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

NCERT Solutions For Class 11 Chemistry Chapter 2 PDF Free 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 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 in this page can also be downloaded as a PDF (for free) by clicking the download button provided above.

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 2019-20. The NCERT Solutions for Class 11 chemistry provided on this page (for chapter 2) provide detailed explanations on 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


 

Q.1. (a)Figure the quantity of electrons which will together measure one gram.

(b)Compute the mass and charge of one mole of electrons.

Ans:

(a)Mass of one electron =9.10939×1031kg9.10939\times 10^{-31}\: kg Quantity of electrons that measure 9.10939×1031kg9.10939\times 10^{-31}\: kg=1

Quantity of electrons that will measure 1 g (1×103kg1\times 10^{-3}kg)

=19.10939×1031kg×(1×103kg)=0.1098×103+31=0.1098×1028=1.098×1027=\frac{1}{9.10939\times 10^{-31}kg}\times (1\times 10^{-3}kg)\\ \\ =0.1098\times 10^{-3+31}\\ \\ =0.1098\times 10^{28}\\ \\ =1.098\times 10^{27}

(b)Mass of one electron =9.10939×1031kg9.10939\times 10^{-31}\: kg

Mass of one mole of electron =(6.022×1023)×(9.10939×1031kg)=5.48×107kg(6.022\times 10^{23})\times (9.10939\times 10^{-31}kg)\\ \\ =5.48\times 10^{-7}kg

Charge on one electron =1.6022×1019coulomb1.6022\times 10^{-19}coulomb

Charge on one mole of electron =(1.6022×1019C)(6.022×1023)=9.65×104C(1.6022\times 10^{-19}C)(6.022\times 10^{23})=9.65\times 10^{4}C

 

Q.2. (I)Figure the aggregate number of electrons present in one mole of methane.

(II)Discover (a) the aggregate number and (b) the aggregate mass of neutrons in 7 mg of 14C.

(Assume that mass of a neutron=1.675×1027kg= 1.675 × 10^{-27} kg).

(III)Discover (a) the aggregate number and (b) the aggregate mass of protons in 34 mg of NH3NH_{3} at STP.

Will the appropriate response change if the temperature and weight are changed?

Ans:

(I)Number of electrons present in 1 molecule of methane (CH4CH_{4})

1(6)+4(1)=10{1(6)+4(1)}=10

Number of electrons present in 1 mole i.e., 6.023×10236.023\times 10^{23} molecules of methane

=6.022×1023×10=6.022×1024=6.022\times 10^{23}\times 10=6.022\times 10^{24}

(II)(a) Number of atoms of 14C14_{C} in 1 mole=6.023×10236.023 \times 10^{23}

Since 1 atom of 14C14_{C} contains (14 – 6) i.e., 8 neutrons, the number of neutrons in 14 g of 14C14_{C} is (6.023×1023)×8(6.023 \times 10^{23})\times8.Or, 14 g of 14C14_{C} contains (6.023×1023×8)(6.023 \times 10^{23}\times8)neutrons.

Number of neutrons in 7 mg

6.022×1023×8×7mg1400mg=2.4092×1021\frac{ 6.022 \times 10^{23}\times 8\times 7\, mg}{1400\, mg}\\ \\ =2.4092\times 10^{21}

 

(b)The aggregate mass of one neutron= 1.67493×1027kg1.67493 \times 10^{–27}\, kg

The aggregate mass of total neutrons in 7 g of 14C14_{C} =(2.4092×1021)(1.67493×1027kg)=4.0352×106kg= (2.4092 \times 10^{21}) (1.67493 \times 10^{–27}\, kg) \\ \\ =4.0352\times 10^{-6}\, kg

(III)(a)1 mole of NH3=1(14)+3(1)gofNH3=17gofNH3=6.022×1023moleculesofNH3NH_{3}={1(14)+3(1)}\, g\: of\: NH_{3}\\ \\ =17\, g\: of\: NH_{3}\\ \\ =6.022\times 10^{23}\: molecules \: of\: NH_{3}

Total number of protons present in 1 molecule of NH3NH_{3} =1(7)+3(1)=10={1(7)+3(1)}\\ \\ =10

Number of protons in 6.023×1023moleculesofNH36.023\times 10^{23}\: molecules\: of\: NH_{3} =(6.023×1023)(10)=6.023×102417gofNH3=(6.023\times 10^{23})(10)\\ \\ =6.023\times 10^{24}\\ \\ \Rightarrow 17 \, g \: of\: NH_{3} contains (6.023×1024)(6.023\times 10^{24}) protons. Number of protons in 34 mg of NH3NH_{3} 6.022×1024×34mg17000mg=1.2046×1022\frac{6.022\times 10^{24}\times 34\, mg}{17000\, mg}\\ \\ =1.2046\times 10^{22}

(b)Mass of one proton = 1.67493×1027kg1.67493 \times 10^{–27}\, kg

Total mass of protons in 34 mg of NH3NH_{3} =(1.67493×1027kg)(1.2046×1022)=2.0176×105kg=(1.67493\times 10^{-27}\, kg)(1.2046\times 10^{22} )\\ \\ =2.0176\times 10^{-5}\, kg

The quantity of protons, electrons, and neutrons in a particle is free of temperature what’s more, weight conditions. Subsequently, the acquired qualities will stay unaltered if the temperature and weight is changed.

 

Q.3.What number of neutrons and protons are there in the following nuclei ?

613C,816O,1224Mg,1656Fe,3888Sr_{6}^{13}\textrm{C}\: ,\: _{8}^{16}\textrm{O}\: ,\: _{12}^{24}\textrm{Mg}\: ,\: _{16}^{56}\textrm{Fe}\: ,\: _{38}^{88}\textrm{Sr}

Ans:

613C_{6}^{13}\textrm{C}:

Atomic mass = 13

Atomic number = Number of protons = 6

Number of neutrons = (Atomic mass) – (Atomic number) = 13 – 6 = 7

816O_{8}^{16}\textrm{O} :

Atomic mass = 16

Atomic number = 8

Number of protons = 8

Number of neutrons = (Atomic mass) – (Atomic number) = 16 – 8 = 8

1224Mg _{12}^{24}\textrm{Mg} :

Atomic mass = 24

Atomic number = Number of protons = 12

Number of neutrons = (Atomic mass) – (Atomic number) = 24 – 12 = 12

1656Fe _{16}^{56}\textrm{Fe} :

Atomic mass = 56

Atomic number = Number of protons = 26

Number of neutrons = (Atomic mass) – (Atomic number) = 56 – 26 = 30

3888Sr _{38}^{88}\textrm{Sr} :

Atomic mass = 88

Atomic number = Number of protons = 38

Number of neutrons = (Atomic mass) – (Atomic number) = 88 – 38 = 50

 

Q.4. Compose the total image for the atom with the given nuclear number (Z) and Atomic mass (A)

(I)Z = 17, A = 35

(II)Z = 92, A = 233

(III)Z = 4, A = 9

Ans:

(I)1735C_{17}^{35}\textrm{C}

(II)92233U_{92}^{233}\textrm{U}

(III)49Be_{4}^{9}\textrm{Be}

 

Q.5.Yellow light radiated from a sodium light has a wavelength (λ\lambda) of 580 nm. Ascertain the frequency (ν\nu) and wave number (νˉ\bar{\nu } ) of the yellow light.

Ans: From the expression,

λ=cν\lambda =\frac{c}{\nu }

We get,

ν=cλ\nu =\frac{c}{ \lambda }      ……….(1)

Where , ν\nu= frequency of yellow light

c = velocity of light in vacuum = 3×108m/s3\times 10^{8}\, m/s λ\lambda= = wavelength of yellow light = 580 nm = 580×109m/s580\times 10^{-9}\, m/s

Substituting the values in expression (1):

ν=3×108580×109=5.17×1014s1\nu =\frac{3\times 10^{8}}{580\times 10^{-9}}=5.17\times 10^{14}\, s^{-1}

Thus, frequency of yellow light emitted from the sodium lamp =

5.17×1014s15.17\times 10^{14}\, s^{-1}

Wave number of yellow light νˉ=1λ\bar{\nu }=\frac{1}{\lambda } =1580×109=1.72×106m1=\frac{1}{580\times 10^{-9}}=1.72\times 10^{6}\, m^{-1}

 

Q.6.Find energy of each of the photons which

(I)Correspond to light of frequency 3×1015Hz3\times 10^{15}\, Hz.

(II)Have wavelength of 0.50 armstrong.

Ans:

(I)Energy (E) of a photon is given by the expression,

E= hνh\nu

Where, h = Planck’s constant = 6.626×1034Js6.626\times 10^{-34}\, Js ν\nu = frequency of light = 3×10153\times 10^{15}Hz

Substituting the values in the given expression of E:

E=(6.626×1034)(3×1015)E=1.988×1018JE=(6.626\times 10^{-34})(3\times 10^{15})\\ \\ E=1.988\times 10^{-18}\, J

(II) Energy (E) of a photon having wavelength (λ\lambda) is given by the expression,

E= hcνhc\nu

h = Planck’s constant = 6.626×1034Js6.626\times 10^{-34}Js

c = velocity of light in vacuum = 3×108m/s3\times 10^{8}\, m/s

Substituting the values in the given expression of E:

E=(6.626×1034)(3×108)0.50×1010=3.976×1015JE=3.98×1015JE=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{0.50\times 10^{-10}}=3.976\times 10^{-15}J\\ \\ ∴ E=3.98\times 10^{-15}J

 

Q.7.Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0×1010s2.0 \times 10^{–10} \, s

Ans: Frequency (ν\nu) of light = 1Period\frac{1}{Period} 1Period=12.0×1010s=5.0×109s1\frac{1}{Period}\\ \\ =\frac{1}{2.0\times 10^{-10}\, s}\\ \\ =5.0\times 10^{9}\, s^{-1 }

Wavelength(λ\lambda) of light=cνc\nu

Where,

c = velocity of light in vacuum =3×108m/s3\times 10^{8}\, m/s

Substituting the value in the given expression of λ\lambda:

λ=3×1085.0×109=6.0×102m\lambda =\frac{3\times 10^{8}}{5.0\times 10^{9}}=6.0\times 10^{-2}m

Wave number (νˉ)(\bar{\nu }) of light = 1λ=16.0×102=1.66×101m1=16.66m\frac{1}{\lambda }=\frac{1}{6.0\times 10^{-2}}=1.66\times 10^{1}\, m^{-1}=16.66\, m

 

Q.8.What is the quantity of photons of light with a wavelength of 4000 pm that give 1 J of energy?

Ans: Energy (E) of a photon = hνh\nu

Energy (EnE_{n}) of ‘n’ photons =nhνnh\nu n=Enλhc\Rightarrow n=\frac{E_{n}\lambda }{hc}

Where, λ\lambda= wavelength of light = 4000 pm = 4000×1012m4000\times 10^{-12}\, m

c = velocity of light in vacuum =3×108m/s3\times 10^{8}\, m/s

h = Planck’s constant = 6.626×1034Js6.626\times 10^{-34}\, Js

Substituting the values in the given expression of n:

n=1×(4000×1012)(6.626×1034)(3×108)=2.012×1016n=\frac{1\times (4000\times 10^{-12})}{(6.626\times 10^{-34})(3\times 10^{8})}=2.012\times 10^{16}

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012×10162.012\times 10^{16}

 

Q.9. A photon of wavelength 4×107m4\times 10^{-7}\, m strikes on metal surface, the work function of the metal being 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= 1.6020×1019J1.6020\times 10^{-19}\,J)

Ans: Energy (E) of a photon = hν=hcλh\nu =\frac{hc}{\lambda }

Where, h = Planck’s constant =6.626×1034Js6.626\times 10^{-34}\,Js

c = velocity of light in vacuum =3×108m/s3\times 10^{8}\,m/s λ\lambda= wavelength of photon =4×107m/s4\times 10^{-7}\,m/s

Substituting the values in the given expression of E:

E=(6.626×1034)(3×108)4×107=4.9695×1019JE=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{4\times 10^{-7}}=4.9695\times 10^{-19}\, J

Hence, the energy of the photon is 4.97×1019J4.97\times 10^{-19}\, J

(II) The kinetic energy of emission EkE_{k} is given by

=hνhν0=(EW)eV=(4.9695×10191.6020×1019)eV2.13eV=(3.10202.13)eV=0.9720eV=h\nu -h\nu _{0}\\ \\ =(E-W)eV\\ \\ =(\frac{4.9695\times 10^{-19}}{1.6020\times 10^{-19}})eV-2.13\, eV\\ \\ =(3.1020-2.13)eV\\ \\ =0.9720\, eV

Hence, the kinetic energy of emission is 0.97 eV.

(III)The velocity of a photoelectron (v) can be calculated by the expression,

12mv2=hνhν0v=2(hνhν0)m\frac{1}{2}mv^{2} =h\nu -h\nu _{0}\\ \\ \Rightarrow v =\sqrt{\frac{2(h\nu -h\nu _{0})}{m}}

Where (hνhν0)(h\nu -h\nu _{0}) is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron.

Substituting the values in the given expression of v:

v=2×(0.9720×1.6020×1019)J9.10939×1031kg=0.3418×1012m2s2v=5.84×105ms1v=\sqrt{\frac{2\times (0.9720\times 1.6020\times 10^{-19})J}{9.10939\times 10^{-31}kg}}\\ \\ =\sqrt{0.3418\times 10^{12}m^{2}s^{2}}\\ \\ \Rightarrow v=5.84\times 10^{5}ms^{-1}

Hence, the velocity of the photoelectron is 5.84×105ms15.84\times 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 mol1mol^{-1}.

Ans: Energy of sodium (E)= NAhcλ=(6.023×1023mol1)(6.626×1034)Js(3×108)ms1242×109m=4.947×105Jmol1=494.7×103Jmol1=494kJmol1\frac{N_{A}hc}{\lambda }\\ \\ =\frac{(6.023\times 10^{23}\, mol^{-1})(6.626\times 10^{-34})Js(3\times 10^{8})ms^{-1}}{242\times 10^{-9}m}\\ \\ =4.947\times 10^{5}\, J\, mol^{-1}\\ \\ =494.7\times 10^{3}\, J\, mol^{-1}\\ \\ =494\, kJ\, mol^{-1}

 

Q.11. A 25 watt bulb discharges monochromatic yellow light of wavelength of 0.57µm. Ascertain the rate of discharge of quanta every second.

Ans: Power of bulb, P = 25 Watt = 25Js125\, Js^{-1}

Energy of one photon, E = hν=hcνh\nu =\frac{hc}{\nu }

Substituting the values in the given expression of E:

E=(6.626×1034)(3×108)(0.57×106)=34.87×1020JE=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{(0.57\times 10^{-6})}=34.87\times 10^{-20}\, J

E= 34.87×1020J34.87\times 10^{-20}\, J

Rate of emission of quanta per second= E=2534.87×1020=7.169×1019s1E=\frac{25}{34.87\times 10^{-20}}=7.169\times 10^{19}\, s^{-1}

 

Q.12. Electrons are discharged with zero speed from a metal surface when it is presented to radiation of wavelength 6800 Å. Figure limit recurrence (ν0\nu _{0} ) and work work (W0W_{0}) of the metal.

Ans: Threshold wavelength of radiation (λ0)(\lambda _{0})= 6800 Å=6800×1010m6800\times 10^{-10}\, m

Threshold frequency (ν0\nu _{0} ) of the metal

=cλ0=3×108ms16.8×107m=4.41×1014s1\frac{c}{\lambda _{0}}=\frac{3\times 10^{8}ms^{-1}}{6.8\times 10^{-7}m}=4.41\times 10^{14}\, s^{-1}

Thus, the threshold frequency (ν0\nu _{0} ) of the metal = hν0=(6.626×1034Js)(4.41×1014s1)=2.922×1019Jh\nu _{0}\\ \\ =(6.626\times 10^{-34}Js)(4.41\times 10^{14}s^{-1})\\ \\ =2.922\times 10^{-19}\, J

 

Q.13. What is the wavelength of light transmitted when the electron in a hydrogen atom experiences move from an energy level with n = 4 to an energy level with n = 2?

Ans: The nin_{i} = 4 to nfn_{f} = 2 transition will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

E=2.18×1018[1ni21nf2]E=2.18\times 10^{-18}[\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}]

Substituting the values in the given expression of E:

E=2.18×1018[142122]=2.18×1018[1416]=2.18×1018×(316)E=(4.0875×1019J)E=2.18\times 10^{-18}[\frac{1}{4^{2}}-\frac{1}{2^{2}}]\\ \\ =2.18\times 10^{-18}[\frac{1-4}{16}]\\ \\ =2.18\times 10^{-18}\times (-\frac{3}{16})\\ \\ E=-(4.0875\times 10^{-19}J)

The negative sign indicates the energy of emission.

Wavelength of light emitted (λ)=hcE(\lambda )=\frac{hc}{E}

(Since E=hc lambdaE=\frac{hc}{\ lambda })

Substituting the values in the given expression of  lambda{\ lambda }:

λ=(6.626×1034)(3×108)(4.0875×1019)=4.8631×107mλ=486.31×109m=486nm\lambda =\frac{(6.626\times 10^{-34})(3\times 10^{8})}{(4.0875\times 10^{-19})}=4.8631\times 10^{-7}\, m\\ \\ \lambda =486.31\times 10^{-9}\, m\\ \\ =486\, nm

 

Q.14. What amount of energy is required to ionize a H molecule if the electron involves n = 5 circle? Contrast your answer and the ionization enthalpy of H molecule (energy required to evacuate the electron from n =1 circle).

Ans: The expression of energy is given by,

En=(2.18×1018)Z2n2E_{n} =\frac{-(2.18\times 10^{-18})Z^{2}}{n^{2}}

Where,

Z= atomic number of the atom n

= principal quantum number

For ionization from nt=5n_{t}=5 to n2=n_{2} =∞,

ΔE=EE5=[((2.18×1018J)(1)2()2)((2.18×1018J)(1)2(5)2)]=0.0872×1018JΔE=8.72×1020J\Delta E=E_{\infty }-E_{5 }\\ \\ =[(\frac{-(2.18\times 10^{-18}\, J)(1)^{2}}{(\infty )^{2}})-(\frac{-(2.18\times 10^{-18}\, J)(1)^{2}}{(5 )^{2}})]\\ \\ =0.0872\times 10^{-18}\, J\\ \\ \Delta E=8.72\times 10^{-20}\, J

Hence, the energy required for ionization from n = 5 to n = ∞Energy required for n1 = 1 to n = ∞, is 8.72×1020m8.72\times 10^{-20}\, m

 

 

ΔE=EE5=[((2.18×1018J)(1)2()2)((2.18×1018J)(1)2(1)2)]=2.18×1018JΔE=2.18×1018J\Delta E=E_{\infty }-E_{5 }\\ \\ =[(\frac{-(2.18\times 10^{-18}\, J)(1)^{2}}{(\infty )^{2}})-(\frac{-(2.18\times 10^{-18}\, J)(1)^{2}}{(1 )^{2}})]\\ \\ =2.18\times 10^{-18}\, J\\ \\ \Delta E=2.18\times 10^{-18}\, J

Hence, less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.

 

Q.15. When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:

Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

The number of spectral lines produced when an electron in the n th level drops down to the ground state is given by

n(n1)2\frac{n(n-1)}{2}

Given, n = 6

Number of spectral lines =6(61)2=15\frac{6(6-1)}{2}=15

 

Q.16.(I) Energy associated with the fifth orbit of hydrogen atom is calculated as:

E5=(2.18×1018)(5)2=(2.18×1018)25=8.72×1020JE_{5}=\frac{-(2.18\times 10^{-18})}{(5)^{2}}=\frac{-(2.18\times 10^{-18})}{25}=-8.72\times 10^{-20}\, J

 

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

For, n = 5

r5=(0.0529nm)(52)r5=1.3225nmr_{5}=(0.0529\, nm)(5^{2})\\ \\ r_{5}=1.3225\, nm

 

Q.17. Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Ans. For the Balmer series, ni = 2. Thus, the expression of wavenumber(νˉ\bar{\nu}) is given by,

νˉ=[1(2)21nf2](1.097×107m1)\bar{\nu}=[\frac{1}{(2)^{2}}-\frac{1}{n_{f}^{2}}](1.097\times 10^{7}\, m^{-1})

Wave number(νˉ\bar{\nu})  is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, (νˉ\bar{\nu}) has to be the smallest.

For  (νˉ\bar{\nu})   to be minimum, nf should be minimum.

For the Balmer series, a transition from ni = 2 to nf = 3 is allowed.

Hence, taking nf = 3, we get:

νˉ=(1.097×107)[1(2)2132]νˉ=(1.097×107)[1419]=(1.097×107)[9436]=(1.097×107)[536]νˉ=1.5236×106m1\bar{\nu}=(1.097\times 10^{7})[\frac{1}{(2)^{2}}-\frac{1}{{3}^{2}}]\\ \\ \bar{\nu}=(1.097\times 10^{7})[\frac{1}{4}-\frac{1}{9}]\\ \\ =(1.097\times 10^{7})[\frac{9-4}{36}]\\ \\ =(1.097\times 10^{7})[\frac{5}{36}]\\ \\ \bar{\nu}=1.5236\times 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.

Ans. Energy (E) of the nth Bohr orbit of an atom is given by,

E5=(2.18×1018)Z2(n)2E_{5}=\frac{-(2.18\times 10^{-18})Z^{2}}{(n)^{2}}

Where, Z = atomic number of the atom

Ground state energy =2.18×1011ergs=2.18×1011×107J=2.18×1018J=-2.18 \times 10^{–11}\, ergs\\ =-2.18 \times 10^{–11} \times 10^{–7}\, J \\ =-2.18 \times 10^{–18} \, J

Energy required to shift the electron from n = 1 to n = 5 is given as:

ΔE=E5E1=[((2.18×1018J)(1)2(5)2)(2.18×1018)]=(2.18×1018)[1125]=(2.18×1018)[2425]=2.0928×1018J\Delta E=E_{5}-E_{1}\\ \\ =[(\frac{-(2.18\times 10^{-18}\, J)(1)^{2}}{(5 )^{2}})-(-2.18\times 10^{-18})]\\ \\ =(2.18\times 10^{-18})[1-\frac{1}{25}]\\ \\ =(2.18\times 10^{-18})[\frac{24}{25}]\\ \\ =2.0928\times 10^{-18}\, J

Wavelength of emitted light=

hcE \frac{hc}{E } E=(6.626×1034)(3×108)(2.0928×1018)=9.498×108mE=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{(2.0928\times 10^{-18})}\\ \\ =9.498\times 10^{-8}\, m

 

Q.19. The electron energy in hydrogen atom is given by En =2.18×1018,n2J-2.18\times 10^{-18}, n^{2}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?

Ans. En=(2.18×1018)(n)2JE_{n}=\frac{-(2.18\times 10^{-18}) }{(n)^{2} }J

Energy required for ionization from n = 2 is given by,

ΔE=EE2=[((2.18×1018()2)((2.18×1018)(2)2)]J=[(2.18×1018)40]J=0.545×1018J\Delta E=E_{\infty }-E_{2}\\ \\ =[(\frac{-(2.18\times 10^{-18}}{(\infty )^{2}})-(\frac{(-2.18\times 10^{-18})}{(2)^{2}})]J\\ \\ =[\frac{(2.18\times 10^{-18})}{4}-0]J\\ \\ =0.545\times 10^{-18}\, J ΔE=5.45×1019J\Delta E=5.45\times 10^{-19}\, J λ=hcΔE\lambda =\frac{hc}{\Delta E}

Here, λ is the longest wavelength causing the transition.

E=(6.626×1034)(3×108)(0.57×106)=34.87×1020JE=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{(0.57\times 10^{-6})}=34.87\times 10^{-20}\, J E=(6.626×1034)(3×108)(5.45×1019)=3.647×107m=3647×1010E=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{(5.45\times 10^{-19})}=3.647\times 10^{-7}\, m\\ \\ =3647\times 10^{-10}

= 3647 Å

 

Q.20. Calculate the wavelength of an electron moving with a velocity of 2.05×107ms12.05\times 10^{7}\, ms^{-1}

Ans. According to de Broglie’s equation,

λ=hmv\lambda =\frac{h}{mv }

Where, λ = wavelength of moving particle

m = mass of particle

v = velocity of particle

h = Planck’s constant

Substituting the values in the expression of λ:

λ=(6.626×1034)Js(9.10939×1031kg)(2.05×107ms1)=3.548×1011m\lambda =\frac{(6.626\times 10^{-34})Js}{(9.10939\times 10^{-31}kg)(2.05\times 10^{7}ms^{-1})}\\ \\ =3.548\times 10^{-11}\, m

Hence, the wavelength of the electron moving with a velocity of 2.05×107ms1is3.548×1011m2.05\times 10^{7}\, ms^{-1}\: is\:\: 3.548\times 10^{-11}\, m

 

Q.21. The mass of an electron is 9.1×10319.1 \times 10^{-31}kg. If its K.E. is 3.0×10253.0\times 10^{-25}J, calculate its wavelength.

Ans. From de Broglie’s equation,

λ=hmv\lambda =\frac{h}{mv }

Given, Kinetic energy (K.E) of the electron = 3.0×10253.0\times 10^{-25}J

Since K.E.= 12mv2\frac{1}{2}mv^{2} Velocity(v)=2K.Em=2(3.0×1025J)9.10939×1031kg=6.5866×104v=811.579ms1∴ Velocity(v)=\sqrt{\frac{2K.E}{m}}\\ \\ =\sqrt{\frac{2(3.0\times 10^{-25}J)}{9.10939\times 10^{-31}kg}}\\ \\ =\sqrt{6.5866\times 10^{4}}\\ \\ v=811.579\, ms^{-1}

Substituting the value in the expression of λ:

λ=(6.626×1034)Js(9.10939×1031kg)(811.579ms1)=8.9625×107m\lambda =\frac{(6.626\times 10^{-34})Js}{(9.10939\times 10^{-31}kg)(811.579ms^{-1})}\\ \\ =8.9625\times 10^{-7}\, m

 

Q.22. (I)Write the electronic configurations of the following ions:

(a)H

(b)Na+

(c)O2–

(d)F –

(II) What are the atomic numbers of elements whose outermost electrons are represented by

(a)3s1

(b)2p3 and

(c)3p5?

(III) Which atoms are indicated by the following configurations?

(a)[He] 2s1

(b)[Ne] 3s2 3p3

(c)[Ar] 4s2 3d1 .

Ans:

(I)(a)H  ion

The electronic configuration of H atom is 1s1 .

A negative charge on the species indicates the gain of an electron by it.

Electronic configuration of H = 1s2

 

(b)Naion

The electronic configuration of Na atom is 1s2 2s2p6 3s1 .

A positive charge on the species indicates the loss of an electron by it.

Electronic configuration of Na+ = 1s2 2s2 2p6 3s0 or 1s2 2s2 2p6

(c)O2– ion

The electronic configuration of 0 atom is 1s 2 2s 2 2p 4 .

A dinegative charge on the species indicates that two electrons are gained by it. Electronic configuration of O2– ion = 1s 2 2s 6

(d)F – ion

The electronic configuration of F atom is 1s 2 2s 2 2p 5 .

A negative charge on the species indicates the gain of an electron by it.

Electron configuration of F ion = 1s 2 2s 2 2p 6

(II) (a)3s1

Completing the electron configuration of the element as 1s 2 2s 2 2p 6 3s 1 . Number of electrons present in the atom of the element = 2 + 2 + 6 + 1 = 11 Atomic number of the element = 11

(b)2p 3

Completing the electron configuration of the element as 1s 2 2s 2 2p 3 .

Number of electrons present in the atom of the element = 2 + 2 + 3 = 7

Atomic number of the element = 7

(c)3p 5

Completing the electron configuration of the element as 1s 2 2s 2 2p 5 .

Number of electrons present in the atom of the element = 2 + 2 + 5 = 9

Atomic number of the element = 9

(III)(a)[He] 2s 1

The electronic configuration of the element is [He] 2s 1 = 1s 2 2s 1 .

Atomic number of the element = 3

Hence, the element with the electronic configuration [He] 2s 1 is lithium (Li).

(b)[Ne] 3s 2 3p3

The electronic configuration of the element is [Ne] 3s 2 3p 3= 1s 2s 2 2p 6 3s 3p 3 . Atomic number of the element = 15

Hence, the element with the electronic configuration [Ne] 3s 2 3p 3 is phosphorus (P).

(c)[Ar] 4s 2 3d1

The electronic configuration of the element is [Ar] 4s 2 3d 1= 1s 2 2s 2p 6 3s 2 3p 6 4s 3d 1 .

Atomic number of the element = 21

Hence, the element with the electronic configuration [Ar] 4s 2 3d 1 is scandium (Sc).

 

Q.23. What is the most reduced estimation of n that permits g orbitals to exist?

Ans. For g-orbitals, l = 4.

With respect to any value “n” of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).

For l = 4, least value of n = 5.

 

Q.24. An electron is in one of the 3d orbitals. Give the possible values of nl, l and ml for this electron.

Ans: For the 3d orbital:

Principal quantum number (n) = 3

Azimuthal quantum number (l) = 2

Magnetic quantum number (ml) = – 2, – 1, 0, 1, 2

 

Q.25.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.

Ans:

(i)For an atom to be neutral, the number of protons is equal to the number of electrons.

Number of protons in the atom of the given element = 29

(ii)The electronic configuration of the atom is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 .

 

Q.26.Give the number of electrons in the species , H2+ and H2 and O2+

Ans: Number of electrons present in hydrogen molecule (H2+) = 1 + 1 = 2

Number of electrons in = 2 – 1 = 1

H2: Number of electrons in H2 = 1 + 1 = 2

O2+:Number of electrons present in oxygen molecule (O2) = 8 + 8 = 16

Number of electrons in = 16 – 1 = 15

 

Q.27. (I)An atomic orbital has n = 3. What are the possible values of l and ml ?

(II)List the quantum numbers (mand l) of electrons for 3d orbital.

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

Ans.

(I)n = 3 (Given)

For a given value of n, l can have values from 0 to (n – 1).

For n = 3

l = 0, 1, 2

For a given value of l, ml can have (2l + 1) values.

For l = 0, m = 0 l = 1, m = – 1, 0, 1 l = 2, m = – 2, – 1, 0, 1, 2

For n = 3

l = 0, 1, 2

m= 0

m1 = – 1, 0, 1

m2 = – 2, – 1, 0, 1, 2

(II)For 3d orbital, l = 2.

For a given value of l, ml can have (2l + 1) values i.e., 5 values.

For l = 2 m= – 2, – 1, 0, 1, 2

(III)Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist. For p-orbital, l = 1.

For a given value of n, l can have values from zero to (n – 1).

For l is equal to 1, the minimum value of n is 2.

Similarly,

For f-orbital, l = 4.

For l = 4, the minimum value of n is 5.

Hence, 1p and 3f do not exist.

 

Q.28.Using s, p, 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.

Ans:

(a)n = 1, l = 0 (Given) The orbital is 1s.

(b)For n = 3 and l = 1 The orbital is 3p.

(c)For n = 4 and l = 2 The orbital is 4d.

(d)For n = 4 and l = 3 The orbital is 4f.

 

Q.29. Explain, giving reasons, which of the following sets of quantum numbers are not possible.

a) n = 0, l = 0, ml= 0, m=+12\frac{1}{2}

b) n = 1, l = 0, ml= 0, m=-12\frac{1}{2}

c) n = 1, l = 1, ml= 0, m=+12\frac{1}{2}

d) n = 2, l = 0, ml= 1, m=-12\frac{1}{2}

e) n = 3, l = 3, ml= -3, m=+12\frac{1}{2}

f) n = 3, l = 0, ml= 1, m=+12\frac{1}{2}

 

Ans.(a)The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.

(b)The given set of quantum numbers is possible.

(c)The given set of quantum numbers is not possible. For a given value of n, ‘l’ can have values from zero to (n – 1). For n = 1, l = 0 and not 1.

(d)The given set of quantum numbers is possible. (e) The given set of quantum numbers is not possible. For n = 3, l = 0 to (3 – 1) l = 0 to 2 i.e., 0, 1, 2

(f)The given set of quantum numbers is possible.

 

Q.30. How many electrons in an atom may have the following quantum numbers?

a) n = 4, m=-12\frac{1}{2}

b)n = 3, l = 0

Ans.(a)Total number of electrons in an atom for a value of n = 2n 2

For n = 4,

Total number of electrons = 2 (4)2 = 32

The given element has a fully filled orbital as

1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 .

Hence, all the electrons are paired.

Number of electrons (having n = 4 and m=-12\frac{1}{2}) = 16

(b)n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.

 

Q.31. 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.Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:

mvr=nh2π.(1)mvr=n\frac{h}{2\pi }\: \: \: ….(1)

Where, n = 1, 2, 3, …

According to de Broglie’s equation:

λ=hmvormv=hλ.(2)\lambda =\frac{h}{mv }\\ \\ or\: mv=\frac{h}{\lambda }\: \: \: \: …….(2) hrλ=nh2πor2πr=nλ..(3)\frac{hr}{\lambda }=n\frac{h}{2\pi }\\ \\ or\: 2\pi r=n\lambda \: \: \: \: …..(3)

Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

 

Q.32.What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of Hespectrum?

Ans. For He+ ion, the wave number associated with the Balmer transition, n = 4 to n = 2 is given by:

νˉ=1λ=RZ2(1n121n22)\bar{\nu}=\frac{1}{\lambda }=RZ^{2}(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})

Where, n1 = 2 n2 = 4

Z = atomic number of helium

νˉ=1λ=R(2)2(14116)=4R(4116)νˉ=1λ=3R4λ=43R\bar{\nu}=\frac{1}{\lambda }=R(2)^{2}(\frac{1}{4}-\frac{1}{16})\\ \\ =4R(\frac{4-1}{16})\\ \\ \bar{\nu}=\frac{1}{\lambda }=\frac{3R}{4}\\ \\ \Rightarrow \lambda =\frac{4}{3R}

According to the question, the desired transition for hydrogen will have the same wavelength as that of He+.

R(1)2[1n121n22]=3R4[1n121n22]=34.(1)R(1)^{2}[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}]=\frac{3R}{4}\\ \\ \Rightarrow [\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}]=\frac{3}{4}\: \: \: \: \: …….(1)

By hit and trail method, the equality given by equation (1) is true only when n1 = 1and n2 = 2.

The transition for n2 = 2 to n = 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of He+ spectrum.

 

Q.33.Calculate the energy required for the process

He(g)+He(g)2++eHe^{+}_{(g)}\rightarrow He^{2+}_{(g)}+e^{-}

The ionization energy for the H atom in the ground state is 2.18×1018Jatom12.18\times 10^{-18}\, J\,atom^{-1}

Ans. Energy associated with hydrogen-like species is given by,

En=2.18×1018(Z2n2)JE_{n}=-2.18\times 10^{-18}(\frac{Z^{2}}{n^{2}})J

For ground state of hydrogen atom,

ΔE=EE1=0[(2.18×1018(1)2(1)2]JΔE=2.18×1018J\Delta E=E_{\infty }-E_{1}\\ \\ =0-[-(2.18\times 10^{-18}{\frac{(1)^{2}}{(1)^{2}}}]J\\ \\ \Delta E=2.18\times 10^{-18}J\\

For the given process,

He(g)+He(g)2++eHe^{+}_{(g)}\rightarrow He^{2+}_{(g)}+e^{-}

An electron is removed from n = 1 to n = ∞.

ΔE=EE1=0[(2.18×1018(2)2(1)2]JΔE=8.72×1018J\Delta E=E_{\infty }-E_{1}\\ \\ =0-[-(2.18\times 10^{-18}{\frac{(2)^{2}}{(1)^{2}}}]J\\ \\ \Delta E=8.72\times 10^{-18}J\\.

The energy required for the process 8.72×1018J8.72\times 10^{-18}J\\.

 

Q.34. 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 length of scale of length 20 cm long.

Ans. 1 m = 100 cm

1 cm = 10–2 m

Length of the scale = 20 cm

=20×102m20\times 10^{-2}\, m

Diameter of a carbon atom = 0.15 nm

=0.15×109m0.15\times 10^{-9}\, m

One carbon atom occupies 0.15×109m0.15\times 10^{-9}\, m.

Number of carbon atoms that can be placed in a straight line 20×102m0.15×109m=133.33×107=1.33×109\frac{20\times 10^{-2}m}{0.15\times 10^{-9}m}\\ \\ =133.33\times 10^{7}\\ \\ =1.33\times 10^{9}

 

Q.35. 2×108m2\times 10^{8}\, m atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Ans. Length of the given arrangement = 2.4 cm

Number of carbon atoms present = 2 × 108

Diameter of carbon atom

2.4×102m2×108m=1.2×1010m\frac{2.4\times 10^{-2}m}{2\times 10^{8}m}\\ \\ =1.2\times 10^{-10}m Radius of carbon atom=Diameter2=1.2×1010m2=6.0×1011m\frac{Diameter}{2}\\ \\ =\frac{1.2\times 10^{-10}m}{2}\\ \\ =6.0\times 10^{-11}m

 

Q.36.The diameter of 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= Diameter2=2.62=1.3×1010m=130×1012m=130pm\frac{Diameter}{2}\\ \\ =\frac{2.6}{2}\\ \\ =1.3\times 10^{-10}m\\ \\ =130\times 10^{-12}m=130pm

(b) )Length of the arrangement = 1.6 cm

=1.6×102m 1.6 \times 10^{-2}m

Diameter of zinc atom =1.6×1010m 1.6 \times 10^{-10}m Number of zinc atoms present in the arrangement

=1.6×102m2.6×1010m=0.6153×108m=6.153×107=\frac{1.6\times 10^{-2}m}{2.6\times 10^{-10}m}\\ \\ =0.6153\times 10^{8}m\\ \\ =6.153\times 10^{7}

 

Q.37. The diameter of 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= Diameter2=2.62=1.3×1010m=130×1012m=130pm\frac{Diameter}{2}\\ \\ =\frac{2.6}{2}\\ \\ =1.3\times 10^{-10}m\\ \\ =130\times 10^{-12}m=130pm

(b) )Length of the arrangement = 1.6 cm

=1.6×102m 1.6 \times 10^{-2}m

Diameter of zinc atom =1.6×1010m 1.6 \times 10^{-10}m

Therefore,  Number of zinc atoms present in the arrangement

=1.6×102m2.6×1010m=0.6153×108m=6.153×107=\frac{1.6\times 10^{-2}m}{2.6\times 10^{-10}m}\\ \\ =0.6153\times 10^{8}m\\ \\ =6.153\times 10^{7}

 

Q.38. A certain particle carries 2.5×1016C 2.5 \times 10^{-16}C  of static electric charge. Calculate the number of electrons present in it.

Ans.

Charge on one electron = 1.6022×1019C 1.6022 \times 10^{-19}C 1.6022×1019C\Rightarrow 1.6022 \times 10^{-19}C charge is carried by 1 electron.

Therefore, Number of electrons carrying a charge of 2.5×1016C 2.5 \times 10^{-16}C 11.6022×1019C(2.5×1016C)=1.560×103C=1560C\frac{1}{1.6022\times 10^{-19}C}(2.5\times 10^{-16}C)\\ \\ =1.560\times 10^{3}C\\ \\ =1560\: C

 

Q.39. In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is1.282×1018C -1.282 \times 10^{-18}C  , calculate the number of electrons present on it.

Ans.

Charge on the oil drop =1.282×1018C 1.282 \times 10^{-18}C

Charge on one electron =1.6022×1019C 1.6022 \times 10^{-19}C

Therefore, Number of electrons present on the oil drop

1.282×1018C1.6022×1019C=0.8001×101=8.0\frac{1.282\times 10^{-18}C}{1.6022\times 10^{-19}C}\\ \\ =0.8001\times 10^{1}\\ \\ =8.0