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NCERT Exemplar Solutions Class 11 Chemistry Chapter Structure of Atom

Multiple-choice Questions (Type-1)

1. Which of the following conclusions could not be derived from Rutherford’s α -particle scattering experiment?

(i) Most of the space in the atom is empty.

(ii) The radius of the atom is about 10–10 m while that of a nucleus is 10–15 m.

(iii) Electrons move in a circular path of fixed energy called orbits.

(iv) Electrons and the nucleus are held together by electrostatic forces of

attraction.

Solution:

Option (iii) is the answer.

2. Which of the following options does not represent ground state electronic configuration of an atom?

(i) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²

(ii) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s²

(iii) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

(iv)1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

Solution:

Option (ii) is the answer.

3. The probability density plots of 1s and 2s orbitals are given in Fig. 2.1:

The density of dots in a region represents the probability density of finding electrons in the region. Based on the above diagram which of the following statements is incorrect?

(i) 1s and 2s orbitals are spherical.

(ii) The probability of finding the electron is maximum near the nucleus.

(iii) The probability of finding the electron at a given distance is equal in all directions.

(iv) The probability density of electrons for 2s orbital decreases uniformly as the distance from the nucleus increases.

Solution:

Option (iv) is the answer.

4. Which of the following statement is not correct about the characteristics of cathode rays?

(i) They start from the cathode and move towards the anode.

(ii) They travel in a straight line in the absence of an external electrical or magnetic field.

(iii) Characteristics of cathode rays do not depend upon the material of electrodes in a cathode ray tube.

(iv) Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube.

Solution:

Option (iv) is the answer.

5. Which of the following statements about the electron is incorrect?

(i) It is a negatively charged particle.

(ii) The mass of an electron is equal to the mass of a neutron.

(iii) It is a basic constituent of all atoms.

(iv) It is a constituent of cathode rays.

Solution:

Option (ii) is the answer.

6. Which of the following properties of an atom could be explained correctly by Thomson Model of an atom?

(i) Overall neutrality of atom.

(ii) Spectra of a hydrogen atom.

(iii) Position of electrons, protons and neutrons in an atom.

(iv) Stability of atom.

Solution:

Option (i) is the answer.

7. Two atoms are said to be isobars if.

(i) they have the same atomic number but a different mass number.

(ii) they have the same number of electrons but a different number of neutrons.

(iii) they have the same number of neutrons but a different number of electrons.

(iv) the sum of the number of protons and neutrons is the same but the number of protons is different.

Solution:

Option (iv) is the answer.

8. The number of radial nodes for 3p orbital is __________.

(i) 3

(ii) 4

(iii) 2

(iv) 1

Solution:

Option (iv) is the answer.

9. Number of angular nodes for 4d orbital is __________.

(i) 4

(ii) 3

(iii) 2

(iv) 1

Solution:

Option (iii) is the answer.

10. Which of the following is responsible to rule out the existence of definite paths or trajectories of electrons?

(i) Pauli’s exclusion principle.

(ii) Heisenberg’s uncertainty principle.

(iii) Hund’s rule of maximum multiplicity.

(iv) Aufbau principle.

Solution:

Option (ii) is the answer.

11. A total number of orbitals associated with the third shell will be __________.

(i) 2

(ii) 4

(iii) 9

(iv) 3

Solution:

Option (iii) is the answer.

12. Orbital angular momentum depends on __________.

(i) l

(ii) n and l

(iii) n and m

(iv) m and s

Solution:

Option (i) is the answer.

13. Chlorine exists in two isotopic forms, Cl-37 and Cl-35 but its atomic mass is 35.5. This indicates the ratio of Cl-37 and Cl-35 is approximately

(i) 1:2

(ii) 1:1

(iii) 1:3

(iv) 3:1

Solution:

Option (iii) is the answer.

14. The pair of ions having same electronic configuration is __________.

(i) Cr³⁺, Fe³⁺

(ii) Fe³⁺, Mn²⁺

(iii) Fe³⁺, Co³⁺

(iv) Sc³⁺, Cr³⁺

Solution:

Option (ii) is the answer.

15. For the electrons of an oxygen atom, which of the following statements is correct?

(i) Z_eff for an electron in a 2s orbital is the same as Z_eff for an electron in a 2p orbital.

(ii) An electron in the 2s orbital has the same energy as an electron in the 2p orbital.

(iii) Z_eff for an electron in 1s orbital is the same as Z_eff for an electron in a 2s orbital.

(iv) The two electrons present in the 2s orbital have spin quantum numbers ms but of opposite sign.

Solution:

Option (iv) is the answer.

16. If travelling at the same speeds, which of the following matter waves has the shortest wavelength?

(i) Electron

(ii) An alpha particle (He2+)

(iii) Neutron

(iv) Proton

Solution:

Option (ii) is the answer.

II. Multiple Choice Questions (Type-II)

In the following questions, two or more options may be correct.

17. Identify the pairs which are not of isotopes?

(i) ₆X¹², ₆Y¹³

(ii) ₁₇X³⁵, ₆Y³⁷

(iii) ₆X¹⁴, ₇Y¹⁴

(iv) ₄X⁸, ₅Y⁸

Solution:

Option (iii) and (iv) are the answers.

18. Out of the following pairs of electrons, identify the pairs of electrons present in degenerate orbitals :

(i) (a) n = 3, l = 2, ml = –2, ms= − 1/2

(b) n = 3, l = 2, ml = –1, ms= − 1/2

(ii) (a) n = 3, l = 1, ml = 1, ms = +1/2

(b) n = 3, l = 2, ml = 1, ms = +1/2

(iii) (a) n = 4, l = 1, ml = 1, ms = +1/2

(b) n = 3, l = 2, ml = 1, ms = +1/2

(iv) (a) n = 3, l = 2, ml = +2, ms = − 1/2

(b) n = 3, l = 2, ml = +2, ms = +1/2

Solution:

Option (i) and (iv) are the answers.

19. Which of the following sets of quantum numbers is correct?

n l ml

(i) 1 1 +2

(ii) 2 1 +1

(iii) 3 2 –2

(iv) 3 4 –2

Solution:

Option (ii) and (iii) are the answers

20. In which of the following pairs, the ions are iso-electronic?

(i) Na⁺, Mg²⁺

(ii) Al³⁺, O^–

(iii) Na⁺, O^2-

(iv) N^3-, Cl^–

Solution;

Option (i) and (iii) are the answers.

21. Which of the following statements concerning the quantum numbers are correct?

(i) The angular quantum number determines the three-dimensional shape of the orbital.

(ii) The principal quantum number determines the orientation and energy of the orbital.

(iii) The magnetic quantum number determines the size of the orbital.

(iv) Spin quantum number of an electron determines the orientation of the spin of the electron relative to the chosen axis.

Solution:

Option (i) and (iv) are the answers.

III. Short Answer Type

22. Arrange s, p and d sub-shells of a shell in the increasing order of effective nuclear charge (Z_eff) experienced by the electron present in them.

Solution:

d

S orbitals shield the electrons from the nucleus more than p-orbitals which shield more in d.

23. Show the distribution of electrons in oxygen atom (atomic number 8) using orbital diagram.

Solution:

₈O¹⁶ = 1s² 2s² 2p⁴

24. Nickel atom can lose two electrons to form Ni²⁺ ion. The atomic number of nickel is 28. From which orbital will nickel lose two electrons.

Solution:

One Ni atom has 28 electrons and its electronic configuration is : [Ar] 4s² 3d⁸

It becomes Ni²⁺ by losing 2 electrons, hence configuration of Ni²⁺ is : [Ar] 4s⁰ 3d⁸

So, nickel loses two electrons from the 4s orbital, not the 3d orbital as per the Aufbau principle

25. Which of the following orbitals are degenerate?

3d_xy, 4d_xy, 3dz2 , 3d_yx, 4d_yx, 4d_zz

Solution:

The energy of orbitals depends on the principal quantum number or the main shell to a large extent. Hence, orbitals with an equal value of n will have the same levels of energy and will be called degenerate orbitals.

Degenerate orbitals are 3dxy, 3dz2, 3dyx because they have the same main shell n = 3.

And 4dxy, 4dyx, 4dzz because they have the same value of n=4.

26. Calculate the total number of angular nodes and radial nodes present in 3p orbital.

Solution:

Nodes are the region present among the orbitals where the probability density of finding electrons will be zero.

In case of np orbitals , radial nodes = n – l – 1 = 3 –1 – 1 = 1

Angular nodes = l = 1.

27. The arrangement of orbitals based on energy is based upon their (n+l ) value. Lower the value of (n+l ), lower is the energy. For orbitals having the same values of (n+l), the orbital with a lower value of n will have lower energy.

I. Based upon the above information, arrange the following orbitals in the increasing order of energy

(a) 1s, 2s, 3s, 2p

(b) 4s, 3s, 3p, 4d

(c) 5p, 4d, 5d, 4f, 6s

(d) 5f, 6d, 7s, 7p

II. Based upon the above information, solve the questions given below :

(a) Which of the following orbitals has the lowest energy?

4d, 4f, 5s, 5p

(b) Which of the following orbitals has the highest energy?

5p, 5d, 5f, 6s, 6p

Solution:

(i) (a) the increasing order of energy of the given orbital is : 1s >2s >2p> 3s

(b) the increasing order of energy of the given orbital is : 3s<3p<4s<4d

(c) the increasing order of energy of the given orbital is : 4d<5p<6s<4f<5d

(d)the increasing order of energy of the given orbital is: 7s<5f<6d<7p

(ii) (a) among the orbitals, 5s has the lowest energy.

the (n+l) value for 5s is the lowest = 5 + 0 = 5. Other orbitals have (n+l )value more than 5 –

5p= 5 + 1 = 6 , 4f = 4 + 3 = 7 , 4d = 4 + 2 = 6.

(b) among the orbitals , 5f has the highest energy because the (n +l ) value – 5 + 3 = 8 is highest.

5d = 5 + 2 = 7 , 5p = 5 + 1= 6 , 6s =6 + 0 = 6 , 6p =6 + 1 = 7.

28. Which of the following will not show deflection from the path on passing through an electric field?

Proton, cathode rays, electron, neutron

Solution:

Neutron will not show deflection from the path on passing through an electric field.

This is due to the neutral nature of the neutron particles. Therefore, it has no charge and does not get affected by any electric field.

Among other 3 particles proton (positive ), electron (negative), cathode rays (the beam of electrons, negatively charged) all have charges in them so they will get deflected easily by an electric field.

29. An atom having atomic mass number 13 has 7 neutrons. What is the atomic number of the atom?

Solution:

The mass number of an atom = number of protons + number of neutrons

Therefore atomic number ( number of protons ) = mass number – no. Of neutrons.

The atomic number of atom: 13 – 7 = 6.

30. Wavelengths of different radiations are given below :

λ(A) = 300 nm

λ(B) = 300 μm

λ(c) = 3 nm

λ (D) 30 A°

Arrange these radiations in the increasing order of their energies.

Solution:

According to Planck’s quantum theory, energy is related to the frequency of radiation by :

E = h × Frequency

So, E is proportional to 1/ λ

Hence, the relation b/w energy and wavelength are inversely proportional, therefore lesser the wavelength higher will be the energy of the radiation.

For the given wavelengths

λ(A) = 300 nm = 300 x 10-9 m = 3 x 10 -7 m

λ(B) = 300 μm = 3 00 x 10-6 = 3 x 10-4

λ(c) = 3 nm = 3 x 10 – 9

λ (D) 30 A° = 3 X 10 – 9

the increasing order of the given wavelengths : λ(c) = λ (D) <λ(A)< λ(B)

hence the increasing order of energy will be the opposite: λ(B)< λ(A:<λ(c) = λ (D)

31. The electronic configuration of the valence shell of Cu is 3d¹⁰ 4s¹ and not 3d⁹4s². How is this configuration explained?

Solution:

Configuration with filled and half-filled orbitals has extra stability. In 3d104s1, d orbitals are filled and s orbitals are half-filled.

32. The Balmer series in the hydrogen spectrum corresponds to the transition from n₁ = 2 to n₂ = 3,4,………. This series lies in the visible region. Calculate the wavenumber of the line associated with the transition in Balmer series when the electron moves to n = 4 orbit. (R_H= 109677 cm^-1)

Solution:

According to Bohr’s model for the hydrogen atom

ν = R_H(1/n₁²-1/ n₂²)cm^-1

here, n₁ = 2 and n₂ = 4 and R_H = Rydberg’s constant = 109677

Hence, wave number -v= 109677 ( ¼-1/16)

= 20564.44cm^-1

33. According to de Broglie, the matter should exhibit dual behaviour, that is both particle and wave-like properties. However, a cricket ball of mass 100 g does not move like a wave when it is thrown by a bowler at a speed of 100 km/h. Calculate the wavelength of the ball and explain why it does not show wave nature.

Solution:

m= 100g or 0.1kg

ν= 100km/h =100*1000/60*60 = 1000/36m/s

λ =h/mν = 2.387*10-34m

34. What is the experimental evidence in support of the idea that electronic energies in an atom are quantized?

Solution:

The bright-line spectrum shows that the energy levels in an atom are quantized. These lines are obtained as a result of electronic transitions between the energy .and if the electronic energy levels were continuous and not quantized or discrete; the atomic spectra would have shown a continuous absorption(from lower to higher energy level transition) or emission (from higher to lower energy level transition.

35. Out of electron and proton which one will have, a higher velocity to produce matter waves of the same wavelength? Explain it.

Solution:

Out of electron and proton, being the lighter particle electron will have a higher velocity and will also produce matter waves of the same wavelength.

36. A hypothetical electromagnetic wave is shown in Fig. 2.2. Find out the wavelength of the radiation.

Solution:

The wavelength can be defined as the distance between two alike successive points in a wave (usually b/w two maxima s i.e. peaks or two minima s i.e. troughs are shown in the fig.)

So, for the given hypothetical wave, wavelength λ = 4 * 2.16 pm

= 8.64 pm.

37. Chlorophyll present in green leaves of plants absorbs light at 4.620 × 1014 Hz. Calculate the wavelength of radiation in nanometer. Which part of the electromagnetic spectrum does it belong to?

Solution:

Relation b/w wavelength and frequency can be expressed as :

λ = c/ν, where c be the velocity of light and ν is the frequency of the radiation.

For the given problem λ = 3 x 108 ms-1 / 4.620 x 1014 Hz

= 0.6494 times10-6m-1

38. What is the difference between the terms orbit and orbital?

Solution:

The orbit stands for a definite circular path for the electrons to revolve around the nucleus. It represents the two-dimensional motion of the electrons around the nucleus, the orbital is not that well-defined path because it’s a region around the nucleus where the probability of finding an electron is maximum.

39. Table-tennis ball has a mass 10 g and a speed of 90 m/s. If speed can be measured within an accuracy of 4% what will be the uncertainty in speed and position?

Solution:

According to Heisenberg’s uncertainty principle :

“It is fundamentally impossible to determine accurately both the velocity and the position of a particle at the same time.

∆x. ∆p ≥ h/4π

From the given problem,

mass of the ball = 4 g and speed is = 90 m /s

hence,the uncertainty of speed is ∆v = 4/100 × 90 = 3.6 m/s

∆x is given by ∆x = h/4πm∆v

Hence , the uncertainty of postion is ∆x = 6.26 × 10-34 / 4 × 3.14 × 4× 3.6

= 1.46 x 10-33 m

40. The effect of the uncertainty principle is significant only for the motion of microscopic particles and is negligible for the macroscopic particles. Justify the statement with the help of a suitable example.

Solution:

The uncertainty principle is only significantly applicable for microscopic particles and not macroscopic particles this can be concluded from the measurement of uncertainty:

For example, if we take a particle or an object of mass 1 milligram i.e. 10-6 kg )

We calculate the,

∆x. ∆ν = 60626*10-34/ 4*3.14*10-6

= 10-28 m-2 s -1

The value we got is negligible and very insignificant for the uncertainty principle to apply to the particle.

41. The hydrogen atom has only one electron, so mutual repulsion between electrons is absent. However, in multielectron atoms mutual repulsion between the electrons is significant. How does this affect the energy of an electron in the orbitals of the same principal quantum number in multielectron atoms?

Solution:

The hydrogen atom has only one electron, so mutual repulsion between electrons is absent. However, in multielectron atoms mutual repulsion between the electrons is significant. How does this affect the energy of an electron in the orbitals of the same principal quantum number in multielectron atoms?

IV. Matching Type

In some of the following questions, one option of the left column maybe

Correlated to more than one option in the right column.

42. Match the following species with their corresponding ground state electronic

configuration.

Atom / Ion (i) Cu (ii) Cu2+ (iii) Zn2+ (iv) Cr3+

Electronic configuration (a) 1s2 2s2 2p6 3s2 3p6 3d10 (b) 1s2 2s2 2p6 3s2 3p6 (c) 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (d) 1s2 2s2 2p6 3s2 3p63d9 (e) 1s2 2s2 2p6 3s2 3p6 3d3

Solution:

(i) → c

(ii) →d

(iii) →a

(iv) →e

43. Match the quantum numbers with the information provided by these.

Quantum number Information provided

(i) Principal quantum number (ii) Azimuthal quantum number (iii) Magnetic quantum number (iv) Spin quantum number

(a) orientation of the orbital (b) energy and size of orbital (c)spin of an electron (d) shape of the orbital

Solution:

(i) → b

(ii) → d

(iii) → a

(iv) → c

44. Match the following

Rules Statements

(i) Hund’s Rule (ii) Aufbau Principle (iii) Pauli Exclusion Principle (iv) Heisenberg’s Uncertainty Principle

(a) No two electrons in an atom can have the same set of four quantum numbers. (b) Half-filled and filled orbitals have extra stability. (c) The pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital is singly occupied. (d) It is impossible to determine the exact position and exact the momentum of a subatomic particle simultaneously. (e) In the ground state of atoms, orbitals are filled in the order of their increasing energies.

Solution;

(i) → c

(ii) → e

(iii) → a

(iv) → d

45. Match the following

(i) X-rays (ii) UV (iii) Long radio waves (iv) Microwave

(a) ν = 100-0 104 Hz (b) ν = 1010Hz (c) ν = 1016 Hz (d) ν = 1018Hz

Solution:

(i)→ d

(ii)→ c

(iii) →a

(iv) →b

46. Match the following

(i) Photon (ii) Electron (iii) ψ2 (iv) The principal quantum number n

(a)Value is 4 for N shell (b)Probability density (c)Always a positive value (d)Exhibits both momentum and wavelength

Solution:

(i)→ d

(ii)→ d

(iii) →b, c

(iv)→ c, a

47. Match species are given in Column I with the electronic configuration given in Column II.

Column I (i) Cr (ii) Fe2+ (iii) Ni2+ (iv) Cu

(a) [Ar]3d84s0 (b) [Ar]3d104s1 (c) [Ar]3d64s0 (d) [Ar] 3d54s1 (e) [Ar]3d64s2

Solution:

(i) →d

(ii) →c

(iii) →a

(iv) →b

V. Assertion and Reason Type

In the following questions, a statement of Assertion (A) followed by a statement

Reason (R) is given. Choose the correct option out of the choices given

below each question.

48. . Assertion (A): All isotopes of a given element show the same type of chemical behaviour.

Reason (R): The chemical properties of an atom are controlled by the number of electrons in the atom.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

Solution:

Option (i) is correct

49. Assertion (A): The black body is an ideal body that emits and absorbs radiations of all frequencies.

Reason (R): The frequency of radiation emitted by a body goes from a lower frequency to higher frequency with an increase in temperature.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the explanation of A.

(iii) A is true and R is false.

(iv) Both A and R are false.

Solution:

Option (ii) is the answer.

50. Assertion (A): It is impossible to determine the exact position and exact

the momentum of an electron simultaneously.

Reason (R): The path of an electron in an atom is clearly defined.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true and R is not the correct explanation of A.

(iii) A is true and R is false.

(iv) Both A and R are false.

Solution:

Option (iii) is correct

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Frequently Asked Questions on NCERT Exemplar Solutions for Class 11 Chemistry Chapter Structure of Atom

Q1

What is the experimental evidence in support of the idea that electronic energies in an atom are quantized in Chapter 2 of NCERT Exemplar Solutions for Class 11 Chemistry?

The bright-line spectrum shows that the energy levels in an atom are quantized. These lines are obtained as a result of electronic transitions between the energy, and if the electronic energy levels were continuous and not quantized or discrete, the atomic spectra would have shown a continuous absorption (from lower to higher energy level transition) or emission (from higher to lower energy level transition).

Q2

Can I get the NCERT Exemplar Solutions for Class 11 Chemistry Chapter 2 PDF online?

The NCERT Exemplar Solutions for Class 11 Chemistry Chapter 2 PDF is available on ANAND CLASSES (A School Of Competitions) website. These solutions are provided with a free download option which can be accessed by the students effortlessly. Students can download the PDF and refer to them while answering the textbook questions to get their doubts cleared instantly. The main advantage of using these solutions is that they can be used by students anywhere and at any time without any difficulty.

Q3

Do the NCERT Exemplar Solutions for Class 11 Chemistry Chapter 2 help students with their annual exam preparation?

The NCERT Exemplar Solutions for Class 11 Chemistry Chapter 2 are very helpful for the students for various purposes. The expert faculty at ANAND CLASSES (A School Of Competitions) has used simple and understandable language, which makes it easy for the students to understand them. These solutions provide students with a strong foundation of fundamental concepts which are important from the exam point of view. By using these solutions, they can save a lot of time in searching for the correct answer as per the CBSE syllabus and guidelines.

Neeraj Anand, Param Anand

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.

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JEE Syllabus for Chapter - ATOMIC STRUCTURE

According to the JEE syllabus, the "Atomic Structure" chapter covers key concepts like: Nature of electromagnetic radiation, photoelectric effect, spectrum of the hydrogen atom, Bohr model of ahydrogen atom - its postulates, derivation of the relations for the energy of the electron and radii of the different orbits, limitations of Bohr's model, dual nature of matter, de Broglie's relationship, Heisenberguncertainty principle, elementary ideas of quantum mechanics, the quantum mechanical model of the atomand its important features, concept of atomic orbitals as one-electron wave functions, variation of Ψ and Ψ² with r for 1s and 2s orbitals, various quantum numbers (principal, angular momentumand magneticquantum numbers) and their significance, shapes of s, p and d - orbitals, electron spin and spin quantumnumber, rules for filling electrons in orbitals – Aufbau principle, Pauli's exclusion principle and Hund'srule, electronic configuration of elements and extra stability of half-filled and completely filled orbitals.

NEET Syllabus for Chapter - ATOMIC STRUCTURE

According to the NEET syllabus, the "Atomic Structure" chapter covers key concepts like: subatomic particles (protons, electrons, neutrons), atomic number and mass number, various atomic models (Dalton's, Thomson's, Rutherford's, Bohr's), quantum mechanical model (Schrödinger's equation, quantum numbers - principal, azimuthal, magnetic, and spin), shapes of orbitals (s, p, d), electronic configuration of elements based on Aufbau principle, Pauli exclusion principle, and Hund's rule; including the limitations of Bohr's model and the concept of dual nature of matter with de Broglie's relationship and Heisenberg's uncertainty principle. 

MCQs on Structure of Atom for Class 11 CBSE Board Exam

Here are some multiple-choice questions (MCQs) on Structure of Atom for Class 11 CBSE along with detailed explanations:

1. Which of the following statements about the nucleus of an atom is correct?

A) It contains protons and neutrons
B) It has a negative charge
C) It occupies most of the volume of the atom
D) It is responsible for chemical properties of an atom

Answer: A) It contains protons and neutrons

Explanation:
The nucleus of an atom contains protons (positively charged) and neutrons (neutral). The electrons revolve around the nucleus in different energy levels. The nucleus has a positive charge due to the presence of protons. It occupies a very small volume but contributes to almost the entire mass of the atom.

2. The total number of electrons that can be accommodated in the second shell (L-shell) is:

A) 2
B) 8
C) 18
D) 32

Answer: B) 8

Explanation:
The maximum number of electrons in a shell is given by 2n², where n is the shell number.
For the second shell (n = 2):
Max electrons = 2(2²) = 8.

3. Which of the following is NOT a postulate of Bohr’s atomic model?

A) Electrons revolve around the nucleus in fixed orbits
B) Electrons emit or absorb energy when they jump between orbits
C) Energy levels are quantized
D) Electrons can have any random energy value

Answer: D) Electrons can have any random energy value

Explanation:
Bohr's model states that electrons revolve in fixed energy levels, and they cannot have arbitrary energy. Electrons only gain or lose energy when they transition between these discrete orbits.

4. The isotope of hydrogen that contains one proton and two neutrons is:

A) Protium
B) Deuterium
C) Tritium
D) None of these

Answer: C) Tritium

Explanation:

• Protium (¹H) → 1 proton, 0 neutrons
• Deuterium (²H) → 1 proton, 1 neutron
• Tritium (³H) → 1 proton, 2 neutrons

5. The wave nature of electrons was proposed by:

A) Bohr
B) Heisenberg
C) de Broglie
D) Rutherford

Answer: C) de Broglie

Explanation:
Louis de Broglie proposed that electrons exhibit both particle and wave nature (wave-particle duality). His equation λ = h/mv relates the wavelength (λ) of a moving particle to its momentum.

6. The uncertainty principle was proposed by:

A) Bohr
B) Heisenberg
C) Rutherford
D) Schrodinger

Answer: B) Heisenberg

Explanation:
Heisenberg's Uncertainty Principle states that it is impossible to simultaneously determine the exact position and momentum of an electron.

7. The quantum number that describes the shape of an orbital is:

A) Principal quantum number (n)
B) Azimuthal quantum number (l)
C) Magnetic quantum number (m)
D) Spin quantum number (s)

Answer: B) Azimuthal quantum number (l)

Explanation:
The Azimuthal quantum number (l) determines the shape of orbitals:

• s-orbital (l = 0) → Spherical
• p-orbital (l = 1) → Dumbbell
• d-orbital (l = 2) → Complex
• f-orbital (l = 3) → More complex

8. Which of the following orbitals cannot exist?

A) 1s
B) 2p
C) 3f
D) 4d

Answer: C) 3f

Explanation:
For an orbital to exist, the Azimuthal quantum number (l) must satisfy:
l = 0 to (n-1), where n is the principal quantum number.
For n = 3, possible l values: 0 (s), 1 (p), 2 (d) → No f-orbital.

9. Which of the following elements has the electronic configuration: 1s² 2s² 2p⁶ 3s¹?

A) Sodium (Na)
B) Magnesium (Mg)
C) Aluminium (Al)
D) Potassium (K)

Answer: A) Sodium (Na)

Explanation:

• 1s² 2s² 2p⁶ 3s¹ is the electronic configuration of sodium (Na) (Atomic number 11).
• Magnesium (Mg) = 1s² 2s² 2p⁶ 3s²
• Aluminium (Al) = 1s² 2s² 2p⁶ 3s² 3p¹

10. The shape of an s-orbital is:

A) Spherical
B) Dumbbell
C) Double dumbbell
D) Complex

Answer: A) Spherical

Explanation:
The s-orbital is spherically symmetric around the nucleus. p-orbitals are dumbbell-shaped.

11. The number of orbitals present in the third energy level (n = 3) is:

A) 3
B) 9
C) 18
D) 5

Answer: B) 9

Explanation:
Total orbitals in an energy level = n²
For n = 3, orbitals = 3² = 9
(1 s-orbital, 3 p-orbitals, 5 d-orbitals)

12. If an electron has quantum numbers n = 3, l = 2, what type of orbital is it in?

A) 3s
B) 3p
C) 3d
D) 3f

Answer: C) 3d

Explanation:

• n = 3 (Third shell)
• l = 2 corresponds to d-orbital → 3d orbital

13. The maximum number of electrons that can be accommodated in an orbital is:

A) 1
B) 2
C) 4
D) 6

Answer: B) 2

Explanation:
Each orbital can hold a maximum of 2 electrons with opposite spins, as per Pauli's exclusion principle.

14. Which quantum number determines the energy of an electron in a hydrogen atom?

A) Principal quantum number (n)
B) Azimuthal quantum number (l)
C) Magnetic quantum number (m)
D) Spin quantum number (s)

Answer: A) Principal quantum number (n)

Explanation:
For hydrogen-like atoms, the energy of an electron depends only on n.

15. The concept of orbitals was introduced by:

A) Bohr
B) Rutherford
C) Schrodinger
D) Heisenberg

Answer: C) Schrodinger

Explanation:
Schrodinger’s wave equation introduced orbitals (regions of high probability of finding an electron), replacing Bohr’s fixed orbits.

Assertion and Reason (A-R) type questions on Structure of Atom for Class 11 CBSE Board Exam

Here are some Assertion and Reason (A-R) type questions on Structure of Atom for Class 11 CBSE, along with detailed explanations.

How to Answer Assertion-Reason Questions?

Each question consists of two statements:

• Assertion (A): A statement of fact.
• Reason (R): An explanation for the assertion.

You must choose the correct option:

1. Both A and R are true, and R is the correct explanation of A.
2. Both A and R are true, but R is NOT the correct explanation of A.
3. A is true, but R is false.
4. A is false, but R is true.

1. Assertion (A): The nucleus of an atom contains protons and neutrons.

Reason (R): The electrons in an atom revolve around the nucleus in fixed orbits.

✅ Answer: Option (2) (Both A and R are true, but R is not the correct explanation of A.)

🔹 Explanation:

• The nucleus is composed of protons and neutrons, which was confirmed by Rutherford’s experiment.
• Electrons revolve around the nucleus in fixed orbits, as described in Bohr’s atomic model.
• However, the presence of electrons in orbits does not explain why the nucleus contains protons and neutrons.

2. Assertion (A): The maximum number of electrons in a shell is given by 2n².

Reason (R): Electrons in an atom are arranged in different shells around the nucleus.

✅ Answer: Option (1) (Both A and R are true, and R is the correct explanation of A.)

🔹 Explanation:

• The Bohr-Bury rule states that the maximum number of electrons in a shell = 2n², where n is the shell number.
• Electrons are distributed in different shells, and this follows the energy level distribution principle.

Example:

• K shell (n = 1): 2(1²) = 2 electrons
• L shell (n = 2): 2(2²) = 8 electrons
• M shell (n = 3): 2(3²) = 18 electrons

Thus, R correctly explains A.

3. Assertion (A): The mass of an atom is concentrated in the nucleus.

Reason (R): The electrons have negligible mass compared to protons and neutrons.

✅ Answer: Option (1) (Both A and R are true, and R is the correct explanation of A.)

🔹 Explanation:

• The nucleus contains protons and neutrons, which are heavy particles.
• Electrons are much lighter (mass = 1/1836 of a proton), so their contribution to atomic mass is negligible.
• Thus, the nucleus contains almost all the mass of an atom, and R explains A correctly.

4. Assertion (A): The energy of electrons in an atom is quantized.

Reason (R): Electrons can exist at any random energy level.

✅ Answer: Option (3) (A is true, but R is false.)

🔹 Explanation:

• According to Bohr’s model, electrons can only occupy specific, discrete energy levels (quantized energy states).
• Electrons cannot have arbitrary energy values.
• Hence, A is true, but R is false because electrons follow quantized energy levels.

5. Assertion (A): The Heisenberg Uncertainty Principle states that the exact position and momentum of an electron cannot be simultaneously determined.

Reason (R): Electrons move in fixed circular orbits around the nucleus.

✅ Answer: Option (3) (A is true, but R is false.)

🔹 Explanation:

• Heisenberg’s Uncertainty Principle states:
  Δx × Δp ≥ h/4π
  (where Δx = uncertainty in position, Δp = uncertainty in momentum)
• This means we cannot know both the exact position and momentum of an electron at the same time.
• Bohr’s model (fixed orbits) was later replaced by Schrodinger’s model (probability orbitals).
• R is false because electrons do not move in fixed orbits but in probabilistic regions (orbitals).

6. Assertion (A): The azimuthal quantum number (l) determines the shape of an orbital.

Reason (R): The principal quantum number (n) determines the size of the orbital.

✅ Answer: Option (1) (Both A and R are true, and R is the correct explanation of A.)

🔹 Explanation:

• The Azimuthal quantum number (l) determines the shape of an orbital (s, p, d, f).
• The Principal quantum number (n) determines the size and energy of an orbital.
• Since both statements are correct and R correctly explains A, the answer is Option (1).

7. Assertion (A): The 3f orbital exists.

Reason (R): The azimuthal quantum number (l) for an f-orbital is 3.

✅ Answer: Option (4) (A is false, but R is true.)

🔹 Explanation:

• l = 3 represents an f-orbital, but for an orbital to exist, n must be greater than l.
• For n = 3, the possible values of l are 0 (s), 1 (p), and 2 (d).
• f-orbital (l = 3) is possible only when n ≥ 4, so 3f does not exist.
• Hence, A is false, but R is true.

8. Assertion (A): The Pauli Exclusion Principle states that no two electrons in an atom can have the same set of four quantum numbers.

Reason (R): An orbital can accommodate a maximum of two electrons with opposite spins.

✅ Answer: Option (1) (Both A and R are true, and R is the correct explanation of A.)

🔹 Explanation:

• Pauli's Exclusion Principle states that no two electrons in an atom can have identical quantum numbers.
• This is because each orbital can hold only two electrons with opposite spins.
• Since R explains A correctly, the answer is Option (1).

9. Assertion (A): The probability of finding an electron is maximum near the nucleus.

Reason (R): The probability density of an electron is given by the wave function (ψ²).

✅ Answer: Option (1) (Both A and R are true, and R is the correct explanation of A.)

🔹 Explanation:

• The Schrodinger wave equation gives ψ², which represents the probability density of finding an electron in a region.
• For s-orbitals, the highest probability is near the nucleus.
• Since R explains A correctly, the answer is Option (1).

FAQs (Important Questions & Answers) on Structure of Atom – Class 11

1. What is the structure of an atom?

The structure of an atom consists of a nucleus containing protons and neutrons, surrounded by electrons moving in discrete energy levels or shells around the nucleus.

2. Who discovered the atom?

The concept of the atom was first proposed by John Dalton in his Atomic Theory (1808). Later, significant contributions were made by J.J. Thomson, Rutherford, and Bohr to explain its structure.

3. What are the fundamental particles of an atom?

The three fundamental particles of an atom are:

• Proton (p⁺) – Positively charged, found in the nucleus.
• Neutron (n⁰) – Neutral, found in the nucleus.
• Electron (e⁻) – Negatively charged, revolves around the nucleus.

4. What is Dalton’s Atomic Theory?

Dalton’s Atomic Theory (1808) states that:

• Matter is made of indivisible atoms.
• Atoms of an element are identical in mass and properties.
• Atoms combine in whole-number ratios to form compounds.
• Atoms cannot be created or destroyed in a chemical reaction.

5. What was J.J. Thomson’s Model of the Atom?

J.J. Thomson proposed the Plum Pudding Model (1897), where an atom was visualized as a positively charged sphere with negatively charged electrons embedded in it, like plums in a pudding. However, this model was later disproven by Rutherford’s experiment.

6. What is Rutherford’s Atomic Model?

Rutherford’s Gold Foil Experiment (1911) led to the discovery that:

• An atom has a dense positively charged nucleus.
• Electrons revolve around the nucleus.
• Most of the atom is empty space.
  This model could not explain atomic stability, which was later addressed by Bohr’s model.

7. What is Bohr’s Atomic Model?

Niels Bohr (1913) proposed that:

• Electrons revolve in fixed energy levels (shells).
• They do not lose energy while in stable orbits.
• Energy is absorbed or emitted when an electron jumps between energy levels.

8. What are Quantum Numbers?

Quantum numbers define the position and energy of an electron in an atom:

1. Principal Quantum Number (n) – Represents the main energy level.
2. Azimuthal Quantum Number (l) – Defines the shape of the orbital.
3. Magnetic Quantum Number (mₗ) – Indicates the orientation of the orbital.
4. Spin Quantum Number (mₛ) – Describes the spin of an electron (+½ or -½).

9. What is the Heisenberg Uncertainty Principle?

Proposed by Werner Heisenberg, it states that it is impossible to simultaneously determine the exact position and momentum of an electron in an atom.

10. What is the difference between an orbit and an orbital?

• Orbit – A fixed path in which electrons revolve (Bohr’s Model).
• Orbital – A 3D region in space where the probability of finding an electron is highest (Quantum Mechanical Model).

11. What is the Aufbau Principle?

The Aufbau Principle states that electrons fill atomic orbitals in order of increasing energy levels, i.e., lower-energy orbitals are filled first before higher ones.

12. What is Hund’s Rule?

Hund’s Rule states that in degenerate (equal energy) orbitals, electrons fill each orbital singly before pairing up.

13. What is Pauli’s Exclusion Principle?

Pauli’s Exclusion Principle states that no two electrons in an atom can have the same set of four quantum numbers. This means an atomic orbital can hold a maximum of two electrons with opposite spins.

14. What is the Quantum Mechanical Model of the Atom?

Developed by Schrödinger, this model describes electrons as wave-like particles with a probability distribution around the nucleus instead of fixed orbits. It introduces atomic orbitals (s, p, d, f) as probable electron locations.

15. What are the types of orbitals in an atom?

There are four types of orbitals:

• s-orbital – Spherical shape (holds max 2 electrons).
• p-orbital – Dumbbell shape (holds max 6 electrons).
• d-orbital – Complex shape (holds max 10 electrons).
• f-orbital – Complex shape (holds max 14 electrons).

16. What is the significance of the atomic number and mass number?

• Atomic Number (Z) – Number of protons in an atom.
• Mass Number (A) – Sum of protons and neutrons in an atom’s nucleus.

17. What are Isotopes, Isobars, and Isotones?

• Isotopes – Same atomic number, different mass number (e.g., Hydrogen: ¹H, ²H, ³H).
• Isobars – Same mass number, different atomic number (e.g., ¹⁴C and ¹⁴N).
• Isotones – Same number of neutrons, different atomic and mass numbers (e.g., ¹⁴C and ¹⁵N).

18. What is the Dual Nature of Electrons?

Proposed by de Broglie, it states that electrons exhibit both particle and wave-like properties, known as wave-particle duality.

19. What is the Electronic Configuration of an Atom?

The electronic configuration of an atom describes how electrons are distributed in different orbitals. It follows the Aufbau principle, Hund’s rule, and Pauli’s exclusion principle (e.g., Oxygen (O) = 1s² 2s² 2p⁴).

20. Why is the Bohr model still used despite its limitations?

Although the Bohr model fails for multi-electron atoms and does not explain fine spectral lines, it is still useful because it provides a simple and understandable representation of electron energy levels.

📚 CBSE Class 11 Chemistry: Structure of Atom – Complete Syllabus Overview

Are you a Class 11 CBSE student aiming to master "Structure of Atom"? Here’s a comprehensive breakdown of the chapter to help you focus your studies!

Case Study-Based MCQs on Structure of Atom – Class 11 CBSE

Case study-based questions are designed to test your analytical skills and conceptual understanding. Below are five case studies with multiple-choice questions (MCQs) and detailed explanations.

Case Study 1: Discovery of Atomic Structure

The structure of an atom has been explored through various experiments. J.J. Thomson discovered electrons using a cathode ray tube experiment, while Goldstein discovered protons using canal rays. Rutherford’s alpha particle scattering experiment led to the discovery of a dense nucleus at the center of the atom. Later, Bohr’s model refined our understanding by proposing that electrons move in discrete energy levels.

1.1 What was the key observation in Rutherford’s experiment?

A) Most alpha particles were deflected back
B) Alpha particles passed through the gold foil without deflection
C) All alpha particles got absorbed by the gold foil
D) Electrons were ejected from the gold foil

✅ Answer: B) Alpha particles passed through the gold foil without deflection

🔹 Explanation: Rutherford’s experiment showed that most of the space in an atom is empty, as most alpha particles passed straight through the foil. However, a few were deflected, indicating the presence of a dense, positively charged nucleus.

1.2 What did J.J. Thomson’s experiment conclude?

A) Atoms have a nucleus
B) Atoms are indivisible
C) Atoms contain negatively charged particles
D) Atoms are mostly empty space

✅ Answer: C) Atoms contain negatively charged particles

🔹 Explanation: J.J. Thomson’s cathode ray experiment showed the presence of negatively charged electrons, leading to the "plum pudding model" of the atom.

Case Study 2: Bohr’s Atomic Model

Niels Bohr proposed that electrons move in fixed orbits (energy levels) around the nucleus without losing energy. He introduced the quantization of energy levels and explained the emission spectra of hydrogen.

2.1 According to Bohr’s model, what happens when an electron jumps from a higher to a lower energy level?

A) The atom becomes unstable
B) The electron absorbs energy
C) The electron loses energy in the form of radiation
D) The electron disappears

✅ Answer: C) The electron loses energy in the form of radiation

🔹 Explanation: When an electron moves from a higher energy level to a lower one, it releases energy in the form of light (photon), which creates the atomic emission spectrum.

2.2 Which of the following is NOT a postulate of Bohr’s model?

A) Electrons revolve around the nucleus in circular orbits
B) Electrons can have any energy within an orbit
C) Energy levels are quantized
D) Electrons do not lose energy while revolving

✅ Answer: B) Electrons can have any energy within an orbit

🔹 Explanation: According to Bohr, electrons can only exist in specific, quantized energy levels, meaning they cannot have any arbitrary energy.

Case Study 3: Quantum Mechanical Model

The quantum mechanical model, developed by Schrodinger, replaced Bohr’s model. It introduced the concept of orbitals, where electrons are found as a probability distribution rather than fixed orbits. The model is based on wave-particle duality and Heisenberg’s Uncertainty Principle.

3.1 What does Heisenberg’s Uncertainty Principle state?

A) Electrons are present in fixed orbits
B) We cannot determine both position and momentum of an electron simultaneously
C) Electrons revolve around the nucleus like planets
D) Electrons follow a predictable path

✅ Answer: B) We cannot determine both position and momentum of an electron simultaneously

🔹 Explanation: Heisenberg’s Uncertainty Principle states that it is impossible to simultaneously know the exact position and momentum of an electron, leading to the concept of probability orbitals.

3.2 Which scientist introduced the wave equation to describe electron behavior?

A) Bohr
B) Heisenberg
C) Schrodinger
D) Rutherford

✅ Answer: C) Schrodinger

🔹 Explanation: Erwin Schrodinger developed the wave equation, which describes the probability of finding an electron in a given region of space, known as an orbital.

Case Study 4: Quantum Numbers

Each electron in an atom is described by four quantum numbers:

1. Principal quantum number (n) → Determines the energy level
2. Azimuthal quantum number (l) → Determines the shape of orbitals
3. Magnetic quantum number (m) → Determines the orientation of orbitals
4. Spin quantum number (s) → Determines electron spin

4.1 What does the principal quantum number (n) determine?

A) Shape of the orbital
B) Orientation of the orbital
C) Size and energy of the orbital
D) Spin of the electron

✅ Answer: C) Size and energy of the orbital

🔹 Explanation: The principal quantum number (n) determines the size of the electron cloud and the energy level where the electron resides.

4.2 How many orbitals are present in the third energy level (n = 3)?

A) 3
B) 9
C) 18
D) 5

✅ Answer: B) 9

🔹 Explanation: The number of orbitals in an energy level is given by n².
For n = 3 → 3² = 9 orbitals (1s, 3p, 5d).

Case Study 5: Electronic Configuration and Periodicity

The electronic configuration of an atom follows the Aufbau principle, Pauli Exclusion Principle, and Hund’s Rule. Elements in the periodic table are arranged based on their atomic number and valence electron configuration.

5.1 Which rule states that electrons fill orbitals in order of increasing energy?

A) Hund’s Rule
B) Aufbau Principle
C) Pauli’s Exclusion Principle
D) Heisenberg’s Principle

✅ Answer: B) Aufbau Principle

🔹 Explanation: The Aufbau Principle states that electrons occupy the lowest energy orbitals first before filling higher ones.

5.2 The electronic configuration of an element is 1s² 2s² 2p⁶ 3s². What is the element?

A) Magnesium (Mg)
B) Sodium (Na)
C) Aluminium (Al)
D) Oxygen (O)

✅ Answer: A) Magnesium (Mg)

🔹 Explanation:

• 1s² 2s² 2p⁶ 3s² corresponds to an atomic number of 12, which is magnesium (Mg).

CBSE Board Exam Syllabus for Chapter 2: Structure of Atom

Key Topics

1. Discovery of Subatomic Particles
   • Electron, Proton, and Neutron
   • Experiments by J.J. Thomson and Rutherford
2. Atomic Models
   • Thomson’s Model – The "Plum Pudding" model
   • Rutherford’s Nuclear Model – Gold foil experiment
   • Bohr’s Model of the Hydrogen Atom
3. Dual Nature of Matter and Radiation
   • Photoelectric Effect – Einstein’s explanation
   • de Broglie’s Hypothesis – Matter waves
4. Heisenberg’s Uncertainty Principle
   • Limitation in determining position and momentum simultaneously
5. Quantum Mechanical Model of Atom
   • Introduction to Schrödinger’s Wave Equation
   • Concept of orbitals and shapes of s, p, and d orbitals
6. Quantum Numbers
   • Principal (n), Azimuthal (l), Magnetic (m), and Spin (s)
   • Significance and rules of electron filling
7. Electronic Configuration of Atoms
   • Aufbau Principle – Electrons fill lower energy orbitals first
   • Pauli’s Exclusion Principle – No two electrons can have identical quantum numbers
   • Hund’s Rule of Maximum Multiplicity
8. Hydrogen Spectrum
   • Explanation of line spectra based on Bohr’s theory
9. Limitations of Bohr’s Model
   • Transition to modern quantum mechanics

Tips to Master The Chapter Structure of Atom

• Understand and visualize atomic models and orbitals for better retention.
• Focus on the photoelectric effect and Heisenberg’s principle, as they are conceptually important.
• Practice writing electronic configurations using rules like the Aufbau Principle and Hund’s Rule.
• Solve questions on quantum numbers to strengthen your grasp of the concept.

💡 Structure of Atom forms the foundation for further chapters like Chemical Bonding and Periodic Properties. Mastering it will make future topics easier to understand.

Happy studying, and good luck with your preparations! 😊

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