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NCERT Exemplar Class 9 Science Chapter 4 Structure of the Atom Solutions

NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom covers all the important questions and answers as well as advanced level questions. It helps in learning about the Charged Particles of Matter, where does this charge come from, Structure of an Atom, different models of atoms proposed, Thomson’s Model of an Atom, Rutherford’s Models of an Atom, nucleus of an atom, orbits and Bohr’s Model of an Atom.

The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like neutrons, proton, electron of an atom, isotopes, isobars, how are the electrons distributed in different shells, valence electrons, atomic number and mass number.


Chapter Name

Chapter 4 Structure of the Atom

Book Title

NCERT Exemplar for Class 9 Science

Related Study

  • NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom
  • Revision Notes for Class 9 Science Chapter 4 Structure of the Atom
  • MCQ for Class 9 Science Chapter 4 Structure of the Atom
  • Important Questions for Class 9 Science Chapter 4 Structure of the Atom

Topics Covered

  • MCQ
  • Short Answers Questions
  • Long Answers Questions

NCERT Exemplar Solutions for Chapter 4 Structure of the Atom Class 9 Science

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

Solution

(b) 2, 8, 2

Atomic number of Mg is 12.Therefore the electronic distribution will be 1s22s22p63s2.


2. Rutherford’s ‘alpha (a ) particles scattering experiment’ resulted in to discovery of :
(a) Electron
(b) Proton
(c) Nucleus in the atom
(d) Atomic mass

Solution

(c) Nucleus in the atom

Rutherford’s ‘alpha (α) particles scattering experiment’ showed that few alpha particles returned to their original path. This showed the presence of nucleus in the centre.


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) 15X31
(b) 16X31
(c) 15X16
(d) 16X15

Solution

(a) 15X31

Atomic number is the number of protons in an element. Number of protons and electrons are equal in an element. The atomic number is written as a subscript and the mass number is written as the superscript of the element.


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 proportions
(a) (i), (ii) and (iii)
(b) (i), (iii) and (iv)
(c) (ii), (iii) and (iv)
(d) (i), (ii) and (iv)

Solution

(d) (i), (ii) and (iv)

Dalton’s theory explains the Law of conservation of mass, Law of constant composition, Law of multiple proportion.


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 a-particles are four times 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)

Solution

(a) (i) and (iii)

The positively charged alpha particles were deflected by the nucleus. This indicates that the nucleus is positively charged. Rutherford also postulated that electrons are arranged in an atom around the nucleus like planets arranged around 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) and (iii)
(d) (ii) and (iv)

Solution

(d) (ii) and (iv)

Atomic number Z is the number of proton present in an atom. This is also equal to number of electrons in an atom. As the mass of neutron is negligible, number of protons and electron are added to obtain mass number of an element.


7. In the Thomson’s model of atom, which of the following statements are correct?
(i) The mass of the atom is assumed to be uniformly distributed over the atom.
(ii) The positive charge is assumed to be uniformly distributed over the atom.
(iii) The electrons are uniformly 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)

Solution

(a) (i), (ii) and (iii)

Thomson proposed that negatively charged electron are stabilised by positively charged protons in the nucleus.


8. Rutherford’s a -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)

Solution

(b) (ii) and (iv)


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

Solution

(b) 10

Mass number (A) of the atom = 27
Number of neutron in the atom =14
Number of electrons = Mass number - Number of neutrons
= 27-14 = 13
Since ions of the element has 3 positive charges number of electron in the ion is 13-3 which equal 10.


10. Identify the Mg2+ ion from the fig. where, n and p represent the number of neutrons and protons respectively :

Solution
Number of protons in Mg atom = 2+ 8 + 2 = 12
Number of neutrons in Mg atom = 24 -12 = 12
[As mass number of Mg atom = 24 and number of neutrons = mass number – number of protons]


11. In a sample of ethyl ethanoate (CH3COOC2H5), 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.

Solution

(c) The two oxygen atoms are isotopes.

Two oxygen atoms in CH3COOC2H5 can have different number of neutrons only if the two O-atoms are isotopes. Isotopes of an element have same number of protons (and electrons) but 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

Solution

(c) Either metals or non-metals

If an element shows positive valency then it is a metal and if it shows negative valency then it is a non-metal.


13. The first model of an atom was given by :
(a) N. Bohr
(b) E. Goldstein
(c) Rutherford
(d) J.J. Thomson

Solution

(d) 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

Solution

(c) 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

Solution

(a) 2, 8, 3

Atomic number of Aluminium is 13. Therefore, the first shell can have maximum of 2 electrons and the second shell holds a maximum of 8 electrons.


16. Which of the following fig. do not represent Bohr’s model of an atom correctly?

(a) (i) and (ii)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (i) and (iv)

Solution

(c) (ii) and (iv)

First shell can have a maximum of 2 electrons and the second shell can have a maximum of 8 electrons.


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.

Solution

(a) An atom has equal number of electrons and protons.

In an atom the number of protons is always equal to number of electrons.


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)

Solution

(c) (ii), (i) and (iii)


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.

Solution

Yes, Hydrogen is the element which has only 1 proton and 1 electron and no neutron. Therefore there is no repulsive force in the nucleus and hence it is stable.
1 proton means atomic no. (Z) = 1
1 neutron means mass no. (A) = p + n = 1 + 1 = 2
1 electron and 1 proton mean that atom is electrically neutral.
Hence, the element is 11H (An isotop of Hydrogen – Deuterium)


20. Write any two observations which support the fact that atoms are divisible.

Solution

  1. Discovery of electrons and protons support the fact that atoms are divisible.
  2. During a chemical reaction, there is either transfer or sharing of electrons which leads to the rearrangement of electrons in the atom.


21. Will 35Cl and 37Cl have different valencies? Justify your answer.

Solution

No, 35Cl and 37Cl both are the isotopes of the same element hence they cannot have different valencies.
Both atoms have valency 1.


22. Why did Rutherford select a gold foil in his a-ray scattering experiment?

Solution

Gold is the most malleable metal and Rutherford wanted the thinnest layer. Therefore, Rutherford used gold for his scattering experiment as it is a heavy metal with high atomic number and it is highly malleable.


23. Find out the valency of the atoms represented by the fig. (a) and (b).

Solution

As atom (a) has 8 electron in its valence shell, it has zero valency.

Atom (b) has a valency of +1 as it has 7 electrons in it outermost shell. To achieve the octet configuration, atom (b) will accept 1 electron.


24. One electron is present in the outermost 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 outermost shell?

Solution

A cation is formed if an electron is removed from the outermost shell and the charge of the element will be +1.


25. Write down the electron distribution of chlorine atom. How many electrons are there in the L shell? (Atomic number of chlorine is 17.)

Solution

Atomic number of chlorine atom = 17

So, its electronic configuration is
17Cl = K-2, L-8, M-7

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?

Solution

6 electrons are already present in the outermost orbital. In order to attain noble gas configuration the element has to accept two electrons. Therefore, the charge on an atom of element X is -2.


27. What information do you get from the fig. about the atomic number, mass number and valency of atoms X, Y and Z? Give your answer in a tabular form.

Solution

Sl. No.

Atomic number

Mass number

Valency

X

5

11

3

Y

8

18

2

Z

15

31

3, 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.

Solution

The statement is incorrect. In a neutral atom the number of protons and electrons is always equal. Number of neutron can be greater than the number of electron. Number of neutron can be equal to or greater than the number of protons because mass number is equal to double the atomic number.


29. Calculate the number of neutrons present in the nucleus of an element X which is represented as 1531X.

Solution

Mass number = No. of protons + No. of neutrons = 31
∴ Number of neutrons = 31–number of protons
= 31–15
= 16


30. Match the names of the scientists given in column A with their contributions towards the understanding of the atomic structure as given in column B :

 

Column A

 

Column 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

Solution

(a)

Ernest Rutherford

Concept of nucleus

(b)

J.J. Thomson

Discovery of electrons

(c)

Dalton

Indivisibility of atoms

(d)

Neils Bohr

Stationary orbits

(e)

James Chadwick

Neutron

(f)

E. Goldstein

Canal rays

(g)

Mosley

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?

Solution

Isobars are elements with different atomic numbers but same mass numbers. Calcium and argon are isobars.


32. Complete the table on the basis of information available in the symbols given below :
(a) 17Cl35 (b) 6C12 (c) 35Br85

Solution

Elements

No. of protons

No. of neutrons

17Cl35

17

35 – 17 = 18

6C12

6

12 – 6 = 6

35Br85

35

81 – 35 = 46


33. Helium atom has 2 electrons in its valence shell but its valency is not 2. Explain.

Solution

As Helium atom has 2 electrons in its outermost shell, its duplet is complete. Therefore the valency is zero.


34. Fill in the blanks in the following statements :
(a) Rutherford’s a -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 _____.

Solution

(a) nucleus
(b) same atomic numbers, mass numbers
(c) 0 and –1
(d) Si = 2,8,4 and S = 2,8,6


35. An element X has a mass number 4 and atomic number 2. Write the valency of this element?

Solution

As the K shell is completely filled, valency is zero.


Long Answer Questions

36. Why do helium, neon and argon have a zero valency?

Solution

Helium has two electrons in its only energy shell, while Argon and Neon have 8 electrons in their valence shells. These elements do not have any tendency to combine with other elements as they have maximum number of electrons in their valence shells. Therefore, they have zero valency.


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’ = 6.4 ×106m, estimate the size of the nucleus.

Solution

(i) Volume of the sphere = 4/3 πr3

Let R be the radius of the atom and r be that of the nucleus.

Therefore ,R = 105r

(ii) If the atom is represented by the planet earth (Re = 6.4×106m) then the radius of the nucleus would be


38. Enlist the conclusions drawn by Rutherford from his a-ray scattering experiment.

Solution

The following conclusions were made from the Rutherford α-particle scattering experiment:

  1. As most of the α-particles passed through the gold foil undeflected, most of the space inside the atom is empty.
  2. Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.
  3. A very small fraction of α-particles were deflected by 180°,indicating that all the positive charges and mass of the gold atom were concentrated in a very small volume within the atom.

From the data he also calculated that the radius of the nucleus is about 105 times less than the radius of the atom.


39. In what way is the Rutherford’s atomic model different from that of Thomson’s atomic model?

Solution

Rutherford’s atomic model proposed that the electrons revolve around the nucleus in well-defined orbits. The center of the atom is positively charged and is called the nucleus. The model also proposed that the size of the nucleus is very small as compared to the size of the atom and nearly the entire mass of an atom is centred in the nucleus.

Thomson’s model of an atom was similar to that of a christmas pudding. The electrons are studded like currants in a positively charged sphere like christmas pudding and the mass of the atom was supposed to be uniformly distributed.


40. What were the drawbacks of Rutherford’s model of an atom?

Solution

Rutherford’s model failed to explain the stability of the atom. Any particle in a circular orbit would undergo an acceleration and the charged particles would radiate energy. Thus, the revolving electron would lose energy and finally fall into the nucleus. In other words, the atom should collapse.


41. What are the postulates of Bohr’s model of an atom?

Solution

The postulates of Neils Bohr’s about the model of an atom are as follows:

  1. Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom.
  2. While revolving in discrete orbits the electrons do not radiate energy.

These orbits are called energy levels. Energy levels in an atom are shown by circles. These orbits are represented by the letters K,L,M,N,... or the numbers, n=1,2,3,4,....


42. Show diagrammatically the electron distributions in a sodium atom and a sodium ion and also give their atomic number.

Solution

As the atomic number of sodium atom is 11, it has 11 electrons. A positively charged sodium ion (Na+) is formed by the removal of one electron from a sodium atom. Thus, a sodium ion has 11–1 = 10 electrons. The electronic configuration of sodium ion will be 2, 8. The atomic number of an element is equal to the number of protons in its atom. As, sodium atom and sodium ion have the same number of protons, therefore, the atomic number of both is 11.


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 > 50º. 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 500.

Solution

% of α-particles deflected more than 50°=1% of α-particles.

% of α-particles deflected less than 50° = 100–1% =  99%

Number of α-particles bombarded = 1 mole = 6.022×1023 particles

Number of particles that deflected at an angle less than 50°

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