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Chemistry: Topic-wise Test- 7 - NEET MCQ


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30 Questions MCQ Test NEET Mock Test Series 2025 - Chemistry: Topic-wise Test- 7

Chemistry: Topic-wise Test- 7 for NEET 2024 is part of NEET Mock Test Series 2025 preparation. The Chemistry: Topic-wise Test- 7 questions and answers have been prepared according to the NEET exam syllabus.The Chemistry: Topic-wise Test- 7 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Chemistry: Topic-wise Test- 7 below.
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Chemistry: Topic-wise Test- 7 - Question 1

If the kinetic energy of an electron is increased four times, the wavelength of the de-Broglie wave associated with it would become

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 1

The wavelength λ is inversely proportional to the square root of kinetic energy. So if KE is increased 4 times, the wavelength becomes half.

λ∝1/√KE

Hence Option A is the answer.

Chemistry: Topic-wise Test- 7 - Question 2

Calculate the wavelength (in nanometer) associated with a proton moving at 1.0×103ms-1 (Mass of proton = 1.67×10-27kg and h = 6.63×10-34Js)

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 2

Given mp = 1.67×10-27kg
h = 6.63×10-34Js
v = 1.0×103ms-1

We know wavelength λ = h/mv
∴λ = 6.63×10-34/(1.67×10-27 × 1.0×103)
= 0.40×10-10
≈ 0.40nm

Hence, Option (B) is the Correct Answer.

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Chemistry: Topic-wise Test- 7 - Question 3

The maximum number of atomic orbitals associated with a principal quantum number 5 is

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 3

Number of orbitals in a shell = n2 = (5)= 25

Chemistry: Topic-wise Test- 7 - Question 4

Which model describes that there is no change in the energy of electrons as long as they keep revolving in the same energy level and atoms remains stable?

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 4

Bohr Model of atom:

  • An atom is made up of three particles: Electrons, neutrons and protons.
  • The protons and neutrons are located in a small nucleus at the centre of the atom.
  • The electrons revolve rapidly around the nucleus at the centre of the atom.
  • There is a limit to the number of electrons that each energy level can hold.
  • Each energy level is associated with a fixed amount of energy.
  • There is no change in the energy of electrons as long as they keep revolving in the same energy level.

Bohr explained the stability through the concept of revolution of electrons in different energy levels.


The change in the energy of an electron occurs when it jumps from lower to higher energy levels. When it gains energy, it excites from lower to higher and vice versa.
Thus energy is not lost and the atom remains stable.

Chemistry: Topic-wise Test- 7 - Question 5

The number of radial nodes for 3p orbital is:

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 5
  • Number of radial nodes = n - l – 1

where n = principal quantum number, l = azimuthal quantum number

  • For 3p orbital, n = 3 – 1 – 1 = 1
  • Number of radial nodes = 3 – 1 – 1 = 1. 
Chemistry: Topic-wise Test- 7 - Question 6

A sub-shell with n = 6 , l = 2 can accommodate a maximum of

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 6

n = 6, l = 2 means 6d → will have 5 orbitals. 

∴ max 10 electrons can be accommodate as each orbital can have maximum of 2 electrons. 

Chemistry: Topic-wise Test- 7 - Question 7

Thomson’s plum pudding model explained:

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 7
  • An atom consists of a positively charged sphere with electrons filled into it. The negative and positive charges present inside an atom are equal and as a whole, an atom is electrically neutral.
  • Thomson’s model of the atom as compared to plum pudding and watermelon.
  • He compared the red edible part of the watermelon to a positively charged sphere whereas the seeds of watermelon to negatively charged particles.

electrical world

Chemistry: Topic-wise Test- 7 - Question 8

Passage II

For the following,

Q.

pH of the solution in the half-cell containing 0.02 HA is(HA is a weak monobasic acid)   

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 8

Anodic

Cathodic

Reaction quotient (Q) 





Chemistry: Topic-wise Test- 7 - Question 9

Passage II

For the following,

Q.

pKa of the weak monobasic acid is  

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 9

Anodic

Cathodic

Reaction quotient (Q) 



*Answer can only contain numeric values
Chemistry: Topic-wise Test- 7 - Question 10

One Integer Value Correct Type

This section contains 3 questions, when worked out will result in an integer value from 0 to 9 (both inclusive)

Q.

For the following cell with metal X electrodes, 

Ecell = -0.028 V at 298 K, if there is no liquid juncton potential,valency of X is.......  


Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 10




Chemistry: Topic-wise Test- 7 - Question 11

Passage II

Given molar conductance of 0.001 M NH4OH solution at 298K = 3.0 x 10-3Sm2mol-1.
Limiting molar conductance of

aq. NH4CI = 1.50 x 10-2Sm2mol-1

aq. NaCl = 1.26 x 10-2Sm2 mol-1

and aq. NaOH = 2.48 x 10-2 Sm2 mol-1

Q.

Degree of dissociation of NH4OH at 298 K is 

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 11

By Kohlrausch’s law,

= 1.50 x 10-2 + 2.48 x 10-2 - 1.26 x 10-2
= 2.72 x 10-2 Sm2mol-1
 


Chemistry: Topic-wise Test- 7 - Question 12

Passage II

Given molar conductance of 0.001 M NH4OH solution at 298K = 3.0 x 10-3Sm2mol-1.
Limiting molar conductance of

aq. NH4CI = 1.50 x 10-2Sm2mol-1

aq. NaCl = 1.26 x 10-2Sm2 mol-1

and aq. NaOH = 2.48 x 10-2 Sm2 mol-1

Q.

pKb of NH4OH  is

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 12

By Kohlrausch’s law,

= 1.50 x 10-2 + 2.48 x 10-2 - 1.26 x 10-2
= 2.72 x 10-2 Sm2mol-1
 


*Answer can only contain numeric values
Chemistry: Topic-wise Test- 7 - Question 13

One Integer Value Correct Type

This section contains 2 questions, when worked out will result in an integer value from 0 to 9 (both inclusive)

Q.

0.01 M aqueous solution of a dibasic acid is diluted to 0.004N such that equivalent conductance is x times.What is the value of x?


Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 13

Dibasic acid is H2 2H+ + A2- 
∴ 0.01M = 0.02N
Equivalent conductance at 0.02 N

Equivalent conductance at 0.004 N 

Chemistry: Topic-wise Test- 7 - Question 14

The internationally recommended unit for conductance is

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 14

The internationally recommended unit for conductance is Siemens (S). 1 siemen = 1 ohm-1

Chemistry: Topic-wise Test- 7 - Question 15

The cell constant G* is calculated by the expression:

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 15

The quantity l/A is called cell constant denoted by the symbol, G*.It depends on the distance between the electrodes and their area of cross-section and has the dimension of length–1 and can be calculated if we know l and A. Measurement of l and A is not only inconvenient but also unreliable. The cell constant is usually determined by measuring the resistance of the cell containing a solution whose conductivity is already known. For this purpose, we generally use KCl solutions whose conductivity is known accurately at various concentrations and at different temperatures. The cell constant, G*, is then given by the equation:
G* =l/A= Rκ

Chemistry: Topic-wise Test- 7 - Question 16

 Which of the following relations are not correct?

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 16

The correct answer is Option B 
 
k =G*l/A      (Since, G = 1/R)
   = l/AR 
Therefore, kR = l/A 
So, l/A = kR not k/R

Chemistry: Topic-wise Test- 7 - Question 17

For the reaction ,

Thus  

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 17

for the given reaction

E0cell  = 2.73 V

n(electrons exchanged) = 12

(4 Al → Al3+ + 12e-)

F=96500 C mol-1

ΔG0 (element) = 0

∴ ΔG(reaction) = -nFE0cell

= -12 x 96500 x 2.73

=3161340 J = 3161.34 kJ

thus,

*Answer can only contain numeric values
Chemistry: Topic-wise Test- 7 - Question 18

One Integer Value Correct Type

This section contains 3 questions, when worked out will result in an integer value from 0 to 9 (both inclusive)

Q.

A platinum electrode is immersed in a solution containing 0.1 M Fe2+ and 0.1 M Fe3+.It iscoupled with SHE.Concentration Fe3+ of increased to 0.1 M without change in [Fe2+], then the change in EMF (in centivolt) is 


Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 18



Chemistry: Topic-wise Test- 7 - Question 19

The molecular formula C5H12 contains how many isomeric alkanes?

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 19

n-pentane, 2-ethylpropane, and 2-methylbutane are the 3 isomeric alkanes of C5H12 (pentane).

Chemistry: Topic-wise Test- 7 - Question 20

How many cycloalkene isomers exist for C5H8 which contain at least one methyl locant directly present on the ring? 

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 20

Chemistry: Topic-wise Test- 7 - Question 21

 Which of the following compounds will exhibit cis-trans isomerism? 

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 21

The compounds with each doubly bonded carbon attached to two different groups (like Cab=Cab, Cab=Ccd) exhibit geometrical isomerism i.e., cis and trans forms. The geometrical isomerism arises due to restricted rotation of double bond.

However, even though there is restricted rotation for triple bond, alkynes do not exhibit geometrical isomerism, since the triply bonded carbons are attached to one group each only.

Chemistry: Topic-wise Test- 7 - Question 22

Direction (Q. Nos. 1-18) This section contains 18 multiple choice questions. Each question has four

 choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
 

Q.

The correct statement regarding a chiral compound is

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 22

According to option A chiral compound at least have one chiral carbon but compounds other than carbon and also chiral like NCl Br I is chiral. According to option b a compound containing only one ka kalakaron is always is always correct that is statement is true it is the minimum requirement for a carbon atom to become chiral. Efficiency is not necessary to be true all the time compound will not be superimposable on its Mirror image. And for option d a plane of symmetry is sufficient but not a mandatory to give achiral compound.

Chemistry: Topic-wise Test- 7 - Question 23

The number of optically active optical isomers of the compound is:

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 23

The number of optically active isomers = 2n−1−2n−1/2 ​=23−1−23−1/2 ​=2.

Chemistry: Topic-wise Test- 7 - Question 24

A hydrocarbon X is optically. X upon hydrogenation gives an optically inactive alkane Y. Which of the following pair of compounds can be X and Y respectively?

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 24

The correct answer is Option B.
The optically active hydrocarbon X is 3-methyl-1-pentene CH2=CH−CH(CH3)CH2CH3​. On catalytic hydrogenation, it forms 3-methyl pentane CH3CH2CH(CH3)CH2CH3, which is optically inactive.

Chemistry: Topic-wise Test- 7 - Question 25

Direction (Q. Nos. 1-13) This section contains 13 multiple choice questions. Each question has four
 choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

Q.
Which among the following defines Meso forms of isomers
 

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 25

The correct option is D Molecule in meso form have a plane of symmetry
Meso forms of isomers are single compound and their molecules are achiral and hence they cannot be separated into optically active enantiometric pairs.
Molecule in meso form have a plane of symmetry due to which the optical rotations of upper and lower parts are equal and in the opposite direction which balanced internally and compound become optically inactive. this property is called internal compensation.

Chemistry: Topic-wise Test- 7 - Question 26

In which type of projection we can get staggered and eclipsed conformations?

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 26

A sawhorse projection can reveal staggered and eclipsed conformations.

Chemistry: Topic-wise Test- 7 - Question 27

Direction (Q. Nos. 1 - 6) This section contains 6 multiple choice questions. Each question has four
choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

 

Q.

A pure enantiomer with molecular formula C6H13OBr, when reacted with PBr3, an achiral product C6H12Br2 is obtained that has no chiral carbon. The compound which satisfy this condition could be (no bond to chiral carbon is broken during the reaction)

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 27

 2-bromomethyl-2-methyl-1-butanol: The bromine substitution occurs at the carbon bonded to oxygen but not at a chiral carbon. Thus option b satisfies the condition.

 

Chemistry: Topic-wise Test- 7 - Question 28

Optical rotation of a newly synthesised chiral compound is found to be +60°. Which of the following experiment can be performed to establish that optical rotation is not actually -300°?

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 28

Option b: Decrease concentration of sample four times and measure the optical rotation. Explanation:

  • Optical rotation is directly proportional to concentration (specific rotation = observed rotation/concentration).
  • If you decrease the concentration of the chiral compound in the sample, the observed optical rotation will decrease proportionally.
  • If the initial observed rotation is indeed +60° and not -300°, decreasing the concentration will still result in a positive rotation, but it will be lower.

The other options are less likely to provide useful information in this context.

Chemistry: Topic-wise Test- 7 - Question 29

How many stereoisomers exist for the compound 4-(1- propenyl) cyclohexane ?

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 29

The correct answer is option C
1,4-Dimethylcyclohexane consists of two diastereomeric stereoisomers, although, according to the n2 rule, four stereoisomers should be expected due to the fact that 1,4-dimethylcyclohexane contains two asymmetric carbons.

Chemistry: Topic-wise Test- 7 - Question 30

If the ionisation energy of hydrogen atom is 13.6 eV, the energy required to excite it from ground state to the next higher state is approximately

Detailed Solution for Chemistry: Topic-wise Test- 7 - Question 30

Correct option A.

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