structure of atom assignment class 11

Talk to our experts

1800-120-456-456

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

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 2024-25 PDF

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

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:

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. 

(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 $0$.

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.

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.

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.

CBSE Class 11 Chemsitry Important Questions

Cbse study materials.

• CBSE Class 11 Chemistry Chapter 2 – Structure of Atom Revision Class 11 Notes

Structure of Atom Class 11 Revision Notes 

Structure of Atom Class 11 Notes – This topic talks about atoms. In addition, it also talks about electron, neutron, and protons. Furthermore, the who and how these particles are discovered by the scientists. Moreover, various principles, functions, and structures are also discussed in the chapter.  Most noteworthy, the chapter talks about sub-atomic particles. Also, shell configuration and how elements complete their shells. Besides, the chapter also discusses atom valency.

Download Toppr app for Android and iOS or signup for free.

Sub-topics covered under Structure of Atom

• Introduction: Structure of Atom – This topic explains the atom and discovery of electron, neutron, and proton.
• Atomic Number – Also, the topic talks about the atomic number and how to find them.
• Bohr’s Model of Atom – This topic overview Bohr’s model of an atom.
• Charged Particles in Matter – Furthermore, the topic discusses charged particles and the discovery of sub-atomic particles.
• Isobars – Moreover, this topic highlights isobars and differences between isobars and isotopes.
• Isotopes – The topic defines isotopes in detail.
• Mass Number – Also, this topic describes the mass numbers, its uses, and properties.
• Neutrons – The topic explains the functions, properties, and discovery of the neutron.
• Rutherford’s Model of an Atom – In addition, this topic overviews Rutherford’s model of an atom.
• Thomson’s Model of an Atom – The topic defines Thomson’s model of an atom.
• Valency – This topic highlights what is valency of an atom.
• How are Electrons Distributed in Different Orbits (Shells)? – Furthermore, the topic describes the electron distribution.
• Sub-Atomic Particles – This topic talks about sub-atomic and important properties.
• Atomic Models – Moreover, the topic describes various atomic models.
• Shapes of Atomic Orbitals – This topic explains the shape of atomic orbitals.
• Energies of Orbitals – The topic talks about the energies of orbitals.
• Quantum Numbers – This topic overviews the quantum numbers.
• Development Leading to Bohr’s Model of Atom – The topic highlights the latest development in Bohr’s model.
• Emission and Absorption Spectra – This topic discusses the emission and absorption of spectra.
• Towards Quantum Mechanical Model of Atom – The topic describes the quantum mechanical model of an atom.

You can download CBSE Class 11 Chemistry Structure of Atom Revision Notes by clicking on the download button below

Customize your course in 30 seconds

Which class are you in.

CBSE Class 12 Chemistry Revision Notes

• CBSE Class 12 Chemistry Chapter 16 – Chemistry in Everyday Life Class 12 Notes
• CBSE Class 12 Chemistry Chapter 15 – Polymers Class 12 Notes
• CBSE Class 12 Chemistry Chapter 14 – Biomolecules Class 12 Notes
• CBSE Class 12 Chemistry Chapter 13 – Amines Class 12 Notes
• CBSE Class 12 Chemistry Chapter 12 – Aldehydes, Ketones and Carboxylic Acids Class 12 Notes
• CBSE Class 12 Chemistry Chapter 11 – Alcohols, Phenols and Ethers Class 12 Notes
• CBSE Class 12 Chemistry Chapter 10 | Haloalkanes and Haloarenes Notes
• CBSE Class 12 Chemistry Chapter 9 – Coordination Compounds Class 12 Notes
• CBSE Class 12 Chemistry Chapter 8 – The d and f Block Elements Class 12 Notes

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Download the App

Chemistry Worksheet Class 11 on Chapter 2 Structure of Atom with Answers - Set 1

An element’s atomic structure refers to the structure of its nucleus and the arrangement of electrons around it. The atomic structure of matter is primarily composed of protons, electrons, and neutrons.

The nucleus of the atom is made up of protons and neutrons, which are surrounded by the atom’s electrons. An element’s atomic number describes the total number of protons in its nucleus.

Neutral atoms contain the same number of protons and electrons. On the other hand, atoms can gain or lose electrons to increase their stability, and the resulting charged entity is known as an ion.

Because different elements have different numbers of protons and electrons, their atomic structures differ. This is the cause of the unique characteristics of different elements.

Download PDF of Class 11 Chemistry Chapter 2 Structure of Atom – Set 1

CBSE Class 11 Chemistry Worksheet Chapter 2 Structure of Atom – Set 1

Q1. Which of the following orbitals do not make sense?

Q2. Orbital angular momentum depends on:

b.) n and l

c.) n and m

d.) m and s

Q3. In the manganese atom, Mn (Z = 25), the total number of orbitals populated by one or more electrons (in the ground state) is:

Q4. The following quantum numbers are possible for how many orbitals?

n = 3, l = 2, m l = +2

Q5. In Bohr’s orbit, the ratio of total kinetic energy and the total energy of the electron is:

Q6. Define Hund’s rule of maximum multiplicity.

Q7. Define electromagnetic spectrum.

Q8. How many protons, electrons, and neutrons are there in the following nuclei?

ii.) 25 12 Mg

iii.) 80 35 Br

Q9. State the drawbacks of the Rutherford Model of an atom.

Q10. Calculate and compare the energies of two radiations, one with a wavelength of 800 nm and the other with a wavelength of 400 nm.

Q11. State Heisenberg’s Uncertainty principle.

Q12. i.) An atomic orbital has n = 3. What are the possible values of l and m l ?

ii.) List the quantum numbers (m l and l) of electrons for the 3-d orbital.

iii.) Which of the following orbitals are possible 1p, 2s, 2p and 3f?

Q13. The radius of the fourth orbit in a hydrogen atom is 0.85 nm. Calculate the velocity of the electron in this orbit (mass of electron = 9.1 × 10 –31 kg).

Q14. State and explain Pauli’s exclusion principle. Write the electronic configuration of the element with atomic number 24.

Q15. a.) What is the Aufbau principle? Write electronic configurations of the elements with atomic numbers 16, 20 and 35.

b.) Explain why half-filled and completely filled orbitals have extra stability?

Q16. How many unpaired electrons are present in the ground state of

i.) P (Z = 15)

ii.) Fe 2+ (Z = 26)

iii.) Cl – (Z = 17)

Q17. On the basis of Heisenberg’s uncertainty principle, show that an electron cannot exist within the atomic nucleus of a radius of 10 –15 m.

Q18. An electron is moving with a kinetic energy of 2.275 × 10 –25 J. Calculate its de-Broglie wavelength.

(Mass of electron = 9.1 × 10 –31 kg, h = 6.6 × 10 –34 J s)

Q19. In the Rydberg equation, a spectral line corresponds to n 1 = 3 and n 2 = 5.

i.) Calculate the wavelength and frequency of this spectral line.

ii) To which spectral series does this line belong?

iii.) In which region of the electromagnetic spectrum will this line fall?

Q20. What is the photoelectric effect? State the results of the photoelectric effect experiment that could not be explained based on the laws of classical physics. Explain this effect based on the quantum theory of electromagnetic radiations.

Download PDF to access answers of the Chemistry Worksheet for Class 11 Chemistry Chapter 2 Structure of Atom Set – 1.

Download PDF

• Class 11 Chemistry Chapter 2 Structure of Atom MCQs
• Important Questions for Class 11 Chemistry Chapter 2 Structure of Atom
• Chemistry Concept Questions and Answers
• CBSE Class 11 Chemistry Practical
• Aufbau Principle
• Subatomic Particles
• Rutherford Atomic Model
• Planck’s Quantum Theory: Quantisation of Energy
• Bohr’s Model of An Atom

• Share Share

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

AssignmentsBag.com

Assignments Class 11 Chemistry Structure of Atom

Please refer to Assignments Class 11 Chemistry Structure of Atom Chapter 2 with solved questions and answers. We have provided Class 11 Chemistry Assignments for all chapters on our website. These problems and solutions for Chapter 2 Structure of Atom Class 11 Chemistry have been prepared as per the latest syllabus and books issued for the current academic year. Learn these solved important questions to get more marks in your class tests and examinations.

Structure of Atom Assignments Class 11 Chemistry

Question. A hydrogen atom in its ground state absorbs 10.2 eV of energy. The orbital angular momentum is increased by (Given Planck constant h = 6.6 × 10 -34 Jsec) (a) 1.05 × 10 -34 Jsec (b) 3.16 × 10 -34 Jsec (c) 2.11 × 10 -34 Jsec (d) 4.22 × 10 -34 Jsec   

Question. In chromium atom, in ground state, the number of occupied orbitals is (a) 14 (b) 15 (c) 7 (d) 12 

Question. The value of Plancks constant is 6.63 × 10 -34 Js. The velocity of light is 3.0 × 10 8 ms -1 . Which  value is closest to the wavelength in nanometres of a quantum of light with frequency of 8 ×  10 15 s -1 (a) 3 × 10 7 (b) 2 × 10 -25 (c) 5 × 10 -18 (d) 4 × 10 1

Question. Maximum number of electrons in a subshell with l = 3 and n = 4 is (a) 10 (b) 12 (c) 14 (d) 16   

Question. The magnetic quantum number specifies (a) Size of orbitals (b) Shape of orbitals (c) Orientation of orbitals (d) Nuclear Stability  

Question. In a multi – electron atom, which of the following orbitals described by the three quantum  numbers will have the same energy in the absence of magnetic acid and electric fields? (a) n = 1, l = 0, m = 0 (b) n = 2, l = 0, m = 0 (c) n = 2, l = 1, m = 1 (d) n = 3, l = 2, m = 1 (e) n = 3, l = 2, m = 0 (a) (a) and (b) (b) (b) and (c) (c) (c) and (d) (d) (d) and (e) 

Question. For principal quantum number n = 4, the total number of orbitals having l = 3 is (a) 3 (b) 7 (c) 5 (d) 9 

Question. In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inner-orbit jumps of the electron for Bohr orbits in an atom of hydrogen? (a) 3 → 2 (b) 5 → 2 (c) 4 → 1 (d) 2 → 5 

Question. The electronic transitions from n = 2 to n = 1 will produce shortest wavelength in (where n = principal quantum state) (a) Li 2+ (b) He + (c) H (d) H +

Question. The ionization enthalpy of hydrogen atom is 1.312 × 10 6 J mol -1 . The energy required to excite the electron in the atom from n = 1 to n = 2 is (a) 8.51 × 10 5 Jmol -1 (b) 6.56 × 10 5 Jmol -1 (c) 7.56 × 10 5 Jmol -1 (d) 9.84 × 10 5 Jmol -1 /sup>   

Related Posts

Assignments class 11 mathematics principle of mathematical induction.

Assignments For Class 11 Mathematics Limits And Derivatives

Assignments Class 11 Chemistry The s-Block Elements

FAQ|Structure of Atom

Faq | solutions| cbse|isc|jee|neet, chemical kinetics |jee main|jee advanced, solutions|jee main |jee advanced|mcq question and answers.

ALL ABOUT CHEMISTRY

Explore the Mystery of Chemistry

• Disclaimer New
• Privacy Policy
• Terms and Conditions
• Quora Space

What is Matter?

Chemistry Terms

Protonation

Mole Concept|JEE Main|JEE Advanced|MCQ question and answers

Structure of atom|class-11.

Describe the discharge tube experiment

The discharge tube filled with a gas is connected to a vacuum pump in order to reduce the pressure. The electrodes are connected to a source of high voltage of 10000 volts.

• Under normal pressure (1 atm), nothing is observed.
• At 10 -2 atm, the gas is found to emit light and the colour of the light depends upon the nature of the gas.
• At 10 -4 atm, the emission of light ceases but the wall of the discharge tube opposite to the cathode starts glowing with a faint greenish light called fluorescence. This is due to the bombardment on the wall by cathode rays.

Mention some properties of cathode rays

• It travels in a straight line, can caste shadow, consists of negatively charged particles, and can exert mechanical pressure.
• It can produce a heating effect, can ionize the gas through which they pass, can produce X-rays when striking on the surface of hard metals like tungsten, copper, molybdenum, etc.
• It is deflected by electric and magnetic field and can pass through thin foils of metals like Al.
• It affects the photographic plate. This is called fogging.
• The e/m ratio of the particles constituting the cathode rays does not depend on the nature of the gas or the material of the cathode.

Mention some properties of anode rays

• It travels in a straight line and can caste shadow.
• It can exert mechanical pressure and is deflected by electric and magnetic field.
• The anode rays are formed by the expulsion of electrons from the atoms of the gas in the space between anode and cathode.
• The e/m ratio of the particles constituting the anode rays depends on the nature of the gas.

Mention the observations of Rutherford’s Experiment

• Most of the a- particles passed through the foil without any deflection from their path.
• A few of them were deflected at some angles.
• Very few (1 in 10000) turned back on their original path.

Mention the conclusions of Rutherford’s experiment

• Most part of the atom is empty.
• The entire mass of the atom is concentrated in a small area that carries positive charge.
• Electrons revolve around the nucleus at a large distance from it.

Different types of atomic species

Atoms of the same elements with different mass numbers but the same atomic number are called isotopes. Ex: 1 H 1 , 1 H 2 , 1 H 3 .

Atoms of different elements which possess same mass number but a different atomic number. Ex: 18 Ar 40 and 20 Ca 40 , 1 H 3 and 2 He 3 , 7 N 14 and 8 O 17 .

Atoms of different elements which possess different atomic and mass number but same number of neutrons. Ex: 6 C 13 and 7 N 14 , 4 Be 9 and 5 B 10 .

Isodiaphers:

Nuclides which have different atomic and mass number but same difference between neutron and proton numbers. Ex: 90 Th 234 and 92 U 238

Mirror nuclei:

Nuclei having the same mass number but with the proton and neutron number interchanged. Ex : 1 H 3 and 2 He 3 , 3 Li 7 and 4 Be 7 .

Isoelectronic species:

Species having the same number of electrons. Ex: N 3- , O 2- , F – , Ne.

Electromagnetic Radiation

The invisible radiation caused by the combined effect of electric and magnetic field.

Photoelectric effect:

The phenomenon of ejection of electrons from the metal surface when light of suitable frequency strikes the metal. The electrons ejected are called photo-electrons.

hν= hν 0 + ½ mv 2

hν = Energy of the incident ray

hν 0 = Threshold energy= work function = Ionisation potential

ν 0 = Threshold frequency

½ mv 2 = kinetic energy of the emitted electrons

Planck’s Quantum Theory:

• Radiant energy is not emitted or absorbed continuously; rather it is absorbed or emitted by a body discontinuously in the form of small packets of energy known as quanta. For light one quantum of energy is called photon.
• The amount of energy associated with a quantum of radiation is directly proportional to its frequency.                    E α ν      or       E = hν
• A body can emit or absorb energy only in terms of integral multiples of quantum.E = nhν

Different types of spectrum

Mention the defects of Rutherford’s Model:

• According to Classical electromagnetic theory of Maxwell, whenever a charged body revolves it gives out radiations and loses energy. Thus electron which carries negative charge, and revolves around the nucleus should continuously emit radiations and loses energy. As a result, the electron follows a spiral path and collides with the nucleus.
• When gases under pressure are subjected to electric discharge, line spectrum is produced. This cannot be explained by Rutherford.

Hydrogen spectra:

Mention the main postulates of Bohr’s Theory

• The electrons move around the nucleus in certain specifically permitted circular orbits called energy levels or energy states. An electron in a particular energy level is associated with a definite amount of energy. While moving in a particular level, it neither absorbs nor loses energy.
• Only those energy levels are permitted in which the angular momentum of an electron is an integral multiple of h/2Π(The angular momentum are quantized)

• The electron absorbs energy when it jumps from lower to higher energy levels and emits energy when it moves from higher to lower levels. The energy gained or lost is equal to the difference between the energies of the two transitions.

Mention the significance of negative sign of energy:

• The energy of the electron in the atom is less than the energy of a free electron and thus the atom is a stable entity.
• Electron is attracted by the nucleus.
• To remove an electron, energy must be supplied from outside.

Mention the merits and demerits of Bohr’s Theory

Difference between rutherford’s and bohr’s concept:, de- broglie equation:.

De- Broglie suggested that electrons shows dual nature of particles (mass) and wave ( wave length).

From Planck’s Equation, E = hc/λ

From Einstein equation, E=mc 2

Where λ is the wave length

h is Planck’s Constant= 6.626 x 10 -27 erg sec

m and v are the mass and velocity respectively

Heisenberg’s Uncertainty Principle:

It states that it is impossible to measure simultaneously the position and exact velocity of a sub-atomic particle.

∆x.∆v≥ h/ 4Πm

∆x= Uncertainty in position

∆v = Uncertainty in velocity

Schrodinger’s Wave Equation:

Quantum numbers:

Set of four numbers which can depict the shape, location, energy, orientation etc of the electrons.

Principal Quantum Number(n):

• It represents the number of shells
• It can have values like 1,2,3,4….
• Determines the energy of the shell
• Provides the maximum number of electrons a shell can hold(2n 2 )

Azimuthal Quantum Number(l):

• The line spectrum when highly resolved shows that each such line is made up of finer lines called the fine spectrum. These fine lines are due to the elliptical orbits.
• The angular momentum of a moving electron in an elliptical path is
• The minimum value of l is 0 and the maximum is n-1. Spectroscopic terms-(s- Sharp,p- Principal,d-Diffuse,f- fundamental)
• The penetrating power and screening effect of an elliptical orbital is: s> p>d>f
• The order of energy: s< p< d <f

Magnetic Quantum Number(m):

• The splitting of fine lines under the magnetic field is called the Zeeman effect.
• Under the influence of the external magnetic field, the electrons present in the subshells orient themselves in certain regions of space called orbitals.
• The value of m is –L to +L including zero.

• The negative values of the magnetic quantum numbers signify that these orbitals are inclined in the direction opposite to the magnetic field and positive value indicates that these orbitals are inclined in the direction of the magnetic field.

The three dimensional region or space around the nucleus of an atom where there is maximum probability of finding an electron having certain energy.

• 1s, 2s and 3s orbitals are all spherical shaped.
• Size and energy of the orbitals -1s<2s<3s
• Nodes are the region where the probability of getting the electrons is zero.

• p-orbitals are dumble shaped.
• Each p-orbitals possess two lobes, where the probability of getting the electron is max.

• All the 3 orbitals possess the same energy and thus are degenerate.

                    Total no of nodes= Radial + angular nodes = n-1

d-Orbitals:

• d-orbitals are double dumble shaped with four lobes.
• dz 2 has two lobes. The annular portion orbital is called doughnut or belly band.
• All the 5 orbitals are degenerate.

Spin Quantum Number(s):

• It expresses the two opposite types of spinning motions of each electron.

• Its values are +1/2 and -1/2

• It signifies the mode of electron spin- clockwise or anticlockwise.
• It is the only quantum no which is not obtained from Schrodinger’s wave equation.

Some important graphs:

• Variation of radial part of wave function with distance from the nucleus.

• Variation of the square of radial wave function(radial probability density) with distance from the nucleus:

• Variation of radial probability distribution function with distance from the nucleus:

Pauli’s Exclusion Principle:

In an atom , no two electrons can have four identical quantum numbers.

Rules for filling up of orbitals:

1.aufbau principle:.

The orbitals are filled up with electrons in the increasing order of their energy.

2. (n+l) rule:

The electron comes in the vacant sub-shell with lowest value of (n+l). If the value of (n+l) is same, then electron enters the sub shell having lowest value of n.

Lanthanides and actinides are exception to the (n+l) rule.

3. Hund’s Multiplicity Rule:

• Electrons continue to enter different orbitals as long as possible. Electron pairing in any orbital cannot take place until each orbital of the same sub-shell contains 1 electron.
•  Two or more electrons each occupying different orbitals must have parallel spins.    (Same direction of spin ­­­↑ or ↓).
• Half filled and full filled subshells possess maximum stability. i.e., s 1 , s 2 , p 3 ,p 6 ,d 5 , d 10, f 7 , f 14 are stable configurations.

Why half-filled and full filled subshells possess maximum stability?

Symmetrical distribution of electrons:.

Subshells with half filled or completely filled electrons are found to have a more symmetrical distribution of electrons. Consequently, they have less energy and more stability.

Exchange energy:

The electrons occupying degenerate orbitals can exchange their positions with other electrons with the same spin. In this process, exchange energy is released. Greater the exchange energy, more stable the configuration is.

Elements with exceptional E.C-Cr(24), Cu(29), Nb(41), Mo(42),Ru(44), Rh(45), Pd(46), Ag(47), La(57), Ce(58), Gd(64), Pt(78), Au(79)

Magnetic nature of atoms and ions:

• Atoms with at least one unpaired electron are paramagnetic in nature( attracted by magnet).
• Atoms with no unpaired electron(s) are diamagnetic in nature(repelled by magnet).
• More the number of unpaired electrons, more the magnetic moment.

Related Posts