Courses
Courses for Kids
Free study material
Free LIVE classes
More

Important Questions for CBSE Class 11 Chemistry Chapter 2 - Structure of Atom

ffImage
LIVE
Join Vedantu’s FREE Mastercalss
VSAT 2023

CBSE Class 11 Chemistry Chapter-2 Important Questions - Free PDF Download

The Important Questions For Class 11 Chemistry Chapter 2 are provided to the students so that they can have some help when it comes to the preparation for their examinations in the best way. The Class 11 Chemistry Chapter 2 important questions deals with the questions in the chapter Structure of Atom and hence is a really good addition to the syllabus for the students of class 11. These are some of the questions which will be able to help in testing the knowledge of the students about the chapter. Hence, we here at Vedantu are going to provide them with all the knowledge and important information that they need.

With the help of the structure of atom class 11 important questions, students can finally delve into the preparation and then make sure that they have all the important answers to the questions for their exams.


Download CBSE Class 11 Chemistry Important Questions 2023-24 PDF

Also, check CBSE Class 11 Chemistry Important Questions for other chapters:

CBSE Class 11 Chemistry Important Questions

Sl.No

Chapter No

Chapter Name

1

Chapter 1

Some Basic Concepts of Chemistry

2

Chapter 2

Structure of Atom

3

Chapter 3

Classification of Elements and Periodicity in Properties

4

Chapter 4

Chemical Bonding and Molecular Structure

5

Chapter 5

States of Matter

6

Chapter 6

Thermodynamics

7

Chapter 7

Equilibrium

8

Chapter 8

Redox Reactions

9

Chapter 9

Hydrogen

10

Chapter 10

The s-Block Elements

11

Chapter 11

The p-Block Elements

12

Chapter 12

Organic Chemistry - Some Basic Principles and Techniques

13

Chapter 13

Hydrocarbons

14

Chapter 14

Environmental Chemistry

Last updated date: 23rd Sep 2023
Total views: 595.8k
Views today: 9.95k
Competitive Exams after 12th Science

CBSE Class 11 Chemistry Important Questions Chapter 2: Structure of Atom

1 Mark Questions

1. Name the sub – atomic particles of an atom.

Ans: The subatomic particles of an atom are:

  • Proton

  • Neutron

  • Electron

2. Name the scientist who first formulated the atomic structure.

Ans: The atomic structure was formulated by John Dalton, a British teacher in 1808 .  He first proposed a firm scientific basis named Dalton’s atomic theory.

3. What is the e/m ratio of an electron?

Ans: According to Thomson’s experiment, e/m ratio for an electron is $1.76\times {{10}^{3}}\text{c}{{\text{g}}^{-1}}$ . It means that every gram of an electron carries charge of $1.76\times {{10}^{3}}C$

4. What is the charge (e) of an electron?

Ans: From Millikan’s experiment, the charge of an electron (e) is $-1.602\times {{10}^{-19}}\text{C}$ . This is universal, and is the same for each and every electron.

5. (i) What is the mass of a proton?

(ii) What is the charge of a proton?

Ans: (i) The mass of a proton is $1.676\times {{10}^{-27}}\text{kg}$

(ii) The charge of a proton is $+1.602\times {{10}^{-19}}\text{C}$.

6. (i) What is the mass of a neutron?

(ii) What is the charge of a neutron?

Ans: (i) The mass of a neutron is $1.676\times {{10}^{-24}}\text{g}$ .

(ii) Neutrons are electrically neutral i.e. it has a charge of 0.

7. Name the scientist who first gave the atomic model.

Ans: J.J. Thomson, in 1898 first proposed the atomic model called raising-pudding model. The negative charges(raisins) were spread around in the plum pudding.

8. What is an isotope?

Ans: Atoms of the same elements which have the same atomic number but different mass number are called isotopes. Chemical properties of isotopes are almost similar.

e.g. ${{^{1}}_{1}}H{{.}^{2}}_{1}H\text{ and}{{\text{ }}^{3}}_{1}H$ ,${{^{35}}_{17}}Cl{{,}^{37}}_{17}Cl{{,}^{12}}_{6}C{{,}^{13}}_{6}C{{,}^{14}}_{6}C$

9. What are isobars?

Ans: Atoms of different elements which have the same mass number but different atomic numbers are called isobars. Isobars differ in chemical property but have same physical properties

e.g. ${{^{14}}_{6}}C{{,}^{14}}_{7}N$ , ${{^{40}}_{18}}Ar{{,}^{40}}_{19}K{{,}^{40}}_{20}Ca$

10. What are isotones?

Ans: Atoms of different elements which contain the same number of neutrons. Isotones have no similarity, when it comes to chemical properties.

e.g. ${{^{14}}_{6}}C{{,}^{15}}_{7}N{{,}^{16}}_{8}O$

11. What is an atomic number?

Ans: Atomic number is the measure of the number of protons which is present in the nucleus of an atom. For example, ${}_{1}^{2}H$ has an atomic number of 1.

12. What is a mass number?

Ans: Mass number of an element is the sum of the number of protons and neutrons present in the nucleus of an atom. For example, ${}_{1}^{2}H$ has a mass number of 2, as it has 1 proton and 1 electron.

13. Give the drawbacks of J.J. Thomson’s experiment.

Ans: The drawbacks are:

(i) It failed to explain the origin of the spectral lines of hydrogen and other atoms. 

(ii) It failed to explain scattering of $\alpha -$ particles in Rutherford’s scattering experiment.

14. Why Rutherford’s model could not explain the stability of an atom?

Ans: According to the electromagnetic theory of Maxwell, when charged particles are accelerated then they should emit electromagnetic radiation. Therefore, an electron in an orbit will continue to emit radiation for infinite time ; the orbit will then continue to shrink which is not the case in an atom.

15. Define photoelectric effect.

Ans: The phenomenon in which the surface of alkali metals like potassium and calcium emit electrons when a beam of light with high frequency is projected on them is called the photoelectric effect.

16. How does the intensity of light affect photoelectrons?

Ans: The number of electrons ejected and kinetic energy associated with them is directly proportional to the intensity of light projected towards the metal.

17. What is the threshold frequency?

Ans: The minimum frequency below which electrons are not ejected is called threshold frequency (${{v}_{0}}$ ). Threshold frequency is different for different metals.

18. Name the scientist who demonstrated the photoelectric effect experiment.  

Ans: In 1887, H. Hertz demonstrated photo electric effect. He observed the photoelectric effect, while working on radio waves.

19. What did Einstein explain about the photoelectric effect?

Ans: Einstein was able to explain the photoelectric effect using Planck’s quantum theory of electromagnetic radiation in 1905. Energy in each quantum of light is equal to a constant multiplied with the speed of light.

20. Calculate energy if 2mole of photons of radiation whose frequency is $5\times {{10}^{14}}\text{Hz}$. 1 Mark

Ans: Energy (E) of one photon $E=hv$

Where $h=6.626\times {{10}^{-34}}\text{Js}$ 

$v=5\times {{10}^{14}}{{s}^{-1}}$ 

$\therefore E=(6.626\times {{10}^{-34}}\times 5\times {{10}^{14}})$ 

$=3.313\times {{10}^{-19}}J$ 

Energy of 2 mole of photon $=(3.313\times {{10}^{-19}}J)\times (2\times 6.022\times {{10}^{23}}mo{{l}^{-1}})$ 

$=3990.2klmo{{l}^{-1}}$


21. States Heisenberg’s Uncertainty Principle.

Ans: It states that it is impossible to determine simultaneously the exact position and exact momentum (or velocity) of an electron. Mathematically, $\Delta x\Delta p\ge \frac{h}{4\pi }$.

22. How would the velocity be affected if the position is known?

Ans: If the position of the electron is known with high degree of accuracy ($\Delta x$ is small), then, according to the equation, the velocity of the electron will be uncertain $(\Delta ({{V}_{x}})$ is large).

23. We don’t see a car moving as a wave on the road why?

Ans: According to de Broglie’s relation, $\lambda ={}^{h}/{}_{mv}\text{ i}\text{.e}\text{. }\lambda \alpha \frac{1}{m}$ the mass of the car is very large and its wavelength $(\lambda )$ or wave character is negligible. Therefore, we do not see a car moving like a wave.

24. Give the de-Broglie’s relation.

Ans: Every  particle in motion is associated with a wavelength and other wave characteristics. The wavelength $(\lambda )$ of a particle in motion is equal to the Planck’s constant (h) divided by the momentum (p) of the particle. 

i.e. $\lambda =\frac{h}{p}=\frac{1}{mv}$ 

Where $m$ is the mass, $v$ is the velocity of the particles.

25. Calculate the uncertainty in the velocity of a wagon of mass $4000\text{ kg}$ whose position is known accurately of $\pm 10\text{m}$.

Ans: $\Delta v\ge \frac{h}{4\pi m\Delta x}$ 

$=\frac{6.6\times {{10}^{-34}}\text{kg}{{\text{m}}^{2}}{{s}^{-1}}}{4\times \frac{22}{7}\times 4\times {{10}^{3}}\text{kg}\times (\pm 10\text{m)}}$ 

$=1.3\times {{10}^{-39}}m{{s}^{-1}}$ 

$\therefore $ The uncertainty in the velocity of the wagon is $=1.3\times {{10}^{-39}}m{{s}^{-1}}$.


26. What is the physical significance of ${{\Psi }^{2}}$ up?

Ans: ${{\Psi }^{2}}$ represents the probability of finding an electron. It is the probability of finding a particle specified by a particular wave function.

27. Which orbital is non-directional?

Ans: S- orbital is spherically symmetrical i.e. it is non-directional.  It has a spherical shape, like a hollow ball.

28. What is the meaning of quantization of energy?

Ans: Quantization of energy means the energy is distributed and transmitted in the form of packets. These packets are called photons.

29. Why is the energy of 1s electron lower than 2s electron?

Ans: 1s electron being close to the nucleus experiences more force of attraction than 2s-electron which is away from the nucleus. Force of attraction is inversely proportional to the square of distance between the particles.

30. What is a nodal surface or nodes?

Ans: The region where the probability of finding an electron is zero i.e. ${{\text{Y}}^{2}}=0$ . This means that the value of the wave function is also zero.

31. How many spherical nodal surfaces are there in 4s – subshell?

Ans: In ns orbital, the number of spherical nodal surfaces are $(n-1)$, hence is $4s(4-1)=3$ nodal surfaces are present.

2 Marks Questions

1. What is the mass (m) of an electron?

Ans: mass of an electron $(m)=\frac{e}{({}^{e}/{}_{m})}$ 

$=\frac{1.602\times {{10}^{-19}}C}{1.76\times {{10}^{3}}C{{g}^{-1}}}$ 

$=9.10\times {{10}^{-28}}g$ 

$=9.1\times {{10}^{-31}}kg$ 

So, the mass of an electron is $=9.1\times {{10}^{-31}}kg$or ${}^{1}/{}_{1837}$  of the mass of a hydrogen atom.

2. Which experiment let to the discovery of electrons and how?

Ans: The beam discharge tube experiment performed by J.J. Thomson led to the invention of negativity charged particles called electron.

A beam tube consists of two thin pieces of metals called electrodes sealed inside a glass tube with sealed ends. The glass tube is attached to a pump and therefore the pressure inside the tube is reduced to 0.01mm . When a fairly high voltage of 10,000 volts is applied across the electrodes, invisible rays are emitted from the cathode called cathode rays. Analysis of this rays led to the invention electrons.

3. Give the main properties of the canal ray experiment.

Ans: The canal ray experiment led to the discovery of:

(i) The canal rays travel in a straight line. 

(ii) They have the ability to penetrate through small openings. 

(iii) They are positively charged as they get deflected from electric and magnetic fields.

4. Find out atomic number, mass number, number of electrons and neutrons in an element $\frac{40}{20}\times $ ?

Ans: The mass no. of $\times $ is $40$ 

The atomic no. of $\times $ is $20$ 

No. of proton is $=Z-A=40-20=20$ 

No. of electron is $(\text{A)}=20$ 

No. of proton is $(\text{A)}=20$

5. Give the main features of Thomson’s Model for an atom.

Ans: The features are:

  • An atom hosts a sphere, which is positively charged, and the electrons are present in it and spread all over.

  • There is balancing of charges, total positive charge is equal to the total negative charge.

6. What did Rutherford conclude from the observations of the $\alpha -ray$ scattering experiment?

Ans: Rutherford proposed the nuclear model of an atom as

(i) The positive charge was concentrated in a small space at the centre, which is the nucleus. 

(ii) The nucleus is surrounded by electrons that move around it in orbits with a very high speed.

(iii) Most of the space inside the atom is empty as most of the rays pass undeflected.

7. What is the relation between kinetic energy and frequency of the photoelectrons?

Ans: Kinetic energy of the ejected electron is proportional to the frequency of the electromagnetic radiation. As the kinetic energy is increased, the light incident on the metal has more energy, and for that, the time period of the electrons ejected reduces, which increases the frequency.

8. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, $\text{n}=4$ to $\text{n}=2$ of $\text{H}{{\text{e}}^{+}}$ spectrum?

Ans: For the Balmer transition, $\text{n}=4$ to $\text{n}=2$ in a $\text{H}{{\text{e}}^{+}}$ ion, we can write.

$\frac{1}{\lambda }={{\text{Z}}^{2}}{{\text{R}}_{H}}\left[ \frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}} \right]$ 

$={{\text{Z}}^{2}}{{\text{R}}_{H}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right]$ 

$=\frac{3}{4}{{\text{R}}_{H}}$  ……(i)

For a hydrogen atom

$\frac{1}{\lambda }={{\text{R}}_{H}}\left[ \frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}} \right]$ ……(ii)

Equating equation (ii) and (i), we get

$\frac{1}{{{\text{n}}_{1}}^{2}}-\frac{1}{{{\text{n}}_{2}}^{2}}=\frac{3}{4}$ 

This equation gives ${{\text{n}}_{1}}=1$ and $\text{n=2}$ . Thus the transition $\text{n=2}$ to $\text{n=1}$ in hydrogen atom will have same wavelength as transition, $\text{n}=4$to $\text{n=2}$ in $\text{H}{{\text{e}}^{+}}$.

9. Spectral lines are regarded as the fingerprints of the elements. Why?   

Ans: Spectral lines are regarded as the fingerprints of the elements because identification of the elements can be done from these lines. Just like fingerprints, the spectral lines of no two elements resemble each other.  Spectral lines are only observed when electrons jump from one energy level to another.

10. Why cannot the motion of an electron around the nucleus be determined accurately?
Ans: We cannot determine the motion of an electron around the nucleus accurately, due to Heisenberg's uncertainty principle. The mass of an electron is fixed, but it is almost impossible to predict the velocity of the electron, for which its position cannot be determined easily, at a specific time.

11. Calculate the uncertainty in the momentum of an electron if it is confined to a linear region of length $1\times {{10}^{-10}}$.

Ans: According to the Uncertainty Principle.

$\Delta x.\Delta p=\frac{h}{4\pi }$ 

or $\Delta p=\frac{h}{4\pi \Delta x}$ 

or $\Delta p=\frac{6.626\times {{10}^{34}}\text{kg}{{\text{m}}^{2}}{{s}^{-1}}}{4\times 3.143\times {{10}^{10}}\text{m)}}$ 

$=5.27\times {{10}^{-26}}\text{kgm}{{\text{s}}^{-1}}$

12. Give the mathematical expression of uncertainty principle.

Ans: Mathematically, it can be given as

$\Delta x\times \Delta {{p}_{x}}\ge {}^{h}/{}_{4\pi }$

or, $\Delta x\times \Delta (m{{v}_{z}})\ge {}^{h}/{}_{4\pi }$

or, $\Delta x\times \Delta {{V}_{x}}\ge {}^{h}/{}_{4\pi m}$ 

Where $\Delta x$ is the uncertainty in position and $\Delta {{p}_{x}}(\Delta {{v}_{x}})$ is the uncertainty in momentum (or velocity) of the particle.

13. Which quantum number determines.

(i) energy of electron

(ii) Orientation of orbitals. 

Ans: 

(i) Principal quantum number (n), and 

(ii) Magnetic quantum number (m).

14. Arrange the electrons represented by the following sets of quantum numbers in decreasing order of energy.

1. $\text{n}=4,\text{I}=0,\text{m}=0,\text{s}=+1/2$ 

2. $\text{n}=3,\text{I}=1,\text{m}=1,\text{s}=-1/2$ 

3. $\text{n}=3,\text{I}=2,\text{m}=0,\text{s}-+1/2$ 

Ans: (i) Represents 4s orbital

(ii) Represents 3p orbital

(iii) Represents 3d orbital

(iv) Represents 3s orbital

The decreasing order of energy $3\text{d4s3p3s}$ 

$\text{n}=3,\text{I}=0,\text{m}=0,\text{s}=-1/2$

3 Marks Questions

1. What designations are given to the orbitals having

(i) $\text{n}=2,\text{I}=1$  (ii) $\text{n}=2,\text{I}=0$  (iii) $\text{n}=4,\text{I}=3$ 

(iv) $\text{n}=4,\text{I}=2$ (v) $\text{n}=1,\text{I}=1$ ?

Ans: (i) Here $\text{n}=2$ and $\text{I}=1$ 

Since $\text{I}=1$ it means a p=orbital, hence the given orbital is designated as 2p.

(ii) Here, $n=2$ and $\text{I}=0$ 

Since $\text{I}=0$ means s-orbital, hence the given orbital is 2s.

(iii) Here, $\text{n}=4$ and $\text{I}=3$ 

Since, $\text{I}=3$ represents f-orbital, hence the given orbital is a 4f orbital.

(iv) Here, $\text{n}=4$ and $\text{I}=2$ 

Since, $\text{I}=2$ represents a d-orbital, hence the given orbital is a 4d-orbital. 

(v) $\text{n}=4$ and $\text{I}=1$ 

Since, $\text{I}=1$ means it is a p-orbital, hence the given orbital can be designated as -4p orbital.

2. Write the electronic configuration of (i) $\text{M}{{\text{n}}^{4+}}$ ,(ii)$\text{F}{{\text{e}}^{3+}}$ ,(iii)$\text{C}{{\text{r}}^{2+}}$ and $\text{Z}{{\text{n}}^{2+}}$ . Mention the number of unpaired electrons in each case. 3 Marks

Ans.(i) $\text{Mn(z}=25),\text{ M}{{\text{n}}^{4+}}(\text{z}=21)$ 

The electronic configuration of $\text{M}{{\text{n}}^{4+}}$ is given by

$1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}\text{3}{{\text{p}}^{6}}3{{\text{d}}^{3}}$ 

As the outermost shell $3\text{d}$ has 3 electrons, thus the number of unpaired electrons is 3.

(ii) $\text{Fe (z}=26),\text{ F}{{\text{e}}^{3+}}(\text{z}=23)$  

The electronic configuration of $\text{F}{{\text{e}}^{3+}}$ is given by

$1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}3{{\text{d}}^{5}}$ 

The number of unpaired electrons is $5$ .

(iii) $\text{Cr(z}=24),\text{ C}{{\text{r}}^{2+}}(\text{z}=22)$ 

The electronic configuration of $\text{C}{{\text{r}}^{2+}}$ is

$1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}\text{3}{{\text{d}}^{4}}$ 

The number of unpaired electrons is $4$ .

(iv) $\text{Zn(z}=30),\text{Z}{{\text{n}}^{2+}}(\text{z}=28)$ 

The electronic configuration of $\text{Z}{{\text{n}}^{2+}}$ is

$1{{\text{s}}^{2}}2{{\text{s}}^{2}}2{{\text{p}}^{6}}3{{\text{s}}^{2}}3{{\text{p}}^{6}}3{{\text{d}}^{10}}$ 

The number of unpaired electrons is $0$.

CBSE Class 11 Chemistry Chapter-2 Important Questions - Free PDF Download

Choose Class 11 Chemistry Chapter 2 Important Questions Right Now 

Having the important questions of structure of atom class 11 can be a huge deal for all the students who want to score some high marks in the examination. One of the most important things that students need to know about these questions is that there are many different types. The exercises are divided into different questions and students need to pick out the important ones if they want to do well in the examinations. So, having the chemistry class 11 chapter 2 important questions is basically a boon for the students.

In the important questions of Ch 2 chemistry class 11, students will get to know about the atomic structures, isotones, isotopes. They will also get their knowledge of Bohr’s Atomic Model and much more information about the atomic theory. These questions are meant to provide an insight to the students about the chapter and with regular practice, they will surely get some more information about the chapter in the best way. For the students who want to make sure that they have a little bit of extra preparation for the exams, choosing to answer the class 11 chemistry chapter 2 extra questions is surely going to be a great help in the near future. We are pretty sure that with the help of these questions, the preparation for exams might become one of the easiest things for the students.

Why Choose Us To Get Structure Of Atom Class 11 Important Questions 

Having a little bit of help when it comes to preparing for exams can be a pretty good thing for students. That is why they are always looking for the important questions class 11 chemistry chapter 2 for their preparation. This is where Vedantu can be a great help. We have some of the best solutions to the CBSE class 11 chemistry chapter 2 important questions. These questions are selected by our team of experts in order to make sure that students get to have the highest possible marks in the exams.

Not to mention that the class 11th chemistry chapter 2 important questions are available for free so that students can download them any time without having to pay a single penny. For those who have any mobile devices such as a phone, laptop, or computer, downloading the PDF format of important questions for class 11 chemistry chapter 2 will definitely be easy to do. Our expert teachers have made sure to include the questions that will give the students all the information that they need about the chapter and hence is the most helpful thing ever.

Conclusion

The class 11 chemistry chapter 2 important questions can be used for the preparation of exams and for revision purposes. However, the students also need to make sure that they are paying attention in their class so that they can get an idea about the chapter, and hence they will be able to answer the questions in the easiest way possible. The structure of atom class 11 important questions will also help the students in preparing the study routine that they have and they can easily prioritize their time for better results. That is one of the main reasons why Vedantu is considered to be one of the leading names for tutoring sites for students who need just a little bit of help for their class 11 exam preparations.


Conclusion

The class 11 chemistry chapter 2 essential questions can be used to help with exam planning and review. To get an understanding of the chapter and, as a result, be able to respond to questions as simply as possible, students must also make sure that they are paying attention in class. Students can simply prioritise their time for better performance by using the structure of the atom class 11 crucial questions to help them plan their study regimen. One of the main reasons Vedantu is regarded as one of the top tutoring services for students who need just a little assistance with their academics is because of this.


Important Related Links for CBSE Class 11 Chemistry

FAQs on Important Questions for CBSE Class 11 Chemistry Chapter 2 - Structure of Atom

1. What are Isotopes?

Answer: Isotopes are the variant atoms of the same element, which have the same number of protons and electrons but a different number of neutrons. Hence, isotopes of the same element have different nucleon numbers. In other words, isotopes are atoms of the same element, having the same atomic number and different mass numbers. Also, the isotopes of an element have similar physical properties but different chemical properties.


For example, Carbon has three isotopes, Carbon-12, Carbon-13, and Carbon-14. These atoms of carbon have mass numbers of 12, 13, and 14 respectively, whereas the atomic number of all three is 6.

2. What are Isobars?

Answer: Isobars are the atoms of various chemical elements, which have the same mass number but differ in atomic numbers. In other words, isobars have the same number of nucleons but have a different number of electrons. For example, Sulphur, Calcium, Potassium, Argon, Chlorine, the atoms of all these elements contain 40 nucleons each but have different numbers of electrons. The existence of isobars in the periodic table helps to determine the stability and radioactivity of various elements. Also, it may be so that the isotopes of different elements having the same atomic weights are isobars.

3. What are Isotones?

Answer: Isotones are the atoms of different elements, which have the same number of neutrons but different numbers of protons. For example, Boron and Carbon are isotones, having 7 neutrons in the nuclei, and a different number of electrons and protons. Again, chlorine and potassium make a pair of isotones, having 20 neutrons each, in their nuclei.

4. Can I Find Important Questions for CBSE Class 11 Chemistry Chapter 2- Structure of Atom Online?

Ans: On Vedantu, you may access key questions for CBSE Class 11 Chemistry Chapter 2: Atomic Structure. Our subject matter specialists have carefully selected and prepared these crucial questions using the NCERT Chemistry textbook for Class 11 and a variety of other reference study tools. Moreover, you may obtain the critical questions for the chapter on Atomic Structure in PDF format for free. You can review all the significant issues and their main points from this chapter by going through these crucial questions. Thus, these crucial questions serve as a very helpful study guide for your exam preparation.

5. What are the main concepts given in Chapter 2 Structure Of Atom of Class 11 Chemistry?


Ans: Chapter 2 of Class 11 Chemistry is based on the structure of atoms and different constituents of atoms. Students will study the concept of isotopes, isotones, models of atoms suggested by different scientists, and other important information related to the atomic theory. Students can understand the main concepts of Class 11 Chemistry Chapter 2 using important questions given on Vedantu. Students can download all important questions for Class 11 Chemistry Chapter 2 by clicking here.

6. How were electrons discovered?


Ans: J. J. Thomson experimented on the cathode ray discharge tube. The experiment helped to discover electrons which are stated to be negatively charged particles present on the outer side of an atom. When high voltage is passed through electrodes present inside a glass tube, invisible rays are emitted which were named cathode rays by the scientist. The scientist analyzed the rays and this led to the discovery of electrons.

7. How important is Chapter 2 Structure Of Atom of Class 11 Chemistry for JEE Mains preparation?


Ans: Atomic structure in Chapter 2 of Class 11 Chemistry is crucial for preparing for the board exams and JEE Mains. The many scientists' descriptions of the atom's structure must be understood by the students. In the board exams and JEE Main paper, they can find questions from the Chapter. As a result, in order to be fully prepared for their tests, pupils must carefully review and comprehend the Chapter's concepts. To master the key concepts of Chapter 2, they might refer to the Chapter 2 critical questions for Class 11 Chemistry.

8. What do you know about the atomic number and mass number of Chapter 2 Structure Of Atom of Class 11 Chemistry?

Ans: The atomic number corresponds to the number of positive particles in it. It helps to differentiate the two given elements from each other. The protons and neutrons present in the nucleus of an atom form the mass number. In the periodic table, the elements are arranged according to the increasing atomic and mass number. The protons and neutrons are the two main particles that are found in the nucleus of an atom.