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BITSAT Chemistry Test - 3 - JEE MCQ


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30 Questions MCQ Test - BITSAT Chemistry Test - 3

BITSAT Chemistry Test - 3 for JEE 2024 is part of JEE preparation. The BITSAT Chemistry Test - 3 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Chemistry Test - 3 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Chemistry Test - 3 below.
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BITSAT Chemistry Test - 3 - Question 1

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

Detailed Solution for BITSAT Chemistry Test - 3 - Question 1

KIO4 + H2O2 → KIO3 + H2O + O2 In this reaction H2O2 act as reducing agent
NH2OH + 3H2O2 → HNO3 + 4H2O In this reaction H2O2 act as oxidising agent.

BITSAT Chemistry Test - 3 - Question 2

Which of the following is correct about heavy water?

Detailed Solution for BITSAT Chemistry Test - 3 - Question 2

Heavy water is formed by the combination of heavier isotope 1Hor D with oxygen.

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BITSAT Chemistry Test - 3 - Question 3

The products of the reaction Na2 O2 + dil . H2 SO4 are

Detailed Solution for BITSAT Chemistry Test - 3 - Question 3

When sodium peroxide is treated with dilute sulphuric acid, we get sodium sulfate and hydrogen peroxide as a product. It is the preparation method of hydrogen peroxide. It is an example of acid-base reaction.
Na2O2+H2SO4→Na2SO4+H2O2

BITSAT Chemistry Test - 3 - Question 4

There are three diatomic species with each having a bond order of three. What is true regarding them?
(a) All of them can be isoelectronic.
(b) They are isostructural.
(c) They may include cationic/anionic species.
(d) They may include neutral species.

Detailed Solution for BITSAT Chemistry Test - 3 - Question 4

NO + , CN - and N2 all have a bond order = 3.
These are isoelectronic means have same number of electrons, in this case have 14 e - and isostructural as they are all linear in shape. These can be ions or neutral species.
All the options are correct.

BITSAT Chemistry Test - 3 - Question 5

Chelation can occur in:

Detailed Solution for BITSAT Chemistry Test - 3 - Question 5

Complex with bidentate ligand form chelate ring and chelation is caused by intramolecular hydrogen bonding also.i.e. (1) Ethylenediamine ligand form ring when it is attached to metal in a complex .(2) Chelation takes place in ortho nitro phenol due to intra molecular H bonding.

BITSAT Chemistry Test - 3 - Question 6

HCl gas is covalent and NaCl is an ionic compound. This is because:

Detailed Solution for BITSAT Chemistry Test - 3 - Question 6

HCl g is covalent whereas NaCl is an ionic solid due to small difference in their electronegativities.
In HCl , the difference in electronegativity of H and Cl atoms is less than 2.10 , whereas in NaCl , the difference in electronegativities of Na and Cl atoms is more than 2.10 .
Electronegativity of Cl = 3.1
Electronegativity of H = 2.2
Electronegativity of Na = 0 .93
Electronegativity difference between H and Cl = 0 . 95
Electronegativity difference between Na and Cl = 2.22

BITSAT Chemistry Test - 3 - Question 7

The correct order of decreasing s-character in hybrid orbitals among the following molecules

Detailed Solution for BITSAT Chemistry Test - 3 - Question 7


BITSAT Chemistry Test - 3 - Question 8

Which one of the following halides has the highest melting point?

Detailed Solution for BITSAT Chemistry Test - 3 - Question 8

NaF has the highest melting point because F- has the smallest size, lattice energy is highest.

Lattice Energy ∝ Melting point

BITSAT Chemistry Test - 3 - Question 9

What is the formula of a metal phosphate, if formula of its chloride is MCl 2 ?

Detailed Solution for BITSAT Chemistry Test - 3 - Question 9

The given formula of chloride is MCl 2.
Since chloride has uni-negative charge, the valency of metal will be 2.
The phosphate ion PO4 - 3 has a valency of 3 .
Hence, by cross valency method for ions of M+2 and PO4-3 , the metal phosphate formula will be M3 ( PO4 )2 .

BITSAT Chemistry Test - 3 - Question 10

The AsF5 molecule is trigonal bipyramidal. The hybrid orbitals used by the As atoms for bonding are

Detailed Solution for BITSAT Chemistry Test - 3 - Question 10

As F5 has sp3d  hybridization. In sp3d hybridization, it is  dz2 orbitals which takes part

BITSAT Chemistry Test - 3 - Question 11

What is formed when calcium carbide reacts with heavy water?

Detailed Solution for BITSAT Chemistry Test - 3 - Question 11

When heavy water reacts with calcium carbide, the products formed are deutero acetylene (C2D2​) and calcium deuteroxide (Ca(OD)2 
CaC2+2D2O→Ca(OD)2+C2D2
These videos will help you learn the above concepts

BITSAT Chemistry Test - 3 - Question 12

Determine which of the following reaction would produce hydrogen gas (H2).

Detailed Solution for BITSAT Chemistry Test - 3 - Question 12

Mn + extremely dilute HNO3 (2%)  → Mn(NO3) + H2 ↑⏐

BITSAT Chemistry Test - 3 - Question 13

Which one of the following reactions does not form gaseous product ?

Detailed Solution for BITSAT Chemistry Test - 3 - Question 13

BITSAT Chemistry Test - 3 - Question 14

Dihydrogen of high purity (>99.95 %) is obtained through :

Detailed Solution for BITSAT Chemistry Test - 3 - Question 14

Dihydrogen of high degree of purity (>99.95%) is obtained by the electrolysis of warm aqueous barium hydroxide solution between nickel electrodes.

BITSAT Chemistry Test - 3 - Question 15

On hydrolysis, peroxodisulphate produce-

Detailed Solution for BITSAT Chemistry Test - 3 - Question 15

Peroxodisulphate obtained by electrolytic oxidation of acidified sulphate solutions at high current density, on hydrolysis yields hydrogen peroxide.

BITSAT Chemistry Test - 3 - Question 16

Which of the following is the true structure of H2 O2?

Detailed Solution for BITSAT Chemistry Test - 3 - Question 16

Hydrogen peroxide has a non–planar structure. The molecular dimensions in the gas phase and solid phase are shown below.


H2O2  structure in solid phase at 110 K, dihedral angle is  90.2° .

BITSAT Chemistry Test - 3 - Question 17

Hydrogen peroxide when added to a solution of potassium permanganate acidified with sulphuric acid

Detailed Solution for BITSAT Chemistry Test - 3 - Question 17


In this reaction hydrogen peroxide acts as a reducing agent and it reduces KMnO4 to Mn2+ ions.
Potassium permanganate is used as an oxidizing agent to determine the hydrogen peroxide amount in a solution. Hydrogen peroxide reduces the permanganate to a colorless product. 

BITSAT Chemistry Test - 3 - Question 18

Mg and Li are similar in their properties due to

Detailed Solution for BITSAT Chemistry Test - 3 - Question 18

Li+ and Mg2+ ions have similar polarising power or ionic potential and therefore, have similar properties. This type of relationship of the first element of a group with the second element of the next group is known as diagonal relationship.

BITSAT Chemistry Test - 3 - Question 19

How will you separate mixture of two gases?

Detailed Solution for BITSAT Chemistry Test - 3 - Question 19

A mixture of two gases can be separated by fractional distillation technique. Fractional distillation is a process by which components in a chemical mixture are separated into different parts called fractions according to their different boiling points.

BITSAT Chemistry Test - 3 - Question 20

 “The addition of unsymmetrical reagents to unsymmetrical alkenes occurs in such a way that the negative part of the addendum goes to that carbon atom of the double bond which carries lesser number of hydrogen atoms” is called by

Detailed Solution for BITSAT Chemistry Test - 3 - Question 20

According to Markownikoff’s rule, the negative part of the unsymmetrical reagent adds to less hydrogenated (more substituted) carbon atom of the double bond, 

BITSAT Chemistry Test - 3 - Question 21

Which of the following gas mixture is used by the divers inside the sea?

Detailed Solution for BITSAT Chemistry Test - 3 - Question 21

Helium is not soluble in blood even under high pressure, a mixture of 80% helium and 20% oxygen is used instead of ordinary air, by sea divers for respiration.

BITSAT Chemistry Test - 3 - Question 22

Which of the following cuts ultraviolet rays?

Detailed Solution for BITSAT Chemistry Test - 3 - Question 22

Crooke's glass is a special type of glass containing cerium oxide. It does not allow the passage of ultraviolet rays and is used for making lenses.

BITSAT Chemistry Test - 3 - Question 23

Acetone is mixed with bleaching powder to give

Detailed Solution for BITSAT Chemistry Test - 3 - Question 23

Acetone forms chloroform when heated with bleaching powder.

BITSAT Chemistry Test - 3 - Question 24

4.4 g of CO2 contains how many litres of CO2 at STP?

Detailed Solution for BITSAT Chemistry Test - 3 - Question 24

At STP,44 g of CO2 weighs = 22.4 litre

BITSAT Chemistry Test - 3 - Question 25

Octane number can be changed by 

Detailed Solution for BITSAT Chemistry Test - 3 - Question 25

The octane number of fuel can be improved by increasing the percentage of branched chain alkanes, alkenes and aromatic hydrocarbons. Thus octane number can be changed by isomerisation (reforming), alkylation and aromatisation (cyclisation) etc.

BITSAT Chemistry Test - 3 - Question 26

Oxygen molecule is

Detailed Solution for BITSAT Chemistry Test - 3 - Question 26

BITSAT Chemistry Test - 3 - Question 27

Triple point of water is

Detailed Solution for BITSAT Chemistry Test - 3 - Question 27

The triple point of any substance is that temperature and pressure at which the material can coexist in all three phases (solid, liquid and gas) in equilibrium. Specifically, the triple point of water is 273.16 K at 611.2 Pa.

BITSAT Chemistry Test - 3 - Question 28

Lanthanides and actinides resemble in

Detailed Solution for BITSAT Chemistry Test - 3 - Question 28

Lanthanides and actinides resemble in electronic configuration:

(n−2)f1−14(n−1)d1−10ns1−2

BITSAT Chemistry Test - 3 - Question 29

Which of these have no unit?

Detailed Solution for BITSAT Chemistry Test - 3 - Question 29

Electronegativity is the property of a bonded atom. The relative tendency of an atom to attract the shared pair of electron towards itself is called electro negativity.

BITSAT Chemistry Test - 3 - Question 30

Photoelectric effect is maximum in

Detailed Solution for BITSAT Chemistry Test - 3 - Question 30

Group I elements are so highly electropositive that they emit electrons even when exposed to light (photoelectric effect) and this character increases on moving down the group from lithium towards cesium.

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