NCERT Solutions for Class 11 Chemistry: Chapter 2 (Structure of Atom)

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 solutions aim to provide students with comprehensive answers to all questions asked in Chapter 2 of the NCERT Class 11 Chemistry textbook. The NCERT Solutions 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 Chemistry – Class 11, 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 CBSE syllabus 2020-21. 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 examinations. The subtopics covered under the chapter are listed below.

Subtopics Of Class 11 Chemistry Chapter 2 Structure Of The Atom

1. Sub-atomic Particles
   • Discovery Of Electron
   • Charge To Mass Ratio Of Electron
   • Charge On The Electron
   • Discovery Of Protons And Neutrons
2. Atomic Models
   • Thomson Model Of Atom
   • Rutherford’s Nuclear Model Of Atom
   • Atomic Number And Mass Number
   • Isobars And Isotopes
   • Drawbacks Of Rutherford Model
3. 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
4. Bohr’s Model For Hydrogen Atom
   • Explanation Of Line Spectrum Of Hydrogen
   • Limitations Of Bohr’s Model
5. Towards Quantum Mechanical Model Of The Atom
   • Dual Behaviour Of Matter
   • Heisenberg’s Uncertainty Principle
6. 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.

NCERT Solutions for Class 11 Chemistry Chapter 2

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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.

Ans:

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

(ii)

Mass of one mole of electrons = N_A* mass of one electron

= (6.022*10²³)*(9.109*10^-31 kg) = 5.48*10^-7 kg

Charge on one mole of electrons = N_A* 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
at
STP.
Will the answer change if the temperature and pressure are changed?

Ans:

(i) 1 molecule of methane contains 10 electrons (6 from carbon, 4 from hydrogen)

Therefore, 1 mole of methane contains 10*N_A = 6.022*10²⁴ electrons.

(ii) Number of neutrons in 14g (1 mol) of ¹⁴C = 8*N_A = 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*N_A = 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?

$_{6}^{13} extrm{C}: ,: _{8}^{16} extrm{O}: ,: _{12}^{24} extrm{Mg}: ,: _{26}^{56} extrm{Fe}: ,: _{38}^{88} extrm{Sr}$

Ans:

$_{6}^{13} extrm{C}$:

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

$_{8}^{16} extrm{O}$:

Mass number of oxygen-16 = 16

Atomic number of oxygen = Number of protons = 8

Therefore, No. neutrons = Mass number – Atomic number = 16 – 8 = 8

$_{12}^{24} extrm{Mg}$:

Mass number = 24

Atomic number = No. protons = 12

No. neutrons = Mass number – Atomic number = 24 – 12 = 12

$_{26}^{56} extrm{Fe}$:

Mass nubmer = 56

Atomic number of iron = No. protons in iron= 26

No. neutrons = Mass number – Atomic number = 56 – 26 = 30

$_{38}^{88} extrm{Sr}$:

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

Ans:

(I)$_{17}^{35} extrm{C}$

(II)$_{92}^{233} extrm{U}$

(III)$_{4}^{9} extrm{Be}$

 

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,

$lambda =frac{c}{ u }$

the following expression can be obtained,

$u =frac{c}{ lambda }$      ……….(1)

Here, $u$ denotes the frequency of the yellow light

c denotes the speed of light ( $3 imes 10^{8}, m/s$ )

$lambda$ denotes the wavelength of the yellow light (580 nm, $580 imes 10^{-9}, m/s$ )

Substituting these values in eq. (1):

$u =frac{3 imes 10^{8}}{580 imes 10^{-9}}=5.17 imes 10^{14}, s^{-1}$

Therefore, the frequency of the yellow light which is emitted by the sodium lamp is:

$5.17 imes 10^{14}, s^{-1}$

The wave number of the yellow light is $ar{ u }=frac{1}{lambda }$ $=frac{1}{580 imes 10^{-9}}=1.72 imes 10^{6}, m^{-1}$

 

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.

Ans:

(i)

The energy of a photon (E) can be calculated by using the following expression:

E= $h u$

Where, ‘h’ denotes Planck’s constant, which is equal to $6.626 imes 10^{-34}, Js$ $u$ (frequency of the light) = $3 imes 10^{15}$Hz

Substituting these values in the expression for the energy of a photon, E:

$E=(6.626 imes 10^{-34})(3 imes 10^{15}) E=1.988 imes 10^{-18}, J$

(ii)

The energy of a photon whose wavelength is ($lambda$) is:

E= $hc u$

Where,

h (Planck’s constant) = $6.626 imes 10^{-34}Js$

c (speed of light) = $3 imes 10^{8}, m/s$

Substituting these values in the equation for ‘E’:

$E=frac{(6.626 imes 10^{-34})(3 imes 10^{8})}{0.50 imes 10^{-10}}=3.976 imes 10^{-15}J ∴ E=3.98 imes 10^{-15}J$

 

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 ($u$) = $frac{1}{Period}$ $frac{1}{Period} =frac{1}{2.0 imes 10^{-10}, s} =5.0 imes 10^{9}, s^{-1 }$

Wavelength of the light wave($lambda$) =$c u$

Where,

c denotes the speed of light,  $3 imes 10^{8}, m/s$

Substituting the value of ‘c’ in the previous expression for $lambda$:

$lambda =frac{3 imes 10^{8}}{5.0 imes 10^{9}}=6.0 imes 10^{-2}m$

Wave number $(ar{ u })$ of light = $frac{1}{lambda }=frac{1}{6.0 imes 10^{-2}}=1.66 imes 10^{1}, m^{-1}=16.66, m^{-1}$

 

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) = $h u$

Energy of ‘n’ photons ($E_{n}$) = $nh u$ $Rightarrow n=frac{E_{n}lambda }{hc}$

Where, $lambda$ is the wavelength of the photons = 4000 pm = $4000 imes 10^{-12}, m$

c denotes the speed of light in vacuum =$3 imes 10^{8}, m/s$

h is Planck’s constant, whose value is $6.626 imes 10^{-34}, Js$

Substituting these values in the expression for n:

$n=frac{1 imes (4000 imes 10^{-12})}{(6.626 imes 10^{-34})(3 imes 10^{8})}=2.012 imes 10^{16}$

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are $2.012 imes 10^{16}$

 

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).

Ans:

(i)

Energy of the photon (E)= $h u =frac{hc}{lambda }$

Where, h denotes Planck’s constant, whose value is $6.626 imes 10^{-34},Js$

c denotes the speed of light = $3 imes 10^{8},m/s$ $lambda$= wavelength of the photon =$4 imes 10^{-7},m/s$

Substituting these values in the expression for E:

$E=frac{(6.626 imes 10^{-34})(3 imes 10^{8})}{4 imes 10^{-7}}=4.9695 imes 10^{-19}, J$

Therefore, energy of the photon = $4.97 imes 10^{-19}, J$

(ii)

The kinetic energy of the emission $E_{k}$ can be calculated as follows:

$=h u -h u _{0} =(E-W)eV =(frac{4.9695 imes 10^{-19}}{1.6020 imes 10^{-19}})eV-2.13, eV =(3.1020-2.13)eV =0.9720, eV$

Therefore, the kinetic energy of the emission = 0.97 eV.

(iii)

The velocity of the photoelectron (v) can be determined using the following expression:

$frac{1}{2}mv^{2} =h u -h u _{0} Rightarrow v =sqrt{frac{2(h u -h u _{0})}{m}}$

Where $(h u -h u _{0})$ is the K.El of the emission (in Joules) and ‘m’ denotes the mass of the photoelectron.

Substituting thes values in the expression for v:

$v=sqrt{frac{2 imes (0.9720 imes 1.6020 imes 10^{-19})J}{9.10939 imes 10^{-31}kg}} =sqrt{0.3418 imes 10^{12}m^{2}s^{2}} Rightarrow v=5.84 imes 10^{5}ms^{-1}$

Therefore, the velocity of the ejected photoelectron is $5.84 imes 10^{5}ms^{-1}$

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 = $frac{N_{A}hc}{lambda } =frac{(6.023 imes 10^{23}, mol^{-1})(6.626 imes 10^{-34})Js(3 imes 10^{8})ms^{-1}}{242 imes 10^{-9}m} =4.947 imes 10^{5}, J, mol^{-1} =494.7 imes 10^{3}, J, mol^{-1} =494, kJ, mol^{-1}$

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 = $25, Js^{-1}$

Energy (E) of one photon= $h u =frac{hc}{ u }$

Substituting these values in the expression for E:

$E=frac{(6.626 imes 10^{-34})(3 imes 10^{8})}{(0.57 imes 10^{-6})}=34.87 imes 10^{-20}, J$

E= $34.87 imes 10^{-20}, J$

Thus, the rate of discharge of quanta (per second) = $E=frac{25}{34.87 imes 10^{-20}}=7.169 imes 10^{19}, s^{-1}$

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 $(lambda _{0})$= 6800 Å=$6800 imes 10^{-10}, m$

Threshold frequency of the metal ($u _{0}$ ) =$frac{c}{lambda _{0}}=frac{3 imes 10^{8}ms^{-1}}{6.8 imes 10^{-7}m}=4.41 imes 10^{14}, s^{-1}$

Therefore, threshold frequency ($u _{0}$ ) of the metal = $h u _{0} =(6.626 imes 10^{-34}Js)(4.41 imes 10^{14}s^{-1}) =2.922 imes 10^{-19}, J$

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?

Ans: The $n_{i}$ = 4 to $n_{f}$ = 2 transition results in a spectral line of the Balmer series. The energy involved in this transition can be calculated using the following expression:

$E=2.18 imes 10^{-18}[frac{1}{n_{i}^{2}}-frac{1}{n_{f}^{2}}]$

Substituting these values in the expression for E:

$E=2.18 imes 10^{-18}[frac{1}{4^{2}}-frac{1}{2^{2}}] =2.18 imes 10^{-18}[frac{1-4}{16}] =2.18 imes 10^{-18} imes (-frac{3}{16}) E=-(4.0875 imes 10^{-19}J)$

Here, the -ve sign denotes the emitted energy.

Wavelength of the emitted light $(lambda )=frac{hc}{E}$

(Since $E=frac{hc}{ lambda }$)

Substituting these values in the expression for ${ lambda }$:

$lambda =frac{(6.626 imes 10^{-34})(3 imes 10^{8})}{(4.0875 imes 10^{-19})}=4.8631 imes 10^{-7}, m lambda =486.31 imes 10^{-9}, m =486, nm$

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,

$E_{n} =frac{-(2.18 imes 10^{-18})Z^{2}}{n^{2}}$

Where Z denotes the atomic number and n is the principal quantum number

For the ionization from $n_{t}=5$ to $n_{2} =∞$,

$Delta E=E_{infty }-E_{5 } =[(frac{-(2.18 imes 10^{-18}, J)(1)^{2}}{(infty )^{2}})-(frac{-(2.18 imes 10^{-18}, J)(1)^{2}}{(5 )^{2}})] =0.0872 imes 10^{-18}, J Delta E=8.72 imes 10^{-20}, J$

Therefore, the required energy for the ionization of hydrogen from n = 5 to n = ∞ = Energy required for n1 = 1 to n = ∞, is $8.72 imes 10^{-20}, m$ $Delta E=E_{infty }-E_{5 } =[(frac{-(2.18 imes 10^{-18}, J)(1)^{2}}{(infty )^{2}})-(frac{-(2.18 imes 10^{-18}, J)(1)^{2}}{(1 )^{2}})] =2.18 imes 10^{-18}, J Delta E=2.18 imes 10^{-18}, J$

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 ‘n^th‘ level drops down to the ground state can be calculated using the following expression:

$frac{n(n-1)}{2}$

Since n = 6, total no. spectral lines = $frac{6(6-1)}{2}=15$

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:

$E_{5}=frac{-(2.18 imes 10^{-18})}{(5)^{2}}=frac{-(2.18 imes 10^{-18})}{25}=-8.72 imes 10^{-20}, J atom^{-1}$

(ii) Radius of Bohr’s n th orbit for hydrogen atom is given by, r_n = (0.0529 nm) n ²

For, n = 5

$r_{5}=(0.0529, nm)(5^{2}) r_{5}=1.3225, nm$

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, n_i = 2. Therefore, the expression for the wavenumber($ar{ u}$) is:

$ar{ u}=[frac{1}{(2)^{2}}-frac{1}{n_{f}^{2}}](1.097 imes 10^{7}, m^{-1})$

Since wave number($ar{ u}$) is inversely proportional to the transition wavelength, the lowest possivle value of ($ar{ u}$) corresponds to the longest wavelength transition.

For ($ar{ u}$) to be of the lowest possible value, n_f should be minimum. In the Balmer series, transitions from n_i = 2 to n_f = 3 are allowed.

Hence, taking n_f = 3, we get:

$ar{ u}=(1.097 imes 10^{7})[frac{1}{(2)^{2}}-frac{1}{{3}^{2}}] ar{ u}=(1.097 imes 10^{7})[frac{1}{4}-frac{1}{9}] =(1.097 imes 10^{7})[frac{9-4}{36}] =(1.097 imes 10^{7})[frac{5}{36}] ar{ u}=1.5236 imes 10^{6}, m^{-1}$

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 n^th Bohr orbit of an atom is:

$E_{5}=frac{-(2.18 imes 10^{-18})Z^{2}}{(n)^{2}}$

Where, Z denotes the atom’s atomic number

Ground state energy $=-2.18 imes 10^{–11}, ergs =-2.18 imes 10^{–11} imes 10^{–7}, J =-2.18 imes 10^{–18} , J$

The required energy for an electron shift from n = 1 to n = 5 is:

$Delta E=E_{5}-E_{1} =[(frac{-(2.18 imes 10^{-18}, J)(1)^{2}}{(5 )^{2}})-(-2.18 imes 10^{-18})] =(2.18 imes 10^{-18})[1-frac{1}{25}] =(2.18 imes 10^{-18})[frac{24}{25}] =2.0928 imes 10^{-18}, J$