Science class 9– Chapter 4 – Structure of Atoms – NCERT Exercise Solution (Question-Answer)

Science Class 9 – Chapter 4 – Structure of Atoms – NCERT Exercise Solution (Question-Answer) is provided below. Total 19 Questions are in this CBSE NCERT Exercise, all are solved here.

Q1. Compare the properties of electrons, protons, and neutrons.

Answer: Properties of electrons, protons, and neutrons:

WeightMass is negligible (1/1800) times of protonsMass is 1 a.m.uMass is 1 a.m.u
AffinityGet attracted towards positive chargeGet attracted  towards negative charge      Do not get attracted
ChargeNegatively chargedPositively chargeNo charge
LocationIt presents outside of nucleusIt presents in the nucleusIt presents in the nucleus of the atom

Q2. What are the limitations of J.J Thomson’s model of the atom?

Answer:   The following are the limitation of J.J Thomson’s model of the atom-

   (According to J.J Thomson’s model of the atom “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 is electrically neutral”)

But experiments done by other scientists showed that protons are present only in the center of the atom and electrons are distributed around it.

This model also fails to explain the outcome of alpha particles scattering which was conducted by Rutherford. The model failed to depict why the majority of these alpha particles pass through gold foil while some are diverted through small and big angles, while some others rebound completely, returning back on their path.

This model did not provide any experimental evidence and was established on imagination.

Q3. What are the limitations of Rutherford’s model of the atom?

Answer: The following are the limitations of Rutherford’s model of the atom:

According to Rutherford’s model of an atom, the electrons are revolving in a circular orbit around the nucleus.

When a charged particle is subjected to the acceleration it radiated energy. Since the electron is charged particle, it will radiate energy and the distance between electron and nucleus will decrease. Hence the motion of the electron will be spiral and finally, the electron should have been collapsed, which does not happen.

Hence, this model is unable to explain the estability of atoms.

Attach Fig

Q4. Describe Bohr’s model of the atom.

Answer: In order to explain the stability of the atom and hydrogen atom spectrum Bohr’s proposed a model known as “Bohr’s atomic model”.  

The following are the postulates of this model:

(i)Electrons move in a certain fixed concentric orbit of constant energy called the stationary orbit. these are also called energy levels.

(ii)So long as electrons move in a particular orbit it neither radiates nor absorbs energy.

(iii)Whenever an electron jump from one orbit to other either it radiated or absorb energy.

(iv)When the electron jumps from lower shell to higher shell, it absorbs energy and when it jumps from higher shell to lower shell it radiated energy.

(v)The orbit or shell are represented by the letters K, L, M, N or the numbers n= 1, 2, ,3, 4

( attach fig from ncert book pg no. 49)

Q5. Compare all the proposed models of an atom given in this chapter.

Answer: There are three models are given in this chapter:

(i)Thomson’s model of atom

(ii)Rutherford’s model of atom

(iii)Bohr’s model of atoms

Thomson’s model of atomRutherford’s model of atomBohr’s model of atoms
Spheres of positive chargeSpheres of positive charge in centre called nucleus. All mass of an atom resides in the nucleusPositive charge in centre called nucleus
Positively charged = negatively chargedIn comparison with the nucleus, the size of an atom is very largeThe distinct orbits are labelled as K, L, M, N
Electrons are spread randomly all over in the sphereElectrons revolve around the nucleus in well defined orbitElectron revolves in discrete orbits and do not radiate energy.
Electrons are negatively chargedElectrons are negatively chargedElectrons are negatively charged

( attach fig of all three models from ncert book .. pg no = 47 for Thomson, 48 for  Rutherford,and 49 for Bohrs)

Q6. Summaries the rules for the writing of distribution of electrons in various shells for the first eighteen elements.

Answer: The following rules are followed for writing the number of electrons in different energy levels or shells:

(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

The 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 accommodates in a given shell unless the inner shell is filled. That is, the shells are filled in a step – wise manner.

Q7. Define valency by taking examples of silicon and oxygen.

Answer: Valency is the combining capacity of an atom.

Valency is defined as the definite combining capacity of the atoms of each element, wherein electrons are lost, gained or shared to make the octet of electrons present in the outermost shell.

To determine the valency, we can figure out the number of electrons that are required to complete the shell in which it is contained or losing excess electrons. if present, once the filling is complete.

Example: To find the valency of silicon

The atomic number of silicon = 14

Electronic configuration of silicon = 2   8   4

The valence electron of silicon is 4, so to fill the orbit 4 electrons are required. Hence, the combining capacity of silicon is 4

Hence the valency of silicon is 4

To find valency of oxygen

The atomic number of oxygen = 8

Electronic configuration of oxygen = 2   6

The valence electron of oxygen is 6, so to fill the orbit 2 electrons are required. Hence, the combining capacity of oxygen is 2

Valency = 2

Q8. Explain with examples (i)Atomic number, (ii)Mass number, (iii)Isotopes and (iv)Isobars. Give any two uses of isotopes.


(i)Atomic Number: The atomic number of an element is the same as the number of protons in the nucleus of its atom.

(ii)Mass Number: The mass number of an atom is equal to the number of nucleons in its nucleus.

(iii)Isotopes: Isotopes are atoms of the same element, which have a different mass number.

(iv)Isobars: Isobars are atoms having the same mass number but a different atomic number.

Uses of Isotopes:

(i) The isotope of iodine is used to treat goitre and iodine deficiency disease.

(ii) Isotope of uranium is used as fuel in nuclear reactors.

(iii) Isotope of cobalt is used I treatment of cancer.

Q9. Na+ has completely filed K and L shells. Explain.

Answer: Atomic  number of sodium(Na) = 11

Electronic configuration = 2    8   1 (k shell, L shell and M shell)

When sodium atom Na losses one atom it forms Na+ ions.

Now electronic configuration of Na+ = 2   8 (because it losses one atom)

Hence, t is a filled state. it is very difficult to eliminate the electron from a filled state as it is very stable.

Q10. If bromine atom is available in the form of, say, two isotopes 7935Br (49.7 %) and 8135Br (50%), Calculate the average atomic mass of bromine atom.


      The atomic masses of two isotopic atoms are 79 (49.7 %) and 81 (50.3 %)

The average atomic mass of bromine atom = = (79 * 49.7/100) + (81 * 50.3/100)

                                                                    =39.263 + 40.743

                                                                 = 80.006 u


Q11. The average atomic mass of a sample of an element X is 16.2 u. What is the percentage of isotopes 168X and 188X in the sample?

Answer: Let the percentage of 8X16 be ‘a’ and that of 8X18 be ‘100-a’

According to given data:

16.2 u = 16a/100 + 18 (100-a)/100

16.2u = 16a/100 + 1800-18a/100

16.2 u =16a – 18a + 1800/100

 16.2 * 100 = -2a + 1800

-180 =   2a

a = 180 / 2 = 90 %

Hence, the percentage of isotope in the sample 8X16 is 90 %

 And that of 8X18 = 100 -a = 100-90 = 10 %

Q12. If Z=3, what would be the valency of the element? Also, name the element.


    Given Z =3 

We know the atomic number is represented by Z

 Electronic configuration of the element = 2 ,1

Hence, its valency is +1 

The 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?

Answer: Mass number of X = protons + neutrons

                                     =6 + 6 = 12

Mass number of Y = protons + neutrons

                                  = 6 + 8 = 14

 Both X and Y are isotopes of same element because atomic number is same i.e. 6

(atomic number = number of protons)

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.

Answer: False

(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.

Answer: False

(c) The mass of an electron is about 1/2000 times that of proton.

Answer: True

(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.

Answer: False

Put tick () 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

Answer: (a) atomic number

Q16. Isotopes of an element have

(a) The same physical properties

(b) Different chemical properties

(c) Different number of neutrons

(d) Different atomic numbers

Answer: (c) Different number of neutrons

Q17. Number of valence electrons in Cl ions are:

(a) 16

(b) 8

(c) 17

(d) 18

Answer: (b) 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

Answer: (d) 2, 8, 1

Q19. Complete the following table:

Atomic numberMass numberNumber of neutronsNumber of protonsNumber of electronsName of the atomic species
9   _______10_____________________


Atomic numberMass numberNumber of neutronsNumber of protonsNumber of electronsName of the atomic species
12111Hydrogen deuterium

Science Class 9– Chapter 2 – Structure of Atoms- NCERT Exercise solution (Question-Answer) ended👍👍👍

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