NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom

Here we have given NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom.

BoardCBSE
TextbookNCERT
ClassClass 9
SubjectScience
ChapterChapter 4
Chapter NameStructure of the Atom
Number of Questions Solved34
CategoryNCERT Solutions

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom

INTEXT Questions

Question 1.
What are canal rays?
Solution:
The beam of rays which travel in a direction away from anode towards cathode when a gas,taken in a discharge tube is subjected to the action of high voltage under low pressure are known as canal rays. It is also called anode rays. It was discovered by E. Goldstein in 1886.

Question 2.
If an atom contains one electron and one proton, will it carry any charge or not?
Solution:
No, the atom will not carry any charge because the electron has one negative charge (-1) and the proton has one positive charge (+1). They netutralise each other.

Question 3.
On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Solution:
According to Thomson’s model of an atom:

  • An atom consists of a positively charged sphere and the electrons are embedded like the seeds in a water-melon.
  • The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically neutral.

Question 4.
On the basis of Rutherford’s model of an atom, which sub-atomic particle is present in the nucleus of an atom?
Solution:
Proton, a positively charged sub-atomic particle is present in the nucleus of an atom according to Rutherford’s model of atom.

Question 5.
Draw a sketch of Bohr’s model of an atom with three shells.
Solution:

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom 1

Question 6.
What do you think would be the observation if the a-particle scattering experiment is carried out using a foil of a metal other than gold?
Solution:
On using the foil of heavy metals like gold (e.g., platinum, silver etc.) the observation will be same but if the foil is of light metal (e.g., sodium, magnesium etc.), the massive α-particles may push the nucleus aside and may not be deflected back.

Question 7.
Name the three sub-atomic particles of an atom.
Solution:
The three sub-atomic particles of an atom are :

  • electron
  • proton
  • neutron.

Question 8.
Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?
Solution:
Mass number of helium = 4
Number of protons = 2
Number of neutrons (n)
= Mass number (A) – No. of protons (p) =4-2=2
Thus, no. of neutrons = 2

Question 9.
Write the distribution of electrons in carbon and sodium atoms.
Solution:
Atomic number of carbon = 6 Hence first shell (K-shell) have 2 electrons and the remaining 4 electrons will be present in the second shell, i.e. L-shell. Thus the distribution will be
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom 2
Atomic number of sodium = 11. Hence, first shell (K-shell) will have 2 electrons and second shell (L-shell) will have 8 electrons and third shell (M-shell) will have 1 electron. Thus, the distribution will be
tiwari academy class 9 science Chapter 4 Structure of the Atom 3
Question 10.
lf K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Solution:
The maximum number of electrons . present in a shell = 2n2 where n – shell number Value of n for K shell = 1
∴ Maximum number of electrons in K shell = 2n2 = 2(1)2 = 2 Value of n for L shell = 2
∴ Maximum number of electrons in L shell = 2(2)2 = 8 Thus, total no. of electrons = 2 + 8 = 10

Question 11.
How will you find the valency of chlorine, sulphur and magnesium?
Solution:
Valency of an atom is the number of electrons gained, lost or shared so as to complete the octet of electrons in the valence shell.
Valency of chlorine: It has electronic configuration = 2, 8, 7 Thus, one electron is gained to complete its octet and so its valency is 1. valency of sulphur: it it has electronic configuration = 2, 8, 6. Thus two electrons are gained to complete its octet and hence its valency = 2 Valency of magnesium : it has electronic configuration = 2, 8, 2. Thus, it can lose two electons to attain octet and hence its valency = 2.

Question 12.
If number of electrons in an atom is 8 and number of protons is also 8, then

  1. what is the atomic number of the atom and
  2. what is the charge on the atom?

Solution:
We know that,
No. of electrons (e) = No. of protons (p) = Atomic number (Z)

  1. No. of electrons = 8 ∴ Atomic number = 8
  2. The number of protons is equal to the number of electrons. So the atom will be neutral.

Question 13.
With the help of Table 4.1, find out the mass number of oxygen and sulphur atom.
Solution:
Mass number of oxygen atom
= No. of protons + No. of neutrons = 8 + 8 = 16
Mass number of sulphur atom
= No. of protons + No. of neutrons = 16 + 16 = 32

Question 14.
Write the electronic configuration of any one pair of isotopes and isobars.
Solution:
Isotopes of chlorine: _{ 17 }^{ 35 }{ CI },_{ 17 }^{ 37 }{ CI } Electronic configuration of each of them: 2,8, 7 Isobars : _{ 20 }^{ 40 }{ Ca },_{ 18 }^{ 40 }{ CI }
Electronic configuration of 20Ca → 2, 8, 8, 2 Electronic configuration of 18Ar → 2, 8, 8

Question 15.
For the symbol H, D and T tabulate three sub-atomic particles found in each of them.
Solution:

SymbolIsotopesAtomic no. (Z)Mass no. (A)No. of protons (p)No. of electrons (e)No. of neutrons (n)
HHydrogen11110
DDeuterium12111
TTritium13112

NCERT Exercises

Question 1.
Compare the properties of electrons, protons and neutrons.
Solution:

PropertyElectronProtonNeutron
1Symbole or _{ 1 }^{ 0 }{ e }Por _{ 1 }^{ 1 }{ p }n or _{ 1 }^{ 0 }{ n }
2Charge– 1 unit or – 1.6 × 10-19 C+ 1 unit or + 1.6 × 10-19 Czero
3Mass9.1 × 10-31 kg1.67 × 10-27 kg1.675 × 10-27 kg
4Locationpresent around the nucleuspresent in nucleuspresent in nucleus
5DiscoveryJ.J. ThomsonE. GoldsteinChadwick

Question 2.
What are the limitations of J.J. Thomson’s model of the atom?
Solution:
The limitation of Thomson’s model are described below :

  1. Thomson’s model of an atom considers an atom to lae a sphere of uniform positive charge. Later researches particularly, Rutherford’s a-particle scattering experiment showed that an atom has a positively charged ‘core’ at its centre.
  2. According to Thomson’s atomic model, mass of an atom is considered to be uniformly distributed. Rutherford’s experiment showed that the entire mass of an atom is concentrated inside the core of the atom.

Question 3.
What are the limitations of Rutherford’s model of the atom?
Solution:
The Rutherford model suffers from the following drawbacks :

  1. An electron revolving around the nucleus gets accelerated towards the nucleus. An accelerating charged particle must emit radiation, and lose energy. Thus, the electrons in an atom must continuously emit radiation and lose energy. Because of this loss of energy, the electron would slow down, and will not be able to withstand the attraction of the nucleus. As a result, the electron should follow a spiral path, and ultimately fall into the nucleus (see figure).
    tiwari academy class 9 science Chapter 4 Structure of the Atom 4
    If it happens, then the atom should collapse in about 10-8 second. But, this does not happen — atoms are stable. This indicates that there is something wrong in Rutherford’s nuclear model of atom.
  2. Rutherford’s model of atom does not say anything about the arrangement of electrons in an atom.

Question 4.
Describe Bohr’s model of the atom.
Solution:
The Danish physicist Neils Bohr proposed the following postulates for revising the Rutherford’s model.

  • Atom has central nucleus surrounded by electrons.
  • An atom consists of small heavy positively charged nucleus in the centre and the electrons revolve around it, in circular paths called orbits or shells.
  • Each orbit has fixed energy, so these orbits are called energy levels or energy shells.
  • The order of the energy of these energy shells will be
    K < L < M < N < O <…………. or, 1 < 2 < 3 < 4 < 5 <…………..
  • As long as an electron remains in a particular orbit, it does not lose or gain energy.
  • Energy is neither absorbed nor emitted when electron is moving in an orbit. But energy is absorbed when it jumps from lower orbit to higher orbit. Whereas energy is emitted when it jumps from higher orbit to lower orbit.

Question 5.
Compare all the proposed models of an atom given in this chapter.
Solution:

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom 001
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom 002
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom 0000001



Question 6.

Summarise the rules for writing the distribution of electrons in various shells for the first eighteen elements.
Solution:
The distribution of elements in different orbits is governed by a scheme called Bohr- Bury scheme. There are following rules :

  • The maximum number of electrons present in any shell is given by the formula 2n2. Where n = no. of orbit.
  • The maximum number of electrons that can be accommodated in the outermost shell is 8.
  • Electrons in an atom do not occupy a new shell unless all the inner shells are completely filled.

Question 7.
Define valency by taking examples of silicon and oxygen.
Solution:
The number of electrons gained, lost or shared so as to complete the octet of electrons in valence shell is called valency.
Valency of silicon : It has electronic configuration → 2, 8, 4 Thus, 4 electrons are shared with other atoms to complete the octet and so its valency = 4
Valency of oxygen : It has electronic configuration → 2, 6 Thus, It will gain 2 electrons to complete its octet. So its valency = 2

Question 8.
Explain with examples

  1. Atomic number,
  2. Mass number,
  3. Isotopes and
  4. Isobars. Give any two uses of isotopes.

Solution:

  1. Atomic number : The number of protons present in the nucleus of an atom is called atomic number. It is denoted by Z. e.g., in _{ 20 }^{ 40 }{ Ca }, atomic number = 20
  2. Mass number : The sum of the number of protons and neutrons present in the nucleus of an atom is called mass number. It is denoted by A. e.g., in _{ 20 }^{ 40 }{ Ca }, mass number = 40
  3. Isotopes : The atoms of the same elements having same atomic number but different mass numbers are called isotopes, e.g., _{ 17 }^{ 37 }{ Cl } and _{ 17 }^{ 35 }{ Cl }
  4. Isobars : The atoms of the different elements having same mass number but different atomic numbers are called isobars. e.g., _{ 20 }^{ 40 }{ Ca } and _{ 18 }^{ 40 }{ Ar }

Uses of isotopes :

  • As nuclear fuel : An isotope of uranium (U – 235) is used as a nuclear fuel.
  • In medical field : An isotope of cobalt is used in the treatment of cancer.

Question 9.
Na+ has completely filled K and L shells. Explain.
Solution:
Atomic number of sodium (Na) = 11 No. of electrons in Na = 11 No. of electrons in Na+ = 11 — 1 = 10 Electronic configuration of Na+ → 2, 8 K L
For K – shell; 2n2 = 2 × l2 = 2
For L – shell; 2n2 = 2 × 22 = 8
Thus, in Na+, K and L shells are completely filled.

Question 10.
If bromine atom is available in the form of, say two isotopes _{ 35 }^{ 79 }{ Br } (49.70%) and _{ 35 }^{ 81 }{ Br } (50.30%), calculate the average atomic mass of bromine atom.
Solution:

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom 8


Question 11.

The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes _{ 8 }^{ 16 }{ X } and _{ 8 }^{ 18 }{ X } in the sample?
Solution:

tiwari academy class 9 science Chapter 4 Structure of the Atom 9


Question 12.

If Z = 3, what would be the valency of the element? Also, name the element.
Solution:
The electronic configuration of (Z) = 2,1 Thus, outermost shell has 1 electron. So, its valency = 1 Atomic number (Z) = 3, So name of the element is lithium.

Question 13.
Composition of the nuclei of two atomic species X and Y are given as under:
X    Y
Protons =    6     6
Neutrons = 6     8
Give the mass numbers of X and Y. What is the relation between the two species?
Solution:
Mass number of X = No. of protons +
No. of neutrons = 6 + 6 = 12
Mass number of Y = 6 + 8 = 14 The species X and Y are isotopes because their atomic numbers are same and their mass numbers are different i.e. _{ 6 }^{ 12 }C and _{ 6 }^{ 14 }C.

Question 14.
For the following statements, write T for True and F for False.

  1. J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
  2. A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
  3. The mass of an electron is about times that of proton. \cfrac { 1 }{ 2000 }
  4. An isotope of iodine is used for making tincture iodine, which is used as a medicine.

Solution:

  1. F : Because it was not proposed by J.J. Thomson.
  2. F : Because neutron is an independent sub-atomic particle.
  3. T : Because it is a fact known from experiments.
  4. F : Because tincture iodine is a solution of ordinary iodine in alcohol.

Question 15.
Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) atomic nucleus
(b) electron
(c) proton
(d) neutron.
Solution:
(a)

Question 16.
Isotopes of an element have
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons
(d) different atomic numbers.
Solution:
(c)

Question 17.
Number of valence electrons in Cl ion are :
(a) 16
(b) 8
(c) 17
(d) 18
Solution:
(b) : Electronic configuration of Cl (Z = 17) = 2, 8, 7. Thus, it has 7 valence electrons. Cl gains 1 electron to form CL ion. So, number of valence electrons = 7 + 1 = 8.

Question 18.
Which one of the following is a correct electronic configuration of sodium?
(a) 2,8
(b) 8,2,1
(c) 2,1,8
(d) 2,8,1
Solution:
(d) : Atomic number of sodium (Na) = 11 Its electronic configuration = 2, 8,1

Question 19.
Complete the following table.
Solution:

Atomic
number
Mass
number
Number of neutronsNumber of protonsNumber of electronsName of the atomic species
910
1632Sulphur
2412
21
1011

First Row : Mass number = Atomic number (9)+ No. of neutrons (10) = 19 No. of protons = Atomic number = 9 No. of electrons = Atomic number = 9 Name of the species = Fluorine (F)

Second Row :
 No. of neutrons = Mass number (32) – Atomic number (16) = 16 No. of protons = Atomic number = 16 No. of electrons = Atomic number = 16

Third Row :
 Atomic number = No. of protons = 12 No. of neutrons = Mass number (24) – Atomic number (12) = 12
No. of electrons = Atomic number = 12 Name of the species = Magnesium (Mg)

Fourth Row :
 Atomic number=No. of protons=1 No. of neutrons = Mass number (2) – Atomic number (1) = 1
No. of electrons = Atomic number = 1 Name of the species = Deuterium (D).

Fifth Row :
 Atomic number = No. of protons = 1 Name of the species = Protium or Hydrogen (H) The complete table can also be represented as:

Atomic
number
Mass
number
Number of neutronsNumber of protonsNumber of electronsName of the atomic species
9191099Fluorine
1632161616Sulphur
1224121212Magnesium
12111Deuterium
11011Hydrogen

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