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The chapter "Structure of Atom" in Class 11 Chemistry is a fundamental chapter that lays the foundation for understanding the nature of atoms, their structure, and properties. This chapter provides a detailed understanding of the various atomic models and theories proposed over the years. Let's have a look at NCERT Solutions of the chapter in this document:
Q. 2.1
(i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Ans.
(i) Mass of one electron = 9.10939 × 10^{–31} kg
Number of electrons that weigh 9.10939 × 10^{–31} kg = 1
Number of electrons that will weigh 1 g = (1 × 10^{–3} kg)
= 0.1098 × 10^{–3 + 31}
= 0.1098 × 10^{28}
= 1.098 × 10^{27}
(ii) Mass of one electron = 9.10939 × 10^{–31} kg
Mass of one mole of electron = (6.022 × 10^{23}) × (9.10939 ×10^{–31} kg)
= 5.48 × 10^{–7} kg
Charge on one electron = 1.6022 × 10^{–19} coulomb
Charge on one mole of electron = (1.6022 × 10^{–19} C) (6.022 × 10^{23})
= 9.65 × 10^{4} C
EduRev Tip: Memorize these values of mass and charge of subatomic particles to improve your understanding of atomic structure.
Q. 2.2
(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 ^{14}C. (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 NH_{3} at STP.
Will the answer change if the temperature and pressure are changed?
Ans.
(i) Number of electrons present in 1 molecule of methane (CH_{4})
{1(6) + 4(1)} = 10
Number of electrons present in 1 mole i.e., 6.023 × 10^{23} molecules of methane
= 6.022 × 10^{23} × 10 = 6.022 × 10^{24}
(ii)
(a) Number of atoms of ^{14}C in 1 mole= 6.023 × 10^{23}
Since 1 atom of ^{14}C contains (14 – 6) i.e., 8 neutrons, the number of neutrons in 14 g of ^{14}C is (6.023 × 10^{23}) ×8.
Or, 14 g of ^{14}C contains (6.022 × 10^{23} × 8) neutrons.
Number of neutrons in 7 mg:
= 2.4092 × 10^{21}
(b) Mass of one neutron = 1.67493 × 10^{–27} kg
Mass of total neutrons in 7 g of ^{14}C
= (2.4092 × 10^{21}) (1.67493 × 10^{–27} kg)
= 4.0352 × 10^{–6} kg
(iii)
(a) 1 mole of NH_{3} = {1(14) + 3(1)} g of NH_{3}
= 17 g of NH_{3}
= 6.022× 10^{23} molecules of NH_{3}
Total number of protons present in 1 molecule of NH_{3}
= {1(7) + 3(1)}
= 10
Number of protons in 6.023 × 10^{23} molecules of NH_{3}
= (6.023 × 10^{23}) (10)
= 6.023 × 10^{24}
⇒ 17 g of NH_{3} contains (6.023 × 10^{24}) protons.
Number of protons in 34 mg of NH_{3}
= 1.2046 × 10^{22}
(b) Mass of one proton = 1.67493 × 10^{–27} kg
Total mass of protons in 34 mg of NH_{3}
= (1.67493 × 10^{–27} kg) (1.2046 × 10^{22})
= 2.0176 × 10^{–5} kg
The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.
Understanding Atomic Symbols
Q. 2.3 How many neutrons and protons are there in the following nuclei?
, , , ,
Ans.
1. ^{13}_{6}C
Atomic mass = 13
Atomic number = Number of protons = 6
Number of neutrons = (Atomic mass) – (Atomic number)
= 13 – 6 = 7
2.
Atomic mass = 16
Atomic number = 8
Number of protons = 8
Number of neutrons = (Atomic mass) – (Atomic number)
= 16 – 8 = 8
3.
Atomic mass = 24
Atomic number = Number of protons = 12
Number of neutrons = (Atomic mass) – (Atomic number)
= 24 – 12 = 12
4.
Atomic mass = 56
Atomic number = Number of protons = 26
Number of neutrons = (Atomic mass) – (Atomic number)
= 56 – 26 = 30
5.
Atomic mass = 88
Atomic number = Number of protons = 38
Number of neutrons = (Atomic mass) – (Atomic number)
= 88 – 38 = 50
Q. 2.4 Write the complete symbol for the atom with the given atomic number(Z) & Atomic mass(A).
(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9
Ans.
(i)
(ii)
(iii)
Q. 2.5 Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wave number () of the yellow light.
Ans. From the expression,
We get, …….. (i)
Where,
ν = frequency of yellow light
c = velocity of light in vacuum = 3 × 10^{8} m/s
λ = wavelength of yellow light = 580 nm = 580 × 10^{–9} m
Substituting the values in expression (i):
Thus, frequency of yellow light emitted from the sodium lamp = 5.17 × 10^{14} s^{–1}
Wave number of yellow light,
Q. 2.6 Find energy of each of the photons which
(i) correspond to light of frequency 3× 10^{15 }Hz.
(ii) have wavelength of 0.50 Å.
Ans.
(i) Energy (E) of a photon is given by the expression, E =
Where, h = Planck’s constant = 6.626 × 10^{–34} Js
ν = frequency of light = 3 × 10^{15} Hz
Substituting the values in the given expression of E:
E = (6.626 × 10^{–34}) (3 × 10^{15})
E = 1.988 × 10^{–18} J
(ii) Energy (E) of a photon having wavelength (λ) is given by the expression,
h = Planck’s constant = 6.626 × 10^{–34} Js
c = velocity of light in vacuum = 3 × 10^{8} m/s
Substituting the values in the given expression of E:
Q. 2.7 Calculate the wavelength,frequency & wave number of a light wave whose period is 2.0×10^{–10}s.
Ans. Frequency (ν) of light
Wavelength (λ) of light
Where, c = velocity of light in vacuum = 3×10^{8} m/s
Substituting the value in the given expression of λ:
Wave number of light
Q. 2.8 What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
Ans. Energy (E) of a photon = hν
Energy (E_{n}) of ‘n’ photons = nhν
Where,
λ = wavelength of light = 4000 pm = 4000 ×10^{–12} m
c = velocity of light in vacuum = 3 × 10^{8 }m/s
h = Planck’s constant = 6.626 × 10^{–34} Js
Substituting the values in the given expression of n:
Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012 × 10^{16}.
Q. 2.9 A photon of wavelength 4 × 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 photo electron (1 eV= 1.6020 × 10^{–19} J).
Ans.
(i) Energy (E) of a photon = hν = hc/λ
Where,
h = Planck’s constant = 6.626 × 10^{–34} Js
c = velocity of light in vacuum = 3 × 10^{8 }m/s
λ = wavelength of photon = 4 × 10^{–7} m
Substituting the values in the given expression of E:
Hence, the energy of the photon is 4.97 × 10^{–19} J.
(ii) The kinetic energy of emission E_{k} is given by
= (3.1020 – 2.13) eV = 0.9720 eV
Hence, the kinetic energy of emission is 0.97 eV.
(iii) The velocity of a photo electron (ν) can be calculated by the expression,
Where, 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 = 5.84 × 10^{5} ms^{–1}
Hence, the velocity of the photoelectron is 5.84 × 10^{5} ms^{–1}.
Q. 2.10 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. Energy of sodium (E)
= 4.947 × 10^{5} J mol^{–1}
= 494.7 × 10^{3} J mol^{–1 }= 494 kJ mol^{–1}
Q. 2.11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the rate of emission of quanta per second.
Ans. Power of bulb, P = 25 Watt = 25 Js^{–1}
Energy of one photon, E = hν = hc/λ
Substituting the values in the given expression of E:
E = 34.87 × 10^{–20} J
Rate of emission of quanta per second
Q. 2.12 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency () and work function (W_{0}) of the metal.
Ans. Threshold wavelength of radian = 6800 × 10^{–10} m
Threshold frequency of the metal
= 4.41 × 10^{14} s^{–1}
Thus, the threshold frequency of the metal is 4.41 × 10^{14} s^{–1}.
Hence, work function (W_{0}) of the metal = hν_{0}
= (6.626 × 10^{–34} Js) (4.41 × 10^{14} s^{–1})
= 2.922 × 10^{–19} J
Q. 2.13 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes 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 will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,
Substituting the values in the given expression of E:
E = – (4.0875 × 10^{–19} J)
The negative sign indicates the energy of emission.
Wavelength of light emitted
Substituting the values in the given expression of λ:
Q. 2.14 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 of energy is given by,
Where,
Z = atomic number of the atom
n = principal quantum number
For ionization from n_{1} = 5 to ,
Hence, the energy required for ionization from n = 5 to n = is 8.72 × 10^{–20} J.
Energy required for n_{1} = 1 to n = ,
Hence, less energy is required to ionize an electron in the 5^{th} orbital of hydrogen atom as compared to that in the ground state.
Q. 2.15 What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?
Ans. 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
Given, n = 6
Number of spectral lines = 6(61)/2 = 15.
Q. 2.16
(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 hydrogen atom.
Ans.
(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:
E_{5} = –8.72 × 10^{–20} J
(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. 2.17 Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Ans. For the Balmer series, n_{i} = 2. Thus, the expression of wave number is given by,
Wave number is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, has to be the smallest.
For to be minimum, n_{f} should be minimum. For the Balmer series, a transition from n_{i} = 2 to n_{f} = 3 is allowed. Hence, taking n_{f} = 3, we get:
Q. 2.18 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 n^{th} Bohr orbit of an atom is given by,
Where,
Z = atomic number of the atom
Ground state energy = – 2.18 × 10^{–11} ergs
= –2.18 × 10^{–11} × 10^{–7} J
= – 2.18 × 10^{–18} J
Energy required to shift the electron from n = 1 to n = 5 is given as: ΔE = E_{5} – E_{1}
Q. 2.19 The electron energy in hydrogen atom is given by E_{n} = (–2.18 × 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. Given,
Energy required for ionization from n = 2 is given by,
ΔE = 0.545 × 10^{–18} J
= 5.45 × 10^{–19} J
Here, λ is the longest wavelength causing the transition.
= 3647 × 10^{–10} m
= 3647 Å
Q. 2.20 Calculate the wavelength of an electron moving with a velocity of 2.05 × 10^{7} ms^{–1}.
Ans. According to de Broglie’s equation,
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:
Hence, the wavelength of the electron moving with a velocity of 2.05 × 10^{7} ms^{–1} is 3.548 × 10^{–11} m.
Q. 2.21 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. From De Broglie’s equation,
λ = h/mv
Given,
Kinetic energy (K.E) of the electron = 3.0 × 10^{–25} J
Substituting the value in the expression of λ:
Hence, the wavelength of the electron is 8.9625 × 10^{–7} m.
Q. 2.22 Which of the following are isoelectronic species i.e., those having the same number of electrons?
Na^{+}^{ }, K^{+}^{ }, Mg^{2+ }, Ca^{2+ }, S^{2–}, Ar
Ans. Isoelectronic species have the same number of electrons.
Number of electrons in sodium (Na^{+}) = 11
Number of electrons in (Na^{+}^{ }) = 10
A positive charge denotes the loss of an electron.
Similarly,
Number of electrons in K^{+}^{ } = 18
Number of electrons in Mg^{2+ } = 10
Number of electrons in Ca^{2+ } = 18
A negative charge denotes the gain of an electron by a species.
Number of electrons in sulphur (S) = 16
∴ Number of electrons in S^{2} = 18
Number of electrons in argon (Ar) = 18
Hence, the following are isoelectronic species:
1) Na^{+} and Mg^{2+ } (10 electrons each)
2) K^{+}^{ }, Ca^{2+ }, S^{2–} and Ar (18 electrons each)
Q. 2.23
(i) Write the electronic configurations of the following ions:
(a) H^{– }
(b) Na^{+}^{ }
(c) O^{2– }
(d) F^{–}
(ii) What are the atomic numbers of elements whose outermost electrons are represented by:
(a) 3s^{1 }
(b) 2p^{3}
(c) 3p^{5}?
(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s^{1}
(b) [Ne] 3s^{2} 3p^{3}
(c) [Ar] 4s^{2} 3d^{1}.
Ans.
(i) (a) H^{–} ion
The electronic configuration of H atom is 1s^{1}.
A negative charge on the species indicates the gain of an electron by it.
∴ Electronic configuration of H^{–} = 1s^{2}
(b) Na^{+}^{ }ion
The electronic configuration of Na atom is 1s^{2} 2s^{2} 2p^{6} 3s^{1}.
A positive charge on the species indicates the loss of an electron by it.
∴ Electronic configuration of Na^{ } = 1s^{2} 2s^{2} 2p^{6} 3s^{0} or 1s^{2} 2s^{2} 2p^{6}
(c) O^{2–} 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 O^{2–} ion = 1s^{2} 2s^{2} p^{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) 3s^{1}
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} 3p^{3}
The electronic configuration of the element is [Ne] 3s^{2} 3p^{3}= 1s^{2} 2s^{2} 2p^{6} 3s^{2} 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} 3d^{1}
The electronic configuration of the element is [Ar] 4s^{2} 3d^{1}= 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{1}.
∴ Atomic number of the element = 21
Hence, the element with the electronic configuration [Ar] 4s^{2} 3d^{1 }is scandium (Sc).
Q. 2.24 What is the lowest value of n that allows g orbitals to exist?
Ans. For gorbitals, l = 4.
As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).
∴ For l = 4, minimum value of n = 5
Q. 2.25 An electron is in one of the 3d orbitals. Give the possible values of n, l, and m_{l} for this electron.
Ans. For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (l) = 2
Magnetic quantum number (m_{l}) = – 2, – 1, 0, 1, 2
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199 videos330 docs229 tests
