4. Structure of the Atom Class 9 NCERT Exemplar Solutions Chapter 4 (CBSE Science)
CBSE Class 9 Science Chapter 4: Structure of the Atom – NCERT Exemplar Questions with Answers
Chapter 4: Structure of the Atom
Multiple Choice Questions
1. Which of the following correctly represent the electronic distribution in the Mg atom?
(a) 3, 8, 1 (b) 2, 8, 2 (c) 1, 8, 3 (d) 8, 2, 2
Answer: (b) 2,8,2
[Atomic number of Mg is 12 .
Electronic configuration of Mg is 2.8.2 ]
2. Rutherford’s ‘alpha (α) particles scattering experiment’ resulted in to discovery of
(a) Electron (b) Proton (c) Nucleus in the atom (d) Atomic mass
Answer: (c) Nucleus in the atom .
[ Rutherford’s experiment led to the discovery of the nucleus of the atom.]
3. The number of electrons in an element X is 15 and the number of neutrons is 16. Which of the following is the correct representation of the element?
(a) 
(b) 
(b) 
(d) 
Answer: (a) 1531X .
[Number of electrons = Number of protons = 15
and Mass number (A) = Protons + Neutrons = 15 + 16 =31 ]
4. Dalton’s atomic theory successfully explained
(i) Law of conservation of mass (ii) Law of constant composition
(iii) Law of radioactivity (iv) Law of multiple proportion
(a) (i), (ii) and (iii) (b) (i), (iii) and (iv)
(c) (ii), (iii) and (iv) (d) (i), (ii) and (iv)
Answer: (d) (i), (ii) and (iv) .
[ Dalton’s atomic theory explained:
(i) Law of conservation of mass .
(ii) Law of constant composition (atoms combine in fixed ratios).
(iv) Law of multiple proportions (different simple ratios when elements combine to form different compounds) ]
5. Which of the following statements about Rutherford’s model of atom are correct?
(i) considered the nucleus as positively charged
(ii) established that the α–particles are four times as heavy as a hydrogen atom
(iii) can be compared to solar system
(iv) was in agreement with Thomson’s model
(a) (i) and (iii) (b) (ii) and (iii) (c) (i) and (iv) (d) only (i)
Answer: (a) (i) and (iii) .
[ (i) Considered the nucleus as positively charged.
Rutherford concluded from the alpha scattering experiment that the nucleus is small, dense, and positively charged.
(iii) Can be compared to solar system.
Rutherford’s model had a central nucleus and electrons revolving around it, similar to planets around the Sun. ]
6. Which of the following are true for an element?
(i) Atomic number = number of protons + number of electrons
(ii) Mass number = number of protons + number of neutrons
(iii) Atomic mass = number of protons = number of neutrons
(iv) Atomic number = number of protons = number of electrons
(a) (i) and (ii) (b) (i) and (iii) (c) (ii) and (iii) (d) (ii) and (iv)
Answer: (d) (ii) and (iv)
[ (ii) Mass number = number of protons + number of neutrons.
(iv) Atomic number = number of protons = number of electrons. ]
7. In the Thomson’s model of atom, which of the following statments are correct?
(i) the mass of the atom is assumed to be uniformaly distributed over the atom
(ii) the positive charge is assumed to be uniformaly distributed over the atom
(iii) the electrons are uniformaly distributed in the positively charged sphere
(iv) the electrons attract each other to stabilise the atom
(a) (i), (ii) and (iii) (b) (i) and (iii) (c) (i) and (iv) (d) (i), (iii) and (iv)
Answer: (a) (i), (ii) and (iii)
[ In J. J. Thomson’s model:
(i) Mass of the atom was considered uniformly distributed.
(ii) Positive charge was spread uniformly throughout the atom.
(iii) Electrons were embedded in the positive sphere (like plums in pudding). ]
8. Rutherford’s α–particle scattering experiment showed that
(i) electrons have negative charge
(ii) the mass and positive charge of the atom is concentrated in the nucleus
(iii) neutron exists in the nucleus
(iv) most of the space in atom is empty
Which of the above statements are correct?
(a) (i) and (iii) (b) (ii) and (iv) (c) (i) and (iv) (d) (iii) and (iv)
Answer: (b) (ii) and (iv)
[ (ii) Rutherford showed that mass and positive charge are concentrated in the nucleus.
(iv) Most of the atom is empty space. ]
9. The ion of an element has 3 positive charges. Mass number of the atom is 27 and the number of neutrons is 14. What is the number of electrons in the ion?
(a) 13 (b) 10 (c) 14 (d) 16
Answer: (b) 10 .
[ Mass number (A) = 27
Number of neutrons (N) = 14
We know, Atomic number (Z) = A − N = 27 – 14 = 13
So, number of protons = number of electron = 13
But the ion has 3 positive charges .
The number of electrons in the ion = 13 – 3 = 10 ]
10. Identify the 
ion from the Fig.4.1 where, n and p represent the number of neutrons and protons respectively
Answer: (d)
[ Atomic number of 
.
Electronic configuration of Mg = 2.8.2
And also, Electronic configuration of 
. So, (d) is correct option. ]
11. In a sample of ethyl ethanoate (
) the two oxygen atoms have the same number of electrons but different number of neutrons. Which of the following is the correct reason for it?
(a) One of the oxygen atoms has gained electrons
(b) One of the oxygen atoms has gained two neutrons
(c) The two oxygen atoms are isotopes
(d) The two oxygen atoms are isobars.
Answer: (c) The two oxygen atoms are isotopes .
[Two oxygen atoms have the same number of electrons . But they have different number of neutrons.]
12. Elements with valency 1 are
(a) always metals (b) always metalloids (c) either metals or non-metals (d) always non-metals
Answer: (c) either metals or non-metals .
[ Valency 1 means the element can lose or gain one electron.
Example : (i) Metals like sodium (Na) lose 1 electron → valency 1.
(ii) Non-metals like chlorine (Cl) gain 1 electron → valency 1.
So, elements with valency 1 can be either metals or non-metals. ]
13. The first model of an atom was given by
(a) N. Bohr (b) E. Goldstein (c) Rutherford (d) J.J. Thomson
Answer: (d) J.J. Thomson .
[The first scientific model of an atom was given by J. J. Thomson. ]
14. An atom with 3 protons and 4 neutrons will have a valency of
(a) 3 (b) 7 (c) 1 (d) 4
Answer: (c) 1
[ Number of protons = 3 . So, atomic number = 3
Name of the element is Lithium (Li) .
Electronic configuration of lithium = 2. 1
Then, the valency of Li is 1 . ]
15. The electron distribution in an aluminium atom is
(a) 2, 8, 3 (b) 2, 8, 2 (c) 8, 2, 3 (d) 2, 3, 8
Answer: (a) 2 , 8 , 3
[ Atomic number of aluminium = 13
Electronic configuration of lithium = 2.8.3 ]
16. Which of the following in Fig. 4.2 do not represent Bohr’s model of an atom correctly?
(i) in figure electron first shell 2 second shell 5 (ii) (iii) (iv)
(a) (i) and (ii) (b) (ii) and (iii) (c) (ii) and (iv) (d) (i) and (iv)
Answer: (c) (ii) and (iv)
[ We know that, maximum electrons in a shell = 2n²
For first shell (n = 1): 2 × 1² = 2 electrons (maximum 2)
For second shell (n = 2): 2 × 2² = 8 electrons (maximum 8)
(i) For first shell 2 electrons and second shell 5 electrons – correct .
(ii) For first shell 4 electrons and second shell 6 electrons – wrong .
(iii) For first shell 2 electrons and second shell 7 electrons – correct .
(iv) For first shell 2 electrons and second shell 9 electrons – wrong . ]
17. Which of the following statement is always correct?
(a) An atom has equal number of electrons and protons.
(b) An atom has equal number of electrons and neutrons.
(c) An atom has equal number of protons and neutrons.
(d) An atom has equal number of electrons, protons and neutrons.
Answer: (a) An atom has equal number of electrons and protons.
[We know that, the number of electrons = number of protons.]
18. Atomic models have been improved over the years. Arrange the following atomic models in the order of their chronological order
(i) Rutherford’s atomic model (ii) Thomson’s atomic model (iii) Bohr’s atomic model
(a) (i), (ii) and (iii) (b) (ii), (iii) and (i) (c) (ii), (i) and (iii) (d) (iii), (ii) and (i)
Answer: (c) (ii), (i) and (iii)
[ J. J. Thomson – Thomson’s atomic model (Plum pudding model)
Ernest Rutherford – Nuclear model
Niels Bohr – Bohr’s model with fixed energy levels ]
Short Answer Questions
19. Is it possible for the atom of an element to have one electron, one proton and no neutron. If so, name the element.
Answer: Yes, it is possible. An atom with one proton, one electron and no neutron is hydrogen. The name of element is hydrogen.
20. Write any two observations which support the fact that atoms are divisible.
Answer: The observations that support the fact that atoms are divisible are:
(i) J. J. Thomson – He discovered electrons in cathode ray experiments. This showed that atoms contain smaller negatively charged particles, so atoms are divisible.
(ii) Ernest Rutherford – His gold foil experiment proved that atoms have a small nucleus with protons, showing atoms are made of smaller particles and are divisible.
21. Will ³⁵Cl and ³⁷Cl have different valencies? Justify your answer.
Answer: No, ³⁵Cl and ³⁷Cl will not have different valencies. Both are isotopes of chlorine. They have the same atomic number (17), so their electronic configuration is 2. 8. 7 . Hence, both have valency 1.
22. Why did Rutherford select a gold foil in his α–ray scattering experiment?
Answer: Ernest Rutherford selected a gold foil because it could be made extremely thin (about 1000 atoms thick). Fast-moving α-particles (mass 4 u) could pass through it. He expected only small deflections since α-particles are much heavier than protons.
23. Find out the valency of the atoms represented by the Fig. 4.3 (a) and (b).
Answer: (a) The electronic configuration of the atom is 2.8.8 .
So, the valency of the atom is 0 .
(b) The electronic configuration of the atom is 2.8.7 .
So, the valency of the atom is 1 (= 8 – 7) .
24. One electron is present in the outer most shell of the atom of an element X. What would be the nature and value of charge on the ion formed if this electron is removed from the outer most shell?
Answer: If one electron is present in the outermost shell of element X, it will easily lose that electron to get a stable configuration. After losing one electron, it will form a positive ion (cation) with a +1 charge, because the number of protons becomes one more than electrons.
25. Write down the electron distribution of chlorine atom. How many electrons are there in the L shell? (Atomic number of chlorine is 17).
Answer: Atomic number of chlorine = 17
Therefore, electronic configuration of chlorine = 2 . 8 . 7
K L M
Number of electrons in the L shell = 8 electrons.
26. In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed?
Answer: The element X has 6 electrons in its outermost shell, so it needs 2 more electrons to complete its octet and attain noble gas configuration . So, it will gain 2 electrons.
When it gains 2 electrons, it forms a negative ion (anion) with –2 charge.
27. What information do you get from the Fig. 4.4 about the atomic number, mass number and valency of atoms X, Y and Z? Give your answer in a tabular form.
Answer: We have,
|
Atom |
X |
Y |
Z |
|
Protons (p) |
5 |
8 |
15 |
|
Neutrons (n) |
6 |
10 |
16 |
|
Atomic number |
5 |
8 |
15 |
|
Mass number |
11 |
18 |
31 |
|
Electronic Configuration |
2.3 |
2.6 |
2.8.5 |
|
Valency |
3 |
2 (= 8 – 6) |
3 (= 8 – 5) |
28. In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement? Justify your answer.
Answer: No . In a neutral atom, the number of protons is equal to the number of electrons. The number of neutrons may be equal to or greater than protons, but protons are never greater than electrons in a neutral atom.
29. Calculate the number of neutrons present in the nucleus of an element X which is represented as
Answer: Given , Mass number (A) = 31 and atomic number (Z) = 15
Number of neutrons = Mass number – Atomic number = 31 – 15 = 16
So, the number of neutrons present in the nucleus of element 
is 16.
30. Match the names of the Scientists given in columnA with their contributions towards the understanding of the atomic structure as given in column B
(A) (B)
(a) Ernest Rutherford (i) Indivisibility of atoms
(b) J.J.Thomson (ii) Stationary orbits
(c) Dalton (iii) Concept of nucleus
(d) Neils Bohr (iv) Discovery of electrons
(e) James Chadwick (v) Atomic number
(f) E. Goldstein (vi) Neutron
(g) Mosley (vii) Canal rays
Answer: (a) Ernest Rutherford → (iii) Concept of nucleus
(b) J. J. Thomson → (iv) Discovery of electrons
(c) John Dalton → (i) Indivisibility of atoms
(d) Niels Bohr → (ii) Stationary orbits
(e) James Chadwick → (vi) Neutron
(f) Eugen Goldstein → (vii) Canal rays
(g) Henry Moseley → (v) Atomic number
31. The atomic number of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements?
Answer: Calcium (atomic number 20) and argon (atomic number 18) have the same mass number (40) but different atomic numbers. Such a pair of elements is called isobars.
32. Complete the Table 4.1 on the basis of information available in the symbols given below
(a) 
(b) 
(c) 
|
Element |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Table 4.1
Answer: We have,
|
|
Symbols |
Number protons ( |
Number neutrons ( |
|
(a) |
|
17 |
18 (= 35 – 17) |
|
(b) |
|
6 |
6 (= 12 – 6) |
|
(c) |
|
35 |
46 (= 81 – 35) |
33. Helium atom has 2 electrons in its valence shell but its valency is not 2, Explain.
Answer: Helium has 2 electrons in its only shell (K shell). The maximum capacity of K shell is also 2 electrons. So, helium already has a completely filled outermost shell. Because of this stable duplet configuration, it does not gain or lose electrons . Hence, its valency is zero, not 2.
34. Fill in the blanks in the following statements
(a) Rutherford’s α-particle scattering experiment led to the discovery of the ———
(b) Isotopes have same ———but different———.
(c) Neon and chlorine have atomic numbers 10 and 17 respectively. Their valencies will be———and———respectively.
(d) The electronic configuration of silicon is ———and that of sulphur is ———.
Answer: (a) Rutherford’s α-particle scattering experiment led to the discovery of the nucleus.
(b) Isotopes have same atomic number but different mass number.
(c) Neon and chlorine have atomic numbers 10 and 17 respectively. Their valencies will be 0 and 1 respectively.
(d) The electronic configuration of silicon is 2, 8, 4 and that of sulphur is 2, 8, 6.
35. An element X has a mass number 4 and atomic number 2. Write the valency of this element?
Answer: Mass number of X = 4
Atomic number of X = 2
Atomic number 2 means the element is helium (He).
Electronic configuration of helium = 2
So, the valency is 2 .
Long Answer Questions
36. Why do Helium, Neon and Argon have a zero valency?
Answer: Helium, Neon and Argon have completely filled outermost shells.
Helium has a duplet (2 electrons) and Neon and Argon have octet (8 electrons). Since their valence shells are full, they do not gain, lose or share electrons. Therefore, their valency is zero.
37. The ratio of the radii of hydrogen atom and its nucleus is ~ 105 . Assuming the atom and the nucleus to be spherical, (i) what will be the ratio of their sizes? (ii) If atom is represented by planet earth ‘Re
, estimate the size of the nucleus.
Answer: Given, 
(i) We have, 
(ii) Radius of atom is represented by radius of Earth 
Radius of nucleus 
So, the radius of the nucleus is 64 metres.
38. Enlist the conclusions drawn by Rutherford from his α-ray scattering experiment.
Answer: From the α–ray scattering experiment, Ernest Rutherford concluded that:
(i) Most of the space inside the atom is empty, because most α–particles passed through the gold foil without any deflection.
(ii) Very few particles were deflected, showing that the positive charge occupies very little space in the atom.
(iii) A very small number of α–particles were deflected by 180°, which proved that almost all the positive charge and mass are concentrated in a tiny central region called the nucleus.
39. In what way is the Rutherford’s atomic model different from that of Thomson’s atomic model?
Answer: Rutherford’s atomic model explained that the positive charge and most of the mass are concentrated in a small, dense nucleus at the centre, and electrons revolve around it in mostly empty space. On the other hand, Thomson’s model stated that the positive charge is spread uniformly throughout the atom, with electrons embedded in it like seeds in a watermelon.
40. What were the drawbacks of Rutherford’s model of an atom?
Answer: The main drawback of Rutherford’s model was that an electron revolving in a circular orbit should continuously radiate energy because it is accelerating. By losing energy, the electron would finally fall into the nucleus. This means the atom should be unstable, but actually atoms are stable.
41. What are the postulates of Bohr’s model of an atom?
Answer: The postulates of Bohr’s model of an atom are :
(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.
42. Show diagramatically the electron distributions in a sodium atom and a sodium ion and also give their atomic number.
Answer: Sodium Atom (Na) : Atomic number of sodium = 11 .
Electron configuration= 2. 8. 1
Sodium Ion (Na⁺) : Atomic number of sodium = 11 – 1 = 10 .
Electron configuration= 2. 8
43. In the Gold foil experiment of Geiger and Marsden, that paved the way for Rutherford’s model of an atom, ~ 1.00% of the α-particles were found to deflect at angles > 50° . If one mole of α-particles were bombarded on the gold foil, compute the number of α-particles that would deflect at angles less than 50° .
Answer: The percentage of α-particles deflecting at angles greater than 50° is 1.00%.
Therefore, the percentage deflecting at angles less than 50° = 100% – 1.00% = 99 % = 0.99 .
The number of α-particles deflecting at angles less than 
.
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