NCERT Class 9 Science Chapter 4 Structure of the Atom Exercise Question Answer Solution For School Students of Class 9th. We also Provides Notes and Important Questions for Class 9 Science. NCERT Class 9th Science book is applied in mostly boards like CBSE, HBSE, RBSE, Up Board, MP Board and also some other state boards.
Also Read:- Class 9 Science NCERT Solution
NCERT Class 9 Science Chapter 4 Structure of the Atom Question and Answer for CBSE, HBSE and Other Boards Solution.
Structure of the Atom Class 9 Science Question Answer
Q1. Compare the properties of electrons, protons and neutrons.
Ans.
| Protons | Electrons | Neutrons |
| Discovered by E. Goldstein in 1886 | Discovered by J.J. Thomson in 1897 | Discovered by James Chadwick in 1932. |
| Have positive charge | Have negative charge | Have no charge |
| Found inside the atom’s nucleus | Revolves around the atom in orbit | Also found inside the atom in nucleus. |
Q2. What are the limitations of J.J. Thomson’s model of the atom?
Ans. The limitations of J.J. Thomson’s model of the atom are as follows :-
(i) J.J. Thomson failed to explain how the positive charge holds on the electrons inside the atom.
(ii) He also failed to explain about the nucleus of an atom which was later discovered by Rutherford.
Q3. What are the limitations of Rutherford’s model of the atom?
Ans. Limitations of Rutherford’s model of atom are as :-
(i) Rutherford’s model of atom could not explains the stability of an atom. As electrons are moving in circular path means they are accelerated. During this they radiate energy and finally they fall into the nucleus. This explains instability of an atom. But we know that atoms are stable.
(ii) It didn’t explained the arrangements of electrons inside the atom.
Q4. Describe Bohr’s model of the atom.
Ans. Bohr proposed many postulates in his atomic model are as :-
(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.
Q5. Compare all the proposed models of an atom given in the chapter.
Ans. Different models of atom are as :-
(i) Thomson’s Model of atom :- Thomson described that an atom consists of a positively charged sphere and electrons are embedded in it. The negative and positive charges are equal in magnitude. So, the atom as a whole in electrically neutral.
(ii) Rutherford’s Model of atom :- His model is based on alpha particles gold foil experiment. From this experiment he concluded that there is a positively charged center of an atom called nucleus. Nearly all the mass of an atom resides in the nucleus. The electron revolve around the nucleus in circular paths. The size of the nucleus is very small as compared to the size of the atom.
(iii) Bohr’s Model of atom :- Bohr’s model is that only certain special orbits known as discrete orbits of electrons, are allowed inside the atom. While revolving in discrete orbits the electrons do not radiate energy. These orbits or shells are called energy levels.
Q6. Summarize the rules for writing of distribution of electrons various shells for the first eighteen elements.
Ans. Rules for writing of distribution of electrons in various shell given by Bohr and are as follows :-
i. The maximum number of electrons present in a shell is equal to 2n2 , where n is the orbit no 1, 2, 3 …. .Hence the maximum no of electrons in different shells are as –
K-shell or first orbit = 2
L-shell or second orbit = 8
M-shell or third orbit = 18
N-shell or fourth orbit = 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.
Q7. Define valency by taking examples of silicon and oxygen.
Ans. Valency :- It is defined as the combining capacity of an atom. It generally ranges from 0 to 4.
Silicon :- Atomic number of silicon is 14. Its electronic configuration is 2, 8, 4. Silicon has 4 electrons in its outermost shell. To make compound it has either loss 4 electrons or to gain 4 electrons. Therefore, valency of silicon is 4.
Oxygen :- Atomic number of oxygen is 8. Its electronic configuration is 2, 6. Oxygen has 6 electrons in its outermost shell. To make compound it has to gain 2 electrons. Therefore, valency of oxygen is 2.
Q8. Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. Give any two uses of isotopes.
Ans.
(i) Atomic number :- It is the number of protons in the nucleus of an atom, denoted by Z.
For example – atomic number of Hydrogen atom is 1.
(ii) Mass number :- The mass number of an atom is equal to the number of nucleons( Number of Protons + Number of Neutrons) present in its nucleus. Mass number is denoted by A.
For example – Mass number of Carbon atom is 12 ( number of Protons = 6 , and also number of Neutrons = 6)
(iii) Isotopes :- These are the atoms of the same element, which have different mass numbers.
For example – Hydrogen has three isotopes. First Protium with Z = 1 and A = 1, Second Deuterium with Z = 1 and A = 2 and Third Tritium with Z = 1 and A = 3.
(iv) Isobars : – These are the atoms of different elements having same mass number but different atomic number.
For example – Argon and Calcium have same atomic mass 40 u . But having different atomic numbers as Argon as 18 and Calcium as 20.
Q9. Na+ has completely filled K and L shells. Explain.
Ans. Electronic configuration of Na = 2, 8, 1
As in symbol Na+ , Na has given 1 electron to other element.
Now the electronic configuration of Na+ = 2, 8.
From the given configuration it is clear that K shell is filled with at most 2 electrons and also L shell with 8 electrons.
Q10. If bromine atom is available in the form of, say, two isotopes 7935Br (49.7%) and 8135Br (50.3%), calculate the average atomic mass of bromine atom.
Ans. The Average Atomic mass of Bromine Atom is :
= 
= 39.263 + 40.743
= 80.006 u
Q11. The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 168X and 188X in the sample?
Ans. Let the percentage of 168X be x and the percentage of 168X be 100 – x.
-2x + 1800 = 16.2 x 100
-2x = 1620-1800
-2x = -180
x = 180/2 = 90
so 
= 90%
and = 10%
Q12. If Z = 3, what would be the valency of the element? Also, name the element.
Ans. Electronic configuration = 2, 1
Number of electrons in outermost shell = 1
Therefore, valency is also 1.
Name of element with atomic number 3 is Lithium.
Q13. 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?
Ans.
As we know that Mass number = Atomic no + number of Neutrons
Therefore, Mass number of element X = 6 + 6 = 12
Mass Number of element Y = 6 + 8 = 14
Both the element have same atomic number but different mass number. Therefore, X and Y are the isotopes ( of carbon as atomic number 6).
Q14. 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. False.
J.J. Thomson proposed that an atom consists of a positively charged sphere and the electrons are embedded in it.
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
Ans. False.
(c) The mass of an electron is about 1/2000 times that of proton.
Ans. True.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Ans. True.
Put the (✓) against correct choice and cross (×) against wrong choice in questions 15, 16 and 17.
Q15. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic Nucleus
(b) Electron
(c) Proton
(d) Neutron
Ans. (a) Atomic Nucleus
Q16. Isotopes of an element have
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons
(d) different atomic numbers.
Ans. (c) different number of neutrons
Isotopes of an element have same atomic number, same chemical properties but have different mass number and different physical properties.
Q17. Number of valence electrons in Cl– ion are :
(a) 16
(b) 8
(c) 17
(d) 18
Ans. (b) 8
As the atomic number of Cl is 17. Therefore, number of electrons in outermost shell is 7. But given is Cl– in which 1 extra electron is carried to complete octet. So valence electrons are 8.
Q18. 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
Ans. (d) 2, 8, 1
As the atomic number of sodium is 11. So, in K shell it occupies 2 electrons, in L shell 8 and remaining 1 in M shell.
Q19. Complete the following table.
| Atomic Number | Mass Number | Number of Neutrons | Number of Protons | Number of Electrons | Name of the Atomic Species |
| 9 | – | – | – | – | – |
| 16 | 32 | – | – | – | Sulphur |
| – | 24 | – | 12 | – | – |
| – | 2 | – | 1 | – | – |
| – | 1 | 0 | 1 | 0 | – |
Ans.
As we know that Atomic Number = Number of Protons,
Mass number = Number of Protons + Number of neutrons,
Number of protons = Number of electrons ( except for isotopes)
| 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 | Aluminium |
| 1 | 2 | 1 | 1 | 1 | Deuterium |
| 1 | 1 | 0 | 1 | 0 | Protium |