CBSE Class 11 Chemistry Chapter 2 Structure of Atom Important Questions with Answers

A solid conceptual understanding of Chemistry is required for students to excel in the subject. One can’t excel without the proper orientation of all the chapters associated with it. So, experts advise students to take the proper guidance from mentors and study from authentic and reliable resources.

The structure of atoms is the study of the fundamental outlook of the atom and its related topics. Students will get a clear understanding of the structure of atoms after completing this chapter. 

The questions covered in Chapter 2 Class 11 Chemistry important questions and based on  the topics below: 

• Discovery Of Sub-Atomic Particles
• Atomic Models
• Developments Leading To The Bohr’s Model Of Atom
• Bohr’s Model For Hydrogen Atom
• Towards Quantum Mechanical Model Of The Atom
• Quantum Mechanical Model Of Atom

All the questions from the major topics and sub-topics have been included in the important questions in Class 11 Chemistry Chapter 2 for students to have command over the entire chapter. Thus, ensuring that no topic remains untouched.

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Structure of Atom Class 11 Important Questions with Answers

The following important questions and their solutions are included in the Chemistry Class 11 Chapter 2 important questions:

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

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

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

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

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

Answer 1. Option (iii) is the correct answer.

Question 2. Which options don’t represent an atom’s ground state electronic configuration?

(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

Answer 2. Option (ii) is the correct answer.

Question 3. The probability density plots of the 1s and 2s orbitals have given in Fig. 

The density of dots in the region represents the probability density of finding an electron in the region.

Based on the above diagram, which among the following statements is incorrect?

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

(ii) The probability of the finding the electron is close to the nucleus.

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

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

Answer 3. Option (iv) is the correct answer.

Question 4. Which of the following statements will not correct the characteristics of cathode rays?

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

(ii) They travel in a  straight line without an external electrical or  magnetic field.

(iii) Characteristics of cathode rays don’t depend upon the material of electrodes in the cathode ray tube.

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

Answer 4. Option (iv) is the correct answer.

Question 5. Which among the following statements about the electron is incorrect?

(i) It is the negatively charged particle.

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

(iii) It is the primary constituent of all atoms.

(iv) It is the constituent of cathode rays.

Answer 5. Option (ii) is the correct answer.

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

(i) Overall neutrality of the atoms.

(ii) Spectra of the hydrogen atom.

(iii) Position of the atom’s electrons, protons and neutrons.

(iv) Stability of the atoms.

Answer 6. Option (i) is the correct answer.

Question 7. Two atoms are said to be isobars only if.

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

(ii) they have the equal number of electrons but  different numbers of neutrons.

(iii) they have the equal number of neutrons but  different numbers of electrons.

(iv) the addition of the number of protons and neutrons is equal, but the number of Protons is different.

Answer 7. Option (iv) is the correct answer

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

(i) 3

(ii) 4

(iii) 2

(iv) 1

Answer 8. Option (iv) is the correct answer.

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

(i) 4

(ii) 3

(iii) 2

(iv) 1

Answer 9. Option (iii) is the correct answer.

Question 10. Which among the following is responsible for ruling out the existence of definite paths or trajectories of the electrons?

(i) Pauli’s exclusion principle.

(ii) Heisenberg’s uncertainty principle.

(iii) Hund’s rule of maximum multiplicity.

(iv) Aufbau principle.

Answer 10. Option (ii) is the correct answer.

Question 11. Arrange s, p and d sub-shells of the shell in the increasing order of their effective nuclear charge (Zeff) experienced by the electron present in them.

Answer 11. d<p<s

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

Question 12. Using the orbital diagram, show the distribution of the electrons in the oxygen atom (atomic number 8).

Answer 12. 8O= 1s2 2s2 2p4

Question 13. The electronic configuration of the valence shell for Cu is 3d10 4s1 and not 3d94s2. How is this configuration explained?

Answer 13. Configuration with a filled and half-filled orbital has extra stability. In 3d104s1, d orbitals are completely filled and s orbitals are half-filled.

Question 14. The hydrogen atom possesses only one electron. Hence the mutual repulsion between the electrons is absent. But, in multielectron atoms, the mutual repulsion between the electrons is significant. How does this affect the energy of the given electron in the orbitals of the same principal quantum number in multielectron atoms?

Answer 14. The hydrogen atom possesses only one electron. Thus, the mutual repulsion between electrons is absent. But in the multielectron atoms, the mutual repulsion between the electrons is significant. Incomplete answer

Question  15. Out of electrons and protons, which one will have a higher velocity to produce matter waves of the same wavelength? Explain it.

Answer 15. Out of the electron and proton, being the lighter particle, the electron will have the higher velocity and produce matter waves of the same wavelength.

Question 16. Calculate the total number of required angular nodes and radial nodes present in the 3p orbital.

Answer 16. Nodes are the region present among the particular orbitals where one can find the probability density of finding electrons would be zero.

In case of np orbitals, the radial nodes, 

= n – l – 1 

= 3 –1 – 1 = 1

Angular nodes = l = 1.

Question 17. A nickel atom can lose two electrons to form a Ni2+ ion. The atomic number of element nickel is 28. From which orbital would the nickel lose two electrons.

Answer 17. One Ni atom contains 28 electrons and the electronic configuration is : [Ar] 4s2 3d8

It becomes Ni2+ by losing two given electrons, so the configuration of Ni2+ is : [Ar] 4s0 3d8

Hence, nickel loses two electrons from the 4s orbital, not the 3d orbital, according to the Aufbau principle.

Question 18. Which among the following orbitals are degenerate?

3dxy , 4dxy, 3dz2 , 3dyx , 4dyx, 4dzz

Answer 18. The energy of the orbitals depend upon the principal quantum number or the main shell to a large extent. Thus, the orbitals with the equal value of n will have the same energy levels and will be known as the degenerate orbitals.

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

Also, 4dxy, 4dyx, and 4dzz as they have the same value of n=4.

Question 19. An atom that has atomic mass number 13 has seven neutrons. What is the atomic number of the given atom?

Answer 19. The mass number of the atom = number of protons + number of neutrons

Hence, the atomic number ( number of protons ) = mass number – no. Of neutrons.

The atomic number of the given atom: 

= 13 – 7 = 6.

Question 20. According to de Broglie, the matter would exhibit dual behaviour, that is, both particle and wave-like properties. However, a cricket ball with a mass of 100 g doesn’t move like a wave when a bowler throws it at the speed of 100 km/h. Calculate the ball’s wavelength and explain why it does not show wave nature.

Answer 20. m= 100g or 0.1kg

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

thus, λ =h/mν = 2.387*10-34 m

Question 21. What experimental evidence supports the idea that an atom’s electronic energies are quantised?

Answer 21. The bright-line spectrum shows that the energy level in an atom is quantised. These lines obtained the result of electronic transition between the energy and if the electronic energy level were continuous and not quantised or discrete; the atomic spectra would have shown a constant absorption(from the lower to higher energy level transition) or emission (from higher to lower energy level transition.

Question 22. What is the difference between terms orbit and orbital?

Answer 22. The orbit is the definite circular path for the electrons to revolve around the nucleus. It represents the two-dimensional motion of electrons around the nucleus. The orbital is not a well-defined path because it’s a region around the nucleus where the probability of finding an electron is maximum.

Question 23. A hypothetical electromagnetic wave is shown in Fig. Find out the wavelength of the radiation.

Answer 23. wavelength defined as the distance between two successive alike points in a wave (usually between two maxima s, i.e. peaks, as well as two minima s, i.e. troughs, as shown in Fig.)

thus, for the given hypothetical wave, wavelength, 

λ = 4 * 2.16 pm

= 8.64 pm.

Question 24. Which among the following will not show deflection from the path on passing through the electric field?

Proton, cathode rays, electron, neutron

Answer 24. Neutron would not show deflection from the path on passing through the electric field.

That is because of the neutral nature of the neutron particles. Hence, it has no charge and is not affected by an electric field.

For the following three particles, proton (positive ), electron (negative), and cathode rays (the beam of the electrons, negatively charged) all have charges in them; thus, they will get deflected easily by an electric field.

Question 25. Chlorophyll present in green leaves of the plants absorbs light at 4.620 × 1014 Hz. Calculate the wavelength of the radiation in nanometers. Which part of the electromagnetic spectrum does it belong to?

Answer 25. Relation between wavelength as well as frequency can be expressed as :

λ = c/ν, here c shows the velocity of light as well, and ν shows the frequency for the radiation.

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

= 0.6494 times 10-6m-1

Question 26. 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 wave number of lines associated by the transition in the Balmer series if the electron moves to n = 4 orbit. (RH= 109677 cm-1)

Answer 26. According to the Bohr’s model for the hydrogen atom

ν = RH(1/ n12-1/ n22)cm-1

where, n1 = 2 and n2 = 4 and H = Rydberg’s constant = 109677

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

= 20564.44cm-1

Question 27. The effect of the uncertainty principle is quite significant only for the motion of microscopic particles and is negligible for the given macroscopic particles. Explain the statement with the help of a suitable example.

Answer 27. The uncertainty principle is only significantly applicable for the microscopic particles, not macroscopic particles. That concluded from the measurement of uncertainty:

For example, if we take the particle or an object of mass 1 milligram that is 10-6 kg 

We calculate the following,

∆x. ∆ν = 60626*10-34/ 4*3.14*10-6

= 10-28 m-2 s -1

The value we get in this case is negligible and very insignificant for the concept of the uncertainty principle to apply to the particles.

Question 28. Table-tennis ball has a  mass of 10 g and a speed of 90 m/s. If speed could be measured with the accuracy of 4%, what will be the uncertainty in speed and position?

Answer 28. According to Heisenberg’s uncertainty principle :

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

∆x. ∆p ≥ h/4π

From given problem,

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

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

∆x has given by ∆x = h/4πm∆v

Hence , the uncertainty of position shows as ∆x = 6.26 × 10-34 / 4 × 3.14 × 4× 3.6

= 1.46 x 10-33 m

Question 29. Wavelengths of the following different radiations have given below :

λ(A) = 300 nm

λ(B) = 300 μm

λ(c) = 3 nm

λ (D) 30 A°

Arrange the following radiations in the increasing order of their energies.

Answer 29. As per the Planck’s quantum theory, energy is related to the frequency of the given radiation by :

E = h × Frequency

So, E is proportional to 1/ λ

So, the relation b/w energy and wavelength are inversely proportional. Thus, the lower  the wavelength higher would be the energy of the radiation.

For the following given wavelength

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

So, the increasing order of the given energy would be the opposite: λ(B)< λ(A:<λ(c) = λ (D)

Question 30. The arrangement of the orbitals is based upon the energy based on the (n+l ) value. The lower the value of the (n+l ), the lower energy found. For orbitals with equal values of (n+l), the orbital with the lower value of n will have the lower energy.

1. Based upon the above-given information, arrange the following orbitals in the increasing order of their energy

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

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

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

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

1. Based upon the above-given information, solve the following questions given below :

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

4d, 4f, 5s, 5p

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

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

Answer 30. (i) (a) The increasing order of the energy of the given orbital is: 1s >2s >2p> 3s

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

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

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

(ii) (a) For the following orbitals, 5s have the lowest energy.

The (n+l) value of the 5s is the lowest = 5 + 0 = 5. 

Other orbitals have (n+l )value of more than 5 –

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

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

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

Question 31. The bromine atom possesses 35 electrons that contain 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electrons experiences the lowest effective nuclear charge?

Answer 31. 

The nuclear charge experienced by electrons (which are present in atoms containing multiple electrons) depends on the distance between its orbital and the atom’s nucleus. The greater the distance, the lower the effective nuclear charge. Among p-orbitals, 4p orbitals are the farthest from the nucleus of the bromine atom with a (+35) charge. Hence, the electrons that reside in the 4p orbital are the ones to experience the lowest effective nuclear charge. These electrons are also shielded by electrons that are present in the 2p and 3p orbitals along with the s-orbitals.

Question 32. Which orbital would experience the larger effective nuclear charge among the following pairs of orbitals?

(i) 2s and 3s,

(ii) 4d and 4f,

(iii) 3d and 3p

Answer 32. 

The nuclear charge can be defined as the net positive charge that acts on an electron in the orbital of an atom with more than 1 electron. It is inversely proportional to the distance between the orbital and the nucleus.

(i)Electrons that reside in the 2s orbital are closer to the nucleus than those in the 3s orbital and, therefore, experience greater nuclear charge.

(ii)4d orbital is closer to the nucleus than 4f orbital and will experience greater nuclear charge.

(iii)3p will experience a greater nuclear charge (since it is closer to the nucleus than the 3f orbital).

Question 33. The unpaired electrons in Al and Si are present in the 3p orbital. Which electrons would experience a more effective nuclear charge from the nucleus?

Answer 33. 

The nuclear charge can be defined as the net positive charge that acts on an electron in the orbital of an atom with more than 1 electron. The greater the atomic number, the greater the nuclear charge. Silicon holds 14 protons, while aluminum holds only 13. Therefore, the nuclear charge of silicon is greater than that of aluminum, implying that the electrons in the 3p orbital of silicon would experience a more effective nuclear charge than aluminum.

Question 34. Indicate the number of unpaired electrons in:

(a)P

(b)Si

(c)Cr

(d)Fe

(e)Kr

Answer 34. 

(a)Phosphorus (P):

The atomic number of phosphorus is 15

Electronic configuration of Phosphorus:

1s 2 2s 2 2p 6 3s 2 3p 3

That can be represented as follows:

From the diagram, it could be observed that phosphorus has three unpaired electrons.

(b) Silicon (Si):

The atomic number of silicon is 14

Electronic configuration of Silicon:

1s 2 2s 2 2p 6 3s 2 3p 2

That can be represented as follows:

From the diagram, it could be observed that silicon has two unpaired electrons.

(c) Chromium (Cr):

The atomic number of Cr is 24

Electronic configuration of Chromium:

1s 2 2s2 2p 6 3s 2 3p 6 4s 1 3d 5

That can be represented as follows:

From the diagram, it could be observed that chromium has six unpaired electrons.

(d) Iron (Fe):

The atomic number of iron is 26

Electronic configuration of Fe:

1s 2 2s2 2p 6 3s 2 3p 6 4s 2 3d 6

That can be represented as follows:

From the diagram, it could be observed that iron has four unpaired electrons.

(e) Krypton (Kr):

The atomic number of Krypton is 36

it is electronic configuration is:

1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d10 4p 6

That can be represented as follows:

From the diagram, it could be observed that Krypton has no unpaired electrons.

Question 35.

(a) How many sub-shells are associated with the n = 4?

(b) How many electrons would be present in the sub-shells having an ms value of –1/2 for n = 4?

Answer 35. 

(a)n = 4 (Given)

For a given value of ‘n’, the values of the ‘l’ range from 0 to (n – 1).

where the possible values of l are 0, 1, 2, and 3

Hence, four subshells are associated with n = 4, which are s, p, d and f.

(b) Number of orbitals in nth shell = n2

For n = 4

Therefore, the number of orbitals when n = 4 is 16

Each orbital has 1 electron with an ms value of -1/2.

Thus, the total number of electrons present in the subshell having an ms value of (-1/2) is 16.

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