CBSE Class IX ( 9th) Science Chapter 4. Structure Of The Atom | Lesson Exercises
Question 1. Compare the properties of electrons, protons and neutrons.
Answer :
| Electrons | Protons | Neutron |
| Electrons are Negatively charged particles Electrons are present in outer shells with in an atom and revolve around the nucleus in well-defined orbits or discrete orbits The mass of an electron is about 1/ 2000 times the mass of an hydrogen atom. A electron is represented as 'e-' | Protons are Positively charged particles. Protons are present in the nucleus of all atoms Number of protons determines the atomic number of an element The mass of a proton is taken as one unit and equals to neutron. A proton is represented as 'p+' | Neutrons do not carry any charge and are neutral. Neutrons are present in the nucleus of all atoms, except hydrogen The mass of a neutron is taken as one unit and equals to that of proton. A neutron is represented as 'n' |
Question 2. What are the limitations of J.J. Thomson's model of the atom?
Answer : Thomson proposed that: an atom consists of a positively charged sphere and the electrons are
embedded in it.The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically neutral.Thomson's model of the atom fails to explain Rutherford's α-particle scattering experiment in which most of the fast moving α-particles passed straight through the gold foil.Only Some of the α-particles were deflected by the foil by small angles.Which clearly established atom has a lot of empty space and positive charge is concentrated in a very small volume within the atom.
Question 3. What are the limitations of Rutherford's model of the atom?
Answer : As per Rutherford's nuclear model of an atom , an atom has a very small sized nucleus with positive charge inside and has electrons revolving around this nucleus in well-defined orbits. Nearly all
the mass of an atom resides in the nucleus.
Rutherford's model of the atom failed to explain the the stability of the atom. As any particle in a
circular orbit would undergo acceleration. During acceleration, revolving electron as charged particles would lose energy and finally fall into the nucleus. This may lead to a collapsed atomic structure, resulting in very unstable atoms.Contrarily atoms are mostly stable.
Question 4. Describe Bohr's model of the atom.
Answer . Neils Bohr put forward the following postulates about the model of an atom:
(i) Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom.
(ii) While revolving in discrete orbits the electrons do not radiate energy.
These orbits or shells are called energy levels.These orbits or shells are represented by the letters K,L,M,N,… or the numbers, n=1,2,3,4,…
.
Question 5. Compare all the proposed models of an atom given in this chapter.
Answer :
| THOMSON'S Model of An Atom | RUTHERFORD'S Model of An Atom | BOHR' S Model of An Atom |
| (i) An atom consists of a positively charged sphere and the electrons are embedded in it. (ii) The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically neutral. | (i) There is a positively charged centre in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus. (ii) The electrons revolve around the nucleus in well-defined orbits. (iii) The size of the nucleus is very small as compared to the size of the atom. | (i) Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom. (ii) While revolving in discrete orbits the electrons do not radiate energy. |
Question 6. Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Answer : The following rules are followed for writing the distribution of electrons in different energy
levels or shells for the first eighteen elements :
(i) The maximum number of electrons present in a shell is given by the formula 2n2, where 'n' is the orbit
number or energy level index, 1,2,3,….Hence the maximum number of electrons in different shells are as
follows:
first orbit or K-shell will be = 2 × 12 = 2,
second orbit or L-shell will be = 2 × 22 = 8,
third orbit or M-shell will be = 2 × 32= 18,
fourth orbit or N-shell will be = 2 × 42= 32,
(ii) The maximum number of electrons that can be accommodated in the outermost orbit is 8.
(iii) Electrons are not accommodated in a given shell, unless the inner shells are filled. That is, the shells are filled in a step-wise manner.
| Distribution of electrons in various shells for the first eighteen elements | ||||||||
| Element | Symbol | Atomic Number | Number of Electron |
Distribution of electrons
| ||||
| Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium Aluminium Silicon Phosphorus Sulphur Chlorine Argon | H He Li Be B C N O F Ne Na Mg Al Si P S Cl A | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
| ||||
Question 7. Define valency by taking examples of silicon and oxygen.
Answer : The valency may be defined as the combining capacity of the atoms of elements, or their tendency to form molecules with atoms of the same or different elements. The electrons in an atom are arranged in different shells/orbits. The electrons present in the outermost shell of an atom are known as the valence electrons. The outermost shell of an atom can accommodate a maximum of 8 electrons. All atoms of various elements have tendency to have a fully-filled outermost shell with 8 electrons or octet . This is achieved through reaction with atoms of other elements, by sharing, gaining or losing electrons.
- In case of Silicon, it has 4 electrons in the outer most shell, thus it is deficient by 4 electrons to have a fully filled outer most shell. Hence Its valency is 4.
- In case of Oxygen, it has 6 electrons in the outer most shell, thus it is deficient by 2 electrons to have a fully filled outer most shell. Hence Its valency is 2.
Question 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 : Atomic number of an element is defined as the number of protons present in its atom. It is denoted by 'Z'.
(ii) Mass number : The mass number is defined as the sum of the total number of protons and neutrons present in the nucleus of an atom.
(iii) Isotopes :Isotopes are defined as the atoms of the same element, having the same atomic number but different mass numbers.
( iv) Isobars : Atoms of different elements with different atomic numbers, which have the same mass number, are known as isobars.
Two uses of isotopes :
(i) An isotope of uranium is used as a fuel in nuclear reactors.
(ii) An isotope of cobalt is used in the treatment of cancer.
Question 9. Na+ has completely filled K and L shells. Explain.
Answer : Na+ represent Sodium with atomic number 11. Which means it has 11 protons. Number of Electrons is equal to number of protons. So Sodium has 11 electrons in the outer shells distributed as 2 electrons in 'k' shell, 8 electron in 'L' shell and 1 electron in 'M' shell. The distribution of electron in the shells is governed by the formula i.e. 2n2. Where 'n' represents the number of shells. The first shell is represented as 'K', second shell as 'L', third shell as 'M', and fourth shell as 'N' and so on.
There for, as per the formula above, the 'K' shell i.e. 1st shell will contain 2 x 12 =2 electrons and 'L' shell will contain 2 x 22 = 8 electrons.Which means it has completely filled 'K' and 'L' shells.
Question 10. If bromine atom is available in the form of, say, two isotopes
and
calculate the average atomic mass of bromine atom.
| 79 | |
| Br (49.7%) | |
| 35 |
| 81 | |
| Br (50.3%). | |
| 35 |
Answer : The average atomic mass of bromine atom (79 × 49.7 ) / 100 + ( 81 × 50.3)/ 100 =(3926.3 + 4074.3 ) / 100 = 8000.6 / 100= 80.06 u
Question 11. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes
and
in the sample?
Answer :
| 16 | |
| X | |
| 8 |
| 18 | |
| X | |
| 8 |
Answer :
Suppose the percentage of
| 16 | |
| X in given sample = x | |
| 8 |
:. The percentage of
| 18 | |
| X in given sample = 100-x | |
| 8 |
As given average atomic mass of a sample of an element X = 16.2 u
=> (16 × x /100) + [18 × (100-x) /100] = 16.2
=> (16x + 1800-18x) / 100 = 16.2
=> (16x + 1800-18x) = 16.2 × 100
=> -2x + 1800 = 1620
=> -2x = 1620-1800
= -2x = -180
=> x = 90
:. The percentage of
| 16 | |
| X in given sample = x = 90 | |
| 8 |
:. The percentage of
| 16 | |
| X in given sample = 100-x =100-90 = 10 | |
| 8 |
Question 12. If Z = 3, what would be the valency of the element? Also, name the element.
Answer :
Answer :
Given Z= 3
we know that Z = Atomic Number = Number of protons = Number of electrons
:. Number of electrons = 3
Distrubution of electrons is given by the formula 2n2 where n is the number of orbit or shell represented by letters K, L, M, N etc
Hence for first orbit i.e. 'k' the number of electrons = 2 × 1 = 2
for second orbit i.e. 'L' the number of electron = total electrons - numbers of electrons in 'K' = 3-2=1
Valency of element is 1 as it will readily lose its outermost valance electrons than to gather 7 more to have a fully filled outermost orbit.
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?
Answer :
We know that :
Mass number of an atom = Number of Protons + number of Neutrons
Atomic Number of an atom= Number of Protons
:. Atomic Number of X = 6
:. Atomic Number of y = 6
:. Mass number of X = 6+6 = 12
:. Mass number of y = 6+8 = 14
X and Y have similar Atomic Number (6) but have different mass numbers, therefor both are isotopes of same element.
Question 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.
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
(c) The mass of an electron is about 1 2000 times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Answer :
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.(F)
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?
Answer :
We know that :
Mass number of an atom = Number of Protons + number of Neutrons
Atomic Number of an atom= Number of Protons
:. Atomic Number of X = 6
:. Atomic Number of y = 6
:. Mass number of X = 6+6 = 12
:. Mass number of y = 6+8 = 14
X and Y have similar Atomic Number (6) but have different mass numbers, therefor both are isotopes of same element.
Question 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.
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
(c) The mass of an electron is about 1 2000 times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Answer :
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.(F)
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.(F)
(c) The mass of an electron is about 1/ 2000 times that of proton.(T)
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.(T)
Put tick (√) against correct choice and cross (×) against wrong choice in questions 15, 16 and 17
Question 15. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic Nucleus
(b) Electron
(c) Proton
(d) Neutron
Answer : (a) Atomic Nucleus (√)
(a) Atomic Nucleus
(b) Electron
(c) Proton
(d) Neutron
Answer : (a) Atomic Nucleus (√)
(b) Electron (×)
(c) Proton (×)
(d) Neutron (×)
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.
Answer : (a) the same physical properties (×)
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.
Answer : (a) the same physical properties (×)
(b) different chemical properties (×)
(c) different number of neutrons (√)
(d) different atomic numbers. (×)
Question 17. Number of valence electrons in Cl– ion are: (a) 16 (b) 8 (c) 17 (d) 18
Answer :(a) 16 (×)
(b) 8 (√)
(c) 17 (×)
(d) 18 (×)
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.
Answer :(d) 2,8,1.
Question 19. Complete the following table.
| Atomic Number | Mass Number | Number of Neutrons | Number of Protons | Number of Electrons | Name of the Atomic Species |
| 9 | - | 10 | - | - | - |
| 10 | 32 | - | - | - | Sulphure |
| - | 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 | Sulphure |
| 12 | 24 | 12 | 12 | 12 | Magnesium |
| 1 | 2 | 1 | 1 | 1 | Deuterium |
| 1 | 1 | 0 | 1 | 1 | Protium |
