NCERT Solutions Class 11 Chemistry Chapter 2 – Free PDF Download
NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom are provided on this page as a free source of educational content for Class 11 students. These NCERT Solutions aim to provide students with comprehensive answers to all questions asked in Chapter 2 of the term – I NCERT Class 11 Chemistry textbook. Chapter 2 will explain completely about the structure of an atom. The main aim of creating NCERT Solutions is to help students focus on important concepts and learn them effectively.
Students will face problems in answering the numericals that would appear in the first term exam. For this purpose, the faculty at BYJU’S have designed the NCERT Solutions for Class 11 Chemistry Chapter 2 in a stepwise manner based on the marks weightage allotted by the CBSE for the 2021-22 term – I syllabus. The solutions also contain explanations for each and every step to provide a quality learning experience for the Class 11 students. The NCERT Solutions for Class 11 Chemistry provided on this page can also be downloaded as a PDF, for free, by clicking the download button provided below.
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NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom
“Structure of Atom” is the second chapter in the NCERT Class 11 Chemistry textbook. The topics covered under this chapter include subatomic particles, Thomson’s atomic model, Rutherford’s atomic model, Bohr’s model and the quantum mechanical model of the atom. The types of questions asked in the NCERT exercise section for this chapter include:
- Basic calculations regarding subatomic particles (protons, electrons and neutrons).
- Numericals based on the relationship between wavelength and frequency.
- Numericals based on calculating the energy associated with electromagnetic radiation.
- Electron transitions to different shells.
- Writing electron configurations.
- Questions related to quantum numbers and their combinations (for electrons).
Students can note that these NCERT Solutions have been prepared and solved by our experienced subject experts, as per the latest term-wise CBSE Syllabus 2021-22 and its guidelines. The NCERT Solutions for Class 11 Chemistry provided on this page (for Chapter 2) provide detailed explanations with the steps to be followed while solving the numerical value questions that are frequently asked in term-wise examinations. The subtopics covered in the chapter are listed below.
Subtopics of NCERT Solutions for Class 11 Chemistry Chapter 2 Structure Of Atom
- Discovery Of Electron
- Charge To Mass Ratio Of Electron
- Charge On The Electron
- Discovery Of Protons And Neutrons
- Thomson Model Of Atom
- Rutherford’s Nuclear Model Of Atom
- Atomic Number And Mass Number
- Isobars And Isotopes
- Drawbacks Of Rutherford Model
Developments Leading To The Bohr’s Model Of Atom
- Wave Nature Of Electromagnetic Radiation
- Particle Nature Of Electromagnetic Radiation: Planck’s Quantum Theory
- Evidence For The Quantized* Electronic Energy Levels: Atomic Spectra
Bohr’s Model For Hydrogen Atom
- Explanation Of Line Spectrum Of Hydrogen
- Limitations Of Bohr’s Model
Towards Quantum Mechanical Model Of The Atom
- Dual Behaviour Of Matter
- Heisenberg’s Uncertainty Principle
Quantum Mechanical Model Of Atom
- Orbitals And Quantum Numbers
- Shapes Of Atomic Orbitals
- Energies Of Orbitals
- Filling Of Orbitals In Atom
- Electronic Configuration Of Atoms
- Stability Of Completely Filled And Half Filled Subshells.
Students can make use of the NCERT Solutions for Class 11 exclusively to comprehend key topics and concepts along with precise answers to exercise questions in NCERT Class 11 Chemistry Textbook.
Access the answers of NCERT Solutions for Class 11 Chemistry Chapter 2
Q.1. (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
1 electron weighs 9.109*10-31 kg. Therefore, number of electrons that weigh 1 g (10-3 kg) =
10-3kg/9.109*10-31 kg = 1.098*1027 electrons
Mass of one mole of electrons = NA* mass of one electron
= (6.022*1023)*(9.109*10-31 kg) = 5.48*10-7 kg
Charge on one mole of electrons = NA* charge of one electron
= (6.022*1023)*(1.6022*10-19 C) = 9.65*104 C
Q.2. (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C.
(Assume that mass of a neutron = 1.675 × 10–27 kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3
Will the answer change if the temperature and pressure are changed?
(i) 1 molecule of methane contains 10 electrons (6 from carbon, 4 from hydrogen)
Therefore, 1 mole of methane contains 10*NA = 6.022*1024 electrons.
(ii) Number of neutrons in 14g (1 mol) of 14C = 8*NA = 4.817*1024 neutrons.
Number of neutrons in 7 mg (0.007g) = (0.007/14)*4.817*1024 = 2.409*1021 neutrons.
Mass of neutrons in 7 mg of 14C = (1.67493*10-27kg)*(2.409*1021) = 4.03*10-6kg
(iii) Molar mass of NH3 = 17g
Number of protons in 1 molecule of NH3 = 7+3 = 10
Therefore, 1 mole (17 grams) of NH3 contains 10*NA = 6.022*1024 protons.
34 mg of NH3 contains (34/1700)*6.022*1024 protons = 1.204*1022 protons.
Total mass accounted for by protons in 34 mg of NH3 = (1.67493*10-27kg)*(1.204*1022) = 2.017*10-5kg.
These values remain constant regardless of any change in temperature and pressure (since these factors do not affect the number of protons in the atom and the mass of each proton).
Q.3. How many neutrons and protons are there in the following nuclei?
Mass number of carbon-13 = 13
Atomic number of carbon = Number of protons in one carbon atom = 6
Therfore, total number of neutrons in 1 carbon atom = Mass number – Atomic number = 13 – 6 = 7
Mass number of oxygen-16 = 16
Atomic number of oxygen = Number of protons = 8
Therefore, No. neutrons = Mass number – Atomic number = 16 – 8 = 8
Mass number = 24
Atomic number = No. protons = 12
No. neutrons = Mass number – Atomic number = 24 – 12 = 12
Mass nubmer = 56
Atomic number of iron = No. protons in iron= 26
No. neutrons = Mass number – Atomic number = 56 – 26 = 30
Mass number = 88
Atomic number = No. protons = 38
No. neutrons = Mass number – Atomic number = 88 – 38 = 50
Q.4. Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(I)Z = 17, A = 35
(II)Z = 92, A = 233
(III)Z = 4, A = 9
Q.5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wavenumber (ν ) of the yellow light.
Ans: Rearranging the expression,
the following expression can be obtained,
c denotes the speed of light (
Substituting these values in eq. (1):
Therefore, the frequency of the yellow light which is emitted by the sodium lamp is:
The wave number of the yellow light is
Q.6. Find the energy of each of the photons which
(i) correspond to light of frequency 3×1015 Hz.
(ii) have a wavelength of 0.50 Å.
The energy of a photon (E) can be calculated by using the following expression:
Where, ‘h’ denotes Planck’s constant, which is equal to
Substituting these values in the expression for the energy of a photon, E:
The energy of a photon whose wavelength is (
h (Planck’s constant) =
c (speed of light) =
Substituting these values in the equation for ‘E’:
Q.7. Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 × 10–10 s.
Ans: Frequency of the light wave (
Wavelength of the light wave(
c denotes the speed of light,
Substituting the value of ‘c’ in the previous expression for
Q.8. What is the number of photons of light with a wavelength of 4000 pm that provides 1J of energy?
Ans: Energy of one photon (E) =
Energy of ‘n’ photons (
c denotes the speed of light in vacuum =
h is Planck’s constant, whose value is
Substituting these values in the expression for n:
Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are
Q.9. A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal is 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J).
Energy of the photon (E)=
Where, h denotes Planck’s constant, whose value is
c denotes the speed of light =
Substituting these values in the expression for E:
Therefore, energy of the photon =
The kinetic energy of the emission
Therefore, the kinetic energy of the emission = 0.97 eV.
The velocity of the photoelectron (v) can be determined using the following expression:
Substituting thes values in the expression for v:
Therefore, the velocity of the ejected photoelectron is
Q.10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the
sodium atom. Calculate the ionisation energy of sodium in kJ mol–1.
Ans: Ionization energy (E) of sodium =
Q.11. A 25-watt bulb emits monochromatic yellow light of the wavelength of 0.57µm. Calculate the rate of emission of quanta per second.
Ans: Power of the bulb, P = 25 Watt =
Energy (E) of one photon=
Substituting these values in the expression for E:
Thus, the rate of discharge of quanta (per second) =
Q.12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0 ) and work function (W0 ) of the metal.
Ans: Threshold wavelength of the radiation
Threshold frequency of the metal (
Therefore, threshold frequency (
Q.13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes the transition from an energy level with n = 4 to an energy level with n = 2?
Substituting these values in the expression for E:
Here, the -ve sign denotes the emitted energy.
Wavelength of the emitted light
Substituting these values in the expression for
Q.14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit).
Ans: The expression for the ionization energy is given by,
Where Z denotes the atomic number and n is the principal quantum number
For the ionization from
Therefore, the required energy for the ionization of hydrogen from n = 5 to n = ∞ = Energy required for n1 = 1 to n = ∞, is
Hence, a lower amount of energy is required in order to ionize electrons in the 5th orbital of a hydrogen atom when compared to do that in the ground state of the atom.
Q.15. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?
A total number of 15 lines (5 + 4 + 3 + 2 + 1) will be obtained in this hydrogen emission spectrum.
Total no. of spectral lines emitted when an electron initially in the ‘nth‘ level drops down to the ground state can be calculated using the following expression:
Since n = 6, total no. spectral lines =
Q.16. (i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18 J atom–1. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for the hydrogen atom.
(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:
(ii) Radius of Bohr’s n th orbit for hydrogen atom is given by, rn = (0.0529 nm) n 2
For, n = 5
Q.17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
Ans. For the Balmer series of the hydrogen emission spectrum, ni = 2. Therefore, the expression for the wavenumber(
Since wave number(
Hence, taking nf = 3, we get:
Q.18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10–11 ergs.
The ground-state electron energy is –2.18 × 10–11 ergs.
Ans. Energy (E) associated with the nth Bohr orbit of an atom is:
Where, Z denotes the atom’s atomic number
Ground state energy
The required energy for an electron shift from n = 1 to n = 5 is:
The wavelength of the emitted light =
Q.19. The electron energy in hydrogen atom is given by En= (–2.18 × 10–18 )/n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Required energy for the ionization from n = 2 is:
If λ is the longest wavelength that can cause this transition,
= 3647 Å
Q.20. Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m s–1
Ans. As per de Broglie’s equation,
Where, λ denotes thr wavelength of the moving particle
m is the mass of the particle
v denotes the velocity of the particle
h is Planck’s constant
Substituting these values in the expression for λ:
Therefore, the wavelength associated with the electron which is moving with a velocity of
Q.21. The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.
Ans. As per de Broglie’s equation,
Given, K.E of electron =
Substituting these values in the expression for λ:
Q.23. (I)Write the electronic configurations of the following ions:
(II) What are the atomic numbers of elements whose outermost electrons are represented by
(b) 2p3 and
(III) Which atoms are indicated by the following configurations?
(b)[Ne] 3s2 3p3
(c)[Ar] 4s2 3d1 .
(I) (a) H– ion
The electronic configuration of the Hydrogen atom (in its ground state) 1s1. The single negative charge on this atom indicates that it has gained an electron. Thus, the electronic configuration of H– = 1s2
(b) Na+ ion
Electron configuration of Na = 1s2 2s2 2p6 3s1 . Here, the +ve charge indicates the loss of an electron.
(c) O2– ion
Electronic configuration of 0xygen = 1s 2 2s 2 2p 4. The ‘-2 ‘ charge suggests that it has gained 2 electrons.
(d) F – ion
Electronic configuration of Fluorine = 1s22s22p5. The species has gained one electron (accounted for by the -1 charge).
(II) (a) 3s1
Complete electronic configuration: 1s 2 2s 2 2p 6 3s 1.
Total no. electrons in the atom = 2 + 2 + 6 + 1 = 11
(b) 2p 3
Complete electronic configuration: 1s 2 2s 2 2p 3 .
Total no. electrons in the atom =2 + 2 + 3 = 7
(c) 3p 5
Complete electronic configuration: 1s 2 2s 2 2p 6 3s2 3p5 .
Total no. electrons in the atom = 2 + 2 + 6 + 2 + 5 = 17
(III)(a)[He] 2s 1
Complete electronic configuration: 1s 2 2s 1 .
(b)[Ne] 3s 2 3p3
Complete electronic configuration: 1s 2 2s 2 2p 6 3s 2 3p 3 .
(c)[Ar] 4s 2 3d1
Complete electronic configuration: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 1 .
Q.24. What is the lowest value of n that allows g orbitals to exist?
Ans. For g-orbitals, l = 4.
For any given value of ‘n’, the possible values of ‘l’ range from 0 to (n-1).
Q.25. An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.
Ans: For the 3d orbital:
Possible values of the Principal quantum number (n) = 3
Possible values of the Azimuthal quantum number (l) = 2
Possible values of the Magnetic quantum number (ml) = – 2, – 1, 0, 1, 2
Q.26. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
In a neutral atom, no.protons = no.electrons.
The electronic configuration of this element (atomic number 29) is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 . The element is copper (Cu).
Q.27.Give the number of electrons in the species , H2+ and H2 and O2+
Ans: No. electrons present in H2 = 1 + 1 = 2.
H2: No. electrons in H2 = 1 + 1 = 2
No. electrons O2 = 8 + 8 = 16.
Q.28. (I)An atomic orbital has n = 3. What are the possible values of l and ml ?
(II)List the quantum numbers (ml and l) of electrons for 3d orbital.
(III) Which of the following orbitals are possible? 1p, 2s, 2p and 3f
The possible values of ‘l’ range from 0 to (n – 1). Thus, for n = 3, the possible values of l are 0, 1, and 2.
The total number of possible values for ml = (2l + 1). Its values range from -l to l.
For n = 3 and l = 0, 1, 2:
m0 = 0
m1 = – 1, 0, 1
m2 = – 2, – 1, 0, 1, 2
For 3d orbitals, n = 3 and l = 2. For l = 2 , possible values of m2 = –2, –1, 0, 1, 2
It is possible for the 2s and 2p orbitals to exist. The 1p and 3f cannot exist.
For the 1p orbital, n=1 and l=1, which is not possible since the value of l must always be lower than that of n.
Similarly, for the 3f orbital, n =3 and l = 3, which is not possible.
Q.29. Using s, p and d notations, describe the orbital with the following quantum numbers.
(a)n = 1, l = 0;
(b)n = 3; l =1
(c) n = 4; l = 2;
(d) n = 4; l =3.
(a)n = 1, l = 0 implies a 1s orbital.
(b)n = 3 and l = 1 implies a 3p orbital.
(c)n = 4 and l = 2 implies a 4d orbital.
(d)n = 4 and l = 3 implies a 4f orbital.
Q.30. Explain, giving reasons, which of the following sets of quantum numbers are not possible.
a) n = 0, l = 0, ml= 0, ms =+
b) n = 1, l = 0, ml= 0, ms =-
c) n = 1, l = 1, ml= 0, ms =+
d) n = 2, l = 0, ml= 1, ms =-
e) n = 3, l = 3, ml= -3, ms =+
f) n = 3, l = 0, ml= 1, ms =+
Ans. (a) Not possible. The value of n cannot be 0.
(c) Not possible. The value of l cannot be equal to that of n.
(d) Not possible. Since, when l = 0, mt cannot be 1.
(e) Not possible. As when n = 3, l cannot be 3
Q.31. How many electrons in an atom may have the following quantum numbers?
a) n = 4, ms =-
b)n = 3, l = 0
Ans.(a)If n is the principal quantum number, the total number of electrons in the atom = 2n 2
Electron configuration for an atom with 32 electrons: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 .
Hence, all the electrons are paired.
(b)n = 3, l = 0 indicates the 3s orbital. Therefore, no. electrons with n = 3 and l = 0 is 2.
Q.32. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Ans. Hydrogen atoms have only one electron. As per Bohr’s postulates, the angular momentum of this electron is:
Where, n = 1, 2, 3, …
As per de Broglie’s equation:
But ‘2πr’ is the Bohr orbit’s circumference. Therefore, equation (3) proves that the Bohr orbit’s circumference for the hydrogen atom is an integral multiple of the de Broglie wavelength of the electron, which is revolving around the orbit.
Q.33. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?
Ans. The wave number associated with the Balmer transition for the He+ ion (n = 4 to n = 2 ) is given by:
Where, n1 = 2 n2 = 4
As per the question, the desired transition in the hydrogen spectrum must have the same wavelength as that of He+ spectrum.
This equality is true only when the value of n1 = 1 and that of n2 = 2.
The transition for n2 = 2 to n = 1 in the hydrogen spectrum would, therefore, have the same wavelength as the Balmer transition from n = 4 to n = 2 of the He+ spectrum.
Q.34. Calculate the energy required for the process
The ionization energy for the H atom in the ground state is
Ans. The energy associated with hydrogen-like species is:
For the ground state of the hydrogen atom,
For the process given by:
An electron is moved from n = 1 to n = ∞.
Q.35. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across the length of the scale of length 20 cm long.
Ans. 1 m = 100 cm
1 cm = 10–2 m
Length of the scale = 20 cm =
Diameter of one carbon atom = 0.15 nm =
Space occupied by one carbon atom
Ans. Length of the arrangement = 2.4 cm
No. carbon atoms present = 2 × 108
The diameter of the carbon atom =
Q.37. The diameter of the zinc atom is 2.6Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.
Ans. (a)Radius of carbon atom=
(b) Length of the arrangement = 1.6 cm
Diameter of a zinc atom =
Q.38. A certain particle carries
Charge held by one electron =
Therefore, No. electrons that carry a charge of
Q.39. In Milikan’s experiment, the static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is
Charge held by the oil drop =
Charge held by one electron =
Therefore, No. electrons present in the drop of oil
Q.40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like Aluminium etc. is used, what difference would be observed from the above results?
The results obtained when a foil of heavy atoms will be different from the results obtained when relatively light atoms are used in the foil. The lighter the atom, the lower the magnitude of positive charge in its nucleus. Therefore, lighter atoms will not cause enough deflection of the positively charged α-particles.
The general convention followed while representing elements along with their atomic masses (A), and their atomic numbers (Z) is
Q.42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Let the No. protons in the element be x.
Therefore, No. neutrons in the element = x + 31.7% of x
= x + 0.317 x
= 1.317 x
Given, Mass number of the element = 81, which implies that (No. protons + No. neutrons) = 81
Therefore, total no. protons = 35, which implies that atomic number = 35.
Therefore, the element is
Q.43. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.
Let the no. electrons in the negatively charged ion be x.
Then, no. neutrons present = x + 11.1% of x = x + 0.111 x = 1.111 x
No. electrons present in the neutral atom = (x – 1)
(When an ion carries a negative charge, it carries an extra electron)
No. protons present in the neutral atom = x – 1
Given, mass number of the ion = 37
(x – 1) + 1.111x = 37
2.111x = 38
x = 18
Therefore, The symbol of the ion is
Q.44. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.
Let the total no. electrons present in
Since the ion has a charge of +3 ,
Therefore, no. protons in neutral atom = x + 3
The Mass number of the ion is 56 (Given)
Therefore, (x+3) + (1.304x) = 56
2.304x = 53
X = 23
Therefore, no. protons = x + 3 = 23 + 3 = 26
The ion is
Q.45. Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.
The increasing order of frequency is as follows:
Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays
The increasing order of a wavelength is as follows:
Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio
Q.46. Nitrogen laser produces radiation at a wavelength of 337.1 nm. If the number of photons emitted is
Power of laser = Energy with which it emits photons
Where, N = number of photons emitted
h = Planck’s constant
c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression of Energy (E):
Hence, the power of the laser is