NCERT Solutions Structure of Atom Class 9 |Chapter – 4th|?
Exercise
1. Compare the properties of electrons, protons and neutrons.
Answer
| Particle | Nature of Charge | Mass | Location |
| Electron | Electrons are negatively charged. | 9 x 10–31 kg | Extra nuclear part distributed in different shell or orbits. |
| Proton | Protons are positively charged. | 1.672 x 10–27 kg (1 µ) (approx. 2000 times that of the electron) | Nucleus |
| Neutron | Neutrons are neutral. | Equal to mass of proton | Nucleus |
2. What are the limitations of J.J. Thomson’s model of the atom?
Answer
The limitations of J.J. Thomson’s model of the atom are:
→ It could not explain the result of scattering experiment performed by rutherford.
→ It did not have any experiment support.
3. What are the limitations of Rutherford’s model of the atom?
Answer
The limitations of Rutherford’s model of the atom are
→ It failed to explain the stability of an atom.
→ It doesn’t explain the spectrum of hydrogen and other atoms.
4. Describe Bohr’s model of the atom.
Answer
→ The atom consists of a small positively charged nucleus at its center.
→ The whole mass of the atom is concentrated at the nucleus and the volume of the nucleus is much smaller than the volume of the atom.
→ All the protons and neutrons of the atom are contained in the nucleus.
→ Only certain orbits known as discrete orbits of electrons are allowed inside the atom.
→ While revolving in these discrete orbits electrons do not radiate energy. These orbits or cells are represented by the letters K, L, M, N etc. or the numbers, n = 1, 2, 3, 4, . . as shown in below figure.
5. Compare all the proposed models of an atom given in this chapter.
Answer
| Thomson’s model | Rutherford’s model | Bohr’s model |
| → An atom consists of a positively charged sphere and the electrons are embedded in it. → The negative and positive charges are equal in magnitude. As a result the atom is electrically neutral. | → An atom consists of a positively charged center in the atom called the nucleus. The mass of the atom is contributed mainly by the nucleus. → The size of the nucleus is very small as compared to the size of the atom. → The electrons revolve around the nucleus in well-defined orbits. | → Bohr agreed with almost all points as said by Rutherford except regarding the revolution of electrons for which he added that there are only certain orbits known as discrete orbits inside the atom in which electrons revolve around the nucleus. → While revolving in its discrete orbits the electrons do not radiate energy. |
6. Summarize the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Answer
The rules for writing of the distribution of electrons in various shells for the first eighteen elements are given below.
→ If n gives the number of orbit or energy level, then 2n2 gives the maximum number of electrons possible in a given orbit or energy level. Thus,
First orbit or K-shell will have 2 electrons,
Second orbit or L-shell will have 8 electrons,
Third orbit or M-shell will have 18 electrons.
→ If it is the outermost orbit, then it should have not more than 8 electrons.
→ There should be step-wise filling of electrons in different orbits, i.e., electrons are not accompanied in a given orbit if the earlier orbits or shells are incompletely filled.
7. Define valency by taking examples of silicon and oxygen.
Answer
The valency of an element is the combining capacity of that element. The valency of an element is determined by the number of valence electrons present in the atom of that element.→ Valency of Silicon: It has electronic configuration: 2,8,4
Thus, the valency of silicon is 4 as these electrons can be shared with others to complete octet.
→ Valency of Oxygen: It has electronic configuration: 2,6
Thus, the valency of oxygen is 2 as it will gain 2 electrons to complete its octet.
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8. Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. Give any two uses of isotopes.
Answer
(i) Atomic number: The atomic number of an element is the total number of protons present in the atom of that element. For example, nitrogen has 7 protons in its atom. Thus, the atomic number of nitrogen is 7.
(ii) Mass number: The mass number of an element is the sum of the number of protons and neutrons present in the atom of that element. For example, the atom of boron has 5 protons and 6 neutrons. So, the mass number of boron is 5 + 6 = 11.
(iii) Isotopes: These are atoms of the same element having the same atomic number, but different mass numbers. For example, chlorine has two isotopes with atomic number 17 but mass numbers 35 and 37 represented by – 17Cl35, 17Cl37
(iv) Isobars: These are atoms having the same mass number, but different atomic numbers i.e., isobars are atoms of different elements having the same mass number. For example, Ne has atomic number 10 and sodium has atomic number 11 but both of them have mass numbers as 22 represented by – 10Ne22, 11Na22
Two uses of isotopes:
→ One isotope of uranium is used as a fuel in nuclear reactors.
→ One isotope of cobalt is used in the treatment of cancer.
9. Na+ has completely filled K and L shells. Explain.
Answer
The atomic number of sodium is 11. So, neutral sodium atom has 11 electrons and its electronic configuration is 2, 8, 1. But Na+ has 10 electrons. Out of 10, K-shell contains 2 and L-shell 8 electrons respectively. Thus, Na+ has completely filled K and L shells.
10. If bromine atom is available in the form of, say, two isotopes 79/35 Br (49.7%) and 81/35 Br (50.3%), calculate the average atomic mass of bromine atom.
Answer
It is given that two isotopes of bromine are 79/35 Br (49.7%) and 81/35 Br (50.3%). Then, the average atomic mass of bromine atom is given by:
11. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 16 / 8 X and 18 / 8 X in the sample?
Answer
It is given that the average atomic mass of the sample of element X is 16.2 u.
Let the percentage of isotope 18 / 8 X be y%. Thus, the percentage of isotope 16 / 8 X will be (100 – y) %.
Therefore,
18y + 1600 – 16y = 1620
2y + 1600 = 1620
2y = 1620 – 1600
y= 10
Therefore, the percentage of isotope 18 / 8 X is 10%.
And, the percentage of isotope 16 / 8 X is (100 – 10) % = 90%.
12. If Z = 3, what would be the valency of the element? Also, name the element.
Answer
By Z = 3, we mean that the atomic number of the element is 3. Its electronic configuration is 2, 1. Hence, the valency of the element is 1 (since the outermost shell has only one electron).
Therefore, the element with Z = 3 is lithium.
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?
Answer
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
These two atomic species X and Y have the same atomic number, but different mass numbers. Hence, they are isotopes.
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.
► False
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
► False
(c) The mass of an electron is about 1 / 2000times that of proton.
► True
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
► False
15. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic nucleus
(b) Electron
(c) Proton
(d) Neutron
► (a) Atomic nucleus
16. Isotopes of an element have
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons
(d) different atomic numbers
► (c) different number of neutrons
17. Number of valence electrons in Cl –ion are:
(a) 16
(b) 8
(c) 17
(d) 18
► (b) 8
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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
► (d) 2, 8, 1
19. Fill the table.
| Atomic number | Mass number | Number of Neutrons | Number of protons | Number of electrons | Name of the Atomic species |
| 9 | − | 10 | − | − | − |
| 16 | 32 | − | − | − | Sulphur |
| − | 24 | − | 12 | − | − |
| − | 2 | − | 1 | − | − |
| − | 1 | 0 | 1 | 0 | − |
Answer
| Atomic number | Mass number | Number of Neutrons | Number of protons | Number of electrons | Name of the Atomic species |
| 9 | 19 | 10 | 9 | 9 | Fluorine |
| 16 | 32 | 16 | 16 | 16 | Sulphur |
| 12 | 24 | 12 | 12 | 12 | Magnesium |
| 1 | 2 | 1 | 1 | 1 | Deuterium |
| 1 | 1 | 0 | 1 | 0 | Hydrogen ion |
NCERT Solutions Structure of Atom Class 9 |Chapter – 4th|?
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