s-Block Elements ‌–‌ ‌JEE‌ ‌Advanced‌ ‌Previous‌ ‌Year‌ ‌Questions with Solutions

s-Block elements JEE Advanced Questions with Solutions given here is a collection of previous year questions that will help every JEE aspirant to gain important insights about the question types, difficulty level and important topics. The detailed and correct solutions that we have provided along with the questions will further enable students to prepare in the right manner. We have used a very simple approach that will help each learner to not only understand the concepts covered in the questions in a clear manner but they will be able to revise the topics before the final exams.

JEE aspirants will benefit more as they can download the s-Block elements JEE Advanced Questions with Solutions PDF for free. They can use this study material to further make a proper self-assessment by solving the questions given in this chapter. As they engage in, solving the questions they will get a better grip on the topic and also develop better problem-solving skills. All in all, they will be ready to face any type of questions that may be asked in the JEE Advanced entrance exam.

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JEE Advanced Previous Year Questions on s-Block Elements

Question 1. Which of the following liberates O₂ upon hydrolysis?

A. Pb₃O₄

B. KO₂

C. Na₂O₂

D. Li₂O₂

Solution: (B)

The reactions are as follows;

A) Pb₃O₄ + H₂O → No reaction (due to insoluble in water)

B) Superoxides liberate oxygen with water.

2KO₂ + 2H₂O → 2KOH + H₂O₂ + O₂

C) Na₂O₂ + 2H₂O → 2NaOH + H₂O₂

D) Li₂O₂ + 2H₂O → 2LiOH + H₂O₂

Question 2. The compound (s) formed upon combustion of sodium metal in excess air is (are)

A. Na₂O₂

B. Na₂O

C. NaO₂

D. NaOH

Solution: (A and B)

Upon combustion,

4Na + O₂ ⟶ 2Na₂O

And traces of peroxide is also formed.

Na + O₂ ⟶ Na₂O₂

The superoxide and NaOH are not formed.

Question 3. Read the following questions and answer as per the direction given below:

Statement I: Alkali metals dissolve in liquid ammonia to give blue solutions.

Statement-II: Alkali metals in liquid ammonia give solvated species of the type [M(NH₃)n] + (M = alkali metals).

A. Statement I is true; Statement II is true; Statement II is a correct explanation of Statement I.

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

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

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

Solution: (B)

Na⁺ + (x + y)NH₃ → [Na(NH₃) x]⁺ + [e(NH₃)y]^–

Ammoniated electrons

Alkali metals dissolve in liquid ammonia giving deep blue solutions which are conducting in nature. The blue colour of the solution is due to ammoniated electrons which absorb energy in the visible region of light. The second statement describes the solvated cations (positively charged species) whereas the colour is due to the solvated electrons. Do note however that both the statements are accurate.

Question 4. A sodium salt when treated with MgCl₂ gives white precipitate only on heating. The anion is:

A. SO₄^2-

B. HCO₃^–

C. CO₃^2-

D. NO₃^–

Solution: (B)

It must be a bicarbonate ion because first magnesium bicarbonate is formed which is soluble. Then on heating, magnesium carbonate is formed which is insoluble and forms a precipitate.

The anion is HCO₃^–.

Question 5. The species that do not contain peroxide ions, is

A. PbO₂

B. H₂O₂

C. SrO₂

D. BaO₂

Solution: (A)

PbO₂ is the Lead (IV) oxide. It is not a peroxide whereas the H₂O₂, SrO₂, and BaO₂ are all peroxides. O₂^-2 is the peroxide anion.

Question 6. The set representing the correct order of first ionisation potential is

A. K > Na > Li

B. Be > Mg > Ca

C. B > C > N

D. Ge > Si > C

Solution: (B)

The ionisation energy decreases down the group with an increase in size. These elements ( Be > Mg > Ca ) are 2nd group elements and the order correctly reflects the trend in ionization energy.

Question 7. Give reason for the following:

BeCl₂ can be easily hydrolysed.

Solution:

BeCl₂ is electron-deficient species so it can be easily hydrolysed.

BeCl₂ + 4H₂O → [Be(H₂O)₄]Cl₂

​Question 8. The following compounds have been arranged in order of their increasing thermal stabilities. Identify the correct order.

K₂CO₃(I), MgCO₃(II), CaCO₃(III), BeCO₃(IV)

A. I < II< III< IV

B. IV < II< III< I

C. IV < II< I< III

D. II< IV < III< I

Solution: (B)

On increasing the size of alkaline earth metal ions, stability of their carbonate salts increases due to increasing polarity, Alkali metal carbonates are more stable than carbonates of alkaline earth metals due to their higher polar characters, Thus, the correct order of stability is IV < II< III< I.

Question 9. A substance absorbs CO₂ and violently reacts with water. The substance is:

A. CaCO₃

B. CaO

C. H₂SO₄

D. ZnO

Solution: (B)

The reactions are as follows.

Ca + CO₂ → CaCO₃

CaO + H₂O → Ca (OH)₂

Question 10. MgSO₄ on reaction with NH₄OH and Na₂HPO₄ forms a white crystalline precipitate. What is its formula?

A. Mg(NH₄)PO₄

B. Mg₃(PO₄)₂

C. MgCl₂ . MgSO₄

D. MgSO₄

Solution: (A)

There is a formation of a white precipitate of magnesium ammonium phosphate when MgSO₄ reacts with NH₄OH and Na₂HPO₄.

MgSO₄ + Na₂HPO₄ + NH₄OH → Mg(NH₄)PO₄↓ + Na₂SO₄ + H₂O

White ppt.

Also Read:-

s-block Elements – Important Topics

s-block Elements – Important Questions