Class 9 Science Chapter 4 Structure of the Atom

NCERT Class 9 Science Chapter 4 Structure of the Atom, Question Answer to each chapter is provided in the list so that you can easily browse throughout different chapters Class 9 Science Chapter 4 Structure of the Atom and select need one.

NCERT Class 9 Science Chapter 4 Structure of the Atom

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Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. These solutions are part of SCERT All Subject Solutions. Here we have given SCERT Class 9 Science Chapter 4 Structure of the Atom Solutions for All Subjects, You can practice these here.

Structure of the Atom

Chapter – 4

GENERAL SCIENCE

TEXTUAL QUESTIONS AND ANSWERS

INTEX QUESTIONS AND ANSWERS

Textbook Page No. 47

1. What are canal rays?

Ans.These are positively charged radiations that consist of particles having charges equal in magnitude. Canal rays can also be described as the beam of positive ions obtained by gas-discharge tubes. The mass of canal particles is extremely more than that of electrons, it is about times larger.

2. If an atom contains one electron and one proton, will it carry any charge or not? 

Ans. An electron is a negatively charged particle, whereas a proton is a positively charged particle. The magnitude of their charges is equal. Therefore, an atom containing one electron and one proton will not carry any charge. 

Textbook Page No. 49.

1. On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.

Ans. On the basis of Thomson’s model of an atom, the positive charges present in an atom due to the presence of protons are equal in magnitude to the negative charges present in the atom due to the presence of electrons. Therefore, an atom is neutral as a whole.

2. On the basis of Rutherford’s model of an atom, which subatomic particle is present in the nucleus of an atom?

Ans. According to Rutherford’s model of an atom, protons are present in the nucleus of an atom.

3. Draw a sketch of Bohr’s model of an atom with three shells.

Ans. 

4. What do you think would be the observation if the a-particle scattering experiment is Bohr’s model of an atom with three shells carried out using a foil of a metal other than gold? 

Ans. To carry out the a-particle scattering experiment, a foil as thin as possible is required. Since no other metal is as malleable as gold, it would be difficult to obtain such a thin foil. If a thick foil is used, more a-particles would bounce back giving some other observations. The gold foil used by Rutherford was about 1000 atoms thick.

Textbook Page No. 49

1. Name the three subatomic particles of an atom.

Ans. The three subatomic particles of an atom are electron, proton and neutron.

2. Helium atom has an atomic mass of 4u and two protons in its nucleus. How many neutrons does it have?

Ans. Atomic mass of a helium atom = 4u

Mass number = Number of protons + Number of neutrons

∴ Number of neutrons = Mass number – Number of protons

= 4 – 2

= 2

Therefore, a helium atom has 2 neutrons.

Textbook Page No. 50

1. Write the distribution of electrons in carbon and sodium atoms.

Ans. Atomic number of carbon = 6

∴ Number of electrons = 6

Number of electrons in the K-shell = 2

Number of electrons in the L-shell = 4

Therefore, distribution of electrons in carbon atom is 2, 4 Atomic number of sodium = 11

∴ Number of electrons = 11

Number of electrons in the K-shell = 2

Number of electrons in the L-shell = 8

Number of electrons in the M-shell = 1

Therefore, distribution of electrons in sodium atom is 2, 8, 1.

2. If K and L shells of an atom are full, then what would be the total number of electrons in the atom?

Ans. Number of electrons in K-shell (n = 1) = 2n² = 2×12 = 2×1 = 2

Number of electrons in L-shell (n = 2) = 2×2² = 2×4 = 8 

Therefore, the total number of electrons in the atom = 2 + 8 = 10

Textbook Page No. 52

1. How will you find the valency of chlorine, sulphur and magnesium?

Ans. The valency of chlorin , sulphur and magnesium are 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 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 electrons to attain octet and hence its valency = 2.

Textbook Page No. 52

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

(i) what is the atomic number of the atom? and

Ans. (i) Atomic number of the atom = Number of protons

= 80.

(ii) what is the charge on the atom?

Ans: As the number of electrons is equal to the number of protons, the charge on the atom is zero. 

2. With the help of the following table, find out the mass number of oxygen and sulphur atom. 

Table: Composition of atoms of the first eighteen elements with electron distribution in various shells 

Ans. Oxygen:

Number of protons = 8

Number of neutrons = 8

∴ Mass number of oxygen = Number of protons + Number of neutrons

= 8 + 8

= 16

Sulphur:

Number of protons = 16

Number of neutrons = 16

∴ Mass number of sulphur = Number of protons + number of neutrons

= 16 + 16

= 32.

Textbook Page No. 53

1. For the symbol H, D and T tabulate three subatomic particles found in each of them.

Ans. 

IsotopeSymbolNo. of protonsNo. of electronsNo. of neutrons
Protium¹₁H110
Deuterium²₁H or D111
Thritium³₁H or T112

2. Write the electrons configuration of any one pair of isotopes and isobars.

Ans. The electrons configuration of one pair of isotopes and isobars are mentioned below: wo isotopes of carbon are 6C12 and 6C14 The electronic configuration of 6C12 is 2 4. The electronic configuration of 6C14 is 2 4.

Isotopes have same electronic configuration]29Ca40 and 18Ar40 are a pair of isobars.The electronic configuration of 29Ca40 is 2 8 8 2. The electronic configuration of 18Ar40 is 2 8 [Isobars have different electronic configuration].

EXERCISE (Textbook Page No. 54-56)

1. Compare the properties of electrons, protons and neutrons. 

Ans. A comparison of the properties of electrons, protons and neutrons are shown below:

ParticleSymbolNature of chargeRelative chargeAbsolute chargeRelative massAbsolute massLocation in the atom
Electrone⁻Negatively charged-11.602×10⁻¹⁹C9.1×10⁻³¹kgOutside the nucleus
Protonp⁺Positively charged+11.602×10⁻¹⁹C1u1.67×10⁻²⁷kgIn the nucleus
NeutronnNo charge001u1.67×10⁻²⁷kgIn the nucleus

2. What are the limitations of J.J. Thomson’s model of the atom? 

Ans. According to J. J. Thomson’s model, an atom is a uniform sphere of positive charges in which negatively charged electrons are embedded. In other words, the electrons and protons in an atom are spread throughout the atom. However later on, Rutherford’s a-particle scattering experiment led to the conclusion that the mass and positive charge of an atom are concentrated in a very small space called nucleus, situated at the centre of the atom. Thus, Thomson’s model of the atom could not explain the results of the experiments done by other scientists.

3. What are the limitations of Rutherford’s model of the atom?

Ans. Rutherford’s model was unable to explain the stability of an atom. According to Rutherford’s postulate, electrons revolve at a very high speed around a nucleus of an atom in a fixed orbit. However, Maxwell explained accelerated charged particles release electromagnetic radiations.

4. Describe Bohr’s model of the atom. 

Ans. Borh’s model of the atom are mentioned below: 

(i) An atom consists of a small positively charged nucleus at the centre. The entire mass of the atom is concentrated at the nucleus.

(ii) The electrons revolve around the nucleus only in certain selected circular paths called orbits. 

(iii) These orbits are associated with definite energies and are numbered as 1; 2, 3, 4, etc. or designated as K, L, M, N, etc. outward from the nucleus.

(iv) The energy of an electron remains constant in a particular orbit. Therefore, these orbits are also called energy levels or energy shells. 

(v) The energy is emitted or absorbed only when the electrons jump from one orbit to another. The energy emitted or absorbed is equal to the difference between the energies of the higher and lower energy shells.

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

Ans. The proposed models of an atom giving in this chapter are mentioned below: 

(i) J.J. Thomson’s model of atom:

(a)  An atom consists of a sphere of positive charge in which electrons are embedded just like seeds in the water melon.

(b) Total positive charge on the sphere is equal to the total negative charge present on the electrons so that atom as a whole is electrically neutral.        

(iii) It could not explain the results of Rutherford’s scatter experiments.

(ii) Rutherford’s model of atom:

(a) An atom consists of a small positively charged nucleus in the centre and the electrons are revolving around it. 

(b) There is very large empty space between the nucleus and the electrons. 

(c) All the mass of the atom is mainly concentrated in the nucleus.                             

(d) It could not explain the stability of the atom because the revolving electron will be accelerated towards nucleus. Hence it will lose energy. Its orbit will become smaller and smaller and ultimately the electron will fall into the nucleus.

(iii) Bohr’s model of atom: 

(a) An atom consists of a small heavy positively charged nucleus in the centre and the electrons revolve around it in circular paths called orbits. 

(b) As long as an electron is revolving in a particular orbit, it can neither lose energy nor gain energy. Thus, the atom is stable and does not collapse. This state of the atom with lowest energy is called ground state of the atom. 

(c) Energy is lost or gained by an electron only when it jumps from one orbit to the other. The energy falls on an electron and it absorbs this energy, it will jump to some outer shell. The atom is then said to be in the excited state. In the excited state, the atom is not stable. It loses or emits energy and jumps back to some inner energy level. In other words, an electron jumps from inner shell to outer shell by absorbing energy whereas energy is emitted when an electron jumps from an outer shell to an inner

6. Summarise the rules for writing of distillation of electrons in various shells for the first eighteen elements.

Ans. The rules for writing of distribution of electrons in various shells for the first eighteen elements is given by the Bohr-Bury Scheme:

(i) The maximum number of electrons which each shell can hold is 2n² where ‘n’ is the number of shell.

nElectron shellMaximum capacity (2n²)
1K2×1² = 2×1 = 2 electrons
2L2×2² = 2×4 = 8 electrons
3M2×3² = 2×9 = 18 electrons

(i) The outermost shell of an atom cannot hold more than 8 electrons. 

(ii) It is necessary that a shell has its full quota of electrons before beginning to fill the next higher shell. In other words, the electrons in the different shells should be filled in a step- wise manner.

7. Define valency by taking examples of silicon and oxygen.

Ans. The combining capacity of an atom is known as its valency. The valence electrons (electrons present in the outermost shell) of an atom determine the valency of the element. The valency of the elements having 1, 2, 3, or 4 electrons respectively in the outermost shell of the respective atoms are 1, 2, 3 and respectively, i.e. the valency of the element is equal to the number of valence electrons. However, if the number of valence electrons of an atom is close to its full capacity then the valency of the element is obtained by subtracting the number of valence electrons from 8. This is done in case of the elements having 5, 6, 7 or 8 valence electrons. For example, the element chlorine (2, 8, 7) has valency 1 (8-7=1) as the chlorine mom has 7 valence electrons. The valency of the inert elements is zero.

Valency of silicon:

The atomic number of silicon is 14. Its electronic configuration is 2, 8, 4. It has 4 valence electrons. It needs four more electrons to fulfil its octet. Therefore, silicon atom shares its four valence electrons with other atoms. Thus, the valency of silicon is 4.

Valency of oxygen:

The atomic number of oxygen is 8. Its electronic configuration is 2, 6. It has 6 valence electrons. It needs two more electrons to fulfil its octet. Therefore, oxygen atom accepts two electrons from other atoms. Thun, the valency of oxygen is 2.

8. Explain with examples 

(i) Atomic number.

Ans. Atomic number: The total number of protons present in the nucleus of an atom of an element is known as atomic number of that element. The atomic number is denoted by Z. For example, there are 8 protons present in the nucleus of an oxygen atom. Therefore, the atomic number of oxygen is 8. All the atoms of the same element have the same atomic number.

(ii) Mass number.

Ans. Mass number: The sum of the total number of neutrons and protons present in the nucleus of an atom of an element is known as its mass number. For example, there are 8 protons and 8 neutrons present in the nucleus of an oxygen atom. Therefore, the mass number of oxygen is 16 (8 + 8 = 16). 

(iii) Isotopes.

Isotopes: Atoms of the same element having same atomic number, but different mass numbers are called isotopes. The physical properties of the isotopes of an element are different, but they have identical chemical properties.

The isotopes of oxygen are ¹⁶₈O, ¹⁷₈O and ¹⁸₈O. Each isotope has same atomic number 8, but the mass numbers of the isotopes are 16, 17 and 18 respectively.

(iv) Isobars. 

Ans. Isobars: The atoms of different elements with different atomic numbers, but having the same mass number are called isobars. For example, Argon (Ar) has atomic number 18 and calcium (Ca) has atomic number 20 but both of these have mass numbers as 40. 

9. Na⁺ has completely filled K and L shells. Explain.

Ans. The Na+ ion possesses one less electron than the Na atom, resulting in a total of ten electrons.

As a result, 2 electrons are sent to the K-shell and 8 electrons are sent to the L-shell, entirely filling the K and L shells.

10. Of bromine atom is available in the form of, say, two isotopes ⁷⁹₃₅ Br (49.7%) and ⁸¹₃₅ Br (50.3%), calculate the average atomic mass of bromine atom.

Ans. Percentage of ⁷⁹₃₅ Br isotope = 49.7

                                                                = 80.006 u

Thus, the average atomic mass of bromine atom is 80.006 u.

11. The average atomic mass of a simple of an element X is 16.2 u. What are the percentage of isotopes ¹⁶₈X and ¹⁸₈X in the sample?

Ans. Let the percentage of ¹⁶₈X isotope in the sample be m%

Therefore, the percentage of ¹⁸₈X isotope in the sample is (100-m)%

          ⇒ m = 90.

Thus, the percentage of ¹⁶₈X isotope in the sample is 90% and the percentage of ¹⁸₈X isotope in the sample is 100-90 = 10%.

12. If Z = 3, what would be the valency of the element? Also, name the element. 

Ans. Since Z represents atomic number, the atomic number of the element is 3 and its electronic configuration is 2, 1. As the number of valence electrons is 1, the valency of the element is 1. The element is lithium (Li).

13. Composition of the nuclei of two atomic species X and Y are given as under.

Give the mass number of X and Y. What is the relation between the two species?

Ans. Mass number of X = Number of protons + Number of neutrons

= 6 + 6

= 12

Mass number of Y = Number of protons + Number of neutrons

= 6 + 8

= 14

Since X and Y both have equal number of protons, their atomic numbers are same but mass numbers are different, therefore the species X and Y are isotopes to each other.

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

(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons. 

Ans. F.

(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.

Ans. F.

(c) The mass of an electron is about 

times that of proton.

Ans. T.

(d) Au isotope of iodine is used for making tincture iodine, which is used as a medicine.

Ans. F.

Put tick (✓) against correct choice and cross (x) against wrong choice in questions 15, 16 and 17.

15. Rutherford’s alpha-particles scattering experiment was responsible for the discovery of

(a) Atomic Nucleus. 

Ans. T.

(b) Electron [x].

Ans. F.

(c) Proton [x].

Ans. F.

(d) Neutron [x].

Ans. F.

16. Isotopes of an element have.

(a) The same physical properties.

Ans. F.

(b) Different chemical properties.

Ans. T.

(c) Different number of neutrons. 

Ans. F.

(d) Gifferent atomic numbers.

Ans. F.

17. Number of valence electrons in CH- ion are:

(a) 16.

Ans. F.

(b) 8.

Ans. T.

(c) 17.

Ans. F.

(d) 18.

Ans. F.

18. Which one of the following is a correct electronic configuration of sodium?

(a) 2, 8.

Ans. F.

(b) 8, 2, 1.

Ans. F.

(c) 2, 1, 8.

Ans. F.

(d) 2, 8, 1.

Ans. T.

19. Complete the following table.

Atomic NumberMass NumberNumber of NeutronsNumber of ProtonsNumber of ElectronsName of the Atomic Species
910
1632Sulphur
2412
21
1010

Ans. 

Atomic NumberMass NumberNumber of NeutronsNumber of ProtonsNumber of ElectronsName of the Atomic Species
9191099Fluorine
1632161616Sulphur
1224121212Magnesium
12111Deuterium
11010Protium

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