1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

You can Download Chapter 2 Structure of Atom Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

1st PUC Chemistry Structure of Atom One Mark Questions and Answers

Question 1.
Mention the constituents of atom.
Answer:
Electron, Proton & Neutron.

Question 2.
Who discovered electron.
Answer:
J.J. Thomson

Question 3.
What is the mass of electron?
Answer:
9.108 × 1031 kg (0.0005487amu).

Question 4.
Mention the charge of electron in colomb.
Answer:
1.61.6*1019C.

Question 5.
What is the charge of proton ?
Answer:
Positive charge.

KSEEB Solutions

Question 6.
What is the mass of proton ?
Answer:
1.672 x 1027 kg(1.00728amu)

Question 7.
Who discovered proton ?
Answer:
E. Goldstein.

Question 8.
Who invented charge on electron ?
Answer:
Robert Millikan.

Question 9.
Name the experiment by which charge on electron determined.
Answer:
Robert Millikan oil drop experiment.

KSEEB Solutions

Question 10.
Who discovered neutron ?
Answer:
James Chadwick.

Question 11.
What is atomic number ?
Answer:
Atomic number *Z’ is the number of protons present in the nucleus of an atom of an element or the number of electrons present in an atom of an element.

Question 12.
What is mass number ?
Answer:
Atomic mass ‘A’. The total number of protons and neutrons present in the nucleus of an atom of an element.

Question 13.
Write the relationship between mass number and atomic number.
Answer:
Number of neutrons = Mass number (A) – Atomic number (Z) = A – Z

Question 14.
How do you represent an atom symbolically with atomic number and mass number?
Answer:
ZXA where A = Mass Number, X = atom, Z = Atomic number.

Question 15.
What is the number of proton and neutron in 92X235 ?
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 1

Question 16.
Give the number of Protons, Electrons and Neutrons present in the atom – having atomic number 27 and mass number 56.
Answer:
Number of Protons 27, Number of Electrons 27, Number of Neturons 29.

Question 17.
Mention the proton, neutron and electons 17C35.
Answer:
Neutron = 35 – 17 = 18, Electron = Proton = 17 .

Question 18.
Name the species which has no electron.
Answer:
Proton

KSEEB Solutions

Question 19.
Name the atom which has no neutron.
Answer:
Hydrogen

Question 20.
What is the ratio between mass of proton and electron ?
Answer:
Mass of proton is 1837 times the mass of electron.

Question 21.
Name the particles which constitute cathode rays.
Answer:
Electrons

Question 22.
Who demonstrated the particle property of an electron ?
Answer:
J. J. Thomson

KSEEB Solutions

Question 23.
Who showed the wave property of electrons ?
Answer:
G.P. Thomson

Question 24.
What is the charge on neutrons ?
Answer:
Nil / Zero

Question 25.
Mention the mass of neutron.
Answer:
1.675 × 1027 kg(1.00866 amu)

Question 26.
Name the scientist who proposed the nuclear theory (Solar model of atom) theory and discovered the existence of nucleus of an atom ?
Answer:
Ernest Rutherford.

Question 27.
What is photon ?
Answer:
Electromagnetic radiations are emitted or absorbed or propagated in the form of small pockets of energy called photon.

Question 28.
What is Isotopes ?
Answer:
Atoms of the same element having the same atomic number but different mass numbers are called isotopes.

Question 29.
What is Isobars ?
Answer:
Atoms of different elements having the same mass number but different atomic numbers are called isobars.

Question 30.
What is Isotones ?
Answer:
Atoms of different elements which contain the same number of neutrons, but different mass numbers are known as isotones.

Question 31.
Define the term ‘radiation’
Answer:
Radiation is the emission and transmission of energy through space in the form of waves.

KSEEB Solutions

Question 32.
How are velocity, frequency and wavelength of light radiation related ?
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 2

Question 33.
State Pauli’s exclusion principles.
Answer:
No two electrons in an atom can have the same set of four quantum numbers.

Question 34.
An atom having mass number 40 has 20 neutrons in its nucleus. What is the atomic number of the element ?
Answer:
Mass Number = Number of protons + Number of neutrons
40 = Number of protons + 20
∴Atomic number = Number of protons = 20.

Question 35.
What is aufbau principle ?
Answer:
Aufbau principle is also known as the building up principle. According to this principle, electrons should eneter the subshells in the order of increasing energies.

Question 36.
What is emission spectrum ?
Answer:
The spectrum of radiation emitted by a substance that has absorbed energy is called in emission spectrum.

Question 37.
How is the magnetic moment of paramagnetic species is related to the number of unpaired electrons present in it ?
Answer:
Magnetic moment \(\mu=\sqrt{n(n+2) B M}\) where n is the number of unpaired electrons.

Question 38.
Differentiate between the terms ‘ground state’ and ‘excited state’.
Answer:
Ground state means the least energy state or the most stable state. Excited state means the higher energy state, in which the electrons are in the higher energy level (unstable state).

KSEEB Solutions

Question 39.
What is the expression for the energy of a photon ?
Answer:
E = hu where, h = planks constant = 6.626 × 10-34 Js , γ = frequency or radiation.

Question 40.
Write the unit for frequency of radiation.
Answer:
Cycles per sec (sec-1) or Hertz (Hz)

Question 41.
Name the experiment which shows that light has particle property.
Answer:
Photoelectric effect.

Question 42.
Name the experiment which shows that light has wave property.
Answer:
Refraction.

Question 43.
How is wave number and wavelength of a wave related?
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 3

Question 44.
What is the velocity of light?
Answer:
3 × 108 m/sec

KSEEB Solutions

Question 45.
Define wavelength.
Answer:
It is the distance between two successive crests (peaks) or trough in a wave.

Question 46.
Define Wave number.
Answer:
It is the number of waves present per meter and is equal to the reciprocal of wavelength. \(\left(\frac{1}{\lambda}\right)=\bar{v}\)

Question 47.
Define frequency of light.
Answer:
The number of waves which pass through a given point in one second is called frequency of light.

Question 48.
What type of waves does light constitute?
Answer:
Electromagnetic waves.

Question 49.
What is orbital (atom orbital) ?
Answer:
Orbital is a three dimensional region in space around the nucleus in which the probability of finding an electron is maximum.

Question 50.
How many electrons can be accommodated in an orbital?
Answer:
Two electrons.

KSEEB Solutions

Question 51.
Write the de Broglie’s equation.
Answer:
\(\lambda=\frac{\mathbf{h}}{\mathbf{m} \mathbf{V}}\) where λ – wavelength of wave, h = Plank’s constant, m = mass of the electron, V – velocity of the electron.

Question 52.
Write Rydberg’s formula.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 4

Question 53.
Write the Balmer equation.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 5

Question 54.
What are the four prominent lines in Balmer series of hydrogen spectrum ?
Answer:
Hα, Hβ, Hγ and H????

Question 55.
What is the value of Rydberg’s constant?
Answer:
10.97 × 106 m-1

Question 56.
Give the range of wavelengths of visible light.
Answer:
400 nm to 750 nm.

KSEEB Solutions

Question 57.
Give the Rydberg equation where R is Rydberg constant?
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 6

Question 58.
Name the element whose atom contain six protons in the nucleus.
Answer:
Carbon

Question 59.
Name the different series of hydrogen spectrum.
Answer:
(i) Lyman
(ii) Balmer
(iii) Pachen
(iv) Bracket
(v) Pfund

Question 60.
Name two series of hydrogen spectra which fall in the infra – red region.
Answer:
Brackett series and P-fund series.

Question 61.
Match the following:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 7
Answer:
a-iii, b-iv, c-ii, d-i

1st PUC Chemistry Structure of Atom Two Marks Questions and Answers

Question 1.
Write the difference between isotope & isobars.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 8

Question 2
Explain Planks quantum theory.
Answer:
Electromagnetic radiations are emitted, absorbed and propagated discontinuously in the form of small packet of energy called photon. A body emits (absorbs) radiations in the integral multiples of ‘photon’.
Energy associated with each photon is E = hv, h = 6.626 × 10 -34 Js

KSEEB Solutions

Question 3.
Distinguish between emission spectra & atomic spectra.
Answer:
Emission spectra : When an object is strongly heated, it starts to emit light. When the emitted light is subjected to depression, emission spectrum is obtained)
Atomic spectra : When an atom of an element is strongly heated, it starts emitting light energy in different region. This emitted light when dispersed, a large number of closely spaced lines where formed they are called atomic spectra.

Question 4.
What do you mean by electromagnetic spectra ?
Answer:
The complete range of electromagnetic waves in the increasing order of wavelength (decreasing order of frequency) is known as electromagnetic spectra.

Question 5.
Write the electromagnetic spectra in the increasing order of wave length. (Decreasing order of frequency)
Answer:
Increasing order of wave length > Cosmic rays > Gamma rays < X-rays < U.V. rays < Visible rays < IR rays < Microwave rays < Radio waves (i.e, decreasing order of frequency).

Question 6.
Deduce the de-Broglies matter wave equation.
Answer:
Dual nature of electron – de-Broglie’s matter wave equation
According to Einstein’s mass – energy relation, E = mc2 … (1)
Where m and c are the mass and velocity of an electron respectively.
According to quantum theory, Energy E = hv = (he) / λ … (2)
Where h is Plank’s constant = 6.626 × 10-34 Js, u is the frequency of moving electron.
Comparing(1)and(2)me2 = (hc)/λ >me = h/λ > = λ = h /me … (3)
Equation (3) is called de-Broglie’s equation.
For an electron its velocity is given as V, So the de-Broglie’s equation for an electron is given by,
λ = h / mv (mv = p = momentum of electron)
λ = h / p is called de-Broglie’s matter – wave equation

KSEEB Solutions

Question 7.
Explain the wave nature of light.
Answer:
Wave nature of light: According to Newton, light is a stream of particles which are also known as corpuscles of light. This could not account for the phenomena of interference and deflection (as Rutherford’s model) but justified reflection and refraction.

Hygen suggested in his wave theory that the fight travels in the form of waves, later James Maxwell proposed that fight and other radiations are transmitted, these waves are associated with oscillating electric and magnetic field.

Question 8.
Distinguish between particle and wave.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 9

Question 9.
Explain Werner Heisenberg’s uncertainty principle (qualitative)
Answer:
It is impossible to determine both the momentum (particle nature) and position (wave nature) of a moving sub atomic particle simultaneously with absolute accuracy.
Mathematically ∆x × Ap = h / 4π where ∆x = uncertainty in position :
∆p = uncertainty in momentum ; h = Plank’s constant = 6.626 × 10-34 Js.

Question 10.
Mention the Merits of Bohr’s theory.
Answer:

  1. Explains the formation of hydrogen spectrums
  2. It explains the stability of atom
  3. Ryd berg constant is calculated by it.

KSEEB Solutions

Question 11.
Write any two limitations of Bohr’s theorem.
Answer:

  1. Not explained spectra of atoms containing more than one electrons.
  2. Not explained fine spectra (when spectroscope of high resolving power is used, it is found that each line in the ordinary specrum spilts into number of components differing in frequency).
  3. Not given explanation regarding Stark effect (spiltting of spectral line in electric field) and Zeeman Effect (splitting of spectral fine in magnetic field).

Question 12.
Write the difference between orbit & orbital.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 10

Question 13.
Draw the structure of p-orbitals. (Draw the shape of orbital whose Azimuthal quantum no is 1).
Answer:
p orbital has 3 orientations px,py &pz
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 11

Question 14.
Draw the structure of d-orbital (Orbital whose Azimuthal quantum no = 2).
Answer:
d- orbital has 5 orientations.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 12

KSEEB Solutions

Question 15.
What do you mean by electronic configuration ? With the sequence.
Answer:
Electronic Configuration is the distribution of electrons in the various available orbitals of an atom of an element.
The sequenced in which the electron occupy the various oribitals is as follows.
Is, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p….

Question 16.
Explain the electronic Configuration of cation Fe+.
Answer:
Example for electronic configuration of cation.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 13

Question 17.
Explain electronic configuration of anion using N.
Answer:
Example for electronic configuration of anion.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 14

Question 18.
State & Explain Paul’s Exclusion Principle.
Answer:
Paul’s Exculsion principle : No two electrons in the same atom can have all the four quantum numbers same.
Illustration : In Is orbital there are two electrons, the set of quantum number of one electron differ with another in spin quantum number.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 15

Question 19.
State and explain Hunds Rule of maximum multiplicity.
Answer:
Hund’s rule or maximum multiplicity : Electron pairing does not take place until orbitals of same energy are singly occupied.

Question 20.
What are quantum number and Name them ?
Answer:
In order to define state energy and location of electron a set of four numbers are used.
These numbers are called quantum numbers.

  • Principal quantum number (n)
  • Azimuthal quantum number (1)
  • Magnetic quantum number (m)
  • Spin quantum number (s).

Question 21.
Write all quantum number values for 3s orbital electrons.
Answer:
S – orbital has two electrons.
3s; n = 3; 1 = 0; m = 0; s = + 1/2 for first electron,
n = 3; 1 = 0; m = 0; s = -1/2 for second electron.

KSEEB Solutions

Question 22.
An orbital can contain only two electrons. Why ?
Answer:
An orbital can have only two electrons, provided these two electrons have anti parallel spins. This arrangement is represented, as ↓↑

Question 23.
Write the atomic number at an element with outer configuration,
i) 4s1, ii) 3d3 .
Answer:
(i) Electronic configuration for outer configuration 4s1 = 1s2. 2s2,2p6, 3s2, 3p6, 4s1.
Atomic number =19
(ii) Exact configuration for outer configuration 3d3.
1s2, 2s2, 2p6, 3s2,3p6, 4s2, 3d3 = 23 Atomic number = 23

Question 24.
Write the electronic configuration of
1. Cl 2. Na+ ion
Answer:
Electronic configuration
1) Cl = 1s2, 2s2, 2p6, 3s2, 3p6
2) Na+ = 1s2, 2s2, 2p6

1st PUC Chemistry Structure of Atom Three / Four Marks Questions and Answers

Question 1.
Summarize the Bohr’s Model of an atom,
Answer:
Bohr’s Model of an atom, the postulates are

  1. Electrons revolve around the nucleus of an atom in a certain definite path called Orbit or stationary state of shell.
  2. The shells are having different energy levels denoted as K, L, M, N
  3. As long as the electron remains in an orbit, they neither absorb nor emit energy.
  4. The electron can move only in that orbit in which angular momentum is quantized, i.e., the angular momentum of the electron is an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\)

KSEEB Solutions

Question 2.
Explain the experimental set up and different series of emission spectrum of hydrogen.
Answer:
Hydrogen Spectrum : Hydrogen spectrum is obtained by exciting electrons present in hydrogen gas using a discharge tube under low pressure and high voltage of current. The electrons present in various atoms of Hydrogen absorb energy and jumps to higher energy levels. Later they return back to the lower energy levels by emitting excess of energy in the form of photons. When these electrons are coming to the lower energy state a series of lines were formed called Hydrogen spectrum. Denoted as Hα,Hβ,Hγ and H????Hε

  • Lyman Series : This series is formed when the electron jumps from 2nd, 3rd, 4th, …. Higher energy level to first (K shell) level by emitting excess of energy in UV region.
  • Baimer Series ; This series is formed when the electron jumps from 3rd, 4th, 5th,… higher energy level to 2nd (L shell) level by emitting excess of energy in visible region.
  • Paschen Series : This series is formed when the electron jumps from 4th, 5th, 6th higher energy level to 3rd (M Shell) level by emitting excess of energy in Infra Red (IR Region).
  • Brackett Series : This series is formed when the electron jumps from 5th, 6 th…. higher energy level to 4th (N shell) level by emitting excess of energy in Infra Red (IR) region.
  • Pfund Series : This series is formed when the electron jumps from 6th, 7th …. higher energy level to 5th (0 shell) level by emitting excess of energy in Infra Red (IR) region.

Question 3.
What is Wave number, Frequency and Amplitude ? Give its SI Units.
Answer:
Wave Number ( \(\overrightarrow{\mathrm{v}}\) ) : Number of waves per unit length. Units : m-1
Frequency (v) ; Number of wave peaks that pass a given point in time unit.
Units: Hz
Amplitude (A): Height of the wave above the central line.

1st PUC Chemistry Structure of Atom Four / Five Marks Questions and Answers

Question 1.
Calculate the wave number, wavelength and frequency first line of hydrogen spectrum Or Calculate the maximum wave length of a line in the Lyman series.
Answer:
First line is Lyman Series, where n1 = 1, n2 = 2. Maximum wave length corresponds to minimum frequency i.e., n1 = 1, n2 = 2.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 16
R = Rydberg constant = 1.097 × 10+7 m. n1 = 1 n2 = 2
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 17
Wave length λ = 0.8227 × 107 = 8.227 × 106 m-1
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 18

KSEEB Solutions

Question 2.
Calculate the limiting frequency of Balmer series.
Answer:
Limiting frequency for Balmer series possible when n1 = 2 and n2 = ∞.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 19
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 20

Question 3.
Calculate the wavelength and wave numbers of the first and second lines in the Balmer series of hydrogen spectrum. Given R = 1.096 × 107m-1
Answer:
For Balmer series n1 = 2 , for third line n2 = 3, for fourth line n2 = 4
i. Wave number \(\overrightarrow{\mathbf{V}}\) of the third line is given by
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 21

Question 4.
Calculate the wavelength and wave number of Hγ and H???? line (R = 10.97 × 106 m-1)
Answer:
For Balmer series n1 = 2 , for third line n2 = 5, for fourth line n2 = 6
i. Wave number \(\overrightarrow{\mathbf{V}}\) of the third line is given by
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 22
= 364 × 10-9m = 364.9nm
Wavelength of H line = 364.9nm

KSEEB Solutions

Question 5.
Calculate the wave number and wavelength of the first spectral line of Lyman series of hydrogen spectrum.
Rydberg constant R = 10.97 × 106 m-1
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 23

Question 6.
Calculate the wave number of the spectral line when electron jumps from the seond Bohr orbit to the ground state. R = 1.097 × 107 m-1.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 24
Wavelength of the spectral line = 8.2258 × 106m-1

Question 7.
In a hydrogen atom, an electron jumps from third orbit to the first orbit. Find out the frequency and wavelength of the spectral line. Given R = 1.097 × 107m-1
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 25
Wavelength of the spectral line = 1025.4 A0
Frequency of the spectral line = 2.9257 × 1015 s-1

Question 8.
Calculate the wavelength of 2nd line and limiting line of Balmer series. If wave length of first line of Balmer series is 656 nm.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 26
For first line m = 2, m = 3 & λ = 656 nm
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 27
For secong line n1 = 2,n2 = 4
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 28
∴ Wavelength of the limiting line n1 = 2, n2 = ∞
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 29

KSEEB Solutions

Question 9.
Calculate the wavelength of a wave of frequency 1012 Hz, travelling with the speed of light 3 x 108 ms-1
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 30

Question 10.
Calculate the frequency of electromagnetic radiation having the wavelength 3 μ. Calculate the wave number corresponding to it. (1μ = 10-6 m)
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 31

Question 11.
Calculate the frequency and energy per quantum of a radiation with a wavelength of 200 nm. (c = 3 × 108 ms-1 and h = 6.625 × 10-34 Js)
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 32
= 1.5 × 1015 Hz = 1.55 × 10 15Hz.
E = hv = 6.625 × 10 -34 x 1.5 × 1015= 9.9375 × 10-19 J

Question 12.
Calculate the number of photon of light with a wavelength of 6000A that provide I joule of energy.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 33
= 0.00331 × 10-16J = 3.31 × 10-19 J
Number of photons 3.31 × 10-19 J make 1 photon
1J = \(\frac{1}{3.31 \times 10^{-19}}\) = 3.021 × 1020 photons

Question 13.
A major line in an atomic emission spectrum occurs at 450 nm. Find the energy decrease, as this photon is emitted.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 34

KSEEB Solutions

Question 14.
Calculate the wave number, wavelength and frequency of the first line in the Baimer series.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 35
For Balmer Series
n1 = 2 and n2 = 3 for the first time
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 36

Question 15.
The red light of neon signs has a wavelength of 693 nm. Find the energy difference (per mole of atoms) between the two energy levels involved.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 37
= 2.8684 × 1015 J
Energy difference per mole of atoms = 2.8684 × 1015 × 6.022 × 1023

Question 16.
Calculate the wavelength of an electron moving with a velocity of 2.5 × 10-7 ms-1 h = 6.626 × 10-34 Js ; mass of an electron = 9.11 × 10-31 kg.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 38

Question 17.
Find the mass of an electrically charged particle moving with a velocity of 3 × 106 ms-1 and having a de Broglie wavelength of 2Å.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 39

KSEEB Solutions

Question 18.
Calculate the energy of a photon whose wavelength is 3.864 × 10-7m.
Answer:
\(\mathrm{E}=\mathrm{h} \vartheta=\frac{\mathrm{h} \mathrm{c}}{\lambda}\)
h = 6.626 × 10-34
c = 3.0 × 108 ms-1
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 40
λ = 3.864 × 10-7

Question 19.
Calculate the de Broglie wavelength of (a) an electron of mass 9.11 × 10-31 kg and moving with a velocity of 1.0 × 106 ms-1 (b) a bullet of mass 25g moving with a velocity of 100ms-1
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 41

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