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Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - JEE MCQ


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23 Questions MCQ Test 35 Years Chapter wise Previous Year Solved Papers for JEE - Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced for JEE 2024 is part of 35 Years Chapter wise Previous Year Solved Papers for JEE preparation. The Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced questions and answers have been prepared according to the JEE exam syllabus.The Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced below.
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Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 1

A substance absorbs CO2 and voilently reacts with water.The substance is 

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 1

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 2

HCl is added to following oxides. Which one would give H2O2?

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 2

BaO2.8H2O + H2SO4 → BaSO4 + H2O2 + 8H2O

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 3

Calcium is obtained by 

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 3

Ca is obtained by electrolysis of molten mixture of CaCl2 mixed with CaF2.

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 4

A solution of sodium metal in liquid ammonia is strongly reducing due to the presence of

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 4

The free ammoniated electrons make the solution of Na in liquid NH3 a very powerful reducing agent.
NOTE : The ammonical solution of an alkali metal is rather favoured as a reducing agent than its aqueous solution because in aqueous solution the alkali metal being highly electropositive evolves hydrogen from water (thus H2O acts as an oxidisng agent) while its solution in ammonia is quite stable, provided no catalyst (transition metal) is present.

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 5

Heavy water is

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 5

Heavy water is D2O, deuterium oxide.

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 6

The hydration energy of Mg++ is larger than that of :

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 6

Na+ ion has larger size than Mg2+ ion and hence hydration energy of Mg2+ is larger than that of Na+.

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 7

The oxide that gives hydrogen peroxide on treatment with a dilute acid is :

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 7

Na2O2 + H2SO4 (20% ice cold) → Na2SO4 + H2O2

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 8

Molecular formula of Glauber ’s salt is :

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 8

Glauber ’s salt is Na2SO4.10 H2O.

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 9

Hydrogen gas will not reduce : 

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 9

NOTE : The more electropositive metal will not be reduced by hydrogen.
Among given choices only Al is more electropositive than hydrogen.
∴ It will not be reduced by hydrogen.

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 10

The pair of compounds which can not exist together in solution is : 

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 10

NOTE : Acidic and basic salts cannot exist together.
Since NaHCO3 is an acid salt of H2CO3, it reacts with NaOH to form Na2CO3 and H2O.
NaHCO3 + NaOH → Na2CO3 + H2O

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 11

The metallic lustre exhibited by sodium is explained by

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 11

Oscillation of loose electrons.

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 12

The volume strength of 1.5 N H2O2 solution is

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 12

Volume strength = Normality  × 5.6 = 1.5 × 5.6 = 8.4 L

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 13

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)

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 13

The increasing thermal stability is

NOTE : Increasing size of cation decreases its polarization ability towards carbonate, making the compound more stable.

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 14

The set representing the correct order of first ionization potential is

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 14

In going from top to bottom in a group, the first ionization potential decreases, thus
Be > Mg > Ca

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 15

A sodium salt on treatment with MgCl2 gives white precipitate only on heating. The anion of the sodium salt is

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 15

Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 16

Hydrogen peroxide in its reaction with KIO4 and NH2OH respectively, is acting as a

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 16

KIO4 + H2O2 → KIO3 + H2O + O2
Thus H2O2 is acting as a reducing agent 2NH2OH + H2O2 → N2 + 4H2O
Here H2O2 is acting as an oxidising agent

*Multiple options can be correct
Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 17

When zeolite, which is hydrated sodium aluminium silicate, is treated with hard water the sodium ions are exchanged with

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 17

*Multiple options can be correct
Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 18

The species that do not contain peroxide ions are

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 18

Only H2O2 (hydrogen peroxide) and BaO2 (Barium peroxide) contain peroxide ions. So (a) and (c) are the correct choices.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 19

Highly pure dilute solution of sodium in liquid ammonia

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 19

A small piece of sodium is cut to expose a fresh surface. The sodium is dropped into liquid ammonia at a temperature of approximately -33 degrees Celsius. Some of the sodium dissolves, forming sodium cations surrounded by ammonia molecules and electrons surrounded by ammonia molecules. The solvated electrons give the blue color to the solution.
Because of the mobility of the electrons, the solution is a good electrical conductor. Bubbles of hydrogen gas are formed by a second reaction that also produces sodium amide. More concentrated solutions appear bronze-colored and have a conductivity similar to metals.
2Na+2NH​→ 2NaNH2+H2.
Hence options A,B,C & D are correct

*Multiple options can be correct
Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 20

The species present in solution when CO2 is dissolved in water are

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 20

*Multiple options can be correct
Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 21

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

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 21

*Multiple options can be correct
Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 22

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

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 22

*Multiple options can be correct
Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 23

The reagent(s) used for softening the temporary hardness of water is (are)

Detailed Solution for Test: MCQs (One or More Correct Option): The s-Block Elements | JEE Advanced - Question 23

Temporary hardness is due to bicarbonates of calcium and magnesium. Temporary hardness can be removed by Clark’s process, which involves the addition of slaked lime, Ca(OH)2. Washing soda (Na2CO3) removes both the temporary and permanent hardness by converting soluble calcium and magnesium compounds into insoluble carbonates.

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