1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Students can Download Chemistry Chapter 2 Structure of Atom Questions and Answers, Notes Pdf, 1st PUC Chemistry Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams

Karnataka 1st PUC Chemistry Question Bank Chapter 2 Some Basic Concepts of Chemistry

Question 1.
Give the limitations of Dalton’s atomic theory.
Answer:
This theory fails

  • to account for the stability of atom after the discovery of subatomic particles.
  • to explain the formation of different kinds of molecules by the combination of different atoms.
  • to compare the behavour of one element from other in terms of both physical and chemical properties.
  • to understand the origin and nature of the characteristics of electromagnetic radiation absorbed or emitted by the atoms.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Sub Atomic Particles

Question 1.
Name the fundamental particles (subatomic particles) present in an atom.
Answer:
Electrons, Protons and Neutrons.

Question 2.
Name the person who first proposed the atomic theory of matter on scientific basis.
Answer:
John Dalton.

Question 3.
What is the basic rule regarding the behaviour of charged particles?
Answer:
“Like charges repel each other and unlike charges attract each other.”

Question 4.
Who was the first scientist began to study electrical discharge in cathode ray discharge tubes?
Answer:
Michael Faraday.

Question 5.
Who discovered electrons?
Answer:
J.J. Thomson

Question 6.
How were electrons (cathode rays) discovered?
Answer:
An electric discharge is passed through the gases in cathode ray discharge tube at very low pressure and at very high voltage as shown in figure.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 1
Current starts flowing through a stream of particles moving in the tube from the negative electrode (cathode) to the positive electrode (anode). These were called cathode rays or cathode ray particles or electrons.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 7.
Give any five characteristics of cathode rays.
Answer:

  1. Cathode rays start from cathode and move towards anode.
  2. They are not visible themselves but when they are passed through zinc sulphide coating screen, a bright spot is developed on the coating due to fluorescence.
  3. In presence of electric and magnetic field, cathode rays shows properties of negatively charged particles. Thus, cathode rays are negatively charged particles called electrons.
  4. In the absence of electric and magnetic field, these rays travel in straight lines.
  5. The characteristics of cathode rays do not depend on nature of gas and nature of electrode material.
    Note: Television picture tubes are cathode ray tubes and television pictures results due to fluorescence on the television screen coated with certain fluorescent or phosphorescent materials.

Question 8.
Name the fluorescent or phosphorescent material used to observe the behaviour of invisible rays in the discharge tube.
Answer:
Zinc sulphide

Question 9.
Who was the first scientist measured the specific charge (charge to mass ratio) of electron.
Answer:
J.J. Thomson.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 10.
How was specific charge (charge to mass ratio) of electron measured by J.J. Thomson?
Answer:
J.J. Thomson measured the ratio of electric charge (e) to the mass of electron by using cathode ray tube by applying electric and magnetic field perpendicular to each other as well as to the path of electrons as shown in the figure.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 2
(a) When only electric field is applied, the electrons deviate from their path and hit the cathode ray tube at point A.
(b) When only magnetic field is applied, electron- strikes the cathode ray tube at point C.
(c) By balancing both electric and magnetic field strength, it is possible to bring back the electron to the path as in the absence of electrical and magnetic field and they hit the screen at point B.
By carrying out accurate measurements on the amount of deflections observed by the electrons on the electric field strength or magnetic field strength, Thomson was able to determine the value of e / me as
e/me = 1 – 75882 × 1011C/kg.

Question 11.
On what factors the amount of deviation of the particles (electrons) from their path in the presence of electric and magnetic field depends? Explain.
Answer:
1. Magnitude of the negative charge on the particle: Greater the magnitude of the negative charge on . the particle, greater is the deflection in the presence of electric and magnetic field.

2. Mass of the particle: Lighter the particle, greater the deflection.

3. Strength of the electric and magnetic field: Higher the strength of the electric or magnetic field, higher the deflection of electrons from their original path.

Question 12.
Who discovered the charge on the electron?
Answer:
R.A. Millikan.

Question 13.
Name the experiment used to determine the charge on the electron.
Answer:
Millikan’s oil drop experiment.

Question 14.
What is the absolute charge and mass of electron?
Answer:
Charge on electron = – 1.6022 × 10-19 Coloumbs.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 3

Question 15.
What is the relative charge and unified mass of electron?
Answer:
Relative charge = – 1
Unified mass = 0.00054 u

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 16.
Who discovered protons?
Answer:
Goldstein

Question 17.
How protons were discovered?
Answer:
Since atoms, as a whole are neutral, the presence of negatively charged electrons suggested the presence of positively charged particles. These rays travel in a direction opposite to cathode rays. They were observed behind the cathode in a gas discharge tube when perforated cathode was used. They were named as canal rays or protons.

Question 18.
Give any four characteristics of canal rays or protons.
Answer:

  1. These are positively charged gaseous ions which depend on nature of gas present in the cathode ray tube.
  2. The charge to mass ratio of these rays depends on the gas from which these originate.
  3. Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.
  4. The behaviour of these particles in the magnetic and electrical field is opposite to that observed for electron.

Question 19.
What is the charge and mass of proton?
Answer:
Charge on proton = + 1.6022 × 10-19 C
Mass of proton = 1.67262 × 10-27 kg

Question 20.
What are the relative charge and unified mass (u) of proton?
Answer:
Relative charge of proton = + 1
Unified mass of proton = 1.00727 u

Question 21.
Who discovered neutron?
Answer:
James Chadwick.

Question 22.
How neutrons were discovered?
Answer:
A thin sheet of Beryllium was bombarded by α – particles. Electrically neutral particles were emitted and were named as neutrons.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 23.
What are absolute charge, relative charge, mass and unified mass of neutron?
Answer:
Absolute charge of neutron = 0
Relative charge of neutron = 0
Mass neutron = 1.67493 × 10-27 kg
Unified mass of neutron = 1.00867 u

Atomic Models

Question 1.
Explain Thomson model of atom.
Answer:
According to this model
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 4
1. An atom possess a spherical shape (radius charge is uniformly distributed.

2. Electrons are embedded into it to balance the electrostatic forces. It gives the most stable electrostatic arrangement.

3. This model can be visualised as a pudding or watermelon of positive charge with plums or seeds (electrons) embedded into it. So this model is also named as plum pudding or raisin pudding or watermelon model.

Question 2.
What is the important feature of Thomson model of an atom?
Answer:
“The mass of the atom is assumed to be uniformly distributed over the atom.”

Question 3.
Explain Rutherford’s a-ray scattering experiment.
Answer:
Rutherford bombarded a very thin piece of gold foil with a-particles as shown below.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 5
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 6
Observations: From this experiment Rutherford observed that,

  • Most of the α – particles passed through the gold foil undeflected.
  • A small fraction of the α – particles were deflected by small angles.
  • A very few α – particles (< 1 in 20,000) bounced back, that is, were deflected by nearly 180°.

Conclusions: On the basis of these observations, Rutherford drew the following conclusions regarding the structure of an atom:

  • As majority of the particles went through the foil undeflected, indicates most of the space in an atom is empty.
  • The scattering of a few particles through smaller angles and very few through longer angles from centre of the atom, indicates protons are present at the centre of atom.
  • The heavy positively charged central part of the atom is called nucleus.
  • Nearly all the mass of the atom is concentrated in the nucleus.
    Note: The size of the nucleus is very small as compared with the size of atom. The radius of the atom is about 10-10 m, while that of nucleus is 10-15 m.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 4.
Explain Rutherford’s nuclear model of atom on the basis of observations and conclusions made by α-ray scattering experiment.
Answer:
On the basis of observations and conclusions made by α-ray scattering experiment, Rutherford proposed the nuclear model of atom (after the discovery of protons).

  • The positive charge and most of the mass of the atom was concentrated in extremely small region called nucleus.
  • The nucleus is surrounded by electrons that move around the nucleus with a very high speed in circular paths called orbits.
  • Electrons and the nucleus are held together by electrostatic forces of attraction.
    Note: Thus, Rutherford’s model of atom resembles the solar system in which the nucleus plays the role of sun and the electrons that of revolving planets.

Question 5.
Give any two drawbacks (limitations) of Rutherford model of atom.
Answer:

  1. Rutherford’s model does not suggest specific radii for orbits of electrons.
  2. The model fails to explain emission and absorption spectra of elements.

Question 6.
Define atomic number (Z).
Answer:’
Atomic number (Z) is the number of protons present in the nucleus.
OR The number of electrons in a neutral atom.

Question 7.
What are nucleons?
Answer:
The number of protons and neutrons present in the nucleus of an atom are called nucleons.

Question 8.
What is mass number (A)?
Answer:
The sum of protons and neutrons present in the nucleus of an atom is called mass number.
Mass number = Number of protons + Number of neutrons = Total number of nucleons.

Question 9.
What are isobars? Give examples.
Answer:
Isobars are the atoms with same mass number (A) and different atomic number (Z).
Examples: 146C and 147N

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 10.
What are isotopes? Give examples.
Answer:
The atoms with same atomic number but different atomic mass (mass number) (A) are called isotopes. Examples: Isotopes of hydrogen
1H1 or 11H (Protium)
1H2 or 21D(Deuterium)
1H3 or 31T (Tritium)
Isotopes of carbon: 126C, 136C, 146C
Note: Chemical properties of isotopes are controlled by the number of electrons which are determined by the number of protons in the nucleus. Therefore all the isotopes of a given element show same chemical behaviour.

Question 11.
Calculate the number of protons, neutrons and electrons in 8035Br.
Answer:
In 8035Br, Z = 35, A = 80 and species is neutral
Number of protons = number of electrons = Z = 35
Number of neutrons = A – Z = 80 – 35 = 45

Question 12.
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species.
Answer:
The atomic number is equal to number of protons = 16. The element is sulphur (S).
Mass number = number of protons + number of neutrons = 16 + 16 = 32.
Species is not neutral as the number of protons is not equal to electrons. It is anion (negatively charged) with charge equal to excess electrons
= A – Z = 18 – 16 = 2. or Charge = p- e = 16 – 18 = -2. \(_{16}^{32} \mathrm{S}^{2-}\)
Symbol is \(_{16}^{32} \mathrm{S}^{2-}\) .

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 13.
How many neutrons and protons are there in the following nuclei? 6C13, 8O16, 12Mg24, 26Fe56, 38Sr88
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 7

Question 14.
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
Answer:
(i) 3517Cl
(ii) 23392U
(iii) 94Be .

Question 15.
Which of the following pairs are not isotopes?
(i) 126X, 136Y
(ii) 3517X, 3717Y
(iii) 146X, 147Y
Answer:
(iii) 146X, 147Y are not isotopes because atomic numbers are different.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Developments Leading to the Bohr’s Model of Atom

Question 1.
Give two developments played major role in the in the formation of Bohr’s model of atom.
Answer:

  1. Dual character of the electromagnetic radiations (wave property and particle property)
  2. Atomic spectra explained only by assuming quantized electronic energy levels in atoms.

Question 2.
What is electromagnetic radiation?
Answer:
It is a form of radiant energy which propagates through space in the form of waves which are associated with electric and magnetic fields, or Radiations propagated in space by the disturbance of electric and magnetic fields are called electro magnetic radiations.

Question 3.
What is electromagnetic spectrum?
Answer:
The arrangement of different types of electromagnetic radiations in the order of increasing wavelengths (or decreasing frequencies) is known as electromagnetic spectrum.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 8

Question 4.
Give any four characteristics of electromagnetic radiations.
Answer:
The main characteristics of electromagnetic radiations are,

  1. Electromagnetic radiation propagates through space in the form of waves.
  2. All the electromagnetic radiations travel with a speed equal to the speed of light.
  3. These radiations do not require any medium to travel.
  4. The energy of an electromagnetic radiation is directly proportional to its frequency or inversely proportional to its wavelength. E ∝ γ or E ∝ \(\frac{1}{\lambda}\)

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 5.
Give the approximate frequency region in radio, microwave, UV and visible region.
Answer:
Radio frequency ≈ 106 Hz – used for broadcasting.
Microwave ≈ 1010 Hz – used for radar.
Infrared region ≈ 1013 Hz – used for heating.
UV region ≈ 1016 Hz – component of sun’s radiation.
Visible region ≈ 1015 Hz – our eyes can detect.

Question 6.
Name any two types of properties which characterize electromagnetic radiations.
Answer:

  1. Wavelength (λ)
  2. Frequency (v)

Question 7.
Define frequency (v)
Answer:
It is defined as the number of waves that pass a given point in one second. It is expressed in Hz or s-1

Question 8.
Define wavelength (λ)
Answer:
The distance between two adjacent troughs or two adjacent crests is called wavelength.
The unit of wavelength is m.

Question 9.
What is the speed of light (all types of electromagnetic radiations in vacuum)?
Answer:
c = 3 × 108 ms-1.

Question 10.
Write the relation between velocity of light (c) wavelength (λ) and frequency (v).
Answer:
c = λ or γ = \(\frac{c}{\lambda}\)

Question 11.
Define wave number (v). Write its unit.
Answer:
It is defined as the number of wavelengths per unit length or Reciprocal of wavelength is called wave number i.e.,\(\bar{\gamma}=\frac{1}{\lambda}\) It expressed in m-1.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 12.
The Vividh bharati station of all India Radio, Delhi, broadcasts on a frequency of 1368 kHz. Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 9
This is a characteristic radio wavelength.

Question 13.
The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelength in frequencies (Hurtz) (1 nm = 10-9 m)
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 10
Frequency range of visible spectrum = 4 × 1014 to 7.5 × 1014Hz

Question 14.
Calculate (a) Wave number and (b) Frequency of yellow radiation having wavelength 5800 A.
Answer:
λ = 5800 Å = 5800 × 10-10m
Velocity of radiation, c = 3 × 108m/s
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 11

Question 15.
Calculate the Wave length, frequency and Wave number of a light Wave Whose period is 2.0 × 10-10s.
Answer:
Frequency (v) of wave is inverse of period
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 12

Question 16.
Explain particle nature of electromagnetic radiation by Planck’s quantum theory.
Answer:
According to Planck’s quantum theory,
1. the body which emits or absorbs radiations of all frequencies is called black body and the radiation emitted by this body is called black body radiation.

2. The exact frequency distribution of the emitted radiation (i.e., intensity versus frequency curve of the radiation) from a black body depends upon its temperature.

3. At a given temperature, intensity of radiation emitted increases with decrease of wavelength, reaches a maximum value at a given

4. wavelength and then starts decreasing with further decrease of wavelength, as shown in Fig.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 13
To explain these experimental observations, Planck proposed the following postulates:
1. The atoms and molecules could emit (or absorb) energy only in small discrete quantity called quanta or photon.

2. The energy of a quanta of radiation is proportional to its frequency and is expressed as E α v ; E = hv the proportionality constant, h, is known as Planck constant and has the value 6.626 × 10-34 Js.

3. Energy is always emitted in integral multiples of hv, i.e., E = nhv, where V is an integer.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 17.
What is photoelectric effect?
Answer:
The process of emission of electrons from the surface of metals when they are exposed to light of suitable frequency is called photoelectric effect.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 14
Note: Einstein got Nobel prize for photo electric effect.

Question 18.
Explain experimental results observed in photoelectric effect.
Answer:
1. The number of electrons ejected is directly proportional to the intensity or brightness of light,

2. There is no time lag between the striking of light beam and the ejection of electron from the metal surface.

3. The minimum frequency required to cause photoelectric effect is called threshold frequency (v0). Below threshold frequency (V0), photoelectric effect is not observed. The kinetic energy of emitted electrons is directly proportional to frequency of the light used.

The minimum amount of energy required to eject an electron from the metal surface is called work function (w0) of that metal.
Since the striking photon has energy equal to hv and minimum energy required to eject electron is hv0, then hv – hv0 is transferred as the kinetic energy of photoelectron \(\left(\frac{1}{2} m_{e} v^{2}\right)\)
∴ hv – hv0 = \(\frac { 1 }{ 2 }\)mev2
hv = hv0 + \(\frac { 1 }{ 2 }\)me v2 or K.E = hv – hv0
This equation is called Einstein’s photoelectric equation.
where me is the mass of electron and V is the velocity associated with the ejected electron, h is the Planck’s constant.

Question 19.
What is the value of Planck’s constant (h)?
Answer:
h = 6.626 × 10-34Js

Question 20.
Calculate the energy of one mole of photons of radiation whose frequency is 5 × 1014Hz.
Answer:
Energy of one photon is given by E = hv
= 6 . 626 × 10-34 × 5 × 1014 = 3.313 × 10-19
Energy of one mole of photon is given by
= 3.313 × 10-19 × 6.022 × 1023 = 19.951 × 104 = 199.51 × 103 Jmol-1 = 199.51 kJ mol-1

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 21.
A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.
Answer:
Power of the bulb = 100 watt = 100 Js-1
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 15

Question 21.
When an electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68 × 105Jmol-1 What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?
Answer:
λ = 300 nm, λ = 300 × 10-9m,
KE = 1.68 × 105 J/mol
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 16
The energy of one mole of photons = 6 . 626 × 10-19 × 6 . 022 × 1023 = 3 . 99 × 105
The minimum energy needed to remove an electron from sodium
Eo = 3.99 × 105 – 1.68 × 105 = 2.31 × 105 J/mol
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 17
This corresponds to green light.

Question 22.
The threshold frequency v0 for a metal is 7.0 × 1014 s-1. Calculate the Kinetic energy of an electron emitted when radiation of frequency v = 1.0 × 1015s-1 hits the metal.
Answer:
According to Einstein’s equation
= hv – hv0 = h(v – v0) = 6.626 × 10-34(1 × 1015-7 × 1014)
= 6.626 × 10-34(10 × 1014 – 7 × 1014)
= 6 . 626 × 10-34[(10 – 7) × 1014]
= 6.626 × 10-34 × 3 × 1014
= 1 . 988 × 10-19 J.

Question 23.
A photon of wavelength 4 × 10-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate
(i) energy of the photon (eV)
(ii) kinetic energy of the photon and
(iii) velocity of the photoelectron (leV = 1 . 602 × 10-19 J).
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 18

(ii) Energy of photon = K.E. + Work function.
Kinetic energy = Energy of photon – Work function
= 4.97 × 10-19 J – 2 . 13 × 1 . 602 × 10-19 J = 1.56 × 10-19J = \(\frac{1 \cdot 56 \times 10^{-19} \mathrm{J}}{1 \cdot 602 \times 10^{-19}}=0.97 \mathrm{eV}\)

(iii) \(\frac { 1 }{ 2 }\)mv2 = 1.56 × 10-19J
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 19
v2 = 0.34 × 1012 = 34 × 1010
v = 5.85 × 105ms-1

Question 24.
When a photon of frequency 1.0 × 1015 s-1 was allowed to hit a metal surface, an electron having 1.988 × 10-19 J of kinetic energy was emitted. Calculate the threshold frequency of this metal.
Answer:
K.E. = hv – hv0
hv0 = hv – K.E.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 20

Question 25.
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kj mol-1.
Answer:
Energy of one photon of radiation having wavelength 242 nm is just sufficient to ionise the sodium atom.
Ionisation energy of sodium = Energy of one photon of radiation of wave length
Energy of photon = hv = \(\frac{\mathrm{hc}}{\lambda}\)
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 21
= 8.22 × 10-19J/photon = 8.22 × 10-22kJ/photon
∴ Ionisation energy per mole = 8 .22 × 10-22 × 6 . 022 × 1023kJ / mol = 495 KJ/mol.

Question 26.
A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 μm. Calculate the rate of emission of quanta per second.
Answer:
Energy emitted by the bulb = 25watt = 25Js-1
Energy of one photon, E = hv = \(\mathrm{h} \frac{\mathrm{c}}{\lambda}\)
Here, λ = 0.57 μm = 0.57 × 10-6m
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 22

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 27.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (v0) and work function (W0) of the metal.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 23

Question 28.
What is spectrum?
Answer:
A series of radiations arranged in the increasing order of their wavelength is called a spectrum.

Question 29.
What is continuous spectrum?
Answer:
The spectrum of white light ranging from violet at 7.5 × 1014 Hz to red at 4 × 1014. Hz is called continuous spectrum.

Question 30.
What is emission spectrum?
Answer:
The spectrum of radiation emitted by a substance that has absorbed energy is called emission spectrum.

Question 31.
What is absorption spectra?
Answer:
A record of radiations absorbed by elements from a composite beam of light as dark lines on a photographic film is called absorption spectrum.

Question 32.
What is spectroscopy?
Answer:
The study of emission or absorption spectra is called spectroscopy.

Question 33.
What is atomic spectra or line spectra?
Answer:
The emission spectra of atoms in a gas phase emit only at specific wavelengths as bright lines with dark spaces between them is called line spectra or atomic spectra.

Question 34.
Give any two importance of line emission spectra.
Answer:

  1. Used to study electronic structure. Each element has unique line emission spectrum.
  2. Line spectra can be used in chemical analysis to identify unknown atoms.

Question 35.
Who was the first scientist used line spectra to identify elements?
Answer:
Robert Bunsen.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 36.
Name any two elements discovered by using spectroscopic method.
Answer:
Rubidium (Rb), Caesium (Cs), Thallium (Tl)

Question 37.
Name the noble gas discovered in the sun by spectroscopic method.
Answer:
Helium (He)

Question 38.
Explain line spectrum of hydrogen.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 24
Hydrogen gas is subjected to electrical discharge in a discharge tube. Rose coloured light is emitted by hydrogen. This light is passed through two slits to get a thin strip of light. This strip of light is then passed through a prism. The different radiations bend at different angles and therefore get resolved and get arranged in the increasing order of their wavelengths when these radiations are made to fall on a photographic film. Normally we get a series of straight lines on the film which is called hydrogen spectrum.

Question 39.
Write Rydberg’s equation.
Answer:
\(\bar{v}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right] \mathrm{m}^{-1}\)
where n1 = 1, 2, 3, ………
n2, = n1 + 1, n1 + 2, n1 + 3
RH is the Rydberg constant.

Question 40.
What is the value of Rydberg constant?
Answer:
RH =109677 cm-1 = 1.097 × 107m-1

Question 41.
Write the Rydberg’s equation for Balmer series of hydrogen spectrum. In which region Balmer series of lines were noticed.
Answer:
\(\bar{v}=109677\left[\frac{1}{2^{2}}-\frac{1}{n_{2}^{2}}\right] \mathrm{cm}^{-1}\)
n2 = 3,4,5 ……..
In visible region, Balmer series can be seen.

Question 42.
Give the postulates of Bohr’s theory of atomic model.
Answer:
Main postulates of Bohr’s atomic theory are:
1. An atom consists of positively charged nucleus at the centre in which the most of the mass of atom is concentrated.

2. Electrons revolve around the nucleus in certain selected circular paths called orbits or energy levels or stationary states.

3. The angular momentum of an electron is equal to integral multiple of \(\frac{\mathrm{h}}{2 \pi}\), where h is Planck’s constant.
i.e., mvr = \(\frac{\mathrm{nh}}{2 \pi}\), where n= 1,2, 3, 4,
(The orbitals are also designated as K, L, M, N etc.,),
m – mass of electron, v – velocity of electron, r – radius of orbit.

4. As long as electron revolve in a particular orbit, it does not emit energy or absorb energy.

5. If an electron jumps from higher energy level to a lower energy level, energy is emitted and when electron jumps from lower energy level to higher energy level, energy is absorbed.

6. Energy released or absorbed by an electron is equal to the difference in energies of the two orbits and is equal to energy of a quantum of radiation.
ΔE = E2 – E1 = hv
where h is the Planck’s constant (6.6256 × 10-34 Js), v is the frequency of radiation, and ΔE indicates the . difference in energy.
Note: With the help of these postulates, Bohr derived the relations to calculate the energy and radii of stationary states.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 43.
Write an expression to calculate the radii of the stationary states for electron of hydrogen atom.
Answer:
rn = n2a0
where ao = radius of first stationary state = 52.9 pm = 0.53Å
n = 1, 2, 3 principal quantum number.

Question 44.
What is Bohr orbit?
Answer:
The first stationary state of electron is called Bohr orbit.

Question 45.
Write an expression to calculate energy of electron in the stationary state of hydrogen atom.
Answer:
En = -RH\(\left(\frac{1}{n^{2}}\right)\) Where n = 1, 2, 3, ………

Question 46.
Calculate the energy of electron in the ground state.
Answer:
En = -RH\(\left(\frac{1}{n^{2}}\right)\) ; En = -2.18 × 10-18\(\left(\frac{1}{1^{2}}\right)\) = 2.18 × 10-18 j
Note: The value of Rydberg constant in Joules = 2.18 × 10-18 j.

Question 47.
Write an expression to calculate energy of electron in the stationary state of hydrogen like atom.
Answer:
En = -2.18 × 10-18 \(\left(\frac{Z^{2}}{n^{2}}\right) J\)

Question 48.
Write an expression to calculate radii of the stationary states of electron of hydrogen like atoms.
Answer:
rn = \(\frac{52 \cdot 9 n^{2}}{2}\) where Z is the atomic number.

Question 49.
Explain hydrogen spectrum based on Bohr’s atomic model.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 25
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 26
When hydrogen gas is subjected to electric discharge, hydrogen molecules break down to give hydrogen atoms. Electrons present in different atoms of hydrogen absorb different amount of energy and get excited to different higher energy levels. These electrons will not remain in the higher energy levels instead they fall back to different lower levels emitting radiations of different energies.

When electrons fall back from higher energy levels to first energy level, U.V rays are produced which cause Lyman series of hydrogen spectrum. When electrons fall back from higher levels to the second energy level, visible radiations are emitted which cause B aimer series of lines.

Similarly, when electrons fall back from higher levels to 3rd 4th, and 5th levels, infra-red (IR) radiations are
produced which cause Paschen, Brackett and Pfund series of lines respectively.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 27

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 50.
Give any three limitations of Bohr’s theory.
Answer:

  1. It does not explain the spectra of atoms having more than one electron.
  2. It does not explain the splitting spectral lines in magnetic field (Zeeman Effect or fine structure) and electric field (Stark effect or hyperfine structure).
  3. Quantisation of angular momentum and idea of stationary states have no theoretical basis.
  4. The postulate of Bohr’s theory that “electrons revolve in well defined orbits around the nucleus with well defined velocities” is not acceptable based on Heisenberg’s uncertainty principle.
  5. de Broglie suggested that electrons have dual character. It has particle and wave character. Bohr treated the electron only as particle.
  6. It could not consider relativistic variation of mass and energy of an electron in motion.

Question 51.
What is the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom?
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 28

Question 52.
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?
Answer:
According to Rydberg’s formula, v̄(cm-1) = RH \(\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\)
In the present case n2 = 4 and n1 = 2
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 29
λ = \(\frac{1}{\bar{v}}=\frac{1}{20564}\) cm = 4.86 × 10-5cm = 486 × 10-5 × 10-2m = 486 nm

Question 53.
The energy associated with the first orbit in the hydrogen atom is -2.× 10-18 J atom-1.
(i) What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 30

(ii) For H-atom. rn = 0 . 529 × n2 A
r5 = 0.529 × 52 = 13.225Å = 1.3225 nm.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 54.
Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
Answer:
For Balmer series longest wavelength transition is from n2 = 3 to n1 = 2
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 31
or = 1.5233 × 106m-1

Towards Quantum Mechanical Model of the Atom

Question 1.
Name the scientist who proposed the dual behaviour of matter.
Answer:
de Broglie.

Question 2.
What are the dual properties of matter?
Answer:
Particle property and wave property.

Question 3.
Write de Broglie’s wave equation.
Answer:
λ = \(\frac { h }{ mv }\) or λ = \(\frac { h }{ p }\)
where λ = wavelength, h = Planck’s constant.
m = mass of the particle, v = velocity of the particle and p = momentum of particle.

Question 4.
What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 ms-1?
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 32

Question 5.
Calculate the mass of a photon with wavelength 3.6 Å.
Answer:
m = ?
λ = 36Å = 3. 6 × 10-10m, v = 3 × 108ms-1
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 33

Question 6.
If the velocity of the electron in Bohr’s first orbit ¡s 2.19 × 106 ms-1, calculate de-Broglie wavelength associated with it.
Answer:
v = 2.19 × 106; m = 9.1 × 10-31 Kg
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 34

Question 7.
The mass of an electron is 91 × 10-31. If its K.E. is 3.0 × 10-25 J , calculate its wavelength.
Answer:
Mass of the electron m = 9.1 × 10-31 kg
KE = \(\frac { 1 }{ 2 }\) mv2 = 3 × 10-25 J
\(\frac { 1 }{ 2 }\) × 9.1 × 10-31kg × v2 = 3 × 10-25J
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 35
h = 6.63 × 10-34 Js = 6.63 × 10-34 Kg ms-1
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 36

Question 8.
State Heisenberg’s uncertainty principle. Give Its mathematical expression.
Answer:
It states that “it is impossible to determine simultaneously, both position and momentum of a moving of a moving particle like electron exactly or accurately.”
If we measure the position exactly there is an error in the momentum.
If we measure the momentum accurately, there is an error in the position.
Let Δx be the error in the position, Δpx be the error in the momentum, then
Δx . Δpx ≥ \(\frac{\mathrm{h}}{4 \pi}\)
or Δx .mΔv ≥ \(\frac{h}{4 \pi m}\)
where Δx = uncertainty in position.
Δpx = uncertainty in momentum of particle

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 9.
Give any two significances of uncertainty principle.
Answer:

  1. It rules out the existence of definite paths of electrons and other similar particles.
  2. This principle is significant only for microscopic objects and is negligible for that of macroscopic objects.

Question 10.
What are the main reasons for the failure of the Bohr model?
Answer:

  1. Bohr’s model ignores dual behaviour of matter.
  2. Bohr’s theory goes against to the Heisenberg uncertainty principle.

Question 11.
A microscope using suitable photons is employed to locate an electron in an atom with in distance of 0.1Å. What is the uncertainty involved in the measurement of its velocity.
Answer:
Δx = 0.1Å = 0.1 × 10-10m
Δv = ? mass of electron = 91 × 10-31kg
We know, & Δx × mΔv = \(\frac{\mathrm{h}}{4 \pi}\)
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 37

Question 12.
A golf ball has a mass of 40g, and a speed of 45 ms-1 If the speed can be measured with in accuracy of 2%. Calculate the uncertainty in the position.
Answer:
m = 40g = 0.040kg = 4 × 10-2kg
v’ = 45m/s
v’ = 2%
v’ = 2% of v = \(\frac { 2 }{ 100 }\) × 45 = 0.9m/s
= \(\frac{6 \cdot 626 \times 10^{-34}}{4 \times 3 \cdot 14 \times 4 \times 10^{-2} \times 0.9}\)
= 0.146 × 10-32
= 1.46 × 10-33m

Question 13.
If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is \(\frac{h}{4 \pi}\) × 0.05nm is there any problem in defining this value.
Answer:
Δx = 0.002nm = 2 × 10-33nm = 2 × 10-3 × 10-9 m = 2 × 10-12 m
Δx × Δp = \(\frac{h}{4 \pi}\)
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 38
If cannot be defined as the actual magnitude of the momentum is smaller than the uncertainty in momentum

Question 14.
Calculate the energy associated with the first orbit of He+.What is the radius of this orbit?
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 39

Question 15.
Electron does not exist inside the nucleus. Why?
Answer:
Δx = (Radius of nucleus) = 1 × 10-15 m
Δx . mΔv = \(\frac{h}{4 \pi}\)
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 40
This uncertainty in velocity is even higher than the velocity of light (3x l&ms) and hence is not possible.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Quantum Mechanical Model of Atom

Question 1.
What is quantum mechanics?
Answer:
Quantum mechanics is a theoretical science which deals with the study of the motion of the macroscopic objects which have both observable wave like and particle like properties.

Question 2.
Who developed the quantum mechanics?
Answer:
Heisenberg and Schrodinger.

Question 3.
Explain quantum mechanical model of an atom.
Answer:
1. The quantum mechanical model is based on the idea of wave function. The wave function is a mathematical function whose value depends upon the coordinates of the electron in the atom and does not carry any physical meaning. ,

2. Energy and motion of an electron wave in an atom is governed by the Schrodinger’s wave equation, ĤΨ = EΨ. where Ĥ is Hamiltonians operator. Ψ is called wave function, Ψ is the amplitude of the electron wave and E is the total energy of the electrons.

3. Although Ψ is a mathematical function and unobservable, the probability of finding the electron at a point within the atom is proportional to lΨ|2 at that point.

4. Each wave function has three variables, called quantum numbers denoted as n, l and m.

5. The wave function Ψ with these 3 variable specifies the energy, shape of the orbital and orientation of the orbital of an electron.

Question 4.
Explain the meaning of Ψ and Ψ2.
Answer:
The orbitals are described by their wave function T. The orbital wave function (()) for an electron in an atom has no physical meaning. It is simply a mathematical function of the coordinates of the electron. However, for different orbitals the plots of corresponding wave functions as a function of r (the distance from the nucleus) are different. The square of the wave function (i.e. ()2) at a point gives the maximum probability density of the electron at that point according to the German physicist, Max Bom.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 5.
Explain the significance of four quantum numbers.
Answer:
1. Principal quantum number (n): It defines the shell, It determines size of the orbital. As the n value (n = 1, 2, 3 ) increases, size and energy of orbital also increases. [As the n value increases, the number of allowed orbitals also increases. There are n2 orbitals are allowed for the given value of n],

2. Azimuthal quantum number (l) or orbital angular momentum or subsidiary quantum number: It defines the shape of the orbital. For the given value of n, the possible value of ‘l’ are l = 0, 1, 2, 3, 4 ……….
Each value of l indicates the following sub shell
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 41

3. Magnetic orbital quantum number (ml): It gives the spatial orientation of the orbital and number of orbitals. For any subshell (for the given l value), there are (2l +1) values of m; are possible.
If l = 0, m = (2l +1) = 1, value, m = 0 (s-orbitals)
l = 1, m = 3 values, -1,0,+ 1 (p-orbitals)
l = 2, m = 5 values, -2, -1, 0, +1, +2 (d-orbitals)
l = 3, m = 7 values, -3, -2, -1, 0, + 1, + 2, + 3 (f-orbitals)
Each value of ml indicates an orbital in a subshell.

4. Electron spin quantum number (ms): refers to the spin of the electron. The values of spin quantum number are +\(\frac { 1 }{ 2 }\) and –\(\frac { 1 }{ 2 }\). (clockwise, s = +\(\frac { 1 }{ 2 }\), anticlockwise, s = –\(\frac { 1 }{ 2 }\))

Question 6.
What is the total number of orbitals associated with the principal quantum number n = 3?
Answer:
Total number of orbitals = n2 = 32 = 9.

Question 7.
If n = 3, what are the possible values of l and ml.
Answer:
If n = 3, l = 0, 1, 2
If l = 0, m = 0
l = 1 m = -1,0,+ 1
l = 2 m = -2, -1, 0, + 1, + 2

Question 8.
Using s, p d and f notations describe the orbital with the following quantum numbers
(a) n = 2, l = 1
(b) n = 4, l = 0
(c) n = 5, l = 3 and n = 3, l = 2.
Answer:
(a) when n = 2, l = 1 the orbital is 2p
(b) when n = 4, l = 0 the orbital is 4s
(c) when n = 5, l = 3 the orbital is 5f
(d) when n = 3, l = 2 the orbital is 3d

Question 9.
Define orbital.
Answer:
It is a region in the space around the nucleus where the probability of finding the electron is maximum.

Question 10.
What is the shape of the s-orbital and draw the shape of s-orbital?
Answer:
Spherical shape
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 42

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 10.
Explain probability density of electron in 1 s and 2 s orbital by using the plots of ()(r) vs r and ()2(r) vs r.
Answer:
In Is orbital, the probability is maximum at the nucleus and it decreases sharply as we move away from it.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 43
In 2s orbital, the probability density first decreases sharply to zero and again starts increasing. After reaching a small maxima, it decreases again and approaches zero as the value of r increases further.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 44

Question 11.
What is node or nodal surface? How many nodes are available in each energy level of s-orbital?
Answer:
The region where the probability of finding the electron is zero is called node or nodal surface. It has been found that ns orbitals have (n – 1) nodes. That is number of nodes in 2s orbital is 1, 3s orbital is 2, 4s orbital is 3 and so on.

Question 12.
What are boundary surface diagrams? Give their importance.
Answer:
The shape of orbitals drawn by boundary surface or contour surface in space on which the value of probability density is constant is called boundary surface diagram. They give the good representation of the shapes of the orbitals.

Question 13.
Draw the boundary surface diagram for Is and 2s orbitals or shape of s-orbital.
Answer:
s-orbitals have spherical shape. They are non-directional
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 45

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 14.
Draw the boundary surface diagrams of the three 2p orbitais or shapes of p-orbitais.
Answer:
p-orbitais have dumb-bell shape. They are directional.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 46

Question 15.
Draw the boundary surface diagrams for five 3d orbitals – d-orbitals have double dumb-bell shape. They are also directional.
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 47

Question 16.
How many radial nodes and angular nodes are possible for the given value of n and 11
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 48

Question 17.
How many radial nodes and angular nodes are present in 4d orbital?
Answer:
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 49

Question 18.
Give the difference between orbit and orbitals.
Answer:

 SI.

No.

 Orbit  Orbital
 1.  It is a circular path around the nucleus in   which an electron revolves.  It is a region of space around the   nucleus,  where the electron is most   likely  to be found.
 2.  An orbit of an electron represents its   energy level which includes all sub energy  levels.  An orbital or a set of orbitals of similar   energy comprise a subenergy level of a   main energy level.
 3.  Orbits are circular in shape.  Orbitals have different shapes, e.g. s-   orbitals are spherical whereas, p-orbitals    are dumb-bell shaped.
 4.  Orbits are non-directional in character   and hence they cannot explain shapes of   molecules.  Orbitals (except s-orbitals) have   directional character and hence they can   account for shapes of molecules.
 5.  Concept of well-defined orbit is against   Heisenberg’s uncertainty principle.  Concept of orbitals is in accordance with   Heisenberg’s uncertainty principle.

Question 19.
Explain the stability of atoms in multielectron systems.
Answer:
In multielectron systems, the stability of atom is due to total attractive interactions are more than the repulsive interactions.
Explanation:
1. Due to the presence of electrons in the inner shells, the electron in the outer shell will not experience the full positive charge of the nucleus.

2. The effect will be lowered due to the partial screening of positive charge on the nucleus by the inner shell electrons. This is known as the shielding of the outer shell electrons from the nucleus by the inner shell electrons. The net positive charge experienced by the outer electrons is known as effective nuclear charge.

3. Due to shielding effect, the attractive force experienced increases with increase in nuclear charge. Therefore, the energy of interaction between the nucleus and electron will increase in atomic number so that the atom will get more stability.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 20.
What is electronic configuration of elements?
Answer:
Electronic configuration is the distribution of electrons in various orbitals in the increasing order of energy.

Question 21.
State Aufbau (building up) principle.
Answer:
Aufbau principle states that “the electrons enter into different atomic orbitals in the increasing order of their energies”.
The order of energies of different orbitals is determined by (n + l) rule.

Question 22.
Explain (n + l) rule.
Answer:
According to this rule,

  • Among the orbitals, the one which has a higher value of (n +l) has more energy.
  • If (n + l) values are same, the one having a higher value of n has more energy.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 50
Examples,
1. Consider 4p and 3s orbitals.
For 4p orbital, n = 4 and l = 1
Therefore n + l = 5
For 3s orbital, n = 3 and l = 0
Therefore n + l = 3 ,
Hence, 4p orbital has more energy than 3 s

2. Consider 3p and 4s.
For 3p, n + l = 4, and for 4s, n + l = 4
According to n + l rule 4s has higher energy than 3p
Based on (n + l) rule the increasing order of energies of different orbitals is represented schematically by the following diagram.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 23.
State Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle states that: in an atom no two electrons can have the same set of four quantum numbers.

Pauli’s exclusion principle indicates that two electrons may have three same quantum numbers but the fourth quantum number must be different. In other words, Pauli’s exclusion principle can also be stated as: “The maximum number of electrons that an orbital can accommodate is two”.

Question 24.
State Hund’s rule of maximum multiplicity.
Answer:
This rule deals with the filling of electrons in the equal energy (degenerate) orbitals of the same sub shell (p, d and/). According to this rule, “electron pairing in p, d and f orbitals cannot occur until all orbitals of a given subshell contains one electron each or is singly occupied”. This is due to the fact that electrons being identical in charge, repel each other when present in the same orbital. This repulsion can, however, be minimized if two electrons move as far apart as possible by occupying different degenerate orbitals. All the electrons in a degenerate set of orbitals will have same spin.

According to Hund’s rule of maximum multiplicity for nitrogen (Z = 7), correct and stable electronic configuration
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 51

Question 25.
Wrute the electron configuration of elements from atomic number 1 to 36.
Answer:
Ground state configuration of Atoms (z = 1 to 36)
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 52

Question 26.
Explain stability of half filled and completely filled orbitals.
Answer:
Half-filled subshell and completely filled subshells have high exchange energy and gain some extra stability. Because of this, the following elements have special electronic configurations.
1st PUC Chemistry Question Bank Chapter 2 Structure of Atom - 53
In case of chromium, one electron from 45 jumps to 3d. The energy absorbed during this transition is less than the energy released due to increase in the stability caused by formation of half filled d and s subshells.

In case of copper, one electron jumps from 4s to 3d to give the extra stability due to fully filled subshells 3d and .half filled 45.

1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

Question 27.
Give the causes of stability of completely filled and half filled subshells.
Answer:
There are two causes for the extra stability of half filled and completely filled subshells.

  • The completely filled or half filled subshells have symmetrical distribution of electrons in them and are therefore more stable.
  • The more stability is also due to high exchange energy.

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