p & s- Block Element JEE Main Questions 2026: Important Practice Questions with PYQs
p & s- Block Element JEE Main Questions 2026 is an important section and holds a weightage of 9.9% in the exam, which is equal to 2-3 questions equivalent to 10-12 marks. Some of the important p & s- Block Element JEE Main Questions have been provided in this article with their solutions.
p & s- Block Element JEE Main Questions 2026 hold a weightage of 9.9% in the exam, which is equivalent to 2 to 3 questions. This is an important section of the JEE Main Chemistry syllabus. The important topics from the p & s- Block chapter are General Properties, Reactions and Compounds, Group 13: Boron Family, Group 14: Carbon Family, etc. Candidates need to prioritise p & s- Block Element chapter from JEE Main syllabus 2026 as it is crucial and every year around 10 to 12 marks are asked. Therefore, aspirants must possess strong JEE Main 2026 preparation strategies to score well in the exams. It is recommended that every aspirant preparing for JEE Main 2026 January and April sessions, should practice the previous years' questions along with the sample questions to get a strong foundation of the topic. In this article, candidates will get a detailed idea of the Important Practice Questions with PYQs related to p & s- Block Element topic of JEE Main Chemistry section. The January session will be conducted from January 21 - 30, 2026, and the April session will be conducted from April 1 - 10, 2026.
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p & s- Block Element JEE Main Questions 2026 PDF
The table below contains the JEE Main Questions 2026 PDF from p & s- Block Element topic.
Practice p & s- Block Element JEE Main Questions 2026
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1. The correct statement for the molecule, CsI3, is :
(1) it contains Cs3+ and I– ions
(2) it contains Cs+, I– and lattice I2 molecule
(3) it is a covalent molecule
(4) it contains Cs+ and I3– ions.
Solution:
CsI3 -> Cs+ + I3–
CsI3 contains Cs+ and I3–.
Hence option (4) is the answer.
Also Read:
JEE Main 2026 Preparation and Study Time Table for 60 Days (2 Months)
2. Which of the following statements about Na2O2 is not correct?
(1) Na2O2 oxidises Cr3+ to CrO42- in acid medium
(2) It is diamagnetic in nature
(3) It is the superoxide of sodium
(4) It is a derivative of H2O2
Solution:
Na2O2 is sodium peroxide. It is not superoxide.
Hence option (3) is the answer.
3. The metal that cannot be obtained by electrolysis of an aqueous solution of its salt is
1) Cr
2) Ag
3) Ca
4) Cu
Solution:
Ca cannot be obtained by the electrolysis of an aqueous solution of its salt.
Hence option (3) is the answer.
4. Based on lattice energy and other considerations, which one of the following alkali metal chlorides is expected to have the highest melting point?
(1) RbCl
(2) LiCl
(3) KCl
(4) NaCl
Solution:
NaCl has the highest lattice energy and thus the highest melting point.
Hence option (4) is the answer.
5. A metal M on heating in nitrogen gas gives Y. Y on treatment with H2O gives a colourless gas which when passed through CuSO4 solution gives a blue colour, Y is :
(1) NH3
(2) MgO
(3) Mg3N2
(4) Mg(NO3)2
Solution:
Magnesium, when heated with nitrogen, forms magnesium nitride.
3Mg + N2 → Mg3N2
Magnesium nitride reacts with water to form ammonia and magnesium hydroxide.
Mg3N2 → Mg(OH)2 + NH3
When ammonia gas is passed through CuSO4 solution, a blue colour is obtained.
Hence option (3) is the answer.
6. Which one of the following will react most vigorously with water?
(1) Li
(2) K
(3) Rb
(4) Na
Solution:
The reactivity of s-block elements increases down a group. So Rubidium will react most vigorously with water.
Hence option (3) is the answer.
7. The hydrogen bomb is based on the principle of
(1) Nuclear fission
(2) Natural radioactivity
(3) Nuclear fusion
(4) Artificial radioactivity
Solution:
Hydrogen bomb is based on the fusion of isotopes of hydrogen.
Hence option (3) is the answer.
8. Amongst LiCl, RbCl, BeCl2 and MgCl2 the compounds with the greatest and the least ionic character, respectively are :
(1) RbCl and MgCl2
(2) LiCl and RbCl
(3) MgCl2 and BeCl2
(4) RbCl and BeCl2
Solution:
Because of the smallest cationic size, BeCl2 is the least ionic. Rb+ has the biggest ionic size. So it has the greatest ionic character.
Hence option (4) is the answer.
9. In curing cement plasters water is sprinkled from time to time. This helps in
(1) developing interlocking needle-like crystals of hydrated silicates.
(2) hydrating sand and gravel mixed with cement
(3) converting sand into silicic acid
(4) keeping it cool
Solution:
During the hydration of calcium aluminosilicates, cross-links are developed
Hence option (1) is the answer.
10. K0₂ (potassium superoxide) is used in oxygen cylinders in space and submarines. This is because it
(1) absorbs C02 and increases 02 content
(2) eliminates moisture
(3) absorbs C02
(4) produces ozone
Solution:
2KO, + 2H20 —> 2KOH + H202 + 02. It eliminates moisture.
Hence option (2) is the answer.
11. A metal M readily forms its sulphate MS04 which is water soluble. It forms its oxide MO which becomes inert on heating. It forms an insoluble hydroxide M(OH)2 which is soluble in NaOH solution. Then M is
(1) Mg
(2) Ba
(3) Ca
(4) Be
Solution:
Because of its small size, Beryllium shows anomalous properties with alkaline earth metals.
Hence option (4) is the answer.
12. The substance not likely to contain CaC03 is
(1) calcined gypsum
(2) sea shells
(3) dolomite
(4) a marble statue
Solution:
The composition of gypsum is CaS04. 2H20.
Hence option (1) is the answer.
13. Several blocks of magnesium are fixed to the bottom of a ship to
(1) make the ship lighter
(2) prevent the action of water and salt
(3) prevent puncturing by under-sea rocks
(4) keep away the sharks
Solution:
Magnesium provides cathodic protection and prevents corrosion.
Hence option (2) is the answer.
14. The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+ will be
1) -2.55 eV
2) -5.1 eV
3)-10.2 eV
4) +2.55 eV
Solution:
The ionisation potential of an alkali metal atom and the electron affinity of its monovalent cation are the same in magnitude but with opposite signs.
Hence option (2) is the answer.
15. In Zeolites and synthetic resin method, which will be more efficient in removing permanent hardness of water.(1) Synthetic resin method as it exchanges only cation.
(2) zeolite resin method as it exchanges only cation.
(3) Synthetic resin method as it exchanges only anion.
(4) Synthetic resin is harmful to nature.
Solution:
The synthetic resin method will be more efficient in removing the permanent hardness of water because it exchanges only cation.
Hence option (1) is the answer.
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