solution: (C)

Heavy water is an oxide of heavy hydrogen, ₁D²(ie, D₂O).

Question 4. The volume strength of 1.5 N H₂O₂ solution is

A. 8.8

B. 8.4

C. 4.8

D. 5.2

Solution: (B)

Volume strength of hydrogen peroxide is defined as the term which is used to express the concentration of H₂O₂ with respect to the volumes of oxygen gas based on its decomposition to form water and oxygen.

Normality (N) = 1.5

The equivalent weight of H₂O₂ is 17

Strength of H₂O₂ = Normality x Equivalent weight

= 1.5 ×17 = 25.5

2H₂O₂ → 2H₂O + O₂

(2 × 34 g) (22.4 L)

Since 68 grams of H₂O₂ produces 22.4 litres of oxygen at NTP, therefore, 25.5 grams of H₂O₂ will produce;

= 22.4/68 x 25.5 = 8.4 litre of oxygen

Thus, the volume strength of the given H₂O₂ solution is 8.4.

Question 5. Read the following statements and answer the questions as per the options given below.

Statement-I: The alkali metals can form ionic hydrides which contain the hydride ion, H^–.

Statement-II: The alkali metals have low electronegativity, their hydrides conduct electricity when fused and liberate hydrogen at the anode.

A. Statement-I is true; Statement-II is true; Statement-II is the correct explanation for Statement-I

B. Statement-I is true; Statement-II is true; Statement-II is not the correct explanation for Statement-I

C. Statement-I is true; Statement-II is false

D. Statement-I is false; Statement-II is true

Solution: (A)

Alkali metals such as NaH, KH, LiH and CsH can form ionic hydrides which contain the hydride ion, H^–.

Alternatively, alkali metals have low electronegativity. In a molten state, they can exist as ions and conduct electricity.

On electrolysis,

At anode H^– oxidises and liberates hydrogen as:

2H^– − 2e^– → H₂

Question 6. The isotopes of hydrogen are:

A. Tritium and protium only

B. Deuterium and tritium only

C. Protium and deuterum only

D. Protium, deuterium and tritium

Solution: (D)

Hydrogen has three naturally occurring isotopes: ¹H (protium), ²H (deuterium), and ³H (tritium).

Question 7. Which of the following properties of dihydrogen is incorrect?

A. It is colourless, odourless, tasteless

B. It is combustible gas

C. It is lighter than air

D. It is soluble in water

Solution: (B)

Dihydrogen gas is a highly inflammable gas.

Question 8. Which of the following metal gives hydrogen with all of these reactants: water, acids, alkalis?

A. Fe

B. Zn

C. Mg

D. Na

Solution: (D)

Sodium, being the most reactive metal, easily reacts with water, acid and base to liberate

hydrogen. The reactions are as follows:

2Na + 2H₂O → 2NaOH + H₂

2Na + H₂SO₄ → Na₂SO4 + H₂

2Na + H₂CO₃ → Na₂CO₃ + H₂

Question 9. Hydrogen peroxide is …….

A. An oxidising agent

B. A reducing agent

C. Both an oxidising and a reducing agent

D. Neither oxidising nor reducing agent

Solution: (C)

Hydrogen peroxide acts as an oxidising as well as a reducing agent in both acidic and alkaline media.

Question 10. To a 25 ml H₂O₂ solution, an excess of an acidified solution of KI was added. The iodine liberated required 20 ml of 0.3 N Na2S2O3 solution. The volume strength of H₂O₂ solution is:

A. 1.344 g/L

B. 3.244 g/L

C. 5.4 g.L

D. 4.08 g/L

Solution: (A)

25ml H₂O₂ was reacted with an excess of KI

Also, 20ml 0.3N Na₂S₂O₃ was required to liberate I₂

The reaction involved in this liberation is

2KI + H₂O₂ ⟶ I₂ + 2KOH

Let, Normality of 25ml H₂O₂ be xN

∴ From above titration

N₁ = Normality of H₂O₂

V₁ = Volume of H₂O₂

N₂ = Normality of Na₂S₂O₃ or I₂

V₂ = Volume of Na₂S₂O₃ or I₂

⇒ N₁V₁ = N₂V₂

⇒ x × 25 = 0.3 × 20

⇒ x = 0.3 × 20 / 25

= 0.24

∴ Normality of 25 solution = 0.24N

Now, We know Volume strength = Normality × Equivalent weight

The equivalent weight of 25 in terms of oxygen = 5.6L

⇒ Volume strength = 0.24 × 5.6

= 1.344gL^-1